M.Sc. (Mathematics), SEM- I Paper - III COMPLEX ANALYSIS
PSMT103
CONTENT
Unit No. Title 1 INTRODUCTION TO COMPLEX NUMBER SYSTEM 2 SEQUENCES OF COMPLEX NUMBERS 3 SERIES OF COMPLEX NUMBERS 4 DIFFERENTIABILITY 5 COMPLEX LOGARITHM 6 ANALYTIC FUNCTIONS 7 COMPLEX INTEGRATION 8 CAUCHY THEOREM 9 THEOREMS IN COMPLEX ANALYSIS 10 MAXIMUM AND MINIMUM MODULUSPRINCIPLE SINGULARITIES 11 RESIDUE CALCULUSAND MEROMORPHIC FUNCTIONS 12 MOBIUS TRANSFORMATION
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SYLLABUS
Unit I. Holomorphic functions
Note: A complex differentiable function defined on an open subset of is called a holomorphic function.
Review: Complex numbers, Geometry of the complex plane, Weierstrass’s M-test and its aplication to uniform convergence, Ratio and root test for convergence of series of complex numbers.
Stereographic projection, Sequence and series of complex numbers, Sequence and series of functions in , Complex differential functions, Chain rule for holomorphic function.
Power series of complex numbers, Radius of convergence of power series, Cauchy-Hadamard formula for radius of convergence of power series. Abel's theorem: let be a power series of radius of convergence R > 0: Then the function f(z) defined by
is holomorphic on the open ball and
for all : Trigonometric functions, Applications of Abel's theorem to trigonometric functions.
Applications of the chain rule to define the logarithm as the inverse of exponential, branches of logarithm, principle branch of the logarithm and its derivative on
Unit II. Contour integration, Cauchy-Goursat theorem
Contour integration, Cauchy-Goursat Theorem for a rectangular region or a triangular region. Cauchy’s theorem(general domain), Cauchy integral formula, Cauchy’s estimates, The index(winding number) of a closed curve, Primitives. Existence of primitives, Morera’s theorem. Power series representation of holomorphic function (Taylor’s theorem). Unit III. Properties of Holomorphic functions
Entire functions, Liouville’s theorem. Fundamental theorem of algebra. Zeros of holomorphic functions, Identity theorem. Counting zeros; Open Mapping Theorem, Maximum modulus theorem, Schwarz’s lemma. Automorphisms of unit disc. Isolated singularities: removable singularities and Removable singularity theorem, poles and essential singularities. Laurent Series development. Casorati- Weierstrass’s theorem.
Unit IV. Residue calculus and Mobius transformation
Residue Theorem and evaluation of standard types of integrals by the residue calculus method. Argument principle. Rouch´e’s theorem. Conformal mapping, Mobius Transformation. 1 1
INTRODUCTION TO COMPLEX NUMBER SYSTEM
Unit Structure : 1.0. Objectives 1.1. Introduction 1.2. The Field of Complex Numbers 1.3. Extended Complex Plane, The Point at Infinity, Stereographic Projection 1.4. Summary 1.5. Unit End Exercises
1.0. OBJECTIVES:
After going through this unit you shall come to know about The field of complex numbers denoted by ℂ. Representations of complex numbers in polar forms . 2 The Euclidean two dimensional plane along with the point at infinity forms the extended complex plane. The extended complex plane is in one to one 3 correspondence with the unit sphere in and such a correspondence is known to be the stereographic projection.
1.1. INTRODUCTION :
Numbers of the form z a bi , where a and b are real numbers and i 1are called as Complex Numbers. The identities involving complex numbers lead to solutions to many problems in the theory of real valued functions .The wider acceptance of complex numbers is because of the geometric representation of complex numbers , which was fully developed and studied by Gauss. The first complete and formal definition of complex umbers was given by William Hamilton. We shall begin with this definition and then consider the geometry of complex numbers. 2
1.2. THE FIELD OF COMPLEX NUMBERS :
A complex number z is an ordered pair x, y of real numbers. i.e. z x,, y x y . Complex number system, denoted by is the set of all ordered pairs of real numbers (i.e. ) with the two operations of addition and multiplication ( or ) which satisfy : (i) x1 , y 1 x 2 , y 2 x 1 x 2 y 1 y 2 x1,,, y 1 x 2 y 2 (ii) xy11 , . xy 22 , xx 12121221 yyxy , xy
The word ordered pair means x1, y 1 and y1, x 1 are distinct unless x1 y 1. Let z x,, y x y .‘x’ is called Real part of a complex number z and it is denoted by x Re z , (Real part of z) and ‘y’ is called Imaginary part of z and it is denoted by y Im z .
Two complex numbers z1 x 1, y 1 and z2 x 2, y 2 are said to be equal iff x1 x 2 and y1 y 2 i.e. real part and imaginary part both are equal.
About Symbol ‘i’: The complex number 0, 1 is denoted by ‘i’ and is called the imaginary number. i2 i. i 0,1 . 0,1 0 1, 0 0 by property (ii) abov 1, 0 i2 Similarly, i3 i 2. i 1,0.0,1 00, 10 0,1 i3 i i4 i 3. i 0,1.0,1 01,00 1,0 3 i4 1
Using this symbol i, we can write a complex number x, y as x iy (Since x iy x,0 0,1 y ,0 x ,0 0, y x , y The complex number z x, y can be written as z x iy Note: (The set of all complex numbers) forms a field. 3
Propeties of complex numbers Let z1 x 1, y 1 , z2 x 2, y 2 and z3 x 3, y 3 . 1) Closure Law : z1 z 2 and z1. z 2
2) Commutative Law of addition : z1 z 2 z 2 z 1 zz1211 xy,,,, xy 22 xxyy 1212 xxyy 2121 x2,, y 2 x 1 y 1 z 2 z 1
3) Associative Law of addition : z1 z 2 z 3 z 1 z 2 z 3
z1 z 2 z 3 x 1,,, y 1 x 2 y 2 x 3 y 3 xyxxyy11,,, 2323 xxxyyy 123123 xxyy1212 ,,,, xy 33 xy 11 xy 22 xy 33 z1 z 2 z 3
4) Existence of additive Identity : The Complex Number 0 0, 0 i.e. z 0 0 i is called the identity with respect to addition.
5) Existence of additive Inverse : For each complex number z1 , a unique complex number z s.t. z1 z z z 1 0 i.e. z z1. The complex number z is called the additive inverse of z1 and it is denoted by z z1.
6) Commutative law of Multiplication : z1.. z 2 z 2 z 1 z.,., z x y x y x x y y, x y x y (1) 1 2 1 1 2 2 1 2 1 2 1 2 2 1 ….. and zz21.,.,.,.. xy 22 xy 11 xxyyxyxy 21212112
x1..,.. x 2 y 2 y 1 x 1 y 2 x 2 y 1 z1. z 2 from (1)
7) Associative Law of Multiplication : z1.... z 2 z 3 z 1 z 2 z 3 z1..,,.. z 2 z 3 x 1 y 1 x 2 y 2 x 3 y 3 x1,..,.. y 1 x 2 x 3 y 2 y 3 x 2 y 3 x 3 y 2 xxxyy12323 ...,.... yxxyy 12323 xxxxy 12332 yxxyy 12323
xxxxyyxxyyyyxxxxxy123123131123123132...... ,
x2. x 3 . y 1 y 1 y 2 y 3 (*) 4
zzz123..,.,,,, xyxy 11 22 xy 33 xxyyxyxyxy 1212122133
xxx123 xyy 312 xxy 123 yyy 123 , xxy 132 xyy 213 xxx 123 yyy 123
z1 z 2. z 3 from (*)
8) Existence of Multiplicative Identity : z1.1 1. z 1 z 1 The complex number 1 1, 0 (i.e. z1 0 i ) is called the identity with respect to multiplication.
9) Existence of Multiplicative Inverse : For each complex number z1 0, there exists a unique complex number z in s.t. 1 z1. z z . z 1 1 i.e. z is called the multiplicative inverse of z1 1 1 complex number z1 and it is denoted by z or z . z1 Let z x, y and z1 x 1, y 1
z1. z 2 1 x, y x1 , y 1 1, 0 xx1 yy 1, xy 1 x 1 y 1, 0 x. x1 y . y 1 1 ……..(i) and x. y1 x 1 . y 0 …...... (ii) Equation (ii) x1 - Equation (i) y1, we get 2 xx1 y 1 x 1 y0 2 xx1 y 1 yy 1 y 1 2 2 y x1 y 1 y 1 y y 1 (iii) 2 2 x1 y 1
Substitute equation (iii) in equation (ii) i.e. x. y1 y . x 1 0 y x y 1 x.. y y x 1 x x 1 1 1 1 2 2 1 2 2 x1 y 1 x1 y 1 y1 x x 1 2 2 x1 y 1 x y z 1, 1 2 2 2 2 x1 y 1 x 1 y 2 z is the multiplicative inverse of complex number z1 x 1, y 1 . 5
10) Distributive Law : z1 z 2 z 3 z 1.. z 2 z 1 z 3
Subtraction: The difference of two complex Numbers z1 x 1, y 1 and z2 x 2, y 2 is defined as : z1 z 2 x 1,, y 1 x 2 y 2 x1 x 2, y 1 y 2
z1 1 Division: It is defined by the equality z1. z 2 z2 0 z2
x y x.. x y y x y x y x,,, y 2 2 1 2 1 2 1 2 2 1 1 1 2 2 2 2 2 2 2 2 x2 y 2 x 2 y 2 x 2 y 2 x 2 y 2 Geometrical Representation of a Complex Number : Consider a complex number z x iy . Complex number is defined as an ordered paired of real numbers. i.e. z x, y Y
y P (x ,y)
X x Fig. 1.1 This form of a complex number z suggest that z can be represented by point (say) P whose Cartesian co-ordinates are x and y referred (relating) to rectangular axis X and Y, usually called the Real and Imaginary axis respectively. To each complex number there corresponds points in the plane and conversely, one and only one each point in the plane there exist one and only one complex number. A plane whose points are represented by the complex numbers is called Complex Plane or Gaussian Plane or Argand Plane. Gauss was first who formulated that complex numbers are represented by points in a plane in 1799 then in 1806 it was done by Argand. 6
Vector Representation of a Complex Numbers : If P is the point in the Complex Plane corresponding to complex number z can be considered as vector OP whose initial point is the origin ‘O’ and terminal point is P z(,) x y as shown in the figure 1.2.
Y P=z≡(x,y)
z
X
Fig 1.2
Conjugate: If z x iy then the complex number x iy is called the conjugate of a complex number z or complex conjugate and it is denoted by z . e.g. z4 3 i z 4 3 i w4 5 e3i w 4 5 e 3 i Geometrically: The complex conjugate of a complex number z(,) x y is the image or reflection of z in the real axis. z = x + iy Y
q1
O X q2
z = x - iy
Fig 1.3
Let z x iy z z z z x Re z and zIm z . xRe z and y Im z 2 2 7
Definition: The modulus or absolute value of a complex number z x iy is defined by z x2 y 2 .
Y P ( x, y)= z
z
O X
Fig 1.4
The distance between Two Complex Numbers : Let z1 x 1,,, y 1 z 2 x 2 y 2 in complex plane is given by 2 2 d z1, z 2 z 1 z 2 x1 x 2 y 1 y 2 Y
( y ) z 2 x2 , 2
z 1( x1 ,y1) X
Fig 1.5
Polar form of a Complex Numbers : If P is a point in the Complex Plane corresponding to complex number z x iy x, y and let r, be the polar co- ordinates of point x, y from figure 1.3, x rcos and y rsin , where r x2 y 2 is called the modulus or absolute value of z
1 y (denoted by z and tan is called the argument or x amplitude of z (denoted by argz ) . Here is the angle between the two lines OP and the real axis (x- axis) . 8
Y P= ( x y) y ,
r= z y
q O x x X Fig 1.6
z x iy z rcos ir sin z r(cos i sin ) This form is called the polar form of a the complex number z. Y q- 2p P
q
O X
q + 2p q + 4p
Fig 1.7 Any complex number z0 has an infinite number of distinct arguments. Any two distinct arguments of z differ each other by an integral multiple of 2. If one of the value of argument of z is then argz 2 n where n 0, 1, 2, ... The value of which lies in the interval or 0 2 is called the principal value of argument of z and it is denoted by Arg z . 9
The relation between Arg z and argz is given by Arg z arg z 2 n where n 0, 1, 2,...
Exponential form of complex number: A complex number can be written in the form of z rei , where r x2 y 2 and
1 y tan .This is known as the exponential form. x (Note: ei cos i sin ,known as Euler’s Identity) i500 Note : 1) ei 1 2)e 4 1 , Solved Examples : 1. Let z1 1 i , z 2 1 2 i , z 3 1 3 i . Find i) z1.z2 ii) z1/z2 iii) z 2 iv) z1 v) arg(z1) vi) Express z1 in polar and exponential form. Solution: 2 i) z1 z 2 (1)(12)12 i i i i 2 i 12 i i 233 i 1 i 1 i 1 2 i 1 2 i i 2 1 3 i ii) z1/ z 2 1 2i 1 2 i 1 2 i 1 4i2 5
iii) z 2=1+2i iv) x1 1, y 1 1 2 2 2 2 z1 x y 1 1 2 v) x1 1, y 1 1 y 1 tan1 (1 ) tan 1 ( ) x1 1 4 vi) z1 r 1(cos 1 sin 1 ) z 2(cos sin ) 1 4 4 2. Find the principal value of arg 'i ' Arg z
1 y Arg itan z i z ( o . x iy ) x 1 1 1 tan tan 0 2 3. Find the principal value of arg 1i z1 i x iy x 1, y 1
1 y 1 1 Arg z tan tan x 1 10
4. Express the Complex Number z 1 3 i in polar form. Solution : z rcos i sin (1) and z x iy 1 3 i x 1 and y 3 r x2 y 2 1 3 4 r 2 1 y 1 3 1 1 tan tan tan 3 tan tan x 1 3 z 2 cos i sin 3 3 3
Results : tan45 tan180 45 1 tan 45 tan180 45 1 tan 60 tan 180 60 3 tan 60 tan180 60 3 5. Expressthe ComplexNumber z 1 i in polar form Solution : z rcos i sin z 1 i (given) Comparing with z x iy x 1, y 1 2 2 2 2 r x y 1 1 1 1 = 2 1 y 1 1 1 1 tan tan tan 1 tan tan x 1 4 4 4 3 3 z 2 cos i sin 4 4
Basic Properties of Complex Numbers : 1) z z iff z is purely a real number Proof: Let z x iy, z x iy Let z z, x iy x iy 2iy 0 y 0 z x z is real number.
2) z 0 iff z 0 11
Proof: 0 z x2 y 2 x 2 0 and y 2 0 x0 and y 0 i.e. z0
3) z z Proof: z x iy and z x iy z x2 y 2 (1) z x2 y2 x 2 y 2 (2) z z from (1) & (2) 4) Rez Re z z 5) Imz Im z z 6) Let z , w then z w z w Let z x iy, w u iv z w x iy u iv x u i y v x u i y v x iy u iv z w z w z w
7) zw z. w zw x iy u iv xu yv i xv yu xu yv i xv yu xu ixv yv iyu x u iv iy u iv u iv x iy z. w 8) z 2 z. z 2 2 2 2 2 2 z x y x y (1) z. z x iy x iy x2 ixy ixy i2 y 2 x2 y 2 (2) z 2 z. z from (1) and (2) 9) zw z w 2 2 2 zw zw zw z z ww = z w zw z w z z 10) w w z 1 1 z z z w w w w 12
11) z z z x iy x iy x iy _____ zxiyxi () y xiyz
Addition of two Complex Numbers : Letz x iy ; w u iv Now, z w OA OB OA AC OC Y
B C
W + Z W
Z A
O X
Fig 1.8
Triangle Inequality : 1) z, w then z w z w Proof: 2 z w z w. z w z w.. z w zz zw zw ww z w 2 z 2 w 2 zw z. w (1) Now, zw.. zw xiyuiv xiyuiv z. w z w 2Re z w 2 z . w Re z z 2 z w z w z. w 2 z w (2)
Substitute (2) in equation (1), we get 2 2 2 2 z w z w 2 z w z w z w z w Geometrically, in any triangle, the sum of the two sides of a triangle is greater than or equal to the third side(the points are collinear, in case it is equal). 13
Y
B x+y y
A x
O X
Fig 1.9
2) Let z, w then z w z w Proof: Let z z w w Taking mod on both the sides z z w w z w w z w z w (i) Interchanging z and w, we get w z w z z w (ii) z w z w z w z w (iii) From equation (i) and (ii), we get z w z w z w z w z w
3) Let z, w then z w 2 z 2 2 Re zw w 2
z w 2 z 2 2 Re zw w 2 4) Parallelogram Law: The sum of the squares of the lengths of the diagonals of a parallelogram is equal to the sum of squares of lengths of its sides. i.e. prove that z w 2 z w 2 2 z 2 w 2 Proof: Let z, w 2 2 z w 2 x u i y v2 z2 Re (z w ) w
2 2 2 z w z2 Re z w w z w2 z w 2 2 z 2 w 2 14
Y z B C z
z A
O X Fig 1.10
5) Let z, w then z w z w z w Proof: i) T.P.T. z w z w Case (i) z w z w by triangle inequality Case (ii) z w z w z w z w From above both cases, z w z w (*)
ii) T.P.T. z w z w Consider z z w w z w w z w w z w z w (a) Consider w w z z w z z w z z w z w z z w w z (b) z w z w (**) From (*) and (**), we get z w z w z w
Theorem: The field is not a linearly totally ordered field OR
The field is not partially ordered field (Total ordering or partial ordaring means that if a b then either a b or a b ). Proof: Suppose that such a total (partial) ordering exists. Then for i.e. , we have either i 0 or i 0 if i 0 1 i . i 0 or if it i 0 i 0 1 i i 0 15
We get 1 0 , which is not true in . Our supposition is not true. is not linearly totally ordered field.
Properties of polar form and exponential form i1 i2 1) Let z1 r 1 e r 1cos 1 i sin 1 , z2 r 2 e r 2cos 1 i sin 2 i1 2 then z1. z 2 r 1 . r 2 cos 1 i sin 1 cos 2 i sin 2 r1. r 2 e i i2 n 2 in e e, e 1, n i i z... z r r e 1 2 r. r e 1 2 1 1 2 1 2 1 2 and arg z1 . z 2 arg z 1 arg z 2 mod 2 in the sense that they are same but for an integral multiple of 2.
Note : argz1 . z 2 arg z 1 arg z 2 2 k where k 0,1or 1 i1 i2 2. Let z1 r 1 e and z2 r 2 e and z2 0 z rcos i sin r 1 1 1 1 1 ei1 2 z2 r 2cos 2 i sin 2 r2
z1 arg argz1 arg z 2 (mod 2 ) z2 Let z1 1 and z2 i , z1 1 x iy x 1 and y 0 argz arg 1 tan1 0 tan1 0 tan 1 tan 1 1 z2 i x iy x 0 and y1 argz arg i tan1y tan 1 1 tan1 2 x 0 2 argz . z arg 1. i arg i tan11 tan 1 1 2 0 2 arg z1 . z 2 arg z 1 arg z 2 2 k where k 0 Let z1 1 and z2 i z1 1 z 1 x iy x 1, y 0 argz arg 1 tan1y tan 1 0 tan1 0 1 x 1 z2 i x iy x 0, y 1 argz arg i tan1y tan 11 tan 1 2 x 0 2 16
argz . z arg 1. i arg i tan1y tan 1 1 1 2 x 0 tan 2 arg z1 . z 2 arg z 1 arg z 2 2 k where k 1 In this case, we get correct answer by adding 2 to bring within the interval , .
When principal argument are added together in multiplication problem, the resulting argument need not be the principle value.
De-Moivre’s Theorem : Theorem: If n is any integer or fraction then cosi sinn cos ( n ) i sin ( n )
Proof: LHS= cosi sinn ( ei ) n e i() n cos n i sin n =RHS cos i sinn cos n i sin n e.g. i) cos i sin5 cos 5 i sin5 2 2 ii) (cosi sin )5 cos i sin 5 Note: sin i cosn sin i cos n n n But, sin i cos cos n i sin n 2 2 cosn i sin n (by above thm) 2 2 2 2 3 e.g. 1) sin i cos 3 cos i sin 2 2 2 2 cos i sin 3 2 3 2 2) cos i sinn cos n i sin n 4 4 4 e.g. cos i sin 5 cos i sin 5 5 z x iy x2 y 2 r 2 is equation of circle with centre at the origin & radius equal to r. 17
r O
Fig 1.11 plane
z x iy z x2 y 2 r 2 r x2 y 2 r 2
The equation of the circle with the centre at c a ib and radius equal to r.
r c
o
Fig 1.12 z c r e.g. 1) z 2 i 1 This is equation of the circle with centre 2,1 and radius 1. z 1 3, circle with centre 1, 0 and radius = 3 z i 2 , circle with centre 0, 1 and radius = 2
Roots of Complex Number : Definition: A number w is called the nth root of complex number z if 1 wn z or w z n .
n Theorem: In , given z 0 , the equation expansion w z has n- 2k i n n distinct solution given by wk r. e , k 0,1,..., n 1 where r z and Argz . 18
Proof: Given, z and z 0 The polar form of complex number z is z rcos i sin where r z and arg z . wn z rcos i sin (given) rcos 2 k i sin 2 k OR 1 1 1 n n w z rcos 2 k i sin 2 k n n 2k 2 k r cos i sin (by De-Moiver’s n n theorem) 2k i n n w wk r e where k 0,1, 2,..., n 1
Note : It is sufficient to take k 0,1, 2,... n 1 since all other values of k lead to repeated roots.
Example : Find all the fourth roots of z1 i and locate these roots in plane. Solution : Let w4 z 1 i x 1, y 1 r x2 y 2 1 1 r 2 tan1y tan 11 tan 1 1 x 1 4 4 w 2 cos i sin (polar form) 4 4 2 cos 2k i sin 2 k 4 4 1 1 4 8 8k 2 k w 2 cos i sin 4 4
1 8 8k 2 k w 2 cos i sin where 16 16 Fourth roots of equations are 1 8 For k 0, w0 zcos i sin 16 16 1 8 9 9 k 1, w1 zcos i sin 16 16 19
1 8 17 17 k 2, w2 zcos i sin 16 16 1 8 25 25 k 3 , w3 zcos i sin 16 16 Which are the required four fourth root of z1 i Y
w1
9p 16 w 17p 0 16 O p 16 X
w 25p 2 16
w3
£- plane Fig 1.13
Example : Find all the fifth roots of z 32 and locate these roots in -plane. Solution : Let w5 z 32 x 32 and y 0 r x2 y 2 322 32
tan1y tan 10 tan 1 0 x 32 w5 32 cos i sin 1 1 5 wk 32 5 cos 2 k i sin 2 k 2k 2 k 2 cos i sin 5 5 For k 0, w0 2 cos i sin 5 5 3 3 k 1, w1 2 cos i sin 5 5 5 5 k 2, w2 2 cos i sin 2 cos i sin 5 5 20
7 7 k 3 , w3 2 cos i sin 5 5 9 9 k 4, w4 2 cos i sin 5 5 Y
w1
3p 5 w0 p p w3 5 X 7p 5 9p 5 w5
w4
Fig 1.14
Example : Solve z8 z 5 z 3 1 0 z8 z 5 z 3 1 0 z5 z 31 1 z 3 1 0 z51 z 3 1 0 Consider, z3 1 0 1 z3 1 w3 z 1 x 1, y 0 r x2 y 2 1 r 1 tan10 tan 1 0 and 1 w z3 1 cos i sin in polar 3 w z cos 2 k i sin 2 k 1 3 wk z cos 2 k i sin 2 k by De-Moivre’s theorem 2k 2 k wk z cos i sin where k 0,1, 2 3 3 21
For k 0, w0 cos i sin 3 3 3 3 k 1, w1 cos i sin cos i sin 3 3 5 5 k 2, w2 cos i sin 3 3 Now, consider z5 1 0 z5 1 1 w5 z 1 r 1 and w cos i sin w cos 2 k i sin 2 k 2k 1 wk cos i sin 2 k 5 where k 0,1, 2, 3, 4 5 2k 2 k wk cos i sin 5 5 For k 0, w0 cos i sin 5 5 3 3 k 1, w1 cos i sin 5 5 k 2, w2 cos i sin 7 7 k 3 , w3 cos i sin 5 5 9 9 k 4, w4 cos i sin 5 5 1 Example: Find all roots of 8 8 3i4 and represent them graphically. (2009)
Solution: Let z 8 8 3 i 2 r 8 1 3 16, tan1 3 3 1 4 1 z4 16 cos 2 k i sin 2 k k 0,1,2,3. 3 3 6k 1 6 k 1 2 cos i sin 12 12 22
13 7 for k 0 w0 2 cos i sin 12 12 12 12 7 7 for k 1 w1 2 cos i sin 12 12 12 13 13 19 for k 2 w2 2 cos i sin 12 12 12 19 19 for k 3 w3 2 cos i sin 12 12 1 Q. Find all the roots of 3i3 and locate them graphically.
1.3. EXTENDED COMPLEX PLANE, THE POINT AT INFINITY AND STEREOGRAPHIC PROJECTION :
Construction of the Stereographic Projection Map. (2012) Let be the Complex plane. Consider a unit sphere S (radius 1) tangent to at a point z 0. The diameter NS is perpendicular to and we call points N and S the north and south poles of the sphere S corresponding to any point z on the Complex Plane , we can construct a straight line NZ intersecting sphere S at a point PN .
N
S = The Unit Sphere
P Z O
x S Fig 1.15
Thus to each point of the Complex Plane , there corresponds one and only one point of the sphere S and conversely, to each point of the sphere S (except N), there corresponds one and only one point on the plane. For completeness, we say that the point N itself corresponds to the point at infinity of the plane . This one-to-one correspondence between the points of the plane and the points of the sphere S is called the Stereographic Projection. The sphere is called the Riemann Sphere (because Complex Number can also be represented by point on the Sphere.) 23
Suppose Complex Plane passes through centre of the unit sphere S. 2 2 2 Let x1 x 2 x 3 1 be the equation of unit sphere S. N 0, 0,1
Also, identify with x1, x 2 , 0 : x 1 , x 2 . Put z x, y and p x1,, x 2 x 3 . We will find equations expressing x1,, x 2 x 3 in terms of x and y.
3 The equation of straight line Nz in passing through points N and Z is given by
1 t z tN : t 1 t x , y t 0, 0, : t 1t x , 1 t y , t t ……(1) Straight line Nz intersects sphere S. 1 t2 x2 1 t 2 y 2 t 2 1 1 t2 x2 y 2 1 t 2 1 t2 x2 y 2 1 t 1 t 1 t2 z 2 1 t 1 t 1 t z 1 t
This equation holds if PN …( if PN then t 1 and z ) z 2 t z 2 1 t z 21 1 z 2 t z 2 1 t (for PN ) 1 z 2 2 2 2 z 1 1 z z 1 2 1 t 1 2 2 2 1 z 1 z 1 z Points N, P, Z are collinear. From equation (1), 24
2x z z x1 1 t x 1 z 2 1 z 2
2y i z z x2 1 t y (2) 1 z 2 1 z 2 z 2 1 x3 t z 2 1
Point z x iy corresponds to point P. 2 z z i z z z 1 P ,, 2 2 2 1 z 1 z z 1 Again from equation (2), x x x x x i x x 1 1 y 2 2 z x iy 1 2 1 t 1 x3 , 1 t 1 x3 , 1 x3 Point P x1,, x 2 x 3 S corresponds to point z.
x1 i x 2 z 1 x3
Note : From figure (Fig 1.20) 3 The straight line Nz in intersects sphere and in exactly one point PN . If z 1, then point P is in the Northern hemisphere and if z 1, then point P is in the southern hemisphere. Also, if z 1, then P z and as z , P approaches N.
Distance function : Let z and z be any two points on the Complex Plane . Suppose point z x, y corresponds to point P x1,, x 2 x 3 S . Suppose point z x, y corresponds to point P x1,, x 2 x 3 S . We define distance function as 2 2 2 dzzdPP ,, xx1 1 xx 2 2 xx 3 3 2 2 2 2 d z, z x1 x 1 x 2 x 2 x 3 x 3 2 2 2 2 2 2 xx11 2 xxxx 1122 2 xxxx 1233 2 xx 33 2 2 2 Sincex1 x 2 x 3 1 and x x x 1 x x x x x x 1 1 2 2 d z, z 2 2 x1 x 1 x 2 x 2 x 3 x 3 25
2 z z i z z z 1 Put x1 ,, x 2 x 3 1 z 2 1 z 2 z 2 1 2 z z i z z z 1 x1 ,, x 2 x 3 1 z 2 1 z 2 z 2 1 1 2 i z z 2 z z z z d z, z 2 2 2 2 2 1 z 1 z 1 z
i z z z 21 z 2 1 2 2 2 1 z z 1 z 1 1.4. SUMMARY
1) A Complex Number Z is an ordered pair x, y of real numbers. 2) The distance between Two Complex Numbers : Let Z1 x 1,,, y 1 Z 2 x 2 y 2 be two complex numbers. The distance between them in complex plane is given by 2 2 d Z1, Z 2 Z 1 Z 2 x1 x 2 y 1 y 2 3) If P is a point in the Complex Plane corresponding to Complex Number Z x iy x, y and let r, be the polar co-ordinates of point x, y from figure x rcos and y rsin where
r x2 y 2 is called the modulus or absolute value of Z (denoted by Z and
1 y tan is called the argument or amplitude of Z (denoted by x argZ ) Here is the angle between the two lines OP & the real axis (axis – X) 4) The modulus or absolute value of a Complex Number Z x iy is defined by Z x2 y 2 . 5) De-Moivre’s Theorem : If n is any integer or fraction then cos i sinn cos n i sin n n 6)Theorem : In , given z 0 , the equation expansion w z has 2k i n n n-distinct solution given by wk r. e , k 0,1,..., n 1 where r z and arg z . 26
1.5. UNIT END EXERCISES :
1) Find two square roots of 2i . 2 (Hint: Let x iy be a square root of 2i x+iy zi . x2 y 2 2 xy i 2 i , comparing real and imaginary parts on both the sides, weget two equations in x,y. x2 y 2 0; 2xy 2 . x iy 1 i or x iy 1 i
2) Describe the set z: z 1 1 in the Complex plane ℂ. Solution: Let z x iy; x Re( z ), y Im( z ) z1 x iy 1 ( x 1)2 y 2 . 2 Hence z 1 1 describes all real number pairs (,)x y in such that x12 y2 1. (x 12 ) y 2 1 This is an equation of the open disc with centre at and radius equal to 1,which can be described as follows :
r = 1 B ((-1,0),1) (-1,0)
3) Find polar form of the Complex Number 1 i .
4) Show that the n th roots of 1 satisfy the “ cyclotomic “ equation zn1 z n 2 z 1 0 . (Hint : Use the identity zn1 z 1 z n1 z n 2 ... z 1 . (2009) 27 2
SEQUENCES OF COMPLEX NUMBERS
Unit Structure : 2.0 Objectives 2.1. Introduction 2.2. Convergent Sequences 2.3. Topological Aspects of the Complex Plane (Limits, Continuity, Uniform Continuity) 2.4. Summary 2.5. Unit End Exercises
2.0 OBJECTIVES
This unit shall make you understand : Cauchy and convergent sequences of complex number . The connection between the convergence of real and imaginary parts of a sequence zn x n y n , namely xn and yn with the convergence of zn in . We shall also see that under what conditions a given sequence of complex number zn x n y n is a Cauchy sequence. Can we relate to
our findings for real values sequence xn and yn .
2.1. INTRODUCTION :
We have already associated the meaning to a sequence of real numbers as a function , a: Z , denoted by (a ( n ))n N . On a similar line , we shall define a sequence of complex numbers , where each term of a sequence is a complex number . For 1 example zn is a sequence of Complex Numbers with terms zn 1 1 1 , , ,...etc. In this Unit, we shall consider the topological z1 z 2 z 3 aspects of the Complex plane. The concept of absolute value can be used to define the notion of a limit of a sequence of complex 28 numbers . We shall begin with the definition of a complex valued sequence .
Definition: A function whose domain is a set of natural number and range is a subset of ,is saidtobe Real sequence.
Any function whose domain is a set of nature numbers and range is subset of complex numbers , is said to be complex sequence.
th Generally, we denote it by zn. zn is the n term of the sequence. e.g. 1) The set of numbers i, i2 , i 3 ,..., i 200 . This is finite sequence th n and its n term is zn i , n 1, 2,..., 200 2 3 2 i 2 i 2 i 2) The set of numbers , , , ... 2 3 n th 2 i It is the infinite sequence and its n term is z . n n Sequences : Definition : A function whose domain is a set of natural number and range is a subset of , is said to be Real sequence.
Any function whose domain is a set of nature numbers and Range is subset of , is said to be Complex sequence.
th Generally, we denote it by zn . zn is the n term of the sequence. e.g. 1) The set of numbers i, i2 , i 3 , ..., i 200 .This is finite sequence th n and its n term is zn i , n 1, 2,..., 200 2 3 2 i 2 i 2 i 2) The set of numbers , , , ... 2 3 n th 2 i It is the infinite sequence and its n term is z . n n 2.2 CONVERGENT SEQUENCES :
A sequence zn is said to converge to a point z0 [or a sequence zn has to limit z0 ] if for every 0 , there is an N s.t.
zn z0 n N and we write lim zn z0 . n 29
Geometrically, zn z0 if every -nbd of z0 contains almost all terms of the sequence zn .
z1 e z4 z2 z 5 z3 z0
Fig 2.1
Divergent Sequences: A sequence is said to be divergent if it is not convergent.
Theorem: Prove that any convergent sequence has a unique limit.
Solution : Let lim zn z1 and lim zn z0 n n z z If z z then for 1 0 0 1 0 2
N1 s.t. n, zn z1 / 2 and N2 s.t. n, zn z0 / 2 choose N = max NNNN1, 2 1 and NN 2
zN z1 2 and zN z0 / 2
z1 z 2 z 1 zNN z z 0
0 z1 z 0 a contradiction.
z1 z 0 .
Theorem : Suppose zn x n iy n and z0 x 0 iy 0 then lim zn z0 n iff lim xn x0 and lim yn y0 . n n
Proof : Suppose lim zn z0 0,, an integer N s.t. n zn z0 n N .
Now, zn z0 x n iy n x 0 iy 0 xn x0 i y n y 0 x x n 0 n N and yn y0 30
lim xn x0 and lim yn y0 n n Conversely, suppose lim xn x0 and lim yn y0 0,N1 n n and N s.t. x x n N 2 n 02 1 and y y n N n 02 2 Choose NNN max 1 , 2 if n N then zn z0 x n iy n x 0 iy 0 xn x0 y n y 0 Rez z and Im z z 2 2 zn z0 lim zn z0 n
Theorem : If lim zn z0 , then lim zn z0 and the sequence n n zn is bdd. Proof : Suppose lim zn z0 0 , an integer N s.t. n zn z0 n N . zn z0 z n z 0 zn z0 n N (1)
lim zn z n n from equation (1) z0 zn z 0 Sequence zn is bounded.
Example : If lim zn z0 and lim wn w0 prove that n n i) lim zn w n z0 w 0 n ii) limzn . w n z0 w 0 n
zn z0 iii) lim provided w0 0 n wn w0
Definition : Cauchy Sequence : A sequence zn is said to be a Cauchy sequence for every 0 there is an integer N s.t.
zn z m n N , and m N . 31
Note : From equation (1) zm z n , n N Put m n p for p 1, 2, 3, ...
zn p z n n N and p 1.
Theorem: Every convergent sequence is a Cauchy Sequence. Proof : Suppose zn is a convergent Sequence.
A sequence zn has a limit of z0 . lim zn z0 n For every 0 , there is an integer N s.t. z z n N n 0 2 If m N and n N then z z m N m 0 2 z z n N n 0 2 z z z z z z z z z z n m n0 0 m n0 0 m 2 2 zn z m if n N and m N zm is a Cauchy sequence.
Theorem : is complete. [i.e. T.P.T. every Cauchy sequence in is convergent.] Proof : Let zn x n iy n be a Cauchy sequence in . xn and yn are Cauchy sequence in . is complete. xn x0 and yn y0 for x0, y 0 limzn lim x n iy n limxn i lim y n x0 iy 0 z0 n n n n sequence zn is convergent. Hence is complete. Note : A sequence is convergent iff it is a Cauchy sequence (Cauchy Criteria for convergence of a sequence.)
Theorem: Let ()a be a sequence of positive real numbers. If n n
an1 lim l 1, then lim an 0. n0 n 0 an 32
l 1 Proof: We have l 1, and by data there exist m such that 2 a l 1 l n1 for all n m an 2 l 1 Put r= . Then 0 < r < 1. Then 2 2 am1 a m, a m 2 a m r r a m r r a m r and so on. We get a a a rk k . Put c m . Then 0 a crn n . Since m k m r m n n 0 r 1, cr 0 as n . So an 0 as n .
2.3 TOPOLOGICAL ASPECTS OF THE COMPLEX PLANE
Topology in the -plane : A function , z, z z z has the following properties. i) z z 0 , if z z and z z 0 if z z ii) z z z z iii) z z z w w z z,, z w
Thus, is a metric space with Euclidean metric (distance) defined by d z1,,, z 2 z 1 z 2 z 1 z 2 1) Let z0 and 0 , then the set B z0,: z z z 0 is called an open disk or open ball with centre at z0 and radius (This is also called the -nbd of z0 or nbd z0 ). Geometrically, B z0, is an open disk, consisting of all points at a distance less than from the point z0 .
Y
e
Z 0 e
X O Fig 2.2 33
2) A set of the form, B z0, z 0 z :0 z z 0 is called the deleted neighbourhood of z0 or punctured disk. 3) The set of the form B z0,: r z z z 0 r is the circle with centre at z0 and radius r and is called the boundary of circle. 4) Let G , A set of G is said to be open in if for every z0 G, r 0 , s.t. B z0, r c G
Y
r
Z 0
X O £ - plane
Fig 2.3 e.g. i) Interior of circle is an open set ii) The entire plane is an open set iii) Half planes: Rez 0, Re z 0, Im z 0, Im z 0 are open set. Y Y
X Re z>0 £ - plane Im z<0 X £ - plane
Fig 2.4
Thoerem :.Any open disk is an open set
Proof : Let z0 , r 0 and B z0,: r z z z 0 r be an open disk. 34
Let a B z0, r a z 0 r (i) If z a d then
z z0 z a a z 0 z a a z0 r (From (i)) r z z0 r i.e. z B a,, z B z0 r
B a,, B z0 r Any open disk is an open set
c 5) The complement of a set S is denoted by S , and defined by c S z: z S c 6) A set F is said to be closed if its complement i.e. F is open. OR A set F is said to be closed if it contains all its limit points. 7) A set of the form B z0,: r z z z 0 r is called the closed disk or closed ball. e.g. i) is closed set ii) is closed set iii) E z:Im z 4 iv) S z : z 2 v) S z: z 2 z 8) Interior point : Let S , then the point z S is said to be an interior point of set S if r 0 s.t. B z, r S . 9) The point CS is said to be exterior point of the set S if a B c, r which does not contain any point of set S. 10) A point p S is said to be a boundary point of set S if it is a neither a interior point nor an exterior point. Y
Exterior
Interior Boundary
X O £ - plane
Fig 2.5 35
11) The set of all interior point of the set S is said to be interior of S. 12) A set G is said to be open if each point of G is an interior point of G. 13) Closure set : The closure of the set S , denoted by Cl S Cl S S S (where S is boundary element is always closed.) 14) A subset S of is said to be Dense if Cl S e.g. i) is dense in . ii) x iy x, y is dense in . 15) An open set G is said to be connected if for any two points z1 and z2 can be joined by a curve that lies entirely in G. OR A metric space X, d is said to be connected if the only subset of X which are both open and closed are X and ()the empty set .
Y
B(0,1) a X b
£- plane
Fig 2.6 e.g. 1) Open disk is a connected set. 2) The unit disk B0,1 z z 1 is a connected set. 3) The annulus B z:1 z 2 is connected Fig. 2.6 4) The set S z: z 2 1 or z2 1 is not connected Fig. 2.6(b).
Y Y
z 1 X X -3 -2 -1 0 1 2 3 z 2
- plane £- plane £
fig (a) fig (b) z 2 1 z 2 1 Fig 2.7 36
16) A domain is an open connected set.
17) A domain together with some none or all of its boundary point is referred to as a region. 18) Bounded Set : A set is said to be bounded if R 0 s.t. S B O,: R z z R . 19) A set which is closed as well as bounded is called compact set. 20) A set that cannot be enclosed by any closed disk is called unbounded set. 21) Let z1 and z2 . These we denote the line segment from z1 to z2 by z1, z 2 1 t z 1 t z 2 : 0 t 1
Y
z2
z1
X
Fig 2.8
Function, limits and continuity : Definition : Let A and B be two non-empty subset of complex numbers. A function from A to B is a rule, f, which associates to each z0 x 0 iy 0 A a unique w0 u 0 i v 0 B
The number w0 is the value of f at z0 and we write f z0 w 0 . If z varies in A then f z w varies in B. We say that f is a complex valued function of a complex variable. Here w is the dependent and z is the independent variable. Let f: A B be a function and SA then f( S ) f ( z ) /, z S } where f S is called the image of S under ‘f’ and the set R f()/ z z A is called range of ‘f’.
Single and Multiple Valued Function : Let z 0, then we write the polar form of a complex number z is z rei where r z and , i.e. z z r, rei . 37
Y
z
q X 0
Fig 2.9 If we increase to 2 i2 i 2 i i z r, 2 re re. e re z r, returning to its original value.
Definition : A function f is said to be a single valued if f satisfies f() z f ((,)) z r f (, r 2) . Otherwise, f is said to be a multiple valued function. n e.g. f z z, n is a single valued function. n Solution : f z f z r, rei n i n2 n i n2 n n in 2 in f z r, 2 r e r e r e. e n in 2 in r e e1, n n rei f z r, n Note : If n then f z z is a multiplied valued function. 2i n e 1, when n Let f: A B be a function. i) If the elements of A are complex numbers and those of B are Real Numbers then we say that f is a real valued function of complex variable. ii) If the elements of A are Real Numbers and those of B are complex numbers then we say that f is a complex valued function of real variable.
Let f : be a function then the graph of f is a subset of and it is two dimensional object and we can represent it very well on the two dimensional page. However the graph of the f : is a 38 subset of ( Cartesian product) i.e. a four dimensional object and we cannot represent it on two dimensional plane. In this case we consider two plane, one plane is z-plane and other one is w-plane.
Y V
.Z
X U plane plane z plane plane
Fig 2.10 Limit Point : Let D be a subset of i.e. D then we say that a point z0 is a limit point of D if every neighbourhood of z0 contains a point of
D other than z0 i.e. ((,))B z0 r z 0 for any r 0 .
Definition : Let f be a complex valued function defined on D and let z0 Cl D. We say that a number is a limit of f z as z z0 and we write lim f z . z z0 iff 0, 0 s.t. f z whenever z D and 0 z z0 . f z B, wherever z D B z0, z 0
z d z0
f (z0 )¹ l
Fig 2.11 39
Note : 1) f may not be defined at z z0 2) z0 need not be in D. 3) even if z0 D , f z0 4) In real variable theory, if x0 then x x0 has only two possible ways, either from left or from right. In complex case, z z0 , in any manner in the Complex Plane.
Theorem : Let f be a complex valued function defined on D and let z0 Cl D. If lim f z exists, then this limit is unique. z z0 Proof : Let lim f z 1 and lim f z 2 z z0 z z0 T.P.T. 1 2 By definition for a given 0 , 1 0, 2 0 f z s.t. 1 2 , whenever z D B z0, 1 z 0 and f z , whenever . 2 2 z D B z0, 2 z 0 Let min1 , 2. If z D B z0, z 0 then
1 2 1 f()() z f z 2 1 f z f z 2 2 2 is arbitrary. 1 2 i.e. limit is unique.
Theorem : Let f be a complex valued function defined on D. suppose, f z u x,, y iv x y , z0 x 0 i y 0 , w0 u 0 iv 0 and z0 Cl D.
Then lim w0 iff limu x , y u0 and limv x , y v0 . z z0 z z0 z z0 Proof : Direct part – Let limf ( z ) w0 and w0 u 0 iv 0 z z0 By definition, 0, 0 s.t. f z w0 whenever z D B z0, z 0 . Now, fz w0 uxy ,, ivxy u 0 iv 0
u x,, y u0 i v x y v 0 40
u x, y u Rez Re z z 0 v x, y v0 Imz Im z z If z D B z0, z 0 then u x, y u0 and
v x, y v0
limu x , y u0 and limv x , y v0 z z0 z z0 Conversely, assume that lim u(,) x y u0 & lim v(,) x y v0 z z0 z z0 By definition given 01 , 2 0. s.t. u( x , y ) u0 / 2 and v( x , y ) v0 / 2 whenever z z0 1 and z z0 2 . let min1 , 2.
whenever z z0 ,
Consider fz()(,)(,) w0 uxy ivxy u 0 iv 0 u(,)(,) x y u v x y v 0 0 2 2 lin f() z w0 z z0
iz Examples : If f() z in the open disk B0,1 , prove that 2 iz i lim . z1 2 2 iz Solution : Given f() z 2 iz i We must prove that for every 0, 0 , s.t. 2 2 whenever z B0,1 and 0 z 1 , f z iz i i if 0 z 1 , then z 1 2 2 2 2 iz i choosing 2 , we see that whenever z B0, 1 and 2 2 0 z 1 . iz i lim z1 2 2 41
i f (z)= iz 2 f (z)= 2 1 z 1 d = 2Î 0
z - plane w - plane
Fig 2.12
zi Problem : If f z in the open disk z 1 prove that 2 i lim f z . z1 2 iz Solution : Given f z 2 We must prove that for every 0 , for given any 0 we can find iz i 0 s.t. whenever 0 z 1 . 2 2 If 0 z 1 iz i i z 1 2 z 1 2 2 i 1 iz i Choosing, 2 , we see that , whenever 0 z 1 2 2 iz i lim z1 2 2
f (z) i 2 e
z d = 2Î 0 1
w - plane z - plane
Fig 2.13 42
z Problem : Prove that lim does not exist. z0 z Solution : We know that the function f z (a unique limit) as z z0 in any manner in the -plane. z Let f z z Let z 0, along the real axis. y 0, z x z x iy z x lim f z lim lim 1 z0 z 0z z 0 x Let z 0, along the imaginary axis. x 0, z iy z x iy z iy lim f z lim lim 1 z0 z 0z z 0 iy limit is not unique along real and imaginary axis. z lim does not exist. z0 z
Problem : If f() z z2 , prove that limf ( z ) z2 . z z0 Solution : Let 0 given, to find 0 s.t. z2 z 2 whenever
0 z z0 2 2 consider z z ( z z0 )( z z 0 )
z z0 z z 0 z z0
z z0 2 z 0 z z0 2 z 0 . 2 z0 Choose 0 s.t. min ,1 1 2 z0 z2 z0 2 .
limf ( z ) zo2 . z z0
Theorem : Let f and g be defined in the neighbourhood of z0 except possibly at z z0 . If lim f z and lim g z m z z0 z z0 Then 1) lim f z g z m z z0 2) lim f z . g z m z z0 43
f z 3) lim z z0 g z m
Continuity Definition : A function f: D is said to be continuous at a point z0 D , iff lim f z exists and lim) f ( z ) f ( z0 ) . z z0 z z0
OR Definition :A function f: D is said to be continuous at a point z0 D iff 0, 0 s.t. f z f z0 whenever z D and z z0
OR Definition : A function f is said to be continuous at a point z0 D iff the following 3 conditions hold true : i) f is defined at z0 i.e. f z0 exists. ii) lim f z exists z z0 iii) lim f z f z0 z z0
OR Definition : A function f: D is continuous or f is continuous on D if f is continuous at every point of D.
Example : If f z z2 then prove that f is continuous at a point z i . 2 Solution : Given, f z z, z0 i f i i2 1 2 2 limz 1 z i limz2 1 f i z i f is continuous at a point z i .
2 z z i Example : Let f z prove that f is not continuous at 0z i a point z i . 44
Solution : f i 0 (given) 2 lim f z lim 1 z i z i lim f z 1 f i z i f is not continuous at z z0
z2 i Problem :Discuss the continuity of f() z at z e 4 z4 3 z 2 1 i i Solution : z e4 z2 e 2 i z 4 1 i f() z 1 1 3i 1 3 i the limit exist z e 4 . i f() z is continuous at z e 4
Uniformly Continuous : A function f: D is said to be uniformly continuous on D iff the following conditions holds for every 0, s.t. for any two points z1 and z2 in D.
z1 z 2 f z 1 f z 2
Example : Let f z z2 in the open disk B0,1 , prove that f is uniformly continuous on B0,1 . Solution : Given, f z z2 . We must prove that for a given 0 , we can find 0, s.t. for any two points z1 and 2 in B0,1 and 2 2 z1 z 2 f z 1 f z 2 z1 z 2 z1 z 2 z 1 z 2
z1 z 2 by triangle inequality
2 z1 1 and z 2 1 Choosing , we see that z2 z 2 whenever z z , 2 1 2 1 2 z1, z 2 B 0,1 . f is uniformly continuous on B0,1 .
Definition : Unbounded set : A set E is said to be unbounded if R 0 s.t. z E v z E . 45
Definition : Limit at Infinity Let f be defined on an unbounded set E. If for each 0 , R 0 s.t. f z whenever z E and z R then we say that f z as z and we write lim f z . z 1 1 e.g. lim 0 for given 0 above R 0 s.t. R z z z 1 1 1 0 z R z z
Infinite Limit : Let f be defined out D except possible at z0 D . If for every R 0, 0 s.t. f z R whenever 0 z z0 then we say that f z as z z0 and we write lim f z . z z0 1 e.g. lim z1 z2 1
2.4. SUMMARY
1) Let f be a complex valued function defined on D and let z0 Cl D. If lim f z exists, then this limit is unique. z z0
2) Let f and g be defined in the neighbourhood of z0 e except possibly at If lim f z and lim g z m z z0 z z0 Then 1) lim f z g z m z z0 2) lim f z . g z m z z0 f z 3) lim z z0 g z m
3) A function f is said to be continuous at a point z0 D iff the following 3 conditions holds i) f is defined at z0 i.e. f z0 exists. ii) lim f z exists z z0 iii) lim f z f z0 z z0 46
4) A function f: D is said to be Uniformly continuous iff the following conditions holds for every 0, s.t. for any two points z1 and z2 in D then z1 z 2 f z 1 f z 2
5) Every convergent sequence is a Cauchy Sequence.
6) is complete.
2.5. UNIT END EXCERCISES :
1)Find the limit of a sequence for z 1. n n Solution : Consider zn 0 z 0 z 0 as n , for z 1.
n 2) Check whether the sequence z is convergent or not. n n i n n i Solution : zn , then an z 2 5 , because n i 3 4i n i 1 1 0 as n . n i() n i n2 1
3) Which of the following subsets of are connected , if not connected then what are it’s components ? (a) X z: z 1 Ans : X is connected . (b) X z: z 1 z : z 2 1
Ans: X is not connected , because X z: z 1 z : z 2 1 is a disjoint union of nonempty closed subsets ( Components) of X.
4) Let zn, z be points in and let d be the metric on . Show that zn z 0 if and only if d zn, z 0 as n . 2 z z (Hint: For z, z , d(,) z z and ((1 z2 )(1 z 2 )) 2 d(,) z 1 2 (1 z ) 2
5) Let P() z be a nonconstant polynomial in . Show that P() z as z . 47
6) Suppose f: X is uniformly continuous , show that if
xn is a Cauchy sequence in , then f() xn is a Cauchy sequence in Ω.
7) Show that if f and g are bounded uniformly continuous functions from X into then fg is also bounded and uniformly continuous function from X into .
(Hint: (fgx )( ) ( fgy )( ) fxgx ( ) ( ) fygy ( ) ( ) fx()()()()()() fygx fygx gy .)
8) Verify the continuity of the following function f of the extended 3 complex plane at the point a (2012) 4 3 f() z if z 4 z 1 3 = if z 4z 3 4
48 3
SERIES OF COMPLEX NUMBERS
Unit Structure : 3.0. Objectives 3.1. Introduction 3.2. Convergence of Series 3.3. Tests for determining the Convergence of Power Series 3.4. Summary 3.5. Unit End Exercises
3.0 OBJECTIVES:
This unit shall make you construct a series of complex numbers by understanding the definition of a series of real’s . Basically, we are going to define a power series of the n form n0 an z . We shall check for the conditions , under which the given power series is convergent or not.Hence, we shall employ certain tests in order to determine the convergence of the given power series.
3.1 INTRODUCTION :
An infinite series of real’s is the expression of the form k1ak , where ak is a real number for all k 1 . Similarly we construct an infinite series of complex numbers as n1 zn , where zn s are complex numbers for all n 1. For example k i k1 is an infinite series of complex numbers. To check k2 i whether the sum exists or not, in other words whether a given series of complex numbers is convergent or not, we employ certain tests for convergence and we shall convert the given problem of checking convergence of the series of complex numbers to checking convergence of the series of real k numbers. For example, i is a convergent series of real k1 k2 i 49
k numbers, because i 1 and we know that 1 k1 2 4 k i k 4 1 k 1 converges. Let us start with defining a series of complex numbers.
3.2 COVERGENCE OF SERIES :
Definition: Let zn be a sequence of complex numbers, Form a new sequence defined by S1 z 1, S 2 z 1 z 2 , ..., Sn z 1 z 2 ... z n , th where Sn is called the sequence of n partial sums of sequence zn. The sequence Sn is symbolized by z1 z 2 ... zn called an n1 Infinite series. If lim SSn exists then the series is said to be convergent and S is n its sum i.e. zn lim z k lim S n S n1n k 1 n
A series is said to be divergent if it is not convergent sequence. (The necessary condition for the convergence of the series.)
Theorem : If the series zn is convergent then limzn 0 . n1 n Proof : Given series is zn . n1 Let Sn z1 z 2 ... z n 1 z n ………..(1) be the nth partial sum of series. Given that the series is convergent Let S be the sum zn n1
lim SSn n from equation (1) zn S n S n1 z1 z 2... zn 1 S n 1 Taking limit on both sides, limzn lim S n lim S n1 S S n n n
limzn 0 n Consider the infinite series zn z0 z 1 ... n0 50
If Rn z n z n1 z n 2 ... then Rn is called remainder of infinite series. If S is sum of infinite series then SSR n n or RSSn n .
Theorem : A series zn of complex terms is convergent iff for n0 every 0 , an integer N s.t. zn z n1 ... z n p n N and p 0 . (Cauchy criteria for convergence of series) Proof : Suppose zn is convergent n0 th Let Sn z0 z 1 z 2... z n 1 be the n partial sum of series and let S be a sum of series. lim SSn n 0, an integer N s.t. SSn (1) n N Let Rn z n z n1 z n 2 ... be the remainder of an infinite series. SSR n n or RSSn n From equation (1) Sn S S S n R n n N i.e. zn z n1 z n 2 ... n N
zn z n1 z n 2 ... z n p n N and p 0
Converse : Given 0 , there is an integer N s.t.
Zn z n1 z n 2 ... z n p ………………….(2) n N and p 0 We know that, zn z0 z 1 ... z n ... n0 If S is its sum then we write SSR n n RSSn n RSSn n
But Rn z n z n1 ... z n p given from (2) n N and p 0
Sn S n N lim SSn n zn is convergent. n1 51
Definition : Let zn . For every n 0 the series zn converges n1 n to z0 iff for every 0, an integer N s.t. zk z0 n N . k1 Definition : A series zn converges absolutely if zn converges. n1 n1 Proposition : If the series zn converges absolutely then zn n1 n1 converges. Proof : Let 0 , consider an infinite series zn . n1 Let Sn z1 z 2 ... z n be the partial sum of series given that zn n1 convergent absolutely. For a given 0, an integer N s.t. zk n N (1) k n1 If n m N then n n Sn S m z m1 z m 2 ... z n zk zk k m1 k n1 zk from (1) k n1 Sn is a Cauchy sequence. Sn is a convergent Sequence. ( by Cauchy criteria) z0 s.t. lim Sn z0 n Thus, zn is convergent. n1 Examples : 1 1) Prove that zn 1 z z2 ... if z 1 n0 1 z 2 Solution : Given zn 1 z z ... n0 2n 1 Let Sn 1 z z ... z 2 3 n z Sn z z z ... z (multiplied by z, z 0 ) 52
1 zn S z S1 zn S 1 z 1 z n S n n n n 1 z (z 1) 1 T.P.T. S ……..(1) n 1 z i.e. T.P.T. limzn 0 n Given any 0 , we must find integer N s.t. zn n N [If z 0, then the result is true]. Let z 0 zn z n nlog z log log n { log z is negative when z 1} log z log Choosing N , we see that zn n N log z limzn 0 n From equation (1) n 1 z 1 n limSn lim z 0 as n n n 1 z 1 z 1 Hence, zn when z 1 n0 1 z
Note : 1) Geometric series (G.S.) zn is cgt when z 1 and n0 divergent when z 1.
Uniformly converges for series : For each n , let fn () z be a complex function of complex variable.The series fn () z converges to f z point wise for each z D iff fn ()() z f z and for each z D [This means that for each z D and for each 0, an integer N (depends on z) and ], s.t.
Sn z f z n N .
Definition : The series fn () z is said to be uniformly convergent on D to f(z) if for every 0 , an integer N (depends only on ) s.t. Sn z f z z D , and n N . A power series about 53 z0 is an infinite series of the n 2 form an z z0 a 0 z z 0 a 1 z z 0 a 2 + ………, where n0 constants an and z0 are called complex numbers and z is a complex variable. n Note : If z0 0 then an z a0 a 1 ... n0 This is power series about origin (i.e. z 0)
e.g. Geometric Series (G.S.) zn 1 z z2 ... n0 1 i) If z 1 then lim Sn and the G.S. converges with n 1 z 1 zn . n0 1 z ii) If z 1 then lim Sn and the G.S. diverges. n
3.3. TESTS FOR DETERMINING THE CONVERGENCE OF A POWER SERIES :
Weierstrass M-test : Statement : Let fn : D be a complex function defined on D s.t. fn z M n z D and n N . If Mn is convergent 1 series of positive real numbers then series fn is uniformly 1 convergent. Proof : Given fn z M n v z D and n N (1) Let 0 Given that Mn is convergent 1 an integer N s.t. Mn n N (2) k n1 Given series fn z 1 Let Sn z f1 z f 2 z ... f n z if n m N , then 54
Sn z S m z f m1 z f m 2 z ... f n z n n fk z fn z k m1 k m 1 n Mk from (1) k m1 n Mk from (2) k m1 Sn z S m z , n, m N
Sn z is a Cauchy sequence. Sequence Sn z is a convergent sequence w s.t. lim Sn z w n Define w f z this gives a function f: D for each z D and for each n N Sn z f z f z S n z .
Sn z f z f z S n z Sn1 z f n 2 z ... fk z fk z k n1 k n 1 Mk from (1) and (2) k n1 Hence Sn z f z z D where n Series fn is uniformly convergent on D. 1 zn Examples : 1) Prove that the series is uniformly n n1 n 1 convergent on a set D z: z 1. zn Solution : Given series n n1 n 1 zn Let fn z n n 1 n z 1 fn z z 1 n n 1 n n 1 1 M n n 1 3 n n 2 55
1 Mn is a p-series and it is convergent 3 n 2 3 p 1 2 By Weierstrass M-test, zn The given series is uniformly convergent. n1 n n 1
2) Given series zn 1 z n1 i) Prove that the series converges for z 1 and find its sum. ii) Prove that the series converges uniformly to the sum z for z 1 2 . iii) Does the series converges uniformly for z 1? Explain. Solution : i) The given series is zn 1 z n1 zn 1 z z 1 z z2 1 z z 3 1 z ... n1 z z2 z 2 z 3 z 3 z 4... zn z n 1 z zn1 z1 z n n1 Let Sn z z z We must prove that given any 0 , we can find an integer N s.t. Sn z z n N n1 n1 n1 Sn z z z z z z z n1 log z log log log n1 , z 0 n 1 N log z log z log Choosing, N 1 (1) log z
SZn n N n1 limSn lim z z z n n Hence, the series converges for z 1 and z is its sum. 56 ii) Since from (i) the series converges to some z for z 1 and 1 hence it converges for z 2 log N 1 from (1) log z 1 log log If z , then N 1 is the largest value of 1. 1 2 log log z 2 log Sn z , n N 1 where N depends only on log 1 2 and not on z. 1 The given series converges uniformly to sum z for z . 2 iii) If z 1 log N 1 from (1) log1 Hence, the series does not converges uniformly for z 1.
Ratio Test for series (2012) Statement: Let zk be an infinite series for non-zero complex term s.t.
zN1 z N 2 z n n N zn z N...... z N zN z N1 z n 1 = L then p zN z n 0 z n p1 i) If L 1, the series converges absolutely. ii) If L 1, the series diverges. iii) If L 1, the series may converge or diverge.
Proof: Suppose L<1. Then for with L< <1, there exist an integer N s.t. z n1 < n N so that, zn
zN1 z N 2 z n n N zn z N...... z N zN z N1 z n 1 p zN p z N for p 1. p Since zN is a convergent (geometric series), zN is p1 n1 covergent by comparision test. This proves (i) 57
z If L>1, then there an integer N s.t. L>k>1 and n1 > k for all zn n N
zN1 z N 2 z n n N for all n > N zn z N...... z N k zN z N1 z n 1
Hence, zn 0 as n and so zn diverges. Hence (ii).
Example: zn 1) Prove that the power series converges for all values of z. n0 n! zn Solution : Given power series is n1 n! 1 1 Here, a a nn! n1 n 1 ! 1 a (n 1)! 1 limn1 lim lim 0 1 n an n 1 n n 1 n! . Therefore the the series is convergent.
Comparison Test : If the series vn converges and un v n then un converges absolutely. Also, un converges.
Abels’ theorem : n Statement : If the power series an z converges to particular value z0 0 of z then it converges absolutely z s.t. z z0 . n Proof : Given that the power series an z us converges for a particular value z0 0 of z. n an z0 converges. n liman z0 0 n n sequence an z0 is bounded. n positive number M s.t. an z0 M M a n n z0 58
n n z or an z M n z0 n z z M. is a geometric series and convergent for 1 z0 z0 i.e. z z0 n By comparison test an z converges absolutely for
z z0 .
Cauchy-Hadaward Theorem : n Statement: For a given power series an z define a number R, n0 1 1 0 R , by lim .sup a n then R n i) If z R , then the series converges absolutely ii) If 0 r R , then the series converges uniformly on z: z r iii) If z R , then the series diverges [Here R is radius of converges of power series.] (2008)
1 1 Proof : Given, lim sup a n (1) R n
[Note : A number L is said to be a limit superior of the sequence un if infinitely many terms of the sequence un are greater than L , while finite number of terms greater than L where 0 .] i) Let z R , then r 0 s.t. z r R 1 1 r R By definition of the lim sup and from equation (1) 1 an rn n z an z n n N r n z z is a G.S. and it is convergent for 1 i.e. r r z r . 59
By comparison test, n an z converges for z R . n an z converges absolutely for z R . n0 ii) Let 0 r R choosing r, 0 r r R . By using part (1), we have 1 1 1 an a n v n N n n r 1 an n r n n n z r an z r r n r is a G.S. of positive real numbers and it convergent r for r r . n By Weierstrass M-test, the power series an z converges n0 uniformly on z: z r. iii) Let z R , then r 0 2 r R 1 1 . r R By definition lim sup and from equation (1) 1 1 an n N n r 1 an rn n n n z r an z 1 rn r n n an z1 n N n an z 0 as n { zn converges then limzn 0, z n 0 as n } n n Power series an z is divergent for z R . n0 60
Definition : The radius of convergence R of the power series n an z is defined as R = sup { r the series converges n0 v z satisfying z r }. n i) If R 0 , then power series an z cgs only for z 0. ii) If R , then the power series converges , for all values of z. iii) If 0 R , the power series converges for all z, fn z R and
diverges v z, fn z R . The power series may converge or diverge on the circle z R . The circle z R is then called the circle of convergence.
n Note : If an z is power series with radius of converges R then n0 a R lim n ,provided this limit exist. n an1
Theorem : If an 0 for all but finitely many values of n then the n radius of convergence R of an z is related by following, n0 a1 a lim inf n1 lim sup n 1 . an R a n a In particular, if lim n1 exists, then n an
1 an1 1 lim lim sup an n . (2007,2008,) R n an n Proof : Given R is radius of convergence of the series an z . n0 an1 an1 Suppose, lim sup L and lim inf an an By the definition of limit sup, an N s.t. a n1 L n N an a a a N1 LLL , N 2 ,..., n aN a N1 a n 1 61
Multiplication of these inequalities gives a a a N1 N 2 ... n L n N aN a N1 a n 1 a n L n N aN n N an a N L 1 1 n N a n a L n n N 1 1 N an a L n L n N is arbitary as n , we get 1 lim sup an n L (1) n limn p 1 n a if p 0 Similarly, by the definition of lim in f 1 lim inf an n (2) From equation (1) and (2), we get 1 1 lim inf an n lim sup a n n L 1 1 1 L lim sup an n R R
an1 If lim exists then L n an 1 1 1 lim inf a n lim sup a n R n n 1 an1 1 lim lim sup an n R n an n a Note : R lim n n an1
n n1 Theorem: Let an z be a power and nan z be the power series n obtained by differentiating an z term by term. Then the derived series has same radius of convergence as the original series. (2009) Proof: Suppose for R and R’ be the radii of the convergenve of the n n1 series an z and nan z respectively. Then we have, 62
11 1 1 1 liman and lim nn a n RRn n 1 In order the desired result we have to show that limn n 1 n
1 Suppose n n =1+h. Then we have
n( n 1) 1 n = (1+h)n=1+nh+ h2 ...... hn > n( n 1) h2 i.e. h2 2! 2 2 < n 1
1 Thus ln1 i i so that limn n 1. 2 2 n
Hence R=R’
n Proposition : Let f z an z have radius of convergence R 0 n0 then for each K 1, the series n K n n1 n 2 ... n K 1 an z has radius of convergence R n K . (2009) Proof : Let R be a radius of convergence of the power series n f z an z . n0 n! Let R be a radius at cgs. of a power series . n K n K ! T.P.T. RR 1 1n ! n lim sup an R n k ! 1 n! n 1 limsup .limsup a n n k ! n 1 1n !n 1 lim sup . ….(1) R n k! R 1 1 lim sup an n R 63
1 an1 Now, lim supan n lim n n an 1 n! n n1 ! n k ! lim sup lim n k! n n 1 k n ! n1 n ! n k ! lim . n n1 k n k ! n! 1 n 1 1 lim lim n n n1 k n 1 1 k n n Substituting in equation (1), we get 1 1 1. RR RR
zn Example: 1) Prove that the power series converges for all n0 n! values of z. zn Solution : Given power series is . n!
1 1 Here, a a nn! n1 n 1 ! 1 a ! R lim n lim n n a n 1 ! n1 n 1 n1 ! n 1 n ! lim lim lim n 1 n n! n n! n R zn converges for all values of z. 0 n! 2) Prove that the power series n! zn converges only for z 0. 0 Solution : Given the power series n! zn 0 an n! a n1 n 1 ! 64
a n! R lim n lim n an1 n n 1 ! n! 1 1 lim lim 0 n n1 n ! n n 1 R 0 n! zn converges only for z 0. 0
n 2 3) Find the radius of convergence of series zn . 3n 5 n 2 Solution : The given series is zn . 3n 5 n 2n 1 2 n 3 an a n1 3n 5 3n 1 5 3 n 8 a n 2 n 3 R lim n lim n an1 n 3n 5 3 n 8 2 8 n 2 3 n 8 n1 n 3 lim lim n n n 3n 5 n 3 n n 3 5 n 1 3 n n 2 8 n2 1 3 lim n n n n2 n 3 5 1 3 n n 1 2 3 8 n 1 0 3 0 3 R 1 3 5 1 3 3 0 1 0 3
2n 1 4) Find the radius of convergence of the power series ()zn . 3n 5 2n 1 Solution : The given series is ()zn 3n 5 2n 1 2n 1 1 2 n 3 a a n3n 5 n1 3 n 1 5 3 n 8 a 2n 1 3 n 8 R lim n lim . n an1 n 3 n 5 2 n 3 n2 2 1 3 8 lim n n n n2 3 5 2 3 n n 65
2 1 3 8 2 0 3 0 1 R 1 3 5 2 3 3+0 2 0
5) Find the radius of convergence of the series c idn zn where c, d . Solution : The given series c idn zn n n1 an c id a n1 c id n a c id R lim n n1 n an1 c id n c id 1 lim lim n c idn c id n c id By Rationalizing c id 1 lim 2 2 n c d c2 d 2
1 6) Find the radius of convergence of the power series zn . 4n 1 1 Solution : The given power series is zn 4n 1 1 1 an a n1 4n 1 4 n1 1 1 4n 4 n R lim 4 lim 4 4 n n 1 n 4 1 4n n2 1 n 7) Find the radius of convergence of the series 1 z . n n2 1 n 1 n2 Solution : The given series is 1 z . Let an (1 ) n n 1 n 2 n n 1 1 1 1 lim sup a n lim sup 1 lim sup 1 R n n
n 1 lim 1 e R 1 n n e 66
8) Find the domain of region of convergence of the power series n 1.3.5.... 2n 1 1 z and show the domain or region n1 n! z graphically. n 1.3.5.... 2n 1 1 z Solution : The given power series is . n1 n! z 1 z Put , we get z 1.3.5.... 2n 1 n n1 n! 1.3.5.... 2n 1 Now, a n n! 1.3.5....2 n 1 2 n 1 a n1 n 1 ! 1.3.5.... 2n 1 a R lim n lim n! n a n 1.3.5....2 n 1 2 n 1 n1 n 1 ! 1.3.5....2 n 1 n 1! lim n 1.3.5....2 n 1 2 n 1 n ! n1 ! n 1 n ! lim lim n 2n 1 n ! n 2n 1 n ! 1 n 1 n 1 lim lim n n 2n 1 n n 2 1 n 1 1 1 1 lim n n 2 1 2 1 n 1 R 2 1 Domain of convergence of power series 2 1 z 1 1 i.e. 1 z z z 2 2
Taking square on both sides. 1 1 z 2 z 2 4 67
1 2 1 z 1 z zz z z. z 4 4 1 z z zz zz 4 4z 4 z 4 zz zz 4 4z 4 z 3 zz 0 4 4 zz z z 0 3 3 Put z x iy z x iy 8 4 x2 y 2 0 3 3 8 16 4 16 x2 x y 2 0 3 9 9 2 4 2 4 4 2 4 x y 0 x y 3 9 3 9 2 C 4 , 0 and r 3 3 4 2 Centre = , 0 , Radius = 3 3
2 r = 3 æ 4 ö cç ,0÷ è 3 ø
Fig 3.1
Given series converges inside the circle.
9) Find the domain of convergence of the power series n iz 1 . n0 3 4i Solution : The given power series is n n iz 1 i z i 2 i 1 n0 3 4i n0 3 4i n i n z i n0 3 4i n i an 3 4i 68
1 n n 1 1 i i lim sup an n lim sup lim sup R 3 4i 3 4i i i 3 4i 3 4 i
z x iy z x2 y 2 i 0 iy i 0 1 1 1 and
3 4i 32 4 2 9 16 25 5
1i 1 R 3 4 i 5 (0,1)
Fig 3.2 R 5 Domain of convergence of power series is z i 5
Centre =(0,1) , Radius = 5 The given series converge inside the circle.
10) Find the radius of converges of the series 3n 2 z 2n . Solution : The given series is 3n 2 z 2n . Put z 2 3n 2n an 3 n 2 a n1 3 n 1 2 3 n 5 2 a 3n 2 n3 Now, R lim n lim lim n 1 n a n 3 n 5 n n 3 5 n1 n R 1 Domain of convergence of power series is z 2 5 i.e a circle with C=(2,0) and r=5
11) Find the region of convergence of power series z 2n1 . 3 n n1 n 1 4 69
z 2n1 Solution : The given series is 3 n n1 n 1 4 1 1 an a n1 n13 4n n 2 3 4 n1 1 11 1 n Now, lim sup an n lim sup R n 13 4 n
1 1 lim sup 3 4 n 1 n R 4 The region of convergence of the series is z 2 R i.e. z 2 4 z 2 2 16 2 z 2 z 2 16 z z . z z. z 2 z 2 z 4 16 zz 2 z z 4 16 Put z x iy z x iy 2 2 2 2 x y 2 x . 2 12 0 z. z x y and z z 2 x x2 y 2 4 x 12 0 x2 4 x 4 y 2 12 4 0 x 22 y2 16 0 x 22 y2 16 c ( 2,0) r 4 Y
4 (-2,0) X -2 -1 0
Fig 3.3 The given series converges inside the circle. 1n (12) Find the radius of convergence of (i) (z / 2)2n (2009) (n !)2 z2 j zn (ii) (2009) (iii) nn z n (2008) (iv) (2008) j2 j( j 1) n 70
1 (13) Find the power series for the function f() z about the point z z=2 and find its radius of convergence. (2007) 14) Check for the convergence of the series nzn . n0
n Solution : Here after comparison with an z , n0 an n n1. 1 1 n ∴ limn supa nn lim n n 1 L 1. 1 n ⇒ R 1. ∴ The series 0n z converges for z 1 and L n diverges for z 1. For z 1, n zn n , as n .
n n0 nz diverges for z 1.
15) Find the radius of convergence of the following series . n n n z z z n2 (i) (ii) 2 (iii) (iv) z n1 n n1 n n1 n! n1
1 1 Solution : (i) ∵ an L lim n sup a n n 1. n2 1 ∴ R 1. The radius of convergence of the power series L n z n1 is equal to 1. n2
2 (iv) ∵ an 1 if ∵ n k for some integer k = 0 otherwise. 1 ∴ Consider ∵ Llimn sup a n n sup 1,0 1. 1 ∴ R 1.The radius of convergence of the power series L 2 n n0 z is equal to 1. 71
3.5. SUMMARY
1) If the series zn is convergent then limzn 0 . n1 n
2) A series zn of complex terms is convergent iff for every n0
0 , an integer N s.t. zn z n1 ... z n p n N and p 0 . (Cauchy criteria of convergence of series)
3) The series fn is said to be uniformly converges on D to f if for every 0 , an integer N (depends only on ) s.t. Sn z f z z D , and n N .
4) Let fn: D be a complex function defined on D s.t. fn z M n z D and n N . If Mn is convergent series 1 of positive Real numbers then series fn is uniformly convergent. 1
5) Let un be an infinite series for non-zero complex term s.t. u lim n1 L then n un i) If L 1, the series converges absolutely. ii) If L 1, the series diverge. iii) If L 1, the series may converges or diverge.
6) A power series about z0 is an infinite series of the form n 2 an ( z z0 ) a 0 ( z z 0 ) a 1 ( z z 0 ) a 2 ..... where constants an n0 and z0 are called complex numbers and z is a complex variable.
n 7) The radius of convergence R of the power series an z is n0 defined as R sup r : the series cgs z satisfying z r . 72
3.6. UNIT END EXCERCIES:
n 1)Check for the convergence of the series n0 z . 1 zn1 1 Solution: If z 1 then 1 z ... zn as 1 z 1 z 1 n zn . n0 1 z n n If z 1, then limn z . The series n0 z diverges.
2) Show that the radius of convergence of the power series n( n 1) n1 z is equal to 1. 1 1 Solution : zn( n 1) z 2 z 6 z 12 .. n1 2 3 an 0 if n k( k 1) for some integer ( 1)n a otherwise. n n 1 L lim sup an sup 0,1 1. R 1 n n L The series has the radius of convergence equal to 1.
k i 3) Find whether k1 converges or not. k2 i k i 1 1 Solution : 2 and we know that k1 k i k 2 1 k 2 1 k i converges. The series k1 converges. k 2 1
1 4) Check whether is convergent or not. (Hint : Check k1 k i 1 whether k1Re i is convergent or not.) k
k 5) Show that f() z kz is continuous in z 1.( Hint : k Here the convergence is uniform . Show that k1kz is k convergent in z 1. Let fk () z kz which is uniformly continuous for all k 1) zn 6) Show that the series is convergent everywhere in n0 n! the complex plane. 73
7) Show that the functions f( z ) cos( z ), g ( z ) sin( z ) are analytic in the whole complex plane. (Hint: Show that each of the series ( 1)nz zn ( 1)nz2 n 1 f( z ) cos( z ) and g( z ) sin( z ) n0 (2n )1 n0 (2n 1)! have infinite radius of convergence. )
8) Let ()an be a sequence of positive real’s and an1 limn L . Show that an 1 n limn a n L .
z2n 1 9) Find the radius of convergence of the series . n0 (2n 1) z2n 1 z 3 z 5 Solution: z ... ... n0 (2n 1)! 3! 5! n Comparing with n0 an z , we get an 0 if n 2, 4, 6,... 1 ifn 3, 5, 7 ... n! 1 Llimn sup a n n sup 0,0 0 . 1 z2n 1 R . The power series has infinite radius of 0 n0 (2n 1) convergence.
10) Find the domain of convergence of the series ( 1)n (z 1)n . n0 n!
Solution: Put z 1. ( 1)n ( 1) n ( 1) n (z 1)n n . a . n0n!!! n 0 nn n
an1 1 Llimn lim n 0 an n 1 1 R . The given power series converges for all θ. 0 But z 1. As θ varies over , also varies over . The given power series converges for all complex numbers. 74
11) Find the domain of convergence of the power series n n0 ()z i (Ans : The series converges for z such that z i 1.
1 1 12) Show that n n 1 as n . ( Hint: Put a log n n n for n 1.)
n 13) If an z has radius of convergence R, what is the radius of convergence of
2n 2 n an z and of an z ? n 1 z 2 14) Prove that the series converges for z 2 4 . 3 n n1 n 1 4 15) Find the radius of convergence of the power series 2 an z n, a . n0 1 1.3 1.3.5 16) Find the domain of convergence of the series z z2 z 3...... 2 2.5 2.5.8
17) Find the radii of convergence of the following power series n 2 z1 n 1i
zn 18) Show that the ROC for the power series n is 1. Discuss the convergence of this series of the points on the boundary or the {z / z 1} disc 75 4
DIFFRENTIABILITY
Unit Structure : 4.0. Objectives 4.1. Introduction 4.2 . Differentiability in complex 4.3. Summary 4.4. Unit End Exercises
4.0. OBJECTIVES:
After going through this chapter you shall come to know about : Defining a polynomial with complex coefficients and in an indeterminant z, which can take any complex number value. n An infinite series of the form n 0an z is called a power series.
We shall investigate for the differentiability of a power series as a function of a complex variable , at the same time we shall also check for the condition , under which two power series are one and the same , that is both the power series represent the same complex valued function.
4.1. INTRODUCTION:
Through this Unit , we shall examine the notion of “ a function of “, where is a Complex Number of the form z x iy. A Complex Number z can be viewed as an ordered pair of real numbers x and y as z(,) x y . The point of view taken in this Unit is to understand some functions , which are direct functions of z x iy and not simply functions of the separate parts x and y Consider for example the function x2 y 2 2 ixy is a direct function of x iy , since x2 y 2 2 ixy ( x iy ) 2 . 76
f() z z2 but x2 y 2 2 ixy is not expressible as a polynomial in variable x iy . Therefore we are compelled to consider a special class of functions , given by direct/ analytic expressions in x iy . We shall name such direct functions as the analytic functions . Let us start this Unit by defining an c polynomial p() z in a Complex variable z.
Definition : A polynomial P(x,y) in a Complex variable z =(x, y) is an expression of the form n P(,) x y 0 1(x iy ) ...n ( x iy ) where 0,.., 1 n are complex constants e.g. (i ) P ( x , y ) x2 y 2 2 ixy (ii ) x2 y 2 2 ixy is not a polynomial in z x iy .
4.2 DIFFERENTIABILITY IN COMPLEX
Differentiation: Let G be an open set in and f: G can be a function, we say that f is differentiable at a point z0 in G if the limit f z f z lim 0 ……(1) z z0 z z0 exists, this limit is denoted by f z0 and is called derivative of f at z0 . Put z z0 h , (complex number) then equation (1) becomes f z0 h f z 0 f z0 lim . z z0 h In terms of ‘ ’ notation limit in equation (1) exists iff f z f z0 0, 0, f z0 whenever 0 z z0 . z z0 If f is differentiable at each point of G then f is differentiable on G. Notice that if f is differentiable on G, f z0 defines a function f: G .
If f continuous then we say that f is continuously differentiable. If f is differentiable, then f is twice differentiable continuing, a different function each successive derivative is differentiable is called infinitely differentiable.
Proposition: If f: G is differentiable at z0 G then f is continuous at z0 (2012) 77
Proof : Given f: G is differentiable at z0 G . f z f z0 lim f z0 exists. z z0 z z0 f z f z0 lim f z f z0 lim . z z 0 z z0 z z 0 z z0 f z0 .0 0
lim f z f z0 z z0 f is continuous at z0 .
Theorem : If f and g are differentiable at z0 G then f g, f . g , f g , g 0 are also differentiable at z0 G .
The Increment Theorem: Let f: G be a complex valued function z0 G and r 0 , B z0, r G . Then f is differentiable at z0 iff a complex number and a function :(;)(,,)B o s o s r such that v h B(;) o s , f z0 h f z 0 h h h and limh 0 . h0 Proof : Let f is differentiable at point z0 . f z h f z (Let) put h 0 0 h So that, f z0 h f z 0 h h h Let f z0 (*) f is differentiable at z0 . f z0 h f z 0 limh lim h0 h 0 h f z from (*) = 0 Conversely, Let limh 0 and (**) h0 f z0 h f z 0 h h h f z h f z 0 0 h h Taking lim on both sides. n 78
f z0 h f z 0 lim lim 0 h0 h 0 f z0 h f z 0 lim 0 from (**) h0 n dw f() z a z n n dz f is differentiable at point z0 and f' z0 .
Composite Function : Let G and be open sets. Let f: G and g : be functions f G . Then for each ZG , the association g f defined by g f z g f z is a function called of composite function.
Note : In general f g g f
Chain Rule : Theorem: Let G , be open sets and let f and g be differentiable on G and (respectively). Suppose f G then g f is differentiable on G and g f z g f z. f z z G
Proof : Fix a point z G , choose r 0 B z , r G . Let 0 h and h r z h Given that f is differentiable on G. f is differentiable at a point ZG . By increment theorem, f z h f z hf z h h where h is continuous function and limh 0 h0 Put K f z h f z , where K hf z h h Also g is differentiable at f z by increment them, g f z h g f z K g f z K g f z K K where K is continuous function and limK 0 . k0 gfzh gfz hfzhh .' gfz K gfz hfzgfz .. h hgfz hf z.. K h h K 79
gfz hgfzfz hh Where, h h g f z f z K h K T.P.T. limh ) 0. h0 h 0 as h 0 As h 0, K f z h f z 0 K 0 as h 0 Hence, limh 0 n by increment theorem, g f is differentiable at z G Z was arbitrary . g f is differentiable on G and g f z g f z f z z G Let z x iy G and f: G be defined by, f z u z iv z , where u and v are real valued function
OR f x,,, y u x y iv x y .
u x h,, y u x y Definition : If lim exists then it is called h0 h partial derivative of u w.r.t x as the point x, y and is denoted by u x, y or u x, y . x x
n Theorem : Let f z an z have the radius of convergence n0 R 0 then 1) The function f is infinitely differentiable on B0; r and ()k n k f( z ) n ( n 1)( n 2)...( n k 1) an z for z R and k 1. n k f ()n (0) 2) If n 0 then a n n! n Proof 1) For z R , we will write f z an z S n z R n z n0 n K where Sn z a n z and Rn z a K z n0 K n1 80
n1 Put g z nan1 z lim Sn z n1 n
Fix a point z0 in BR0;
(Choose r 0, z0 r R and z r R z z0
We will prove that f z0 g z 0
Let 0 be arbitary B z0; B 0, r
Let z B z0, then
f()() z f z0 Sn()()()() z R n z S n z0 R n z 0 g()() z0 g z 0 z z0 z z 0
Sn z S n z0 R n z R n z 0 Sn z0 S n z 0 z z0 z z 0 Taking modulus on both the sides.
f z f z0 Sn z S n z 0 g z0 Sn z 0 S n z 0 g z 0 z z0 z z 0 R z R z n n 0 (1) z z0 Let 0 , be given Rn z R n z0 1 KK Now, aK z z0 z z0 z z 0 K n1
1 a z z zKKKK1 z z 2... z z 2 z 1 K 0 0 0 0 z z0 K n1 Rn z R n z0 KKKK1 2 2 1 aK z z0 z... z z 0 z 0 z z0 K n1 KKK1 1 1 K 1 aK r r... r aK . K r K n1 K n1 n1 The derived series n an z is convergent at z r . n1 K 1 The power series aK . K r converges for r R . K 1 81
For the above 0, an integer N1 a .. K rK 1 n N . (by Cauchy criteria) K 3 1 K n1
Rn z R n z0 Thus, n N1 (2) z z0 3
lim Sn z0 g z 0 n
For the above 0 , an integer N2 s.t.
S z g z (3) n 0 0 3
Choose Nmax N1 , N 2 . for n For this n, we can find 0 s.t.
Sn z S n z0 Sn z0 (4) z z0 3 whenever 0 z z0 . From equation (1), (2), (3), (4) we get
f z f z0 g z0 z z0 3 3 3
whenever 0 z z0
f is differentiable at z0 B0, R z is arbitary. f is differentiable on BR0; A repeated application of this argument shows that the heigher derivatives f, f ,..., f K ... exists, so that
K n K f z n n1 ... n K 1 an z exists for z R and n K K 1. f is infinitely differentiable on BR0; .
2) Since K n K n! n K f z n n1 ... n K 1 an z an z n K n K n K ! 82
n! n K K! ak a n z K n1 n K ! Put z 0 K f0 K ! aK 0 f K 0 a k K! Replace K by n f n 0 a n n! n Corollary : If the power series an z has radius of convergence n0 n R 0 , then f() z an z is analytic on BR0; .
Theorem : If G is an open connected set and f: G is differentiable with f z 0 z G , then f is constant.
Proof : Fix a point z0 G and let w0 f z 0 .
Let A z G; f z w0 T.P.T. AG . [i.e. by showing A is both open and closed and A ] T.P.T. A is closed.
Let z G and zn be a sequence in A lim zn z n
f zn w0 , for each n f is differentiable on G (given) f is continuous on G. f z flim zn lim f z n w0 n n z A A contain its limit point. A is closed. Now, T.P.T. A is open. Fix a A , since G is open. r 0 B a ; r G 83
Let z B a; r and set g t f tz 1 t a , 0 t 1 g t g s f tz1 t a f S 1 S a t S z S t a z t s t S z S t a t S
g t g S f tz1 t a f sz 1 S a lim lim z a t S t S t S t S z S t a g' s f ' sz (1 s ) a ( z a ) g'( s ) 0 , 0 s 1 f z 0, z G g s constant, 0 S 1 g1 constant g 0
f z g1 constant g 0 f a w0 z A z B a; R z A B(;) a r cA
A is open and A () z A Hence, by the connectedness of G AG f is constant on G.
4.3 SUMMARY
1) If f: G is differentiable at a point z0 in G, then f is continuous at z0 . 2)The Increment Theorem : Let f: G be a complex valued function z0 G and r 0 , B z0, r G . Then f is differentiable at z0 iff a complex number and a function :B 0; s 0, S , r h B 0; S f z0 h f z 0 h h h and limh 0. h0 3) Chain Rule : Let G , be open sets and let f and g be differentiable on G and (respectively). Suppose f G then g f is differentiable on G and g f z g f z. f z z G 84
u x h,, y u x y 4) If lim exists then it is called partial h0 h derivative of u w.r.t x as the point x, y and is denoted by u x, y or u x, y . x x n 5) Let f z an z has the radius of convergence R 0 then n0 i) The function f is infinitely differentiable on B0; r and ()k n k f( z ) n ( n 1)( n 2)...( n k 1) an z for z R and K 1. n k f n 0 ii) If n 0 then G n n! 6) If G is an open connected set and f: G is differentiable with f z 0 z G , then f is constant.
4.4. UNIT END EXERCISES :
1) Check for the differentiability of the power series zn f() z . n0 n! zn Solution: We know that the series converges for all n1 n! complex numbers. ∴ f() z exist for all z and nzn1 z n 1 z n f()() z f z . n1n! n 1 ( n 1)! n 1 n ! f()() z f z for all z .
n 2) If the series n0 an () z a has the radius of convergence n R 0 , then show that f()() z n0 an z a is analytic in B(;) a R . ( Hint : Use the fact that f is infinitely differentiable on 1 B(;) a R and a fn ( a ) n 1.) n n!
85 5
COMPLEX LOGARITHM
Unit Structure : 5.0. Objectives 5.1. Introduction 5.2. Logarithmic function 5.3. Branches of Logarithmic Function 5.4. Properties of Logarithmic Function 5.5 Trigonometric and hyperbolic functions 5.6. Summary 5.7. Unit End Exercises
5.0. OBJECTIVES:
We are already familiar with a logarithm function, defined for positive real x. In the same manner one can define a complex logarithm Log (z) of a complex number z . We shall study the branches of this complex logarithm function . The complex logarithm Log (z) posseses some branches , which we shall try toinvestigate. We shall also study the properties of a complex logarithm in detail.
5.1. INTRODUCTION:
With the help of order completeness property of, we proved in our earlier course that if y 0 and n 2 is any integer, then there is a unique positive number x such that xn y . x is called nth root of y, since there is a unique positive 1 number x satisfying this, defining y n is justified. We proved that, y x y x y x xy for a 1and x , a., a a x y and a a . x f : 0, defined by f x a is a bijective function and it’s inverse is called as the logarithm of y to the base a, denoted by loga (y ) . We want to discuss these concepts once 86 again but we consider the logarithm of complex numbers with base e , hence we try to identify the nature of inverse of the exponential function of complex variable z , namely z f() z e on some domain D . Here we shall start defining log (z ) for z .
5.2 LOGARITHMIC FUNCTION :
Definition : For z 0 , the logarithmic function of a complex variable z, denoted by logz , is defined as logz ln z i arg z 2 n where argz , or 0, 2 and n. Here, log z is a single valued function.
5.3 BRANCHES OF LOGARITHMIC FUNCTION :
Definition : If 0G is an open connected set in and f: G is a continuous function such that ef z z, z G , then f is branch of logarithm.
Theorem: A branch of the logarithm is analytic and its derivative is 1 . z Proof: Let f z log z ln z i arg z be a branch of logarithm, where z 0 , arg z [ , ]. Let f z u z iv z and z x iy u iv ln x2 y 2 i tan 1 y where z x2 y 2 and x tan1 y x u x, y ln x2 y 2 and v tan1 y x u x, y ln x2 y 2 u 2 x x u y x, y and x, y x 2 x2 y 2 x 2 y 2 y x2 y 2 v tan1 y x 87
v 1 y y y and x y 2 x 2 x2 y 2 x2 y 2 1 x2 x 2 x v x y x2 y 2 Therefore C-R equations are satisfied f z log z z z u v x y z z 1 i i 2 2 2 2 2 x x x y x y z z. z z
Theorem : Let 0G be an open connected set in and suppose that f: G is analytic. Then f is a branch of logarithm iff 1 f z , z G and ef a a for atleast one a G . z Proof : Suppose f is a branch of a logarithm. ef z z z G (1) Differentiate w.r.t. to z on both sides. f z e. f z 1 1 f() z e f() z 1 f z z G z Clearly, from equation (1), ef a a for atleast one a G . Conversely, 1 Suppose f z z G and ef a a for atleast one a G . z T.P.T. f is a branch of the logarithm. Define, g z z. e f z (2) g is analytic. g is differentiable. d dz d g z z e f z e f z. z e f z dz dz dz e f z f z. z e f z f z f z 1 1 e z. e f z z z e f z e f z g z 0 z G 88
g z constant = K (Say) (3) To find K, put z a in equation (2) and (3) g a a. e f a and g a K . f a 1 f a 1 K a.. e a e a a 1 K 1 Put K 1 in equation (3), we get g z 1 Put g z 1 in equation (2), we get 1 z . e f z ef z z z G f is branch of logarithm (by definition)
A single valued function (is branch of logarithm) logz ln z i arg z z 0and arg z , is continuous function in the region or a Domain D z x iy; y 0, x 0 logz ln z i arg z z 0 and arg z , log z is not defined at the point z 0.
Theorem : Prove that logz is not continuous on the negative real axis.
Proof : Let z0 x 0 0 be any point on the negative real axis. For z x iy with x 0, y 0 , we have, lim argz lim arg( x iy ) z z0 x x 0 y0 For z x iy with x 0 and y 0 , We have lim argz lim arg( x iy ) z z0 x x 0 y0 Two limits obtained are different. i.e. argz fails to possess a limit every point of the negative real axis. logz is not continuous along the negative real axis.
Theorem: Let 0G be an open connected set in . If a branch of the logarithm f : is related by g z f z 2 in [for some integer n] with g: G then g is branch of logarithm.(2008)
Proof : Given that f is a branch of the logarithm. ef z z z G 89
Given, g z e z 2 in (for some intn ) eg z e f z2 in ef z. e2in g z f z 2in e z z G e zand e 1 g is a branch of logarithm.
5.4 PROPERTIES OF LOGARITHM FUNCTION :
Theorem: 1) logz1 . z 2 log z 1 log z 2 2 in , where n 1, 0οr1 by definition.
Proof : logz ln z i arg z where z 0 and argz ,.
log z1 . z 2 ln z 1 . z 2 i arg z 1 . z 2 lnz1 ln z 2 i arg z 1 arg z 2 2 in ( arg z1 . z 2 arg z 1 arg z 2 2 in) ln z1 i arg z 1 ln z 2 i arg z 2 2 in logz1 log z 2 2 in
z1 2) log logz1 log z 2 2 in z2 1 3) log logz z
1 1 1 Proof: log ln i arg z z z 1 ln z i arg z arg arg z z ln z i arg z ln z i arg z logz
Evaluate :- 1) logi z x iy i x 0 and y 1 logz ln z i arg z (by definition) logi ln i i arg i ln1 i i 2 2 2) log 1i z1 i x 1 and y 1, z x2 y 2 1 1 2 90
log 1i ln 1 i i arg 1 i ln 2 i tan1 1 i ln 2 i tan1 1 ln 2 i 4
3) log 1 i z 1 i x 1, y 1 z 2 log 1 i ln 1 i i arg 1 i ln 2 i tan1 1 1 ln 2 i tan1 1 ln 2 i 4 4) In unit disk B0,1 z : z 1 prove that power series zn 1 1 log where log is a branch of the logarithm n1 n 1 z 1 z 1 log . 1 z zn Solution : Let f z (1) n1 n 1 and g z log (2) 1 z g is differentiable. d 1 1 1 1 z g z log 1 2 dz 1 z 1 1 z 1 z2 1 z2 1 g z (3) 1 z zn Given, the power series f z n1 n 1 1 Here, a , a nn n1 n 1 1 a 1 n 1 R lim n lim n 1 lim n n an1 n n n n
1 1 0 f is analytic in open disk B0,1 (using corollary) f is differentiable in B0;1 n. zn1 f z zn1 zn n1n n 1 n0 91
1 f z (4) 1z 1 znis a G.S. and cgs to z 1, G n 1 z n0 n 0 From equation (3) and (4), we get 1 f z g z 0 f z g z 0 f z g z = constant = K (5) To find k, put z 0 in equation (1), (2) and (3) f0 0, g 0 0 f0 g 0 K K 0 Put K 0 in equation (3), we get f z g z 0 f z g z zn 1 log for all z B0;1 n1 n 1 z
b th Definition : Given 0 , the principal value of z (i.e. the b b b.log z power of z) is defined by z e, b Here, zb is analytic. ( log z is analytic) b b.log z Consider z e, b Here, zb is multivalued function. argz (and hence logz ) is a multiple valued function.
Case I : If b is an integer then zb e b. log z is a single valued function.
Proof : Let b K . kln z i arg z 2 n zb e b.log z e ln z i arg z ek . e i2Kn klog z i2 Kn e i e1, n eblog z zb e b.log z is a single valued function.
Case II : If b p (real rational) then zb has produces exactly q q values. 92
Case III : If b is an irrational number or imaginary number then zb is infinite valued function.
Example 1 : Find the principal value of ii.
b b.log z Solution : z e, b (by definition) 2 2 iln i i arg i i0 i i . ii e i.log i e e e 2 ii e 2
Example 2 : Find all the values of i2i .
b b. log z Solution : z e b (by definition) 2i 0 i 2 n 2i ln i i arg i 2 n i2i e 2 i . log i e e 2
4n 2i i 2 2 i1 4 n e e 1 1 4n e n e4n 1 Here, the principal value of i2i is e .(All values are not found.)
Example 3 : Find the value of i2 .
2 ln i i arg i 2 n 2 ln1i arg i 2 n Solution : i2 e 2 log i e e 2 0 1 2 4ni i 4ni e. e e e 1, n cos i sin 1 0 i2 1
1i Example 4 : Find all the values of 1 i .
5.5 TRIGNOMETRIC AND HYPERBOLIC FUNCTIONS:
Trigonometric Function : The Complex trigonometric functions sin and cos are defined by eiz e iz ez e iz sinz and cosz (2008) 2! 2 93
eiz e iz 2 Similarly, tanz , sec z i eiz e iz eiz e iz iz iz 2i i e e cosecz , cot z eiz e iz e iz e iz
Note : 1) sin2z cos 2 z 1 d d 2) sinz cos z and cosz sin z dz dz 3) sinz sin z and cosz cos z 4) sinz w sin z .cos w cos z .sin w 5) cos z w cos z .cos w sin z .sin w 6) sin 2z 2 sin z .cos z
Hyperbolic Function : The complex hyperbolic functions sinh and cosh are defined by ez e z ez e z sinhz and cosh . 2 2 ez e z e z e z Similarly, tanhz , cot hz ez e z e z e z 2 2 sechz , cosec hz ez e z e z e z
Note : 1) cos2z sin 2 z 1 d 2) sinhz cos hz dz d 3) coshz sin hz dz 4) sinh z w sin hz .cos hw cos hw .sin hz 5) cosh z w cos hz .cos hw sin hw .sin hz
Relation between Trigonometric and Hyperbolic Function : 1) siniz i sin hz
Proof : 2 z z z z ei iz e i iz ez e z i e e i e e siniz 2i 2! 2i 2 isin hz 94
2) cosiz cos hz eiz e iz ez e z Proof : cos(z ) coshz 2 2 3) taniz i tan hz 4) coshiz cos z 5) sinhiz i sin z 6) tanhiz i tan z
Periodic Function : A function f: G is said to be periodic if a non-zero complex number T f z T f z z G . Here T is a period of the function n.
Periodicity of ez : Let T be the period of ez z T z e e z . To find T, put z 0 eT e01 e 2 in Let T i ei e2 in e 1 and ei e2 in 0 and 2n
T i 0 2 i n is a period of ez . OR T log1 0 (is not possible by definition) Let T i ei 1 e. cos 1 and sin 0
e 0 and 2n T i 0 2 in 2 in 2in is a period of ez ez2 ni e z
Periodicity of sin z : Let T be the period of sinz . sin z T sin z Put z 0 95
sin 0T sin 0 sinT 0 T n, where n 0, 1, 2, ... sin z 2 sin z .cos 2 cos z .sin 2 1n sin z sinz if n is even. The period of sinz is 2n where n.
Periodicity of cosz : Let T be the period of cosz . cosz T cos z Put z 0 cos 0T cos0 cosT 1 T 2 n n The period of cos z is 2n, where n.
5.6. SUMMARY
1) For z 0 , the logarithmic function of a complex variable z, denoted by logz , is defined as logz ln z i arg z 2 n where argz , or 0, 2 and n.
2) If 0G is an open connected set in and f: G is a continuous function ef z z, z G , then f is a branch of the logarithm.
3) Given 0c , the principal value of zb (i.e. the bth power of z) is defined by b b.log z z e, b
4) The Hyperbolic functions sinh and cosh are defined by ez e z ez e z sinhz and cosh . 2 2
5.7. UNIT END EXERCISES:
1) Suppose that f: G is a branch of the logarithm and n is any integer . Prove that zn exp( nf ( z )) for all z G . 96
Solution: Since f: G is a branch of the logarithm ( G is an open connected set .) z exp( f ( z )) for all z G . z2 exp( fz ( ).exp( fz ( )) exp( fzfz ( ) ( )) exp(2 fz ( )) . zn exp( nf ( z )) for all z G . (By induction on power of z .
2) Describe the branches of an analytic function f() z √z .
1 Solution: f( z ) z exp log( z ) ,since 2 2 1 1 1 exp log(z ) exp log( z ) log( z ) exp log( z ) z . 2 2 2 This defines and it is analytic , where the log(z ) is analytic.
Different branches of log(z ) yield different branches of z . log(z ) has infinitely many different branches log(z ) 2 ki for any integer k but there are only two different branches of z . 1 1 Since exp log(z ) exp log z 2 ki whenever k is an even 2 2 integer .
3)Find all values of the complex number ii .
Solution: ii e ilog() i e log() i i arg() i e arg() i . Here we know that 3 5 7 arg(i ) ..., , , , ,... 2 n : n . 2 2 2 2 2 3 5 i i ..., e2 , e 2 , e 2 ,....
4) Find all values of 1 i1i (Hint : 1 i1i e1i log 1 i ).
a b 5) Let f: G and g: G be branches of z and z f respectively. Show that fg is a branch of za b and is a g branch of za b . 97
Solution: f() z za and g() z zb for all z G . a b a b fg: G defined by fg( z ) f ( z ). g ( z ) z . z z fg() z za b for all z G . za b has a branch fg on G. f Similarly, is a branch of za b . g
6) Let z1,,..., z 2 zn be complex numbers such that Re(zk ) 0 and
Re(z1 z 2 ... zk ) 0 , for 1 k n . Then show that log(z1 z 2 ... zk ) log( z 1 ) ... log( z n ) .
Solution : Let f( z ) log( z ) be the principle branch of the logarithm function ef( z ) e log( z ) z .
Take a z1... zn . Since the arguments of each zk and that of n z z... z lies between to for all 1 k n . Therefore 1 2 k 2 2 log(zz1 2 ... zk ) log zz 1 ... k iArgzz ( 1 2 ... z k ) and log(zk ) log z k iArg ( z k ) , for 1 k n .
Argzz(1 2 ...) zn Argz () 1 Argz ()... 2 Argz ()2 n k where k is any integer .
f( a ) log z1 ... zn i A rg( z 1 z 2 ... z n )
n logzk iArgz ()1 Argz ()... 2 Argz ()2 n k k1 n k1logzk iArg ( z k ) 2 k i
n k1log(zk ) 2 k i
f( a )n log( z ) e2k i 1 e k1 k n f( a ) k1 log( zk ) 0
n f( a ) k1 log( zk )
log(z1 z 2 ... zn ) log( z 1 ) ... log( z n ) .
7) Give the principal branch of 1 z ( Hint : log 1z 1 e log(1 z ) ) 2 98
8) Prove that there is no branch of the logarithm defined on G 0. (Hint: Assume the existence of a continuous function L(z) defined on a connected open set G of the complex plane such that L(z) is a logarithm of z for each z in G , compare L(z) with the Principal branch of log(z ) . As α goes from 0 to 2 , since L() eia ia and L being continuous function of α , L e2i 2 i L (1) 0 , a contradiction.)
9) Evaluate ii by taking the logarithm in its principal branch.
10) Prove that sinz2 sin2 x sinh 2 y
Solution: 2 sinz2 sin xiy sin xiy cos cos xiy sin 2 sin x cosh yix cos sinh y 2 sinx cosh y i cos x sinh y (sin x cosh y i cos x sinh y ) (sinx cosh y i cos x sinh y )(sin x cosh y i cos x sinh y ) sin2x cosh 2 y cos 2 x sinh 2 y sin 2 x (1 sinh 2 y ) (1 sin 2 x )sinh 2 y sin2x sin 2 x sinh 2 y sinh 2 y sinh 2 y sin 2 x sin 2 x sinh 2 y
11) Find the principal value of i2i
99 6
ANALYTIC FUNCTIONS
Unit Structure : 6.0. Objectives 6.1. Introduction 6.2. Analytic Functions 6.3. Cauchy Riemann equations 6.4. Harmonic Functions 6.5. The Functions ez ,sin( z ),cos( z ) etc. 6.6. Summary 6.7. Unit End Exercises
6.0. OBJECTIVES:
In this unit we shall characterise the differentiability of a complex valued function in terms of it’s power series expansion, in this case the function is said to be an analytic function about some point z0 . An analytic function f() z satisfies some properties, among these one important property is to satisfy Cauchy-Riemann equations . Further we shall also see the term by term differentiation of a power series function, provided that such term by term differentiation is possible. We shall also study the inverse function theorem then we shall define a class of functions called as harmonic functions .We shall also discuss the differentiability of a complex valued functions like ez ,sin( z ),cos( z ) etc.
6.1.INTRODUCTION :
Given a function of the complex variable z, we wish to examine if f is a differentiable function of z or not. As we saw in the case real valued functions, we look for existence f()() z h f z of the limit lim which should exist regardless n0 n 100 of the manner in which h approaches 0 through complex values. An immediate consequence is that the partial derivatives of f , considered as a function of two real variables x and y(()()(,) f z f x iy f x y must satisfy the Cauchy Riemann equations. Let us define the derivative of a function of complex variable z at the point z z0 . Let z x iy G and f: G be defined by, f z u z iv z , where u and v are real valued function OR f x,,, y u x y iv x y .
u x h,, y u x y Definition : If lim exists then it is called h0 h partial derivative of u w.r.t x as the point x, y and is denoted by u x, y or u x, y . x x
6.2 ANALYTIC FUNCTIONS :
A function f is said to be analytic (or holomorphic or regular) at a point z z0 if f is differentiable at every point of some nbd. of z0 .
Definition : A function f: G is analytic if f is continuous differentiable on G.
A function f is analytic on a closed set S if f is differentiable at every point of some open set containing S. n Theorem : Let f z an z have the radius of converges R 0 n0 then 1) The function f is infinitely differentiable on B0; r and ()K n K f( z ) n ( n 1)...... ( n K 1) an z for z R and K 1. n K f n (0) 2) If n 0 then a n n! 101
Proof : 1) For z R , we will write n n f z an z S n z R n z where Sn z a n z and n0 n0 K Rn z a K z K n1 1 Put g z n an z lim Sn z n1 n
Fix a point z0 in BR0;
(Choose r 0, z0 r R and z r R z z0
We will prove that f z0 g z 0
Let 0 be arbitrary B z0; B 0, r
Let z B z0, then
f z f z0 Sn z R n z S n z0 R n z 0 z z0 g zl0 z z0 z z 0
Sn z S n z0 R n z R n z 0 Sn z0 S n z 0 z z0 z z 0
f z f z0 Sn z S n z 0 g z0 Sn z 0 S n z 0 g z 0 z z0 z z 0 R z R z n n 0 ….(1) z z0 Let 0 , be given Rn z R n z0 1 KK Now, aK z z0 z z0 z z 0 K n1 1 a z z zKKKK1 z z 2... z z 2 z 1 K 0 0 0 0 z z0 K n1 Rn z R n z0 KKKK1 2 2 1 aK z z0 z... z z 0 z 0 z z0 K n1 KKK1 1 1 aK r r... r K n1 K 1 aK . K r K n1 102
n1 The derived series n an z is convergent at z r . n1 K 1 The power series aK . K r converges for r R . K 1
For the above 0, an integer N1 a .. K rK 1 n N . ( by Cauchy criteria) K 3 1 K n1
Rn z R n z0 Thus, n N1 (2) z z0 3
lim Sn z0 g z 0 n
For the above 0 , an integer N2 s.t. S z g z (3) n 0 0 3
Choose NNNmax 1 , 2 For this n, we can find 0 s.t.
Sn z S n z0 Sn z0 (4) z z0 3 whenever 0 z z0 .
From equation (1), (2), (3), (4) we get
f z f z0 g z0 whenever 0 z z0 z z0 3 3 3
f is differentiable at z0 B.; R z is arbitrary. f is differentiable on BR0;
A repeated application of this argument shows that the heigher derivatives f, f ,..., f K ... exists, so that K n K f z n n1 ... n K 1 an z exists for z R and n K K 1. f is infinitely differentiable on BR0; . 103
2) Since K n K n! n K f z n n1 ... n K 1 an z an z n K n K n K ! n! n K K! ak a n z K n1 n K ! Put z 0 K f0 K ! aK 0 f K 0 a k K! Replace K by n f n 0 a n n! n Corollary : If the power series an z has radius of cgs. R 0 , n0 n then f() z an z is analytic on BR0; .
Theorem : If G is an open connected set and f: G is differentiable with f z 0 z G , then f is constant.
Proof : Fix a point z0 G and let w0 f z 0 .
Let A z G; f z w0 T.P.T. AG . [i.e. by showing A is both open and closed and A ] T.P.T. A is closed.
Let z G and zn be a sequence in A lim zn z n
f zn w0 , for each n f is differentiable on G (given) f is continuous on G. f z flim zn lim f z n w0 n n z A A contain its limit point. A is closed. Now, T.P.T. A is open. Fix a A , since G is open. r 0 B a ; r G 104
Let z B a; r and set g t f tz 1 t a , 0 t 1 g t g s f tz1 t a f S 1 S a t S z S t a z t s t S z S t a t S
g t g S f tz1 t a f sz 1 S a lim lim z a t S t S t S t S z S t a
g S f Sz 1 S a z a g S 0 , 0 S 1 f z 0, z G g s constant, 0 S 1 g1 constant g 0
f z g1 constant g 0 f a w0 z A z B a; R z A z b(;) a r A A is open. Hence, by the connectedness of G. AG f is constant on G.
6.3 CAUCHY RIEMANN EQUATIONS (C-R Eq.):
Theorem : Let u and v be real valued function defined on the domain G and suppose that u and v have continuous partial derivatives then f: G defined by f z u z iv z is analytic iff u and v u v u v satisfy Cauchy Riemann equation. i.e. and x y y x (2006, 2007, 2008, 2009) Proof : Let z x iy G and z x i y .
Given, f: G is defined by f u z iv z OR f x,,, y u x y iv x y f is analytic on G. f is differentiable at z G . f z z f z f z (a unique limit) z 0 as z 0 in any manner in . 105
Now, f z z f z ux xy,,,, y ivx xy y uxy ivxy z x i y ux xy,,., yuxy vx xy yvxy i x i y x i y …………………(I) Suppose z 0, along the real axis (x-axis) z x and y 0 fzzfz uxxyuxy ,, lim lim z0z x 0 x i v x x,, y v x y x u v f z x,, y i x y x x …………………(II) Suppose z 0 , along the imaginary axis (y-axis). z i y and x 0 fzzfz uxyyuxy ,, lim lim z0z y 0 i y i v x,, y y v x y i y 1 u v u v f z i i y y y y 1 i i i i i2 i u v i x,, y x y y y ………………..(III) From equation (II) and (III) u v u v i i x x y y Equating Real and imaginary part on both sides. u v u v i.e. and x y y x which are Cauchy Riemann equations. 106
Conversely, Let z G G is open r 0, B z , r G Let z x i y B0; r . Given, u and v have continuous partial derivatives. u u x v u,,,,, v are continuous on G. x y x y The expression u x x,, y y u x y can be written as (by definition of partial derivative.) ux xy,,.,,, yuxy xuxy x yuxy y xy x, y z where, lim 0 OR lim 0 x y0 x i y z0 z …………………(IV) Similarly, vx xy,,,,, yvxy xvxy x yvxy y xy x, y where lim 0 ……………..(V) x i y0 x i y fz z fz ux xy,, yuxy i v x x,, y y v x y from equation (I) xu.,, x yu y xyixv x yv y xy from equation (IV) and (V)
xuiv x x yuiv y y xyi,, xy By Cauchy-Riemann equations u v u v and x y y x 2 2 i.e. ux v y and uy v x i v x i 1 2 fz zfz xuiv x x yiviu x x i
x ux iv x i y iv x u x i
x i y ux iv x i f z z f z x i y i u i v z z x y z 107
u v i i z x i y where x x x i y i lim 0 } x i y0 x i y f is differentiable at z and f z z f z u v lim f z i z0 z x x u v f is continuous ( and are continuous) x x f is continuously differentiable. f is analytic.
Note: If f(z) is analytic then it can be differentiated directly
Example : z (1) Prove that the function f z e is analytic in . Also find its derivative. Solution : Let f z u z iv z and z x iy Given that, f z ez u iv ex iy e x. e iy e x cos y i sin y Equating real and imaginary parts on both sides. T.P.T. f is analytic. By previous theorem, we see that in order to prove and is analytic we have to verify that u and v are satisfy Cauchy-Riemann equations. u v u v i.e. and x y y x Now, u excos y and v exsin y u v ex.cos y , e x sin y x y u v u v and x y y x u and v satisfies the C-R equation. f is analytic. e e sini sin h 2 108
e e and cosi cos h 2 u v f z i excos y i e x sin y ex cos y i sin y ex. e iy x x ex iy ez z (Or f() z is analytic it can be differentiated directly i.e. f() z e )
2) Show that the function f z w sin z is analytic and also find dw . dz
Solution : Let z x iy and u iv w Given that w sin z u iv sin x iy sinx cos iy cos x .sin iy sinx .cos hy i cos x .sin hy Comparing real and imaginary parts u sin x . cos hy , v cos x .sin hy u and v are real valued function of x and y. u v cosx . cos hy , sin x .sin hy x x u v and sinhy .sin x , cos x .cos hy y y u v u v and x y y x Cauchy-Riemann equations are satisfied. f z sin z is an analytic function. w u v f z i cosx cos hy i sin x .sin hy z x x (Or f() z is analytic it can be differentiated directly i.e. f( z ) sin z )
3) Using the Cauchy- Riemann equations , verify that x2 y 2 2 ixy is not analytic. Solution : ∵ P x, y x2 y 2 2 ixy
Py 2 y 2 ixP , x 2 x 2 iy P y iP x P x, y is analytic. 109
4) Using Cauchy-Riemann equations , verify that x2 y 2 2 ixy is not analytic.
2 2 Solution : P x, y x y 2 ixy
Py 2 y 2 ixP , x 2 x 2 iy iP x 2 ix 2 y
Py 2 y 2 ix , P x 2 x 2 iy iP x 2 ix 2 y P y iP x P x , y is not analytic.
(5) Give an example of function which is continuous everywhere but not analytic
Solution: Let f() z xy iy u xy, v y . Since u and v are polynomials , they are continuous everywhere. u u y y y, x , 0, 1 x y x x Now, u v u v , x y y x Therefore f(z) is not analytic
6.4 HARMONIC FUNCTIONS :
If G is an open subset of , then the function UG: (i.e. Real valued function of complex variable) is harmonic if, it has continuous second order partial derivatives and 2u 2 v 0 (This is called Laplace’s equation) x2 y 2 e.g. u x, y ex .cos y is harmonic function ? u u ex.cos y e x sin y x y 2u 2 u ex.cos y e x cos y x2 y 2 2u 2 u excos y e x cos y 0 x2 y 2 Above function is Harmonic function. 110
Proposition : Let f be a analytic function in a region and f z u z iv z . If u and v have continuous second partial derivatives then u, v are harmonic function. OR 1) If f: G defined by f z u z iv z is analytic then, u Re f and v Im f are harmonic functions.
Proof : Given that f z u z iv z is analytic. Cauchy Riemann equations are satisfied. u v i.e. (I) x y u v and (II) y x
Differentiate equation (I) partially w.r.t. x and (II) w.r.t. y. 2u 2 v 2u 2 v and x2 x. y y2 x. y 2u 2 v 2 v 2 v 0 x2 y 2 x.. y x y is harmonic function.
Differentiate equation (I) and (II) partially w.r.t. y and x respectively. 2u 2 v 2u 2 v and x. y y2 x. y y2 2v 2 v 2 u 2 u Consider, 0 x2 y 2 x.. y x y v is harmonic function.
Definition : If f: G is analytic and f z u z iv z then u Re f and v Im f are harmonic conjugate i.e. u and v are harmonic conjugate then u and v are harmonic function and u, v are satisfied C – R equations.
Example : If f: G is analytic and f z u z iv z then prove that harmonic function u satisfies the partial differential 2u equations 0 . z. z 111
Solution : Given f is analytic. Let z x iy z x iy . z z z z Here, u u x, y where x and y . 2 2 u u x u y 1u 1 u u x, y .. z z x z y z 2 x 2 i y 1u u 1 i i i i 2 x y i i2 1 u u x u y 1u 1 u u x, y .. z z xz y z 2 x 2 i y 1 u u i 2 x y
OR 1 i z 2 x y 2u u 1 1 u u i . i z. z z z 2 x y 2 x y
1 2u 2 u 2 u 2u 1 2u 2 u i i =0 4 2 2 y . x y. x 4 2 2 x y x y 2u 0 z. z
Example: Prove that the function u x, y x3 3 xy 2 3 x 2 3 y 2 2 is harmonic. Find its harmonic conjugate and corresponding analytic function f z u z iv z . Solution : Given function u x, y x3 3 xy 2 3 x 2 3 y 2 2 u u 3x2 3 y 2 6 x and 6xy 6 y x y 2u 2u 6x 6 and 6x 6 x2 y2
2u 2 u 6x 6 6 x 6 0 x2 2 y u is a harmonic function. 112
To find v, we use Cauchy Riemann equation. u v u v and x y y x u v 3x2 3 y 2 6 x 3 x 2 3 y 2 6 x x y By integrating v 3 x2 3 y 2 6 x y 3y3 v 3 x2 y 6 xy x 3x2 y y 3 6 xy x 3 (1) where x is an arbitrary function of x. To find x, we use another equation of Cauchy-Riemann. u v y x 6xy 6 y 3 x2 y y 3 6 xy x x x 6xy 6 y 6xy 6 y x x x 0 Integrating, we get x c , where c is constant.
v 3 x2 y y 3 6 xy c i.e. the required harmonic conjugate. Analytic function f()(,)(,) z u x y iv x y x3 3 xy 2 3 x 2 3 y 2 2 ixyy 3 2 3 6 xyc Put x z and y 0 f( z ) z3 3 z 2 c (Alternate method to find harmonic conjugate using Milne Thompson method) Given function u x, y x3 3 xy 2 3 x 2 3 y 2 2 u 2 2 u let 1(,)x y 3x 3 y 6 x and 2 (,)x y 6xy 6 y x y Now, 2 f( z ) 1 ( z ,0) i 2 ( z ,0) (3 z 6z ) i (0) putting x z, y 0 f( z ) (3 z2 6 z ) dz z 3 3 z 2 Put z = x+iy , 113
f( z ) ( x iy )3 3( x iy ) 2 x3 3 xyi 2 3 xyi 2 2 yi 3 3 3 x 2 6 xyi 3 yi 2 2 Separating real and imaginary parts we get v 3x2 y y 3 6 xy
Example : If f: G defined by f z u z u z iv z is analytic and u v ex cos y sin y then find the function f z in terms of z.
Solution : f() z u iv if() z ui v (1 ifz ) ( ) uivuiv uviuv ( ) U iV() say x Ux e(cos y sin y ) 1 ( x , y ) x Uy e( sin y cos y ) 2 ( x , y )
(1)() i f z Ux iU y 1 (,0) z 2 (,0) z (1 ifz ) ( ) ( ez iedz z ) (1 iedz ) z f() z ez c
Proposition: Suppose that f is analytic in a region G. If f() z constant.
Proof : Let z x iy G and f z u z iv z . Given that f is analytic. Cauchy-Riemann equations are satisfied. u v u v and x y y x Here, we given that f z constant K (say). Let k 0 [If k 0 , it is obvious that f z 0]. u iv K u2 v 2 K u2 v 2 K 2 …….(I) Differentiate equation (I) partially w.r.t. x u v 2u 2 v 0 x x u v u v 0 x x u u u v u v 0 ………….(II) x y y x 114
Again, differentiate equation (I) partially w.r.t y u v u v 2u 2 v 0 u v 0 y y y y u u u v u v 0 {} ………….(III) y x x y
Multiplying equation (II) by u and equation (III) by v and add u u u2 uv 0 x y u u v2 uv 0 x y u2 v 2 u 0 dx u u K 2 0 0 x x
Multiply equation II by v and III by u and subtract. u u uv v2 0 x y u u uv u2 0 x y u2 v 2 u 0 dy u K 2 0 y u 0 y Using Cauchy-Riemann equation u v u v 0 and 0 x y y x v v 0 x y f is analytic at z. f is differentiable at z and y v f z i 0i 0 0 x x f z 0, z G f z = constant.
Theorem : Suppose that f is analytic in a domain (region) D then a) If f z 0, z D , f is constant. b) Ifanyoneof f , Re f , Im f , are f is constant in D (2008) 115
Proof : Let z x iy D and f z u z iv z Given that f is analytic in D. Cauchy-Riemann (C.R) equations are satisfied. u v u v i.e. and (1) x y y x a) If f z 0, z D then f is constant. (Already done, last proposition) b) i) Let f = constant.
u2 v 2 K (say) u2 v 2 K (2) where u x, y and v x, y are real valued function. Let K 0 [if K 0 then nothing to prove.] Differentiate equation (2) w.r.t. x u v 2u 2 v 0 x x u v u v 0 x x u v u v u v 0 (3) x y y x Again differentiate equation (2) w.r.t y u v 2u 2 v 0 y y u v u v 0 y y u u u v v u 0 (4) x y x y
Multiplying equation (2) by u and (4) by v and adding u u u2 uv 0 x y u u v2 uv 0 x y u2 v 2 u 0 dx u K 2 0 x u 0 x
Multiply equation (3) by v and (4) by u and subtracting 116
u u uv v2 0 x y u u uv u2 0 x y u2 v 2 u 0 dy u K 2 0 y u 0 y v v From equation (1), 0 and 0 x y u v f z i 0 i 0 0 x x z is arbitrary. f z 0 z D by part (a) f is constant. ii) Let Re f constant K (say) u x, y K (say) u u 0 and 0 x y from equation (1) v v 0 and 0 x y f is analytic. u v f is differentiable and f z i x x f z 0 i .0 0 f z 0 f z constant by (a) iii) Let Im f constant = K v x, y K (say) v v 0 and 0 x y f is analytic. v y f is differentiate and f z i x y 117
u v By Cauchy Riemann equation 0 x y f z 0 i .0 0 f z constant by (a) iv) Let arg f z constant 1 v arg f z tan K (say) u 1 v tan = K u u v tanK u v cot K vtan K u vcot K 0 Put cut K C u v c 0 unless v. c 0 But u cv is a real part of 1 ic f z 1 ic f z constant by (a) constant f z constant 1 ic constant 1 ic f is constant.
Example 1 : Prove that the function f z z is not differentiable anywhere in the complex plane.
Solution : We know that, f is differentiable at z if f z z f z f z (a unique limit) z as z 0 in any manner in -plane. f z z z z z z f z z f z f z lim z0 z z z z z z z lim lim lim z0z z 0 z z0 z x i y lim …………(1) z0 x i y Let z 0 along the Real axis. z x and y 0 118
x f z lim 1 x0 x Let z 0 along the imaginary axis. z i y and x 0 i y f z lim 1 y0 i y Two limits obtained are different. i.e. limit is not unique.
Given function f z z is not differentiable in to check 1 i z 0 . x y , we get . 1 i
2) Show that the function f z z 2 is differentiable only at the origin. (2006 )
Solution : Let f() z x2 y 2 . Since, x2 y 2 is continuous everywhere, f(z) is continuous everywhere. 2 2 f()() z0 z f z 0 z0 z z 0 f( z0 ) lim lim z z0 z z0 z (z zz )( zzz ) ( z zz )( zzz ) lim0 0 0 0 lim 0 0 0 0 z0 z z 0 z zzzz zz zzzz zz lim0 0 lim 0 0 z0 z z 0 z z lim z0 z 0 z z0 z (i) When z is real: Then y =0 and z z x . As z 0, x 0 z f( z0 ) lim z 0 z 0 z lim z 0 z 0 x z 0 z 0 x0 z x 0 (ii) When z is imaginary: Then x 0 and z i y,. z i y As z 0, y 0 z i z fz(0 ) lim z 0 zz 0 lim z 0 ( ) ziyzz 0 0 0 z0 z x 0 z Since the two limits are different along two different paths except at z=0, f() z0 does not exist anywhere except at z=0 Hence, f(z) is not differentiable anywhere except at z=0 119
x3 y 3 i x 3 y 3 x 0, y 0 i.e. z 0 3) Let f z x2 y 2 0x 0, y 0 i.e. z 0 Prove that C.R. equations are satisfied at the origin but f 0 does not exist i.e. f z is not differentiable there.
Solution : Let z x iy and f x u z iv z f z u z iv z x3 y 3 i x 3 y 3 z x, y 0 x2 y 2 0z 0 x3 y 3 i x 3 y 3 z 0 u x, y x2 y 2 0z 0 x3 y 3 i x 3 y 3 z 0 v x, y x2 y 2 0z 0 u u x hy u x, y x, y lim x h0 h u u h, 0 u 0, 0 0, 0 lim xh0 h h3 0 0 2 h lim h 0 lim h0h h 0 h
Similarly, 0 h3 0 u u0, u u 0, 0 2 h 0, 0 lim lim 0 h lim 1 yh0 h h0h h 0 h h3 0 0 v 2 h 0, 0 lim h lim 1 xh0 h h 0 h 120
0 h3 v v x,, y h v x y 2 h3 x, y lim lim 0 h lim 1 yh0 h h0h h 0 h2 u v u v and x y y x Cauchy-Riemann equations are satisfied at the origin. f z z f z f z lim z0 z f z f 0 f 0 lim z0 z 0 where z 0, in any manner in Complex Plane. Along the x-axis, y 0 and z x x3 ix 3 0 f x f 0 2 x3 1 i f0 x lim lim x lim h0x x 0 x x0 x3 1 i Along the y-axis, x 0 and z iy y3 iy 3 f y f 0 y2 f 0 lim lim iy0iy iy 0 iy i2 i i i 1 1 i i i Let z 0 along the line y x , x y , and z x ix x1 i.
f x f 0 i. 2 x3 ix i f 0 lim lim lim x1 i 0 x1 i x0 2x2 1 i x0 1 ix 1 i Hence, f 0 is not unique. The given f z is not differentiable at origin.
4) If f: G defined by f z u z iv z is analytic and suppose that u, v be any continuously differentiable function. 22 2 2 2 Prove that f z . x y u v Solution : Let z x iy and f x,,, y u x y iv x y Given, f z is analytic. Cauchy Riemann equations are satisfied. 121
u v u v and (1) x y y x Now, u, v where u u x, y and v v x, y and exist. x y u, v x x u v .. (2) u x v x Similarly, u v .. y u y v y By C.R. equations v u .. (3) y u x v x Squaring and adding equations (2) and (3) 22 2 2 u u v u v . . 2 . . . x y u x v x u x v x 2 2 v u u u . . 2 . . . u x v x u x v x 2 2 2 2 2 u v u v x u v x 2 2 2 2 u v (*) u v x x f is analytic at z. f is differentiable at z. u v u v f z i and f z x x x x 2 2 u v u v (*) i i u v x x x x 22 2 2 2 f z x y u v
5) Prove that the function Re z is differentiable anywhere in . f z z Solution : f z (a unique limit) z as z 0 in any manner in plane. 122
Here f z Re Z
f z Re z z
6) Prove that Cauchy Riemann equation can be written polar Co- u 1 dv v 1 u ordinates as and . r r d r r
Solution : Let f r,,, u r iv r Let x rcos and y rsin r x2 y 2 and tan1 y x Now, u u r, where r and are functions of x and y. u u r u .. x r x x r x2 y 2 and tan1 y x r 2 x r cos cos x 2 x2 y 2 r and 1 y rsin sin x2 y 2 2 x y x2 y 2 r r 1 x2 x 2 x u u u sin .cos x r r (1) u u r u u1 u Similarly, .. .sin cos . y r y y r r (2) v v sin v Similarly, cos . . x r r (3) v v cos v and sin . . y r r (4) By Cauchy – Riemann Equations u v u v u v and 0 x y y x x y Consider, u v u 1 u v 1 v cos sin sin cos x y r r r r 123
u 1 v v 1 u cos sin 0 r r r r (5) u v and 0 y x u 1 u v sin v sin cos . cos . 0 r r r r u 1 v 1 u v sin cos . 0 r r r r (6) Multiplying equation (3) cos and equation (6) sin and adding. 2 u 1 v v 1 u cos sin .cos 0 r r r r
2 u 1 v v 1 u sin sin .cos 0 r r r r
2 2 u 1 v sin cos 0 r r u 1 v 0 r r u 1 v r r Multiplying equation (3) sin and equation (6) cos and subtracting. u 1 v 2 v 1 u sin .cos sin 0 r r r r
u 1 v 2 v 1 u sin .cos cos 0 r r r r
2 2 v 1 u sin cos 0 r r 2 2 v 1 u sin cos 0 r r v 1 u 0 r r v 1 u r r 124
6.5 THE FUNCTIONS ez,sin( z ),cos( z ) etc :
Exponential Function : The exponential function in Complex z Plane denoted by e , is defined by as zn z2 z 3 ez 1 z ... n0 n! 2! 3! 1) The exponential function f z ez is analytic in the whole Complex Plane and f z f zwith f 0 1.
zn Solution : We have, f z ez . n0 n! 1 1 a a nn! n1 n 1 ! a 1 n+1 ! R lim n lim n an1 n n! 1 lim n 1 n The radius of convergence of the given power series is R . The power series converges for all z and convergence is uniform for each compact subset of . By using corollary. n [If f z an z has a radius of convergence R 0 then f is analytic in BR0, ] f z ez is analytic in whole Complex Plane.
z2n 1 z 2 n Note: Similarly, sinz ,cos z are analytic in n0(2n 1)! n 0 (2 n )! whole Complex Plane.
Definition : Entire Function : If the function f is analytic everywhere in whole Complex Plane (except at ) is called an Entire function or integral function. e.g. ez , cos z , sin z .
6.6. SUMMARY
1) A function f is said to be analytic (or holomorphic or regular) at a point z z0 , if f is differentiable at every point of some nbd of z0 . 125
n 2) If the power series an z has radius of cgs. R 0 , then n0 n f() z an z is analytic on BR0; .
3) Let u and v be real valued function defined on the domain G and suppose that u and v have continuous partial derivatives then f: G defined by f z u z iv z is analytic iff u and v u v u v satisfy Cauchy Riemann equation. i.e. and . x y y x
zn z2 z 3 4) ez 1 z ... n0 n! 2! 3!
6.7. UNIT END EXERCISES:
1) Give an example of a function which is not differentiable at the origin but the partial derivatives exist and satisfy the Cauchy- Riemann equations there .
Solution: Consider f : defined by xy() x iy f(,) x y (,)(0,0) x y x2 y 2 0 (x , y ) (0,0) .
fx (0,0) 0 similarly f y (0,0) 0 f( z ) f (0) xy But limz0 lim ( x , y ) (0,0) does not exist. For, on z x2 y 2 f( z ) f (0) the line y ax for z 0 . The limit depends z 1 a2 on real number a .
2) Check at what points does the function f() z z 2 is differentiable.
Solution : f() z z 2 . Let z x iy . 2 2 f() z x y . fx 2 x , f y 2 y f has continuous partial derivatives for all
f is differentiable at z , provided that fy if x . 2y 2 ix 2 y i 2 x 0 (,)(0,0) x y . f is differentiable only at the origin (0,0) 126
3) Show that f() z x2 iy 2 is differentiable at all points on the line y x .
2 2 Solution: f( z ) x iy . fx 2 x , f y 2 iy
fx, f y exist and are continuous functions of z x iy
fy 2 iy if x 2 xi iff x y 0 . This is possible iff x y . By proposition f is differentiable at all points on the line y x .
4) Suppose f is analytic in a region and ateverypoint, either f 0 or f 2 0 . Show that f is a constant function. (Hint: Consider The derivative of f2() z )
5) Find all analytic functions f u iv with u(,) x y x2 y 2
Solution: f u iv is analytic. f satisfies Cauchy- Riemann equations . ux v y, v x u y. 2 x v y , 2 y v x
vy 2 x , v x 2 y . v( x , y ) 2 xy c , where c is any real constant. f ( x2 y 2 ) 2 ixy ic
6) If f is a analytic in a region and if f is constant there , then show that f is constant.
Solution: If f 0, the proof is immediate , otherwise assume that f 0 . Let f u iv u2 v 2 0. Taking the partial derivatives w.r.t. x and y, we see that
uux vv x 0, uu y vv y 0 . Makinguse of the Cauchy- Riemann equations , we get ,
uux vu y 0, vu x uu y 0 2 2 (u v ) ux 0 u x v y 0 , similarly uy v x 0 . f is constant. 127
d 7) Show that (sin(z )) cos( z ) . (Hint: Use dz 1 that sin(z ) ( eiz e iz ) ) 2i 8) Find a power series representation for cos(z ) .
1 zn Solution: cos(z ) ( eiz e iz ), e z 2n0 n ! ()()izn iz n eiz , e iz n0n!! n 0 n 1 ( 1)nz2 n ()eiz e iz 2n0 (2n )! ( 1)nz2 n z 2 z 4 z 6 cos(z ) 1 ... n0 (2n )! 2! 4! 6!
2 2 9) Show that log(x y ) is harmonic in 0 .
Solution: Let u( x , y ) log( x2 y 2 ) . 2x 2 y ux , u y 0 x2 y 2 x 2 y 2 2y2 2 x 2 2 x 2 2 y 2 uxx , u yy ()()x2 y 2 2 x 2 y 2 2
uxx, u yy exist and are continuous functions of z on 0 , also uxx u yy 0 . u(,) x y is harmonic in 0 .
10) For the function f(z) defined by
()z 2 f z if z 0 z
= 0if z 0
Prove that C-R eq. are not satisfied at the origin , but the function f(z) is not differentiable at the origin(2009)
11) Find the holomorphic function f(z) whose real part is 2xy+2x (2008)
12) Find the analytic function f z u r, i v r, whose real 2 part is r cos2 . 128
2 z , z 0 13) For the function, f z defined by f z . z 0, z 0
Prove that the Cauchy Riemann equations are satisfied at 0, 0 but the function is not differentiable at 0, 0. 2 2 2 2 14) If f: G is analytic, prove that Ref z 2 f ' z 2 2 x y
15) Construct an analytic function f z u z iv() z , whose real part is cos x cosh y. express the result as a function of z.
16) Construct an analytic function f z u z iv() z , whose real part is ex( x cos y y sin y ) .
Express the result as a function of z.
146 8
COMPLEX INTEGRATION
Unit Structure : 8.0. Objectives 8.1. Introduction 8.2. Complex Line integrals 8.3. Integration along piecewise smooth path, The Closed Curve Theorem 8.4. Summary 8.5. Unit End Exercises
8.0 OBJECTIVES:
Through this unit we shall study the concept of complex integration, an integration of the form f() z dz taken over a piecewise smooth path further we shall derive certain properties of this integral. We would like to know further that what can be the integral of an entire function along a boundary of a rectangle in a complex plane, the answer is given in form of a closed curve theorem.
8.1 INTRODUCTION:
We have to recall theorem on differentiability of a power n series that states that a power series f() z an z converges for n0 n1 z CR . Then f() z exists and f() z nan z on the open disc n1 z CR . Therefore an everywhere convergent power series represents an entire function. Our main goal in this unit is to study the converse of this result namely that every entire function can be expanded as an everywhere convergent power series. This result has a consequence that every entire function is infinitely differentiable. We shall also arrive at these results by discussing integrals. Let us start by defining a Line integral. 147
8.2 COMPLEX INTEGRATION :
Definition : Trace of a curve : If x:, a b is a curve, then the set x t is called the trace x and is denoted by x x t: a t b The trace of x is always a compact set.
Definition : Contour : A contour is a piecewise smooth curve.
Definition : A complex valued function f is said to be continuous on a smooth curve x:, a b if, f z f x t u t iv t is continuous.
8.3 INTEGRATION ALONG A PIECEWISE SMOOTH PATH, THE CLOSED CURVE THEOREM :
Definition : Complex Line Integral :
Suppose f is complex valued, continuous and defined on open set G and that x:, a b is a piecewise smooth curve with x G . Then, the expression b n1 j1 f z dz f x t x t dt f x t x t dt x aj1 i where a t0 t 1 t ... tn 1 b is called the complex line integral of f over x.
This curve x is called path of integration of this integral.
Connection between Real and Complex line integral : If f z u z iv z then the complex line integral f z dz x can be expressed as f z dz u dx v dy i u dy v dx x x x
Theorem : Let x be such that x t x t i y t, a t b is a smooth curve and suppose that f and g are continuous function on open set G containing x then ; 148 i) f z dz f z dz where is a complex constant. x x ii) f z g z dz f z dz g z dz x x x
1 Example 1 : Evaluate dz , where x t eit , t 0, 2. z x Solution : By the definition of complex line integral. b f z dz f x t x t dt x a 1 Here, f z , x t eit , a 0, b 2 z 1 it f x t , x t i e eit 2 2 1 1 dz .i eit dt i dt i t2 2i z it 0 x 0 e 0
Definition : Rectifiable Curve :A curve is rectifiable if it has finite length.
Note : Every piecewise smooth curve is Rectifiable.
Definition : If x is s.t. x(). t a t b is rectifiable, then its length L x b us defined by L x x t dt a Example 1 : Find the length of the curve x t 4 eit , t 0, 2. 2 2 Solution : Length of x L x 4 i eit dt 4 i eit dt 0 0 2 2 4dt 4 8 dt 0 0
Example 2 : Find the length of the curve x t 1 i t , t 0, 4.
Solution : x t 1 i t x t 1 i 149
Length of 4 4 4 x L x 1 i dt 12 1 2dt 2dt 2 t 4 4 2 0 0 0 0
Definition : Opposite Curve : If x:, a b is a given curve then, the opposite curve x to x is defined as x t x a b t;, t a b
Example : Let x t eit ; t 0,.
x -x
-1 1 -1 1
Fig. 8.1
Definition : Let x1 :, a b and x2 :, a b be two smooth curves such that x1 b 1 x 2 a 2 . Then we define the path x1 x 2:, a 1 b 1 b 2 a 2 as follows.
x1 x 2 t x 1 t if t a 1 , b 1 OR xxt1 2 xtba 2 1 2 if tbbba 1 , 1 2 2
The path x1 x 2 is called the sum of two curves x1 or x2 or the union of two curves x1 and x2 .
Example 1 : Let x t t, t 0, 3 and x t 3 eit , t 0, 3 . 2 2 150
x 2
0 3 x1
Fig 8.2
Theorem : Let x be s.t. x()()() t x1 t iy 1 t be a smooth curve and suppose that f is a continuous function on an open set G containing x. Then, i) f z dz f z dz x x ii) f z dz s z dz x x iii) If M Max f x t and L L x (Length of x) t a, b
then f z dz ML . x (This property is called standard estimate for the integral.) iv) If x1 and x2 are smooth curves in G then, fzdz fzdz fzdz , where x1 x 2 are sum x1 x 2 x 1 x 2 of 2 curves.
Proof : i) By definition of opposite curve x t x a b t By definition of complex line integral. b f z dz f x t x a b t dt x a 151
Put a b t u dt du , When t a, u b t b, u a
a b f z dz f x u. x u du f x u x u du x b a f z. dz x ii) If f z dz 0 , then there is nothing to prove. x Let f z dz 0 x Put u ei , where arg f z dz x u 1
fzdz ei fzdz u fzdz (1) x x x z Re z f z dz Re f z dz Re u f z dz from (1) x x x Re u f z dz re u f z dz x x u f z dz ( Rez z ) x u . f z dz x
f z dz f z dz ( u 1) x x iii) Given that, M Max f x t and L L x t a, b By using part (ii), f z dz f z dz x x 152
b b fzdz fxt xtdt M x t dt ML x x a a
f z dz ML x iv) Let x1:, a 1 b 1 G and x2:, a 2 b 2 G with x1 b 1 x 2 a 2 .
We define x1 x 2: a 1 b 1 b 2 a 2 G x1 t if t a 1 , b 1 x1 x 2 t x2 t b 1 a 2 if t b 1 , b 1 b 2 a 2
x1 x 2 has derivative x1 t in a1, b 1 and t b1 a 2 in b1, b 1 b 2 a 2 b1 b 2 a 2 fzdz fxxtxx 1 2 1 2 tdt x1 x 2 a 1 b1 b1 b 2 b 3 fxtxtdt 1 1 xtbaxtbadt 2 1 2 2 1 2 a1 b 1 b1 b 2 b 3 fzdz fxtbaxtbadt 2 1 2 2 1 2 x1 b 1 Put t b1 a 2 u dt du , then t b1, u a2 , t b1 b 2 a 2 , u b2 . b2 f z dz f x2 u x 2 u du x1 a 2 f z dz f z dz x1 x 2
Note : i) If x is piecewise smooth then, there is a partition P a t0 t 1 ... tn b of a, b s.t. the restriction xk of curve x to tk1, t k is smooth for 1 k n . x x1 x 2 ... xn fzdz fzdz fzdz fzdz ... fzdz x x1 x 2 xn x 1 x 2 x n ii) Re f z dz Re f z dz x x 153
Example 1 : Let f z 1 and x t it , t 0,1 . b 1 f z dt f x t x t dt 1. i . dt i x a 0 Re f z dt Re i 0 (1) x 1 and Re f z dz 1. i dt 1 (2) x 0 From equation (1) and (2) Re f z dz Re f z dz x x
Change of Parameter : Let x:, a b and :,c d be two smooth curve. Then the curve is equivalent to curve x if, there is a function :,,c d a b which is contain non-decreasing and with c a and d b s.t. x .
Here, we call the function a change of parameter. This new curve x is called the Reparametrization of the curve x.
Theorem : Let x:, a b be a smooth curve and suppose that :,,c d a b is a continuous non-decreasing function with c a and d b . If f is continuous on x then f z dz f z dz x x
Proof : Given that x:, a b is a smooth curve and :,,c d a b is continuous non-decreasing function with c a and d b .
By hypothesis, there is a change of parameter c,, d a b s.t. x s x s and x s x s s for s c, d d f z dz f x s x s ds x c 154
d f x s x s s ds c Put t s dt s when s c , t c a and when s d , t d b b f z dz f x t x t dt x a f z dz f z dz x x
Fundamental theorem of calculus : If f is continuous on a, b and F x f x in a, b then b f()()() x dx F b F a . s
Primitive or Antiderivative of a function : A function f: G is said to be primitive or antiderivative of f in G if, F is analytic in G and F z f z in G.
Theorem : Let G be an open set in and suppose that f: G is a continuous function with primitive FG: . If x:, a b G is a smooth curve, then f z dz F x b F x a . x In particular, if x is closed then f z dz 0 . x Proof : Given that, f: G is a continuous function with primitive f: G . F z f z b fzdz fxt xtdt x a b F x t dt ( F x t f x t a b F x t (by using fundamental) a F x b F x a If x is closed, then x a x b. f z dz F x a F x b 0 x 155
it2 Example : Evaluate z2 dz where x t t , t 0,. x Solution : Given integral is z2 dz . x Here, f z z2 z3 F z is primitive of f. 3 a 0, b . By previous theorem, b f z dz F x t a x 3 it2 it2 1 z2 dz f t t . 3 x 0 0 2 3 1 1 3 i 0 i 3 3 3 1 i3 3
Definition : The index of curve or winding number : If x is a closed rectifiable curve in and if x , then, 1 dz x, is called the index of x w.r.t the point . 2i z x It is also called the winding number about .
x b x c x a x a b c
n(x ; a)=0
x; 1 x; 2 x; c 1 x; 0 x; 1 8.3 156
x x: x t z : z r eit , t 0, 4 a r
z r eit r e it r n(x;a)=2
Fig 8.4
Theorem : If x : 0,1 is a closed rectifiable curve and x then x; is an integer. Proof : Define g : 0,1 by, t x s ds g t t 0,1 x s 0 g 0 0 , g is continuous on 0,1 and dz g 1 (*) z x (put x s z x s ds dz ) To prove that g1 2 in for some integer . from equation (1), x t g t , t 0,1 x t d Now, eg t xt extxt g t e g t gt dt x t g t g t e x t x t e x t x t x t eg t.0 0 eg t. x t constant = K (say) OR x t K. eg t (2) To find K, put t 0 g0 0 0 x0 K e K e K e 1 K x0 157
Putting the value of K in (2), we get x t x0 . eg t Put t 1, x1 eg1 1 ( x is closed x1 x 0 ) x0 g1 2in 2in e e ( e 1) g1 2 in For some integer n substitute the above value in (*), we get dz 2in z x 1 dz n n x; 2i z x
Component of a Metric space: A subset D of a metric space X is a component of X if D is a maximal connected subset of X i.e. D is connected and there is no connected subset of X that properly contains D.
D3 D4
D1
D2
G x Fig 8.5
Note: If G is open then component of G also open.
Simply and multiply connected domains :
Definition : A domain D is said to be simply connected if any simple closed curve which lies in D can be shrunk to a point without leaving domain D.
Definition : A domain which is not simply connected is said to be a multiply connected domain. 158
Green’s Theorem : Let M x, y and N x, y be continuous and have continuous partial derivatives in a domain and on its boundary x, Green’s theorem states that, NM M dx N dy dx dy x y x
Cauchy Theorem : Let G be an open set in and suppose that x:, a b G is a smooth curve. If f is analytic with f continuous inside and on a simple closed curve x then, f z dz 0 . x G x f
W
Fig 8.6
Proof : Let z x iy , f z u z iv z and Int x fzdz uz ivz dx idy x x udx vdy i vdx udy (1) x x Given that, f is analytic in and on its boundary x. f is continuous in and hence u and v are continuous. Also, given that, f is continuous inside and on a simple closed curve x. Partial derivatives of u and v are also continuous in and on its boundary x. By Green’s theorem v u u dx v dy dx dy x y x 159
u v and v dx u dy dx, dy x y x Substituting above values in equations (1), we get v u u v fzdz dxdyi dxdy x y x y x Given that, f is analytic. Cauchy Riemann equations are satisfied. u v v v and x y y x u u v v fzdz dxdyi dxdy y y y y x f z dz 0 x
Note : In Cauchy’s theorem, Cauchy assumed the continuity of derived function f z . It was Goursat who first proved that this condition can removed from the hypothesis in the theorem. The revised form of the theorem is known as Cauchy- Goursat theorem which we shall study in the next chapter.
8.4. SUMMARY
b n1 j1 1) f z dz f x t x t dt f x t x t dt x aj1 i f zdz udx vdy i udy vdx x x x
2) If x:, a b is a given curve then, the opposite curve x to x is defined as x t x a b t;, t a b
3) If x is piecewise smooth then, there is a partition P a t0 t 1 ... tn b of a, b s.t. the restriction xk of curve x to tk1, t k is smooth for fzdz()()() fzdz fzdz h x1 x 2 x 2
z h z z hz0 z h Fzh()()()()()()()() Fz fz fwdw fwdw fwdw fwdw fwdw z0 z 0 z 0 z z 160
F( z h ) F ( z ) 1z h f ( z ) lim f ( z ) f ( w ) dw h h0 h hz h 1z h z h 1 z h fwdwfz()()(()()) dw fw fzdw h z z h z 1z h 1 z h 1 0 f ( w ) f ( z ) . dw dw h hz h z h 1 k n . fzdz fzdz fzdz fzdz ... fzdz x x1 x 2 xn x 1 x 2 x n
x x1 x 2 ... xn
4) Let x:, a b and :,c d be two smooth curve. Then the curve is equivalent to curve x if, there is a function :,,c d a b which is contain non-decreasing and with c a and d b s.t. x .
5) Fundamental theorem of calculus : If f is continuous on a, b and F x f x in a, b then b f()()() x dx F b F a . a
6) Primitive or Antiderivative of a function : A function f : is said to be Primitive or Antiderivative of f in G if, F is analytic in G and F z f z in G.
7) The index of curve or winding number : If x is a closed rectifiable curve in and if x , then, 1 dz x, is called the index of x w.r.t. the point . 2i z x It is also called the winding number about .
8) A domain D is said to be simply connected if any simple closed curve which lies in D can be shrunk to a point without leaving domain D. 161
8.5 UNIT END EXERCISES:
1) Suppose f() z x2 iy 2 where z x iy . Then evaluate f() z dz , where e c: z ( t ) t it , 0 t 1.
Solution: Consider c: z ( t ) t it , 0 t 1. Then z( t ) 1 i , and 1 2 2 2i f( z ) dz ( t it )(1 i ) dt . c 0 3
2) Find the integral of the function taken over a circle of radius .
1 x iy Solution: f() z z x2 y 2 x 2 y 2 c:)) z t R cos() t iR sin(),0 t t 2, R 0 . Then 2 cos(t ) i sin( t ) f( z ) dz 0 R sin( t ) iR cos( t ) 2 i c RR
3) Let c be any smooth curve in. Let f( z ) 1. Then find f() z dz . c
4) Let c be the Unit circle and suppose that f 1 on c . Then prove that f( z ) dz 2 . c (Hint: ML 1, 2 . Apply ML formula.)
5) Let c be any closed curve not passing through the origin , then show that 1 k dz 0 : z dz 0, k 1, k is any integer. cz2 c 1 1 (Hint: g()() z F z where F() z and F() z is analytic z2 z everywhere except at the origin .)
6) Evaluate ()z i dz , where c is the parabolic segment c c: z ( t ) t it2 , 1 t 1
Solution : Let f() z z i . Then f is the derivative of an analytic z2 function F() z iz . 2 162
By proposition , f( z ) dz F ( z ( b )) F ( z ( a )) c z2 z 2 (z i ) dz iz iz 0 (z i ) dz 0 . c 2 2 c 1i 1 i
1 7) Find zz dz . Where (a) is the upper half of the Unit circle y from +1 to -1. (b) is the lower half of the Unit circle from +1 to -1. (Hint: (a) Let (t ) cos( t ) i sin( t ),0 t (b) Let ()t cos() t i sin(), t t 2
8) Let (t ) 2 eit , for t . Find (z2 1) 1 dz .
9) Prove the following integration by parts formula . Let f and g be analytic in G , let be a rectifiable curve from a to b in G . Then show that fg fbgb()()()() faga fg .
z 10) Evaluate the integral z esin z z dz where is the circle z 2 . 163 9
CAUCHY THEOREM
Unit Structure : 9.0. Objectives 9.1. Introduction 9.2. Cauchy Theorem for an Open Star shaped Domain 9.3. Cauchy Integral Formula 9.4. Summary 9.5. Unit End Exercises
9.0. OBJECTIVES:
Our main goal in this unit is toshow that a function analytic in a disc can be represented as a power series. We shall prove Cauchy’s theorem for an open star-shaped domain and Cauchy integral formula for an analytic function in a disc.
9.1. INTRODUCTION :
We have seen that a function is analytic on a closed curve c but f 0 . For example consider the function c 1 f : 0 defined as f() z . In this example z 1 f( z ) dz dz 2 i . Whereas , the closed curve theorem z1 z 1 z states that if f is analytic throughout a disc, the integral around any closed curve is 0. Weshalltry to find the most general type of domain in which the closed curve theorem is 1 valid. We should note that f() z is analytic in the punctured z plane 0 . We shall see that the existence of a hole at z 0 allows us to construct an example above , for which the integral is non-vanishing. The property of a domain, which assures that it has no holes is called simple connectedness. The formal definition is as follows. 164
Definition : A domain D is said to be simply connected if any simple closed curve which lies in D can be shrunk to a point without leaving domain D.
Definition :Singular point : A point at which the function f is not analytic is said to be a singular point or singularity of the function f. z2 e.g. f x z 3 Here, f is not defined at z 3 and hence not analytic at z 3, therefore z 3 is singular point.
9.2 CAUCHY THEOREM FOR AN OPEN STAR SHAPED DOMAIN :
Theorem : Let G be star like w.r.t. point z0 and suppose that f is analytic in G. Then there exists an analytic function F in G s.t. F z f z in G.
In particular, f z dz 0 , for every closed, piecewise smooth curve x x in G. (2008)
Proof : Given that, G is a starlike w.r.t point z0 and f analytic in G. By definition, z0, z G z G Fix a point z in G and define z F z f d f d z0, z z 0
z + h
z z0
Fig 9.1 165
Choose h with h 0 s.t. B z; h G and z, z h G
Since G is starlike w.r.t. point z0 .
The triangle z0,, z 1 z h is contained in G. f is analytic inside and on the boundary of the triangle . By Cauchy Goursat theorem, f()() z dz d z z h z0 f d f d f d 0 z0 z z h z h F z f d F z h 0 z z h F z h F z f d z z h z h FzhFzhfz f d fzd z z z h F z h F z 1 f z f f z d h h z z h F z h E z 1 f z f f z d (1) h h z Given that, F is analytic in G. f is differentiable in G. f is continuous at a point z G . For a given 0, 0 s.t. f f z whenever z Choose z h From equation (1), z h F z h F z d . h h h h z F z h F z By definition lim f z h0 h F z f z in B z; h 166
Since z is fixed but arbitrary. F z f z in G. (2) The derivative F exists and is continuous at every point z in G. F is analytic in G. From equation (2), F is primitive of f. By using the theorem f z dz 0 for every closed, piecewise smooth curve x in G. x
In Cauchy’s theorem, Cauchy assumed the continuity of derived function f z . It was Goursat who first prove that this condition can removed from the hypothesis in the theorem. The revised form of the theorem is known as Cauchy-Goursat theorem.
Cauchy-Goursat Theorem : (Cauchy Triangular Theorem) :
Let f be analytic in an open set GD . Let z1,, z 2 z 3 be points in G. Assume that the triangle with vertices z1,, z 2 z 3 is continuous in G then f z dz 0 where is the boundary of a triangle . x (2007,2008, 2009) Proof : Given that, the triangle with vertices z1,, z 2 z 3 is continued in G. Let m1,, m 2 m 3 be mid points of line segment z1,,, z 2 z 2 z 3 , z3, z 1 respectively. Then we get 4 smallest triangles 1,,, 2 3 4 . z 3
3
m2
m3 4 2
1 z 2 m1 z1 Fig 9.2 167 f z dz f z dz z m z m z m z 1 1 2 2 3 3 1 f z dz f z dz m z m m m z m m 3 1 1 3 1 2 2 1
f z dz f z dz m z m m m m 2 3 3 2 3 1
f z dz f z dz m m m m 1 2 2 3 f z dz f z dz f z dz f z dz 1 2 3 4 4 f z dz K 1 K
Among this 4 triangles, there is one triangle, call it , s.t.
f z dz f z dz K 1, 2, 3, 4 1 K
f z dz 4 f z dz 1 Let L be the perimeter of a triangle , then 1 1 LL and diam. diam. (diameter of triangle 1 2 1 2 means the length of its largest side)
Now, perform the same process on the triangle 1 getting a triangle 2 with analogus properties. (i) 0 1 2
(ii) f z dz 42 f z dz 2 1 (iii) LL2 22 1 (iv) diam2 diam 22 Continue the process and at the nth stage, we get (i) 0 1 2 ... n 168
(ii) f z dz 4n f z dz n 1 (iii) LLn 2n 1 (iv) diamn diam 2n A metric space is complete iff for any sequence Fn of non-empty closed sets with FFF1 2 3 ... and diam Fn 0, F n consists of a single point. n1 Since is complete and n is a sequence of non-empty closed sets with 0 1 ... n 1 and limdiamn lim diam n 2 n n 0 limxn 0 if x 1 n By using cantor’s theorem, n consists of a single point say z0 . n0 In particular z0 G . Given that, f is analytic on G. f is differentiable at a point z0 G for every 0, 0 s.t. f z f z0 f z0 where 0 z z0 z z0 By increment theorem, fzfz 0 zzfz 0 0 zz 0 z with z is continuous and z for z z0 . fzdz fzdz 0 zzfzdz 0 0 zz 0 zdz n n n n 0 0 z z0 z dz ( f z 0 if x is n x closed cure and here n is closed.)
f z dz z z0 z dz n n 169
1 Choose s.t. diamn diam . 2n z0 n n B z 0;
d
Dn
Fig 9.3
f z dz diamn dz n n ( 2 and z z0 diamn ) diamn L n ( Ln d2 ) n 1 1 .. diam L 2n 2 1 1 ( diamn diam and LLn 2n 2n 4n f z dz diam . L n
f z dz diam. L ( by (ii), n
f z dz 4n f z dz ) n Since 0 is arbitrary. f z dz 0 170
Check your progress: 1) State and prove Cauchy Goursat Theorem for a closed quadrilateral. 2) State and prove Cauchy Goursat Theorem for a closed Rectangle.
Theorem : Let G be an open set in . Let f be analytic in G except possibly at a point z0 G . Assume that f is continuous in G and that the triangle with vertex at z0 is contained in G. Then f( z ) dz 0 , where is a boundary of the triangle .
Proof : Let z0,, z 1 z 2 .
Let 1be a point on the line segment z0, z 1 and 2 be a point on the line segment z0, z 2 .
z2
x 2 2
3 1 z1
x1 z3
Fig 9.4
Consider the subtriangles, 1 201,,,,,,,,z 2 z 122 z 3 112 z f z dz f z dz 2z 0 1 z 1 z 2 2 fzdz fzdz fzdz 2012z 1121 z z 1221 z fzdz fzdz fzdz f z dz 0 ... 1 2 3 1 (by Cauchy Goursat theorem) 171
f z dz f z dz 1 Put M max f z z
f z dz M dz ML1 1 As 1 and 2 tend to z0 , perimeter of the triangle 1 tends to zero i.e. L1 0. f z dz 0 x2
1 x z0 1 Fig 9.5
Theorem : Let G be an open set in . Let f beanalytic G for some G . If f iscontinuousonG,then f z dz 0, where is a boundary of the triangle contained in G.
Star Shaped Domains : Definition : A set G in is said to be convex if, given any two points z and w in G, the line segment z, w lies entirely in G .
Fig 9.6 172
Definition : A set G is said to be starlike (or star shaped) w.r.t. points G if for any point z in G, the line segment ,z lies entirely in G.
z a a a
Open sets, closed, half planes The above set is starlike w.r.t. if Rez 0, Im z 0 etc. are
a
Fig 9.7
Punctured disk is not starlike w.r.t .
Note : Every starlike set is not convex but every convex set is starlike.
Question : If f is analytic in a simply connected domain D, then z2 f z dz is independent of path in D, joining any two points z1 z1 and z2 in D.
Solution : Let x1 and x2:, a b G be two smooth paths in G such that x1 a x 2 a z 1 and x1 b x 2 b z 2 . x1 t 1 x 2 t 2,,, t 1 t 2 a b Y D
z2
x2 x1
z1 X s Fig 9.8 173
Form a simple closed curve which moves from z1 to z2 via x1 and z2 to z1 via x2 . f is analytic inside and on a simple closed curve . By Cauchy Goursat theorem, f z dz 0 f z dz f z dz 0 x1 x 2 f z dz f z dz x1 x 2 z2 fzdz fzdz fzdz x1 x 2 z 1 z2 f z dz is independent of path. z1 Cauchy Deformation Theorem : Statement : If f is analytic in c domain bounded by two simple closed curves x1 and x2 (where x2 is inside x1 ) and on these curves, then f z dz f z dz where x1 and x2 are both x1 x 2 traversed in anticlockwise direction.
Proof : Join two curves x1 and x2 by lines AB and CD. Denote, x1 = lower section of x1 from A to D. x1u x1
x x2u 2 C D
A B x 2l
x 1l Fig 9.9 174 x1 u = the upper section of x1 from D to A. x2 = the lower section of x2 from B to C. x2 u = the upper section of x2 from C to B.
Form a simple closed curve 1 by transversing from A to B then from B to C by x2 u , then from C to D and finally from C to D and finally from D back to A by x1 u . f is analytic inside and on the simple closed curve 1. By Cauchy- Goursat theorem. f z dz 0 1 fzdz fzdz fzdz fzdz 0 (1) AB x2 u CD x 1 u
From a simple closed curve 2 by transversing from A to D by x1 then from D to C, then from C to B by x2 and finally from B back to A. f is analytic inside and on the simple closed curve 2 . By Cauchy - Goursat theorem, f z dz 0 2 fzdz fzdz fzdz fzdz 0 (2) x1 DC x 2 BA Adding equations (1) and (2), we get fzdz fzdz fzdz fzdz 0 x1 u x 1 x 2 x 2 u f z dz f z dz x x f z dz f z dz 0 x1 x 2 f z dz f z dz x1 x 2
Generalization of Cauchy Deformation Theorem : Statement : If f is analytic in a domain bounded by non-intersecting simple closed curves x, x1 ,..., xn where x, x1 ,..., xn are inside x and on this curves, then 175 fzdz fzdz fzdz ... fzdz where, x x1 x 2 xn x, x1 , x 2 ,..., xn are traversed in anticlockwise direction.
a1 a3 a2 a4
bn-1 bn
Fig 9.10
8.3 CAUCHY INTEGRAL FORMULA :
Statement : Let f be analytic in a simply connected domain G, GD . If x is a simple closed curve in G and be any point inside x then, f z dz2 i f z 2 i f where, x is traversed in z z x anticlockwise direction.
Proof : Given that, f is analytic in a simply connected domain G. construct a circle with centre at and radius r 0 so that lies entirely inside x .
r z a
G
Fig 9.11 176
f z The function is analytic in a domain which is bounded z by two simple closed curves x and and on these curves. By Cauchy deformation theorem, f z f z f z f f dz dz dz z z z x f z f z f dz dz dz f (1) z z z x dz Consider, z Equation of the circle is, z r or z erit, t 0, 2 dz i reit dt z r z r rei, 0, 2 reit , t 0, 2 a+reiq
q a+ r a r
Fig 9.12
2 2 dz ireit dt i dt 2 i z it 0re 0 Equation (1), becomes, f z f z f dz dz 2 i f z z x 1 f z 1 f z f dz f dz 2i z 2 i z x 1 f z 1 f z f dz f dz (2) 2i z 2 z x 2i 2 i 2 i 1 177
Given that, f is analytic in G. f is differentiable in G. f is continuous in G. f is continuous at a point G . For a given 0 , 0 such that, z f z f Choose, r s.t. r From equation (2) 1 f z dz dz f z 2i z 2 r x dz .LT . 2r 2r 2r 2r 1 f z dz f 2i z x is arbitrary. f z 2i f z x
Theorem : Let f be analytic in a simply connected domain G. G . If x is closed rectifiable curve in G and G x, then 1 f z dz f x; . 2i z x
Proof : Given that, f is analytic in a simply connected domain G. Define F z as follows, f z f ,z z F z (1) f x, z f z f F z is analytic in G . z 1 ( f z f and are analytic in G ) z f()() z f lim()limf z f' () F () by (1) z z z F is continuous at a point G and hence F is continuous in G. F is analytic in G. 178 By Cauchy- Goursat theorem F z dz 0 x f z f dz 0 z x f z dz dz f z z x x f z 1 dz f. 2 in x , ... x ; z 2 i x 1 f z dz f.; x 2i z x z2 2 Example 1 : Use Cauchy integral formula, to evaluate dz z 1 x where x is a circle z 2. z2 2 Solution : Given that, dz z 1 x f z z2 2 F z z1 z 1
x
0 1 2
Fig 9.13 F has singular point at z 1. Given equation of circle is z 2. Centre is origin and radius is 2. The singular point z 1, lies inside the circle. We use Cauchy integral formula f z dz 2 i f z z z x 2 z 2 2 dz 2 i z 2 2i 1 2 2 i 3 6i . z 1 z1 x 179
cotz dz z 1 2) Evaluate where x is a circle 2 . x cos z Solution: cot z dz dz sin z x x For finding singular point put sinz 0 z sin1 0 n z 1 Singular point z=0, lies inside the circle 2 . cos z Hence by Cauchy’s integral formula dz 2 i cos 0 2 i x sin z
sin(z ) 3) Evaluate where C is the unit circle oriented z 1 c 2 clockwise. sin(z ) Solution: Let I= z 1 c 2 z 1 is a point of singularity and lies inside z 1 2 z 1 , f z sin z 0 2 I 2 if z 2 i sin 0 2
Theorem : Let f be analytic in a simply connected domain G, G and suppose x is a simple closed curve in G. If is any point inside x then, 1 f z f dz , where x is traversed in 2i 2 x z anticlockwise direction.
(Note: To prove the theorem we will need the following theorem Boundedness Theorem :Let f be continuous on a compact set S. then, f is bounded on S i.e. there exists a number M s.t. f z M z S )
Proof : Given that, x is a simple closed curve in G and is any point inside x. By Cauchy-integral formula, 1 f z 1f ( z ) f dz and f() h dz 2i z 2ix z ( h ) 180
x z
a+h
a
Fig 9.14 1 f z f z f h f dz dz 2i z z h x x 1 z h f z z f z 2i z z h x 1 hf z dz 2i z h z x f h f 1 f z dz h2 i z h z x 1 z f z z h f z dz 2i 2 x z h z 1 h f z dz 2i 2 x z h z h f z dz 2i 2 x z h z f h f 1 f z h f z dz dz h2 i 2 2 2 x z x z h z (1) Theorem will be proved if L.H.S. of equation (1) tends to zero as h 0 . Choose r inf z : z x r z . r Choose h h 2 z h z h a b a b r r r 2 2 181 f is analytic in G. 2 f is continuous on x x is compact set By boundedness theorem, M 0 s.t. f().V z M z x From equation (1),
f h f 1 f z h M. dz dz h2 i 2 2 r .r 2 x z x 2 h . M h . M dz L x 3 3 r x r L.H.S. tends to zero as h 0 . f h f 1 f z lim dz h0 h2 i 2 x z
1 f z f x dz 2i 2 x z
Generalization of the above theorem : Theorem : Let f be analytic in a simply connected domain G, G and suppose x is a simple closed curve in G. If is any point inside x, then n! f z fn dz 2i n1 x z where n 0,1, x is traversed in anticlockwise direction. Prove this theorem by induction on n.
Note : If a function f is analytic at a point G , then its derivatives of all orders are also analytic at a point .
Example 1 : Use Cauchy integral formula or theorem, to evaluate sinz cos z 3 dz where x is a circle z . z1 z 2 2 x
sinz cos z Solution : Given integral dz z1 z 2 x 182
By partial fraction, sinz cos z sin z cos z sin z cos z dz dz . z1 z 2 z 1 z 2 x x x f z Here, F z F z has singular point at z 1 z1 z 2 and z 2 .
(Note : If the singular points lies inside the circle then we use Cauchy integral formula and it lies outside the circle then we use Cauchy integral theorem.) 9 Given equation of circle, z 3 i.e. x2 y 2 2 4 Singular point z 1 lies inside the circle and z 2 lies outside the circle. For z 1, we use Cauchy integral formula f z dz 2 i f z z z x sinz cos z dz 2 i sin z cos z z 1 z1 x 2i sin cos 2i 0 1 2i ….(a) For z 2 , we use Cauchy integral formula theorem F z dz 0 x sinz cos z 0 (b) z 2 x Substituting (a) and (b) in equation (1), we get sinz cos z dz 0 2 i 2 i z1 z 2 x
e2z 2) dz where x is the rectangle with vertices at i 3 x 1 z 2 and 1 i .
f z Solution : Here, F z has singular point at z 1 of 3 2 1 z 2 order 3. 183
i (1,1) 1+i
0 1/2 -i 1-i Fig 9.15
By Generalization of derivative of an analytic function. n! f z fn dz 2i n1 x z f z 2i dz f n n1 n! x z z 2 e 2i 2 1 d 2 dz f i e z 3 2 ! 2 2 1 x 1 dz z z z 2 1 2 i4 e2z i4 e 2 4ie z 1 2
Exercise : Use Cauchy integral formula to evaluate, cos z i) where x is a rectangle with vertices at 2 i and 2 x z 1 2 i . ii) tan z dz where x is a circle z . 2 2 x ez z iii) dz where x is a circle z 3 . 3 x z 2 z2 1 iv) dz where x is a circle z 3 1. 1 x z i 2 1 Singular point is P i and here centre c 3, 0 2 2 1 1 2 2 p d c, p d 3, i 3 0, 1 1 . 2 2 4 Singular point p lies outside the circle. 184
z2 2 z 3 v) where x is a circle z i 2. z i x
Cauchy estimate or Cauchy inequality : Theorem: If f is analytic in an open disk BR; and f z M z B, R n! M then f n n 0,1, 2,... Rn
Proof : For 0 r R , Construct a circle x with centre at and radius . By generalization of the theorem on Derivative of an analytic function. n! f z fn dz n 0,1, 2,... 2i n1 x z n! f z fn dz (1) 2 n1 x z Given that, f z M z x For any point z on x, we have z r From equation (1),
n n! M n! M n! M f dz dz . 2r 2 n1 2 n1 2 n1 x r r x r n! M f n rn r R is arbitrary. n n! M As r R , we have f n 0,1, 2, ... Rn
Cauchy Integral formula for Multiply connected domains: Theorem: It f is analytic in a domain which is bounded by two simple closed curves x1 and x 2 (where x2 lies inside x1 ) and on these curves and if z0 is any point in G. Then 1f ( z ) 1 f ( z ) f() z0 dz dz 2i z z0 2 i z z 0 x1 x 2 Where x1 and x 2 are transverse in anticlockwise direction 185
Proof: Construct a circle x with centre at z0 and radius r, so that x lies inside x1 f() z The function is analytic in a domain which is bounded by z z0 non-intersecting simple closed curves x1,, x x 2 where ( x1 and x 2 lies inside x1 ) and on these curves x 1 x
r z0 x2
Fig 9.16
By using Cauchy determination theorem, f()()() z f z f z dz dz dz (I) z z0 z z 0 z z 0 x1 x 1 x 2 f() z By using Cauchy integer formula, dz2 i f ( z ) z z 0 x 0 Put this value in equation (I), f()() z f z dz2 i f ( z0 ) dz z z0 z z 0 x1 x 2 1f ( z ) 1 f ( z ) f() z0 dz dz 2i z z0 2 i z z 0 x1 x 2
9.4. SUMMARY :
1) Let G be starlike w.r.t. point z0 and suppose that f is analytic in G. Then there exists an analytic function F in G s.t. F z f z in G. In particular, f z dz 0 , for every closed, piecewise x smooth curve x in G. 186
2) Cauchy-Goursat Theorem : (Cauchy Triangular Theorem) : Let f be analytic in an open set GD . Let z1,, z 2 z 3 be points in G. Assume that the triangle with vertices z1,, z 2 z 3 is continuous in G then f z dz 0 , where is the boundary of a x triangle .
3) Cauchy Deformation Theorem : Statement : If f is analytic in c domain bounded by two simple closed curves x1 and x2 (where x2 is inside x1 ) and on these curves, then f z dz f z dz where x1 and x2 are both x1 x 2 transverse in anticlockwise direction.
4) Statement : Let f be analytic in a simply connected domain G, GD . If x is a simple closed curve in G and be any point inside x f z then, dz2 i f z 2 i f z z x where, x is traversed in anticlockwise direction.
5) Theorem : Let f be analytic in a simply connected domain G, G and suppose x is a simple closed curve in G. If is any 1 f z point inside x then, f dz , where x is traversed 2i 2 x z in anticlockwise direction.
6) Generalization of the above theorem : Theorem : Let f be analytic in a simply connected domain G, G and suppose x is a simple closed curve in G. If is any point inside x, then n! f z fn dz ,where n 0,1, x is 2i n1 x z traversed in anticlockwise direction.
9.5. UNIT END EXCERCISES :
1) Suppose f: G be an analytic function, define : GG by f()() z f w (,)z w if z w z w f() z if z w . 187
Prove that is a continuous and for each fixed w, z (,) z w is analytic function of z.
( Hint: Take z z0 h, w w 0 k and z0 h w 0 k for any h, k .
Consider lim(,)(0,0)h k (z 0 h , w 0 k ) ( z 00 , w ) , similarly lim(,)(0,0)h k (z 0 h , w 0 k ) ( z 00 , w ) for z0 w 0 and h k or h k .
2) Let be a closed rectifiable curve in and a . Then show that for n 2 , (z a )n dz 0 .
Solution: Use the lemma that, if is a rectifiable curve and ϕ is a function defined and continuous on { }.
For each m 1, F( z ) ( w )( w z )m dw for z . Then each m
Fm is analytic on
and Fm mF m1 . Take m 1, 1 on . Then F()() z w z1 dw . 1
F1()() z F 2 z on . Here a .
F1()() a F 2 a F2 ( a ) 0 , since F1() a is constant number independent of a. (z a )2 dz 0 Inductively (z a )n dz 0 for n 2 .
3) Let f be analytic on DB (0,;1) . Suppose f( z ) 1 for z 1. Then show that f (0) 1
( Hint: f is analytic in a simply connected set B(0,;1) . 1 Let boundary B0; , then is a simple closed curve and 2 α=0 is a point inside . 1f ( z ) Consider f(0) dz 2i z1 188
1f ( z ) 1 f ( z ) 1 f(0) dz .2 1. 2i z2 2 z 2 2
n it z 4) Let (t ) 1 e for 0 t 2 .Find dz for all positive z 1 integers n .
n f() z n z n ( Hint : Put f() z z . Then dz f( z ) /( z 1) dz . z2 z 1 Where ()t (1 cos(),sin()) t t for 0 t 2 . Apply the following Cauchy- Integral formula n n!() f z f()(;) a a dz , for n 1. 2i ()z a n1
5) Use Cauchy- Integral formula to evaluate cos(nz ) (i) dz , is a rectangle with vertices at 2 i , 2 i z2 1
ez z (ii) dz , is a circle z 3. (z 2)3
Solution (ii) Let f() z ez z .
By generalisation of Cauchy formula for derivative of analytic function n n!() f z f()(;) a a dz , for n 1. 2i ()z a n1
Here a 2 and we have z 3 and n 2 . z 2 2i e z f(2) ( ;2) dz . 2i (z 2)3 z e z 2 dz if (2) , since ( ;2) 1. (z 2)3 z (2) z 2 z e z 2 f() z e . ϕ f(2) e . dz i e . (z 2)3 sin z (iii) dz where is the cicle z 2 z z 2 189
dz i 6) Evaluate the integral , 2 cos(2 ) e for z2 1 0 2 .
7) Use Cauchy-integral theorem or formula to evaluate cosz sin z dz where is the circle z 2 , taken in z21 r positive sense.
cosez 8) Evaluate dz , where is a unit circle. z z 2
190 10
THEOREMS IN COMPLEX ANALYSIS
Unit Structure : 10.0. Objectives 10.1. Introduction 10.2. Morera’s Theorem 10.3. Liouville’s Theorem 10.4. Taylor’s Theorem 10.5. Fundamental Theorem of Algebra 10.6. Summary 10.7. Unit End Exercises.
9.0 OBJECTIVES:
In this unit we shall prove the important theorems in complex analysis. 1) Morera’s theorem 2) Liouville’s theorem 3) Taylor’s theorem 4) Fundamental theorem of Algebra
9.1. INTRODUCTION:
Given an entire function f , we saw that f has a power f k (0) series representation as a , where each a . In fact k0 k k k! being an entire function the k th order derivative fk () z exists k 0 .
In this unit we propose to prove some important theorems in complex analysis .Let us start with the Taylor’s theorem. 191
9.2 MORERA’S THEOREM :
Note: this is a sort of converse of Cauchy Goursat thm.
Theorem: If f(z) is continuous in a simply connected domain D and f() z dz =0 where x is rectifiable curve in D , then f(z) is analytic in x D
Proof: Suppose z is any variable point and z0 is a fixed in the region D. Also suppose x1 and x2 are any two continuous rectifiable curve in D joining z0 to z and x is the closed continuous rectifiable curve consisting of x1 and -x2. Then we have fzdz()()() fzdz fzdz and f( z ) dz 0 (given) x x1 x 2 x fzdz()()() fzdz fzdz x1 x 2 x 2 i.e. the integral along every rectifiable curve in D joining z0 to z is the same z Now, consider a function F(z) defined by F()() z f w dw ……..(1) z0 As discussed above (1) depends only on the end points z0 and z If z+h is a point in the neighbourhood of z, then we z h have F()() z h f w dw …..(2) z0 From(1)and (2), we have
z h z z h z0 Fzh()()()()()() Fz fwdw fwdw fwdw fwdw z0 z 0 z 0 z z h f() w dw …….(3) z
Since the integral on the RHS of (3) is path independent therefore it may be taken along the straight line joining z to z+h, so that F( z h ) F ( z ) 1z h f ( z ) f()() z f w dw h h hz h 1z h z h 1 z h fwdwfz( ) ( ) dw ( fw ( ) fzdw ( )) …….(4) h z z h z
The function f(w)is given to be continuous at x therefore for a given >0 there exist >0 s.t. f()() w f z s.t. w z 192
Since h is any arbitrary therefore choosing h so that every point w lying on the line joining z to z+h satisfies ….. (5) From (4)and(5), we have F( z h ) F ( z ) 1 z h f()()(). z f w f z dw h h z 1z h 1 dw h hz h Since is small and positive, therefore we have F()() z h F z F()() z h F z f( z ) 0 or lim f ( z ) h h0 h Hence F()() z f z i.e. F(z) is differentiable for all values of z inD. Therefore F(z) is analytic in D. Since the deriviative of an analytic function therefore f(z) is analytic in D
9.3 LIOUVILLE’S THEOREM :
Statement : If f is an entire and bounded function, then f is constant. (2006, 2008)
Proof : Given that, f is an entire and bounded function. M 0 s.t. f z M z f is an entire function and hence f is analytic everywhere in Complex Plane and BR; (say) n! M fn n 0,1, 2,... Rn Put z n! M fn z n 0,1, 2,... Rn Put n 1 1!M f z R M f z 0 as R . R f z 0 z . f is constant.
Aliter : Given that f is an entire and bounded function. Let z1 and z2 be any two points in . 193
Construct a circle x with centre at z1 and radius R so that point z2 lies inside x.
By Cauchy’s integral formula 1 f 1 f f z d and f z d 1 2i z 2 2i z x 1 x 2 1 f f f z f z d d 1 2 2i z z x2 x 1 1 f z z d (1) 2 2 1 z z x 2 1
R Choose R, so large that z z , since is any point on the 2 1 2 circle x. z1 R RR Now, z z z z z z z R 2 1 1 2 1 2 1 2 2
Given that f is bounded function. f M x From equation (1) 1 M f z2 f z 1 z2 z 1 d 2 R x .R 2 z z 2 1 .M d 2 R x z z M2 R 2 z z M 2 1 2 1 R2 R 2 f z f z z z M 0 as R . 2 1R 2 1 f z1 f z 2 0
f z1 f z 2 for any two points z1 and z2 in . f is constant.
Note : If f is a non-constant entire function the f is unbounded.
Example : Let f z u z iv z be an entire function and suppose u z M z . Prove that f is constant. 194
Proof : Given that, f z u z iv z is an entire function. Define g z e f z g is an entire function. g z ef z e u z iv z eu z
i e 1 g z eM constant. g is bounded. u z M z Thus, g is entire and bounded function. By Liouville’s theorem, g is constant. e f z is constant. f z u z iv z is constant. u and v are constant.
n Note : If f z an z has radius of convergence. n0 R 0 , then f is analytic in BR; .
9.4 TAYLOR’S THEOREM :
If f is analytic in a domain G, then for any point z B, R G, f n has Taylor series expansion, f()() z an z where n0 f n a . n n!
Proof : Given that f is analytic in G. For 0 r R , construct a circle x with centre at and radius r so that the point z lies inside x. 1 f By Cauchy integral formula, f z d (1) 2i z x 1 1 1 Now, z z z 1 195
1 1 z 1 n1 n 1 z z z 1 ... ... r n1 n 1 2 x 1 x x ... n1 n 1 z z z 1 1 ... . z n1 n z 1 1 x1 1 x x2 ... xn 1 x n x n 1 ...
1 x x2 ... xn 1 x n 1 x x 2 ...
1 x x2 ... xn 1 x n 1 x1 n1 n 1 z z z 1 ... (2) 2 n n z f Multiplying equation (2) by and integrating w.r.t. over. 2i 1 f f z d 2i z x 1 f z f d d ... 2i 2 i 2 x x f z n f z n1 d d (3) n 2i z x x By generalization of the theorem on derivative of analytic function. n n! f f d n 0,1, 2, ... 2i n1 x n f 1 f OR d n! 2 i n1 x Substituting all these values in equation (3), we get 2 2z n 1 f n1 f z f z f x f ... A 2!n 1 ! n 196
2n 1 a0 a 1 z a 2 z ... an 1 z A n n f z f a and A d nn! n 2 i n x z
The theorem will be proved if limAn 0 n z n f A d n 2i n x z z n f A d (4) n 2 n x z Choose 9 0 s.t. z 9 0 9 r Since is any point on the circle x. Now, z z z r 9 Given that, f is analytic in G. f is continuous on x xis compact set
By boundness theorem, M 0 s.t. f M x From equation (4), n n 9MM 9 An d d 2 n 2 r r xr r 9 9 x n M 9 . 2r 2 r 9 r n Mr 9 An 0 as n r9 r n limx 0,0 x 1andhence09 1 0 9 r n
limAn 0 n Given series is convergent and we write f z f z f x z 2 f ... n n f f z an z where an n0 n! Put 0 in equation (1), we get, n n f 0 f z an z where a n n0 n! 197
Example 1 : Expand sinz in a Taylor series about z . 4 Solution : By Taylor series z 2 z 3 f z f z f f f ... 2! 3! Here, f z sin z and 4 1 f sin sin 4 2 1 f z cos z f cos cos 4 2 1 f z sin z f sin sin 4 2 1 f z cos z f cos cos 4 2 Substituting above values in equation (1), we get
2 3 z z 1 1 4 1 4 1 sinz z ... 2 4 2 2 2 6 2 2 3 1 z 2 1 z 4 4 ... 4 2 6 2 Zeros of an Analytic Function :
Definition : A complex polynomial p z of degree n is an n expression of the form a0 a 1 z a 2 z ... an z , where a0, a 1 ,..., an are complex constants and an 0 .
Definition : Let G be an open set and suppose f: G is a given function. A point z0 G is said to be zero (or root) of f if f z0 0 . e.g. f z z2 5 z 6 z 2 3 z 2 z 6 z 3 z 2 . Here roots or zero of, f are z 2 and z 3.
Definition : If f: G is analytic and in G satisfies f 0, then is a zero of f of order (multiplicity) m 1, if an analytic m function g: G s.t. f z z g z where g 0 . 198
Example : Let f: G be an analytic function then f has a zero of order m 1 at z if f z z m g z where g is analytic on G and g 0 . (2006)
Solution : Let f z z m g z where g is analytic on G and g 0 (1) By Taylor series, For any z B; G , g has Taylor series expansion z 2 g z g z g g ... 2! z m2 f z z m g z g g ... 2! z m2 z m g z m1 g g ... 2! z m z m1 z m 2 f z fm f m1 m! m 1 ! m 2 !
f m2 ... Clearly, this is a Taylor series expansion about z and f f f ... f m1 0 and f m 0 . f has a zero of order m 1 at z .
Conversely, assume that f has a zero of order m 1 at z . f f ... f m1 0 n f z an z n0 f z an z n m m m1 m 2 am z a m1 z a m 2 z ... z m a a z a z 2 ... m m1 m 2 m n m z an m z f z z g z g 0 n0 199
Note : f z m z m1 g z z m g z m m1 m 2 ... 2 g z ... ... fm z m! Each of these terms have z as one form. fm m! g 0 0 ... fm m! g
Definition : A zero of an analytic function f is said to be isolated if it has a neighbourhood in which there is no other zero of f.
Theorem : Any zero of an analytic function is isolated in the set of its zeros. (2009)
Proof : Let f: G be an analytic function. Suppose f has a zero of order m at z . f z z m g z (1) where g is analytic on G and g 0 . Let 0 be given. Put g 2 0 g is analytic on G. g is continuous at G . for the above 0, 0 s.t. z g z g When z i.e. z B, g()()()() z g g g z g g g 2 g z 0 g z 0 z B , g z 0 z B , and z 0 z B , or z 0 z B , From equation (1) f z z m g z 0 z B , is arbitrary. Any zeros of an analytic function is isolated in the set of its zeros. 200
9.4 FUNDAMENTAL THEOREM OF ALGEBRA :
Statement: Every non-constant complex polynomial has a root. (2009) OR If p z is a non-constant complex polynomial then, there is a complex number and with p 0
Proof : Given that p z is a non-constant complex polynomial T.P.T. p 0 . Assume that this is not true. Suppose p z 0 z . p z is an entire function. 1 Let f z (1) p z f is an entire function. p z is non-constant entire function. p z is unbounded (by contra positive statement of Lioville’s theorem) lim p z z 1 1 From equation (1), lim f z lim 0 z z p z Let f be defined on an bounded set E . If for a given 0 , R 0 s.t. f z whenever z R and z E Then, we say that f z as z . lim f z z
E
R
B (O,R) closed disk
(closeddiskB (O,R) iscompactset)
Fig 10.1 201
For given 0,R 0 s.t. f z whenever z R f is an entire function. f is continuous on BR0; . By boundedness theorem. f is bounded on BR0; . M 0 s.t. f z M z B0; R f is entire and bounded on BR0; By Liouville’s theorem, f is constant. From equation (1), 1 1 p z constant f z constant which contradicts the hypothesis that p z is non-constant. Our assumption is wrong. There is a Complex Number with p 0 .
Exercise : Prove that a complex polynomial 2 n p z a0 a 1 z a 2 z... an z has exactly n roots where a0, a 1 ,..., an are complex constant and an 0 . (Use fundamental theorem of Algebra)
Theorem : Suppose that f is analytic in domain G. If z f , the set of zerosof f in G, has limitpoint in G,then f z 0 in G.
Proof : Given that f is analytic in a domain G. zf z: f z 0 and is a limit point of z f .
Let zn be a sequence of zeros of f in G, such that lim zn n f is analytic on G, f is continuous on G. f f lim zn lim f z n 0 n n zn is a zero of f z n 0 f 0 zeros of an analytic function are isolated. either f z 0 z B ; OR f z 0 B ; — x is connected if the only sets of x which are both open and closed are and x. Assume that f z 0 in B;— 202
is a limit point of z f every nbd of continuously infinitely many points of z f .
For sufficiently large n, there is a point zn such that f zn 0 in B;— Our assumption is wrong. Hence, we must have f z 0 z B ;
d a
f (Z )=0
Fig 10.2 Given that, G is a domain. G is open and connected. We split the set G into two sets. A G: is a point of z f
BGA : , where z f is the set of zeros of f in G. AB and ABG A is a limit point of z f in G. f z 0 z B ; z A z B; z A BA, A is an open set and A A Let B , then is not a limit point of z f . By continuity of f at , 0 s.t. f z 0 z B ; z B Thus, z B, z B BB; B is an open set. G is connected. 203
It can not be written as a union of two non-empty disjoint open sets. A or B But A A B AG ABGB and Each point of G is a limit point of z f . f z 0 z G
Theorem: Let f and g be analytic in a domain G. If T is a subset of G having limit point in G and if f z g z z T , then f z g z z G . Fz fzgzT . , zGFz : 0 and use of previous theorem.
Theorem : Let f be analytic in a domain G such that for some n in G and f 0 , n 0,1, 2, ... then f z 0 z G . (Use Taylor’s theorem)
Exercise : Prove that the function f z zez z has a zero of order 2 at origin. f z zez z by Maclaurin expansion f z zez z z z2 z 3 z 4 z 5 z 6 z(1 ) z 1! 2! 3! 4! 5! 6! z2 z 3 z 4 z 5 z 6 z 7 ()z z 1! 2! 3! 4! 5! 6! z2 z 3 z 4 z 5 z 6 z 7 1! 2! 3! 4! 5! 6! 1 z z2 z 3 z 4 z 5 z2 () 1! 2! 3! 4! 5! 6! since lowest power of z is 2
10.5 SUMMARY :
1) Morera’s Theorem : Statement : Let G be a domain in and let f: G be a continuous function s.t. f z dz 0 for any triangle in G then 204 f is analytic in G. (This is a partial converse of Cauchy – Goursat theorem.)
2) Liouville’s Theorem :
Statement : If f is an entire and bounded function, then f is constant.
1) Taylor’s Theorem: If f is analytic in a domain G, then for any point z B, R G, f has taylor series expansion, n n f f()() z an z where an . n0 n!
3) Fundamental Theorem of Algebra :
Statement : Every non-constant complex polynomial has a root. OR If p z is a non-constant complex polynomial then, there is a complex number and with p 0 .
10.6 UNIT END EXERCISES :
1) Show that an entire function is infinitely differentiable .
Solution: If f is entire , by Taylors expansion of f , f has a power series representation. In fact f k (0) exist k 1. k f (0) k f(), z z z k0 k!
We can see that f() z has an infinite radius of convergence . f() z converges for all z .
By the result that Power series are infinitely differentiable within their domain of convergence, f() z is infinitely differentiable. kfk(0) f k (0) f1() z zk 1 z k 1 k1k! k 1 ( k 1)! k) k 1) fk (0) f k (0) f2() z zk 2 z k 2 and so on. k2k! k 2 ( k 2)! 205
2) Find the power series expansion of f() z z2 around z 2 .
f (2) Solution: f() z f (2) f (2)( z 2) ( z 2)2 ... 2! f( z ) 4 4( z 2) ( z 2)2
3)Find the power series expansion for ez about any point a .
4) Suppose an entire function f is bounded by M, along z R . Show that the coefficients in it’s power series expansion about M f k (0) z 0 satisfy ck . ( Hint: ck , Rk k! k k!() f z by Cauchy formula f(0) dz k 1) 2iz Rz k 1
5) Let f be an entire function , if for some integer k 0 , there exist positive constants A and B such that f() z A B z k , then f is a polynomial of degree atmost k . ( Hint: Use Liouville’s Theorem)
6) Using Morera’s theorem show that the function f defined by ezt f() z dt is analytic in the left half plane D: Re( z ) 0 . 0 t 1 ezt xt 1 Solution: dt e dt for Re(z ) x 0 . 0t1 0 x 1 This integral is absolutely convergent and f() z x ezt Consider f() z dz dt dz . Here Γ= The boundary of 0 t 1 some closed rectangle in D. ezt Since, dtdz converges hence we can interchange the order 0 t 1 of integration . zt zt e e f( z ) dz dtdz 0 dt 0 , since is analytic inside 0 t 1 0 t 1 and on a closed curve Γ. By Morera’s Theorem f is analytic in D. 206
1 sin(2t ) 7) Show that dt is an entire function.( Use Morera’s 0 t Theorem. )
8) Show that α is a zero of multiplicity k if and only if P( a ) P ( a ) ... Pk1 ( a ) 0 but Pk ( a ) 0 . ( Hint : Use the Fundamental theorem of Algebra .)
z 9) Find the order of zero at z 0 of the function f z z ze z.
2 10) Find the Maclaurin series expansion of f z sin z .
207 11
MAXIMUM AND MINIMUM MODULUS PRINCIPLE
Unit Structure : 11.0. Objectives 11.1. Introduction 11.2. Maximum Modulus Principle, Schwarz Lemma ,Open Mapping Theorem 11.3. Automorphisms of the Unit Disc 11.4. Summary 11.5. Unit End Exercises
11.0 OBJECTIVES :
After going through this unit, we shall understand 1)Maximum modulus principle and open mapping theorem for analytic functions . 2) Corollaries on open mapping theorem and maximum modulus principle. 3) Possible Automorphisms of the Unit disc B(0;1) . 4)Harmonic functions and their properties.
11.1 INTRODUCTION :
In previous sections, we have studied the connections between everywhere convergent power series and entire functions. We shall now turn our attention to the general relationship between power series and analytic functions. According to a theorem, every power series represents an analytic function inside it’s circle of convergence.
Our first goal is the converse of this theorem. We then turn to the questions of analytic functions in arbitrary open sets and local behaviour of such functions. 208
11.2 THE MAXIMUM MODULUS PRINCIPLE :
Definition : Let G be any subset of . A complex function f defined on G is said to have local maximum modulus at a point in G if, there exists 0 s.t. BG; and f z f z B(,) .
Similarly, f has local minimum modulus at a point in G, if 0 s.t. BG; and f z f z B, .
Theorem : Suppose f is analytic in a domain G and there is a point in G s.t. f z f z G . Then f is constant i.e. if f attains to maximum modulus in G then f is constant.
Proof : Given that, f is analytic in domain G (open and connected set) G and G is open. r 0 s.t. B; r G where x B; r = boundary of closed disk B; r
x
r a
Fig 11.1
By Cauchy integral formla, 1 f z f dz 2i z x
Here x is the circle z r z reit t 0, 2 , dz i reit dt 209
2 f reit 2 1 1 it f .i reit dt f re dt 2 i it 2 0 re 0 (1) Given that, f z f z G f reit f From equation (1), we get 2 2 1 1 f f reit dt f dt 2 2 0 0 f 2 f dt t2 2 2 0 0 f 2 f 2 2 1 f f reit dt f 2 0 2 1 f reit dt 2 0 2 2 f f reit dt 0 2 2 f dt f reit dt 0 0 2 2 f f dt 0 2 f f reit dt 0 0 Here, the integral f f reit is continuous and non- negative. f f reit 0 t 0, 2 f reit f t 0, 2 f z f z x 210
This equation holds on all circles z 0 s r f z is constant in B; r f z is constant in B; r
( If f: G is analytic and f z constant if z G then, f is constant on G .
By using theorem [Let f and g be analytic in a domain G. If T is a subset of G having limit point in G and if f z g z z T then f z g z z G ] f is constant in G.
Maximum Modulus Principle : Suppose f is analytic in a bounded domain D and continuous on D (Closure of D). Then , f attains its maximum on the boundary D of D. (2006, 2012)
Proof : If f is constant, then there is nothing to prove. Let f be a non-constant function. Given that, f is continuous on D ( D is a compact set). f attains its maximum value at same point in D . Maximum modulus principal, f does not attain its maximum in D. f attains its maximum on the boundary D of D .
DDD
Minimum Modulus Principle : Suppose f is a non-constant and analytic function in a domain G. If f attains its local minimum G at , then f 0 or f is constant (2006,2007)
Proof : Given that, f is a non-constant analytic function in a domain G and f attains its local minimum at a point in G. f f z z B; G (1) To prove that f 0 Assume that this is not a true. i.e. f 0 in some open disk B, r G . 211
1 Let g z (2) f z g is analytic in B; r G . From equation (1), 1 1 f f z
OR
g z g z B, r g has a local maximum modulus at a point in G. By maximum modulus principle, G is constant in G. From equation (2), g 1 f z constant g z constant f is constant in G. which is contradicts that f is a non-constant function. Our assumption is wrong. f 0
Theorem : If f is a non-constant analytic function in a bounded domain G and f z 0 for any z G , then f can not attain its minimum in G.
Example : Let f z ez and T B2 3 i ,1 . Find a point in T at which f attains its maximum value. Solution : Given function, f z ez T B2 3 i ,1
B=BU¶B
(3, 3) (2, 3)
Fig 11.2 212
The boundary of T is the circle. z 2 3 i 1 By maximum modulus theorem, f attains its maximum value on B . i z 2 3i 1 e 2 cos i 3 sin 2 f z e e e. e e cos
i e 1 e2 cos We know that, the value of cos is maximum when 0 . ez e2 1 e 3 ez e3 is the maximum value of f at a point 3, 3 or 3 3i in T.
Exercise : Let f z z and TB 0;1 . Prove that the function f cannot attain its minimum value on the boundary of T.
Schwarz’s Lemma Let D z : z 1 be the unit disk and suppose f is analytic in D with, (i) f (0) = 0 and (ii) f z 1 for z D . Then, f z z z D and f 0 1. Moreover, if f z z for some z 0, then there is a constant C with c 1 s.t. f c. D.(2004, 2005, 2008) Proof: Given that, f is analytic in D, with (i) f0 0 and (ii) f z 1 for z D. Define g: D by z =r f z / z z 0 g z , f 0 z 0 r g is analytic in D 0 Choose r s.t. 0 r 1 on the circle z r 1 f z g z z Fig 11.3 213
1 g z on the circle z r r …….. f z 1 and z r by Maximum Modulus theorem, 1 g z z B0; r r As r1, g z 1 z D B 0,1 ------(2) f z 1 ------(by (1)) f z z z D Again, from equation (1) f0 g 0 1 ------(2) f 0 1 If f z z for some z 0 then, f z z g z 1 z z g attains its maximum value of some point z inside D. by maximum Modules principle, g is constant in D i.e. g z c where, c is constant and c 1 f z c z or f z cz z D or f c D
z Example: ; z in D, define the M.T. z 1 z Let f be analytic in D and let f z 1, then f z f z (i) OR f z z , z D 1 f f z 1 z f x f z 1 (ii) z D 2 2 1 f z 1 z
Solution: Fix a pint in D arbitrarily z Put z z 1 z 1 214
f z f Define f z g ------(1) f 1 f f z f f 1 f z f 1 f f 1 g is analytic in D, g(o) = o and g 1 By using Schwarz’s Lemma, g o 1 and g D
f z f z 1 f f z 1 z OR
f x f z z z D From equation (1) d d f z f g dz dz1 f f z 1 z 1 z g 2 1 z 1 f fzfz fzf f fzf f fz 2 1 f f z 2 1 2 1 f f z z g 1 z 2 2 1 z 1 f f z Put z 2 2 1 1 f f g o 2 2 2 2 1 1 f 1 2 f g o 1 g o 1 2 1 f 215
f 1 2 2 1 f 1
f 2 2 1 f f z 1 Put z, z D 2 2 1 f z 1 z
f o z iii) f z 1 f o z
g D f z f o z 1 1 f o f z 1 z Put o f z f o z 1 f o f z If a1 and b 1 a b a b a b 1 a b1 ab 1 a b f z f o f z f o z 1 f z f o 1 f o and f z f z f o z 1 f z f o f z f o z1 f z f o f z f o z z f z f o f z z f z f o f o z f o z f z 1 f o z
Counting zero Definition : A zero of order one is said to be a simple zero. e.g. Let f z z2 3 z 2 Here, f has simple zeros at z 1 and z 2 . 216
f z 2 z 3 f 1 1 0 f 2 1 0 Prove that all zeros of the function sin z are simple. To find zeros at sin z put sinz 0 1 sinz 0 z sin 0 2 , n 0, 1, 2,... n f z sin z f z cos z n f z cos n 1 0, n All zeros of sinz are simple.
Theorem : Let f be analytic in a domain G with zeros 1, 2 , ...,m (repeated according to order) If x is a smooth closed curve in G which does not pass through any k s then 1 f z m dz n x; 2i f z k x k 1 Proof : Given that, f is analytic in a domain G with zeros 1, 2 ,..., n (repeated according to order or multiplicities.) f z z 1 z 2 ... z m , where g is analytic and gk 0, k 1, 2,..., m
Taking log on both sides and differentiating w.r.t. z, we get f z 1 1 1 g z ... f z z 1 z 2 z m g z
1 Multiply this equation by and integrate w.r.t z over {x } on 2i both side. 1 f z 1dz 1 dz 1 g z dz ... dz 2i f z 2 i z 2 i z 2 i g z x x1 xm x g and g are analytic in G. g is analytic in G and x is a smooth closed curve. g g z By Cauchy theorem, dz 0 g z x 217
1 f z dz n x; n x ; ... n x ; 2i f z 1 2 m x m n x;k k 1
Note : 1 f z m dz n x, = Number of zeros of f inside x, 2i f z k x k 1 where each zero is counted according to its order. 2 e.g. for f z z 1 z 2 g z 1 f z dz x; x , 3 x , 1 0 3 4 2i f z 1 2 3 x G x
a1 a3
a2
Fig 11.4
Corollary : Let f, G and x be as in the preceding theorem except that 1, 2 , 3 ,..., n are the points in G that satisfy the equation f z then, 1 f x m dz x; Number of zeros of 2i f z k x k 1 f z inside x. f z z z 12 Example : Evaluate dz where f z and x is f z 3 x z 5 the circle z 1.2 . z z 12 Solution : Given function, f z . z3 5 Here, f z has simple zero at z 0 and z 1 is a zero of order 2. Given, equation of circle, z 1.2 . Zero z 0 and z 1 lies inside x. 218
x
0 1 1.2
£ Fig 11.5 1 f z By theorem, dz = number of zeros of inside x where 2i f z x each zero is counted according to its order. 1 f z f z dz 1 2 3 dz 6 i 2i f z f z x x f z z2 z1 z 3 Exercise: Evaluate dz where f z and f z 3 x z 2 x is the circle z 1.5 .
Note : Let x: 0,1 G be a closed (Smooth) curve in and suppose f: G is an analytic function. Then f x is also a closed curve in w-plane. If is a Complex Number f x , we write, 1dw 1 t ; dt 2i w 2 i t 0 1 f x t x t dt 2i f x t 0 1 f z dz 1 2 3 2i f z x f z dz 6 i f z x f z z2 z1 z 3 Exercise : Evaluate dz where f z and x f z 3 x z 2 is the circle z 1.5 . 219
Note : Let x: 0,1 G be a closed (smooth) curve in and suppose f: G is an analytic function. Then f x is also a closed curve in w-plane. If is a Complex Number f x , we write, 1dw 1 t ; dt 2i w 2 i t x x 1 f x t x t dt 1 f z dz 1 2 3 2i f x t 2i f z 0 x m x,k = numbers at zeros of f z inside x where each k1 zero is counted according to its order.
f G s= f x x o
£- plane z- plane w- plane Fig 11.6
where 1, 2 ,..., n are points in G with f k . x f
[ 0, 1] G £
s= f ox
Fig 11.7 220
Theorem : Suppose that f is analytic in B a; R and let f a has a zero of order m at z a then there is an 0 and 0 s.t. and the equation f z has exactly m simple roots in B a; .
Proof : Given that, f has a zero of order m at z a . Zeros of an analytic function are isolated. We can choose 0 s.t. R . 2 f z has no solution with 0 z a 2 and f z 0 if 0 z a 2 .
R
b1 b2 e
O bm 2 e
z- plane Fig 11.8
Let x be the circle, z a i.e. x t a e2it , t 0,1 . x: 0,1 B a ; R f is an analytic on B a; R f x is also a closed curve in w-plane. 221
s= f ox
d a Ïe
w- plane Fig 11.9
Now, So, 0 s.t. B; (It means open disk B; does not touch trace of . B; is contained in the same component of w . For B; i.e. ;; (1) 1 dw Now, ; f()(). z dw f z dz 2i w 1 f z (;) dz 2i f z x = Number of zeroes of f z inside x, where each zero is counted according to its order. = Thm f() z has a zero of order m at z a ) 1 dw Again, ; 2i w 1 f z ; dz 2i f z x 222
1 f z dz ;; n m By (1) and 2i f z x ; m f z has exactly m-roots in B a; f z 0 for 0 z a The equation f z has exactly m-simple roots in B a, .
Definition : If x and are Metric spaces and f: X Y has the property that f 0 is open in whenever U is open in X, then f is called an open map.
Open Mapping Theorem : Statement : Suppose G is a domain in , f is a non-constant analytic function on G. Then for any open set U in G, f U is an open. (2007, 2009)
Proof : Given that, f is a non-constant analytic function on G. Let a U and f a U is open 0 s.t. B a; U . f is non-constant analytic function on G. by fundamental theorem of algebra, an integer m 1. f z has a zero of order m at z a . by using previous theorem, for the above 0, 0 s.t. for the above 0, 0 s.t. and the equation f z has exactly m simple roots in B;. Thus B; , we an find m points in B; which are mapped to by f. B;; f B B; f U BU; is interior point of f 0 . But is arbitrary. f U is open. 223
11.3 AUTOMORPHISMS OF THE UNIT DISC :
A function f: D D is said to be an Analytic automorphism or Automorphism of the Unit disc D if f is bijective and if both f, f 1 are analytic in D.
Note: Let 0 D z : z 1 z For ,,z in D define the Mobius transformation z 1 z This Mobius transformation is analytic in D and also in D DU D , 1 z is not analytic at a point z . Which lies outside the disk D .( D 1 ) 1 1 1and z z
Note: : 0 D is an analytic automorphism. 1. 0 and 0 2. for any point z D, z D
D
+ a - a
o o
£- plane w- plane Fig 11.10 i.e. z 1 z D 2 z z Since, z 1 z 1 z 2 z z z 2 2 z z z z z z z 2 z z 1 z z zz 1 2z 2 z z z z …. z D z 1, 1 224
2 z 1
z 1
z D
maps D onto itself i.e. 0 D
(3) If D, so is and for any z D, 1 z z z i.e. z z 1 z Now z z 1 z z 1 2 z 2 z z 1 z z 1 z z 2 1 2
Similarly, z z 1 1 maps D onto D in a one-one manner Hence, and are automorphisms of the Unit disc. 4. For z D,z z D i.e. z 1 z D z ¶D Since, z 1 z z D For any point z D 1 z ei, 0, 2 o Fig 12.10 i ei 1 ei e z 11. z ………… 1 ei ei e i ei 11
i i i e1 and e e
z1 z z
z D maps D onto D i.e. DD combining results (2) and (4), we get maps D onto D . 225
z 5. Mobius transformation z is analytic in D, 1 z 1 z 1 z 1 2 z 2 2 1 z 1 2 In particular, 2 o 1 0 D 1 1 0 1 2 21 1 2 0
Proposition: If z 1 then, is one map of D onto itself. The inverse of is . Furthermore, maps D onto D , 2 1 0, 0 1 z and . 1 2
Theorem: Let f: D D be a one-one analytic map of D onto itself, with f 0 and suppose that is a one-one analytic map of D onto itself with analytic inverse . Then, there is a complex no. C with C 1 s.t. f C .
Proof: Given that, f: D D is an one-one analytic map of D onto itself with f o . z Put , 1 1 Define, g f g is a one-one analytic map of D onto itself and g o f o f o ….. f o and 1 and g f f z 1 …… z and f: D D and D is unit Disk z D f z D f z 1 by Schwarz’s Lemma, g D and g o 1 226
let g for some o in D by second part of Schwarz’s Lemma, There is a complex no. C with C 1 s.t. g c. D
f c f z c z f c
11.4 SUMMARY :
1) The Maximum Modulus Principle: Let G be any subset of . A complex function f defined on G is said to have local maximum modulus at a point in G if, there exists 0 s.t. BG; and f z f z B ; . Similarly, f has local minimum modulus at a point in G, if 0 s.t. BG; and f z f z B, .
2) Minimum Modulus Principle : Suppose f is a non-constant and analytic function in a domain G. If f attains its local minimum G at , then f 0.
3) Schwarz’s Lemma: Let D z : z 1 be the unit disk and suppose f is analytic in D with, (i) f (0) = 0 and (ii) f z 1 for z D . Then, f z z z D and f 0 1. Moreover, if f z z for some z 0, then there is a constant C with c 1 s.t. f c. D.
4) Theorem : If f is a non-constant analytic function in a bounded domain G and f z 0 for any z G , then f cannot attain its minimum in G.
5) Theorem : Let f be analytic in a domain G with zeros 1, 2 , ...,m (repeated according to order) If x is a smooth closed curve in G which does not pass through any k s then 1 f z m dz n x; 2i f z k x k 1 227
6) Open Mapping Theorem : Statement : Suppose G is a domain in , f is a non-constant analytic function on G. Then for any open set U in G, f U is an open.
7) A function f: D D is said to be an Analytic automorphism or Automorphism of the unit disc D, if f is bijective and if both f, f 1 are analytic in D.
11.5 UNIT END EXERCISES:
1) Find the maximum modulus of z2 z in the disc z 1. Solution: z2 z z( z 1) The maximum modulus is assumed at the boundary of the disc z 1 That is at z 1. 2 maxz 1 z z 2
2) Show that the maximum modulus of ez is always assumed on the boundary of the compact domain . Solution: Since ez e x where z x iy
ez is maximum at a point inthe domain with maximal x. (At a point farthest to the right. )
3) Suppose f,g both are analytic in a compact domain D. Show that f()() z g z takes it’s maximum on the boundary. ( Hint: Take f(),() z Aei g z Be i then put h()()() z f z g z h()()()()() z f z g z f z g z
Let z0 be an interior point of a compact domainD. Assume that
f g takes maximum values inside D, say f()() z0 g z 0 h()()() z f z g z
f()()()()() z g z f z0 g z 0 h z 0
h()() z h z0 The analytic function h() z assumes it’s maximum at the interior point z0 ( not on the boundary), which is not possible. f()() z g z takes it’s maximum on the boundary. 228
4) Let f be analytic and bounded by 1 in the Unit disc and 1 3 f 0 Estimate f . 2 4 1 Solution: Since f 0 , define g : as follows: 2 f() z 1 g() z for z 1 z 2 z / 1 2 2 3 1 1 f for z 4 2 2 Then g is analytic in z 1 Letting z 1we find that g 1 on the disc . 1 z 3 3 2 f() z 2 for z , f 2 1 4 4 5 2
5) Show that among all functions , which are analytic and 1 bounded by 1, in the Unit disc, Max f is assumed, when 3 1 f 0 . 3 1 f() z f 1 3 Solution: Suppose f 0, consider g() z 3 1 1 f f ( z ) 3 1 w f 3 1 when w 1 and f 1 in z 1, 1 1 f w 3 By Maximum-Modulus Theorem g 1 in z 1 . 1 f 1 3 1 1 By direct calculations g g f 2 3 1 3 3 1 f 3 This is a contradiction.
6) Show that the automorphisms of the Unit Disc are of the i z form g( z ) e , 1. a z 229
z Solution: Let g() z . Then g( z ) 1for z 1. 1 z Since g( ) 0 g is an automorphism of the Unit disc. Assume that f is an automorphism of the Unit Disc with f ( ) 0 . h fog 1 is an automorphism with h(0) 0 . By the lemma that describes automorphisms of the Unit disc , i i z h() z e z or f() z e . 1 z
230 12
SINGULARITIES
Unit Structure : 12.0. Objectives 12.1. Introduction 12.2. Three Types of isolated Singularities 12.3. Laurent’s Theorem 12.4. Classification of Singularities by the Principal Part of Laurent’s Expansion 12.5 Casorati- Weirstrass Theorem 12.6. Summary 12.7. Unit End Exercises
12.0. OBJECTIVES:
After going through this unit , you will understand the concept of continuing an analytic function to another region. We shall also study three types of singularities of a function f() z and the theorems like Casorati-Weirstrass theorem and the Laurent’s theorem.
Given a singularity z0 of a function f() z , we shall try to classify the singularities by finding the principal part of Laurent series expansion of a function f() z .
12.1. INTRODUCTION :
We shall recall the uniqueness theorem that states that if
f is analytic in a region D and znis a sequence of distinct points such that f( zn ) 0 n and znconverges to some z0 , then f is identically zero in a region D. Suppose we are given a function f , which is analytic in regionD. The question is that of continuing f analytically to a region D1 such that g f on DD1 . By the uniqueness theorem such 231 continuation of f is uniquely determined. The Schwarz reflection principle is an example of how, in some cases, an analytic function can be continued beyond it’s original domain of analyticity. In this unit, we shall examine the possibilities of such extensions for functions given by power series.
12.2 THREE TYPES OF ISOLATED SINGULARITIES
Definition : A point at which the function f is not analytic is said to be a singular point or singularity of the function f. z2 e.g. f x z 3 Here, f is not defined at z 3 and hence not analytic at z 3, therefore z 3 is singular point.
Definition : A point at which the function f is analytic is said to be a Regular point.
Definition : A function f has isolated singular point at z z0 if an R 0 s.t. f is defined and analytic in 0 z z0 R but not B z0, R . z2 e.g. 1) f z z1 z 3 z 1and z 3 are points of singularity sin z 2) f z cot z cos z Put cos z = 0 z=n Singular points are n, n .
Definition : Let f be analytic 0 z z0 R . Let z0 be an isolated singular point of f. A point z z0 is said to be a Removable singularity of f, if an analytic function g:, B z0 R g s.t. f z g z for 0 z z0 R .(2007) Or Definition : If a single valued function f z is not defined at a point z z0 but lim f z exists. Then z z0 is said to be a z z0 removable singularity of f. sinz e.g. f z , z 0 z 232
sinz In this case, f is not defined at a point z 0 but lim 1 z0 z exists. z 0 is a Removable singularity of f. OR Define g: B 0; R s.t. sin z z 0 z g z 1z 0 f z g z for 0 z 0 R z2 9 f( z ) , z 3 z 3 Here, f is not defined at z 3 . z2 9 z 3 z 3 But, lim f z lim lim 6 exist. z3 z 3z 3 z 3 z 3 z 3 is a removable singularity of f.
Definition: A singular point which is not isolated is said to be Non- isolated singular point. 1 1 e.g. f z cos ec z sin(z ) In the delta nbd at zeros, there are other singular point of f z = 0 is a non isolated singular point of f. 1 For, Singular points, Put sin 0 z 1 n n z 1 1 z 0 n , n 0 , 1 , 2 ,.... .Since z 0 as n n n Here, z = 0 is a non-isolated singular point, whereas other singular points are isolated.
Theorem: If f has an isolated singular point at z0 , then z z0 is a removable singularity of f iff lim z z0 f z 0 z z0
Proof: Let lim z z0 f z 0 z z0
T.P.T. z = z0 is a removable singularity of f. Given, f has an isolated singular point at z z0 233
there exists R > 0 s.t. f is defined and analytic in B z0;\ R z 0 but not in B z0; R Define
()()z z0 f z z z0 h(), z (I) 0 z z0
h is analytic in B z0;\ R z 0 and
limh ( z ) lim ( z z0 ) f ( z ) 0 h ( z 0 ) from (I) z z0 z z 0
h is continuous in B z0; R
T.P.T. h is analytic in B z0; R i.e. T.P.T. h( z ) dz 0 for every triangle int in B z0; R There are four cases : By Morera’s Theorem, h is analytic in B z0, R from equation (I), h z0 0
z0 is a zero (root) of h
f an analytic function g:;.. B z R s t h()() z z z0 g z where g( z0 ) 0
for 0 z z0 R h z z z0 g z z z0 f() z by (I)
f()() z g z for 0 z z0 R
z z0 is a removable singularity of f. Conversely,
Suppose z z0 is a removable singularity of f.
T.P.T. lim z z0 f ( z ) 0 z z0
By definition, f an analytic function g:, B z0 R s.t. f()() z g z for 0 z z0 R
limf () z lim g () z g ( z0 ) 0 z z0 z z 0
lim z z0 f ( z ) 0 g ( z 0 ) 0 z z0
lim z z0 f ( z ) 0 z z0 234
Definition: If f has an isolated singular point at z0, then z z0 is a pole of f if limf ( z ) i.e. for any M >0, z z0
0 s.t. f() z M where 0 z z0
Definition: If f has a pole at z z0 and m 1 is the smallest positive m integer s.t. ()()z z0 f z has a removable singularity at z z0 then, f has a pole of order m at z z0 .
Definition: A pole of order one is said to be a simple pole. z3 e.g. f() z z 4 Here, z = 4 is an isolated singular point at f. z3 limf ( z ) lim z4 z 4 z 4 z 4 is a simple pole of f. z2 f() z (z z )( z 1)4 Here f has simple pole at z = 2 and z =1 is a pole of order 4.
Essential Singularity: An isolated Singular point which is neither a pole nor a removable singularity is said to be Essential singularity. e.g. 1 f() z e z Here, f has essential singularity at z = 0. T.P.T. z = 0 neither a pole nor a removable singularity.
Theorem: If a function f(z) of analytic for all finite values of z and as z , f z a z k then f(z) is a polynomial of degree k.
Proof: Since f(z) is analytic for all finite values of z therefore it can n be expanded by Taylor’s theorem in the form f() z an z , for n0 z R , where R is large. Let max f() z M on the circle z r r R . Then by Cauchy’s M inequality , we have a for all values of n n r n 235
k Ar k Arn k r n k k , since f() z = A z when z r n
Ar k Ar n k , which tends to zero when r since n>k. r n Thus an=0, n k .
2 k Hence, we have f(z)=a0+ a1z+ a2 z +………. +ak z , which is a polynomial of degree k .
12.3 LAURENT’S THEOREM :
Theorem: If f is analytic in G ann ; R2 , R 1 , R 2 0 then for any point z in G, f has unique representation n bn 1f ( ) f()() z an z where, an d , n ()z n 2i () n1 n0 n 1 x1 1f ( ) =0,1,2..and bn d , n = 1, 2, …… 2i () n1 x2 and x1, x 2 are circles r1, r 2 respectively with R2 r 2 r 1 R 1
Proof: for a given z G, choose r1 and r 2 s.t. R2 r 2 z r 1 R 1 by using Cauchy integral formula for multiply connected domain 1f ( ) 1 f ( ) f() z d d (1) 2i z 2 i 2 x1 x 2 1f ( ) Consider, d 2i 2 x1 For any point on x1 1 1 1 Consider z z z 1 n1 n 1z z z 1 1 ...... n1 n z 1
12 3n 1 n 1 ……. 1 x x x 1.... x x 1 x 1 x 236
n1 n 1 z z z ..... 2 n n f () Multiply the above equation by // and then integrating w.r.t. 2i over x1 1f 1 f z f d d d .... 2i z 2 i z i 2 x1 x 1 x 1 ()()z n1 f z n f ...... d 2i n 2 i n x1 x 1 1 f Given that, an d n = 0, 1, 2, 3 …….. 2i n1 x1 1 f d a a z 0.... a ( z )n1 R (2) 2i z 0 1n 1 n x1 n z f () Where, Rn d 2i n z x1
T.P.T. limRn 0 n z n f d Rn (3) 2 n z x1
Choose 9 0 s.t. z 9 equation of the circle x1 is, r ,
Now, z z z r1 9 Given that, f is analytic in G
f is continuous on x1 (Compact set) By boundedness theorem
M1 0 s.t. f M x1 Put all the above values in equation (3) we get n n 9 MM19 1 Rn d d 2 rn r 2 r n r x11 19 1 1 9 x 1 n n 9 M1 M1 r 1 9 2r1 0as n r1 2 r 1 9 r19 r 1
limRn 0 n 237
from equation (2) 1 f n d an z (4) 2i z x1 n0 Consider, 1 f d 2i z x2
For any point on x2 , 1 1 1 1 Consider z z z z 1 z
n1 n 1 1 1 ...... n1 n z z z z 1 z n1 n 1 ..... 2 n n z z z z z
Solving in the same manner as above, we get, 1 f b d n ______ 2i z z n x2 n1 From equation (1), (4) and , we get n b f() z a z n ______ n n n0 n 1 z n Note: (i) equation can also be written as f z an z n for R2 z R 1
1) Where x is the circle r with R2 r R 1 1 f a d n 0, 1, 2 n 2i n1 x 238
Proof: If x is the circle r s.t. R2 r R 1, then both functions f f and are analytic in RR by using n1 n 1 2 1 Cauchy De-formation Theorem, 1f 1 f an d d 2i n1 2 i n 1 x1 x 1f 1 f bn d d 2i n1 2 i n 1 x2 x x1
F x2
ò F = ò F x x 1 2
Fig 12.1
We observe that, bn a n From n n f z an z a n z n0 n 1 n n an z a n z n0 n 1 n f() z an z n 1 f Where a d n 0, 1, 2...... n 2i n1 x
2. The Laurent Series expansion is Unique.
Proof: Suppose that we have another Laurent series expansion m f() z Am z for R2 z R 1 (1) m 239
We prove that equation (1) is identical with n f() z an z for R2 z R 1 n 1 f Where a d, n 0, 1, 2...... n 2i n1 x To prove that an A m Let x be the circle r with R2 r R 1 1 f 1m 1 a d= A d n 2i n1 2i m n1 x x m 1 m n 1 1 m n 1 A d A d 2i m 2i m x m m x 1 m n1 A rei i re i 2i m m r rei0,2 1 A rm n e i() m n d . This integral 2 ,m n 2i m m 0m n
1 A r02 2i m an A m Laurent series is unique in G.
Note: 1. The Laurent series for given function n n is f z an z a n z R2 z R 1 n1 n 0 n 1) The part an z of Laurent series is called the n1 principal part of f(z) at z . n 2) The part an z of Laurent series is called the n0 Analytic part of f(z) at z . 240
g() z 2. If f has a pole of order m at z , then f z where, g z m is analytic at a point and g 0 z2 e.g. f() z z 33 g() z z2 is analytic at point 3 and g(3) 32 9 0.
12.4 CLASSIFICATION OF SINGULARITIES BY THE PRINCIPAL PART OF LAURRENT’S EXPANSION
Corollary: Let z be an isolated singularity of f() z and let n f() z an z be its Laurent expansion in ann ; 0, R n 0 z R (Punctured disk or deleted nbd of ). Then,
i) z is a removable singularity of f iff an 0 for n 1 i.e. (Principal part is zero) (2008)
ii) z is a pole of order m iff am 0 and a n 0 for n m 1 . i.e (Principal part is finite)
iii) z is an essential singularity of f iff an 0 for infinitely many negative integers n. i.e. (Principal part is infinite) n Proof: Given, f() z an z n n n f() z an z a n z (1) n1 n 0
i) Let an 0 for n 1 T.P.T. z is a removable singularity of f. From equation (1) n f z an z an 0 for n 1 n0 n f() z a0 an z n1 n limf ( z ) lim a0 an z a 0 0 a 0 0 z z n1 241
limz f z lim z lim f z 0 a0 0 z z z z is a removable singularity of f Conversely, suppose, z is a removable singularity of f. an analytic function g:; B R s.t. f z g() z in 0 z R
g is analytic in BR; for any point z B; R g has Taylor series expansion n g()() z an z n0 the Laurent series expansion for f(z) must coincide with the Taylor series expansion for g(z) about z n f() z an z n0
an 0 for n 1 (Compare equation (1) and above equation)
zsin z E.g. f( z ) , z 0 z3 z3 z 5 z 7 z z ...... 3! 5! 7! 1 z2 z 4 f() z ...... z3 3! 5! 7!
Principal part is zero i.e. an 0 for n 1. This is a Laurent series expansion for f(z) but principal part contain no negative power of z. i.e. an 0 for n 1 z 0 is a removable singularity of f. ii) Given am 0 and a n for n ( m 1) n n n f()() z an z an()() z a n z n n1 n 0 a a n f( z ) m ...... 1 a z m n z z n0 m Multiplying above equation by z , 242
mm1 m n z f() z am ..... a1 ( z ) ( z ) a n z n0 m m1 m n limz f ( z ) lim am ...... a1 z z a n z z z n0 am 0 0 .... 0 0 am 0 m1 limz f ( z ) lim z am 0 z z m the function z f() z has a removable singularity at z z is a pole of order m
Converse, (Exercise) ez E.g. f(), z Here f has pole of order 3 at z = 0 z3 z2 z 3 1 z .... 2! 3! 1 z2 z 3 f( z ) 1 z .... 3 3 z z 2! 3! 1 1 1 1 z ...... z3 z 2 2!.z 3! 4!
This is a Laurent series expansion for f (z) but principal part of Laurent series is finite. f has a pole of order 3 at z = 0 iii) Combine part (i) and (ii) and by definition essential singularity, we see that z is an essential singularity of f iff an 0 for infinitely many negative integers n. 1 e.g. f() z e z 1 1 1 1 1 ...... z 2!z2 3! z 3 4! z 4 Prinicpal part This is a Laurent series expansion for f(z) and principal part of Laurent series is infinite. z 0 is an essential singularity of f.
Theorem: Let f be analytic in 0 z R R 0. Then, f has a pole of order m at z iff there exists an analytic function, g() z g:; B R s.t. f() z where g 0 ()z m 243
(Note: We know that the geometric series zn converges for n0 1 z 1 and we write zn 1 z z2 z 3 ...... 1 z n0 Similarly, the geometric series ( 1)nz n converges for z 1 and n0 1 we write ( 1)nz n 1 z z2 z 3 z 4 ...... ) 1 z n0
Example:1 5 Expand f() z in a Laurent series valid for (i) 2 z 3 z2 z 6 (ii) z 2 (iii) z 3 (iv) 0 z 2 4
Solution: Give function 5 5 1 1 f() z (1) z2 z 6 z 2 z 3 z 2 z 3 2 1) For 2 z 3, if 2 z then 1 z n 1 1 1 2 2n z 2 2 z z zn1 z 1 n0 n 0 z z if z 3 then 1 3 n n 1 1 1 n z n z ( 1) ( 1) z 3 z 3 3 3n1 3 1 n0 n 0 3 From equation (1), we get n n 2 n z f( z ) 1 n1 n 1 n0z n 0 3 23 2 2 2 1 1 z z 2 ...... ..... 4 3 2 2 3 z z z 2 3 3 3 23 2 2 2 1 1 z z 2 ...... ...... (In this case, z = 0 z4 z 3 z 2z 3 3 2 3 3 is an essential singularity) This is the required Laurent series for 2 z 3 244 ii) For z 2 z z z 2 3 1 and 1 2 3 n 1 1 1 z zn z 2z 2 2 n1 2 1 2 n0 n 0 2 n n 1 1 1 n z n z 1 ( 1) z 3 z 3 3 3n1 3 1 n0 n 0 3 from equation (1) zn z n f( z ) ( 1)n n1 n 1 n02 n 0 3 This is the required Laurent series for z 2 (In this case, z = 0 is removable singularity) iii) For z 3 2 3 z 3 2 1 and 1 z z n 1 1 1 2 2n 2 n1 z 2 z 1 z zn0 z n 0 2 n 1 1 1 3 3n z 3 3 z z n1 z 1 z n0 n 0 z from equation (1), 2n 3 n f() z n1 n 1 n0z n 0 z This is the required Laurent series (In this case z = 0 is essential singularity) iv) For 0 z 2 4 put z – 2 = u z = u + 2 from equation (1) 5 5 5 f() z z 2 z 3 z 2 z 2 5 u u 5 u 0 u 4 5 1 5 n 5 1 n u f( z ) ( 1) u u 5 4 5 1 5 n0 245
Example: 2 4 Expand f() z in a Laurent series valid for z2 2 z 3 (i) 1 z 3 (ii) z 3 (iii) z 1 (iv) 0 z 1 4
Example 3: cos(z ) 1 Expand f() z in a Laurent series about z = 0 and name z4 the singularity .
Cosz 1 Solution: f() z z4 z2 z 4 z 6 1 ...... 1 2! 4! 6! 1 1 z2 ...... z4 2!z2 4! 6! Z=0isapoleoforder2
1 Example 4:Find the Laurent series of f(z)= in the z z1 z 2 annular region 1 z 2 . (2012) 1 Solution: f(z)= z z1 z 2 1 z By partial fraction for1 z 2, 1 & 1 z 2 1 1 1 f z 2z z 1 1 4 1 z z 2 1 1z 1 1 1 1 z 2 2z z 4 1 1 1 1 1 z z2 1 2 ..... 1 ...... 2z z z z 4 2 4
Definition: A set D in a Metric space X is said to be Dense iff DX , where D, is closure of D. OR Let D be a subset of ()D , we say that D is dense in if, for any 0 and 0 B0, n D 246
12.5 CASORATI -WEIERSTRASS THEOREM :
If f has an essential singularity at z , then for every 0,
f ann ; 0, is dense in . (2005, 2008) OR f ann ;0,
Proof: Let G ann ;0, 0 z B ; \ Given that, f has an essential singularity at z f is analytic in G. T.P.T. f(G) is dense in i.e. f G i.e. T.P.T. for given 0 , 0, 0,z s.t. z and f() z 0
f (2) 0 x 2
plane plane plane plane
Fig 12.1
Assume this is not true i.e. Assume there is 0 and 0 s.t. f() z 0 2G ann ;0, f() z lim 0 z z f z The function 0 has a pole at z z z z0 is a pole of f if limf ( z ) z z0
Note: If limx 0 then limx 2 0 x x 247
m f() z 0 If m 1 is the order of this pole then, z has a z removable singularity at z . m1 limz f ( z ) 0 0 (1) z Now, z m1 f()() z z m 1 f z c c
lim z m1 f ( z ) lim z m 1 f ( z ) c z m 1 c z z = 0 + 0 ------by (1) m 1 limz f ( z ) 0 z m the function z f z has a removable singularity at z f has a pole of order m at z which contradicts the hypothesis that z is an essential singularity of f. Our assumption was wrong
Hence, f() z 0 z G
f() G is dense in i.e. f() G
z z0 is a removable singularity lim z z0 f ( z ) 0 z z0 If z is an essential singularity of f then f has Laurent series expansion about z . If f is analytic in B; than for any z B; and has a Taylor series expansion about z = . m1 m( m 1) d m g()()()()() z z f z g z z f z dzm1 m1 d m g(m 1) lim g ( m 1) ( z ) lim z f z z z dz
12.6. SUMMARY
1) A point at which the function f is not analytic is said to be a singular point or singularity of the function f.
2) A function f has isolated singular point at z z0 if an R 0 s.t. f is defined and analytic in 0 z z0 R but not B z0, R . 248
3) If f has an isolated singular point at z0 , then z z0 is a removable singularity of f iff lim z z0 f z 0 z z0
4) If f has a pole at z z0 and m 1 is the smallest positive integer m s.t. ()()z z0 f z has a removable singularity at z z0 then, f has a pole of order m at z z0
5) Laurent Theorem: If f is analytic in G ann ; R2 , R 1 , R 2 0 then for any point z in G, f has unique representation bn f()() z a z n n n n0 n 1 ()z 1f ( ) Where, an d n = 0, 1, 2, … 2i () n1 x1 1f ( ) , bn d n = 1, 2, 3, …………… 2i () n1 x2 and x1, x 2 are circles r1, r 2 respectively with R2 r 2 r 1 R 1 .
6) Let f be analytic in 0 z R R 0 Then, f has a pole of order m at z iff there exists an analytic function, g:; B R s.t. g() z f() z where g 0 ()z m
7) Casorati Weierstrass Theorem: If f has an essential singularity at z , then for every 0,
f ann ; 0, is dense in .
12.7. UNIT END EXERCISES:
1) Each of the following functions f has an isolated singularity at z 0 . Determine it’s nature , if it is removable singularity , define f (0) , so that f is analytic at z 0 if it is a pole, find the singular part; if it is an essential singularity determine f z: 0 z for arbitrarily small values. 249
sin(z ) cos(z ) 1 (a) f() z (b) f() z z z z2 1 1 (c) f() z ( d) f( z ) z sin . z( z 1) z sin(z ) Solution: (a) limf ( z ) lim 1exists. z0 z 0 z z 0 is a removable singularity of f . sin(z ) Define g: B (0; r ) as g( z ) , z 0 1,z 0 . z Then f()() z g z for 0 z R and g is analytic on B(0; r ) .
z(cos( z ) 1) (b) lim (0)()limz f z 0 z0 z 0 z By theorem on removable singularity, z 0 is a removable singularity of f . Define f (0) 0 . cos(s ) 1 Define g: B (0; r ) as f( z ) z 0 z 0z 0 Then f()() z g z for 0 z R and g is analytic on B(0; r ) .
(c) f has a pole at z 0 . exp(1/z2 ) 2) Classify the singularities of (a) cot(z) (b) z 1 sin z (c) f z 2 (2005) z z1 z 2
1 exp 2 Solution: (b) Let f() z z . z 1 1 exp z2 limz1 f ( z ) lim f ( z ) has a pole at z 1. z1 z 1
A() z 1 Since f() z , where A( z ) exp , B ( z ) z 1. B() z z2 ABB(1) 0, (1) 0 has a zero of order 1 at z 1⇒ f() z has a pole of order 1 at z 1. We know that f is analytic in a deleted neighbourhood BR(0; ) 0 of 0 and f is not analytic at z 0 z 0 is a 1 exp 2 singularity of f. Since we know that lim z does not exist. z1 z 1 250
z 0 is not a removable singularity of f. A()() z A z f() z cant be written as f() z , where A and are B() z z2 analytic at z 0 z 0 is not a pole of f. z 0 is an essential singularity of f. sin z (c) f z z z1 z 22
Solution: f(z) has pole at z=0, z=1, z=2 of order 1,1,2 respectively.
1 3) Find the Laurent series expansion of (a) about z 0 . z2 z 2 1 (b) about z 0 . z2 4
Solution: (a) 1 1 12 1 2 k 2 k Let f() z z z k0 ( 1) z z4 z 2 z 21 z 2 1 ( z 2 ) ( 1)k1z 2 k k1 4) Check whether z 0 is a removable singularity of sin(z ) f() z or not. z
sin(z ) Solution: f() z z sin(z ) z2 z 4 f( z ) 1 ... z 3! 5!
Here all the coefficients ck 0 for k 0 . f(z) has removable singularity at z=0
5) Show that the image of B(0;1) 0 under the function 1 f( z ) cos ec is dense in the Complex plane. (Hint: z 0 is z 1 essential singularity of sin . Make use of the following z theorem: If f is analytic in a deleted neighbourhood D of z0 except for poles at all points of a sequence zn z0 . Then f() D is dense in the Complex plane. 251
1 2 6)Expand f() z ze z in a Laurent series about z = 0 and name the singularity.
e2z 7) Expand f() z in a Laurent series about z = 1 and name z 13 e2 e2 z 1 e2z e 2 z e 2 e 2 the singularity. (Hint: f() z ) 3 3 3 z1 z 1 z 1
8) Determine the number of zeroes, counting multiplicities, of the polynomial z4-2z3+9z2+z-1 inside the circle z 2
z 9) Expand f z in a Laurent series valid for z1 2 z (i) 0 z 2 (ii) z 1 (2008)
e2z 10) Expand f z in a Laurent series about z=1 and name z 13 the singularity. (2007)
z 11) Expand f z in a Laurent series valid for z1 z 2 (i)1 z 2 (ii) z 2 (2007)
1 12) Expand f z in a Laurent series valid for z1 z 2
(i) 1 z 2 , (ii) 2 z .
1 13) Expand f z z 3 sin in a Laurent series about z 2 z 2 and name the singularity. 1 (14) Expand f z in a Laurent series valid for z24 z 3
(i)1 z 3 , (ii) z 3 .
252 13
RESIDUE CALCULUS AND MEROMORPHIC FUNCTIONS
Unit Structure : 13.0. Objectives 13.1. Introduction 13.2. The Residue Theorem and it’s Application 13.3. Evaluation of Standard Types of Integrals by the Residue Calculus Method 13.4. Argument Principle 13.5. Rouche’s Theorem 13.6. Summary 13.7. Unit End Exercises
13.0. OBJECTIVES:
In this unit we shall study the generalisation of the Cauchy closed curve theorem to functions having isolated singularities. We shall prove the Residue theorem and further we shall use it to evaluate the standard types of integrals like sin(x ) dx dx,,() f z dz etc. . We shall also prove the x 0 1 x2 z 1 Argument Principle and Rouche’s theorem for Meromorphic functions in the complex plane .
13.1. INTRODUCTION :
In this unit, we now seek to generalize the Cauchy closed curve theorem to functions, which have isolated singularities. If is a circle surrounding a single isolated singularity z0 and k f()() z k ck z z0 in a deleted neighbourhood of z0 that contains trace of a circle , then f f 2 ic1 . Thus the coefficient c1 is of special significance in this context. We 253 shall see some of the applications of the Residue theorem. Let us start defining the Meromorphic functions.
Residues: Let f has an isolated singularity at z and n Let f() z an z be its Laurent expansion about z Res n 1 f ; = Coefficient of z in Laurent series = a1.
13.2 THE RESIDUE THEOREM AND IT’S APPLICATIONS :
Proposition: If f has a pole of order m at z and m1 g g()()(), z z f z then Res f ; m 1 ! Proof: Given that, f has a pole of order m at z z is an isolated singularity of f by definition, R 0 s.t. f is analytic in 0 z R or B ; R \ f has a pole of order m at z g() z f( z ) , where g 0 and g is analytic in BR; . z m for any z B; R , g has Taylor expansion about z ()n n g g() z an z where an ------(2) n0 n! m1 m m 1 gzaaz( ) 0 1 ( ) ...... azm 1 az m az m 1 g() z aa a f( z ) 0 1 ...... m 1 m m m1 z z z z 2 am a m1 z a m 2( z ) ...... This is a Laurent expansion for f(z) about z by definition of residue, 1 Res f ; Coefficient of z in a Laurent series am1
gm1 g()n () Res f ; ------ a n m 1 ! n! 254
Calculation of Residues: 1) If f has a pole of order mat z then, 1 d m1 Res f; lim ( z )m f ( z ) Where m = order m1 m 1 ! z dz
2) If f has a simple pole at z , then, Res f; lim z f z z e2 Example: Determine the residue of f z at its 2 z 2 z 3 poles ez Solution: Given function, f() z z 2 z 32 Heref has simple poleatz=2and z=3is a poleof order2. For z= 2:
Res f; lim z f z z
e2 e 2 Res f ;2 lim 2 2 lim z2 2 z 2 2 z 2 2 3 2 3 e2 e 2 = e2 2 32 1 2 For z=3 1 d m1 Res f; lim z m f z m1 m 1 ! z dz 3,m 2 1 d 2 e Res f ;3 lim z 3 2 2 1 ! z3 d z 2 z 3 2 d c2 lim 2 z 2 z3 d 1 1 1 e3 lim ez e2 e3 e 3 e 3 0 z3 2 z 2 2 3 2 z 2 3 2 255
Example: Determine the residues of f( z ) cot( z ) at its poles Solution: Here fzdz fzdz fzdz .... fzdz x x1 x 2 xm 2 fzdz a() rei n ired i n k k n xk 0
0 z z0 0 z 1 e2z f z z 13 cos z f z cot z sin z f has simple pole at n, where n is an integer Res f ; lim z f z z cos z 0 Res f , n lim z n . form z n sin z 0 (z n )sin z cos z By using L’hospital rule, lim 1 z n cos z
sin z f z Example: Compute the residue of z 4 at the pole z=0. (2012) Solution: sinz 1 z3 z 5 z 7 f z 4 4 z ...... z z 3! 5! 7! 1 1 z z3 ..... z3 6 z 5! 7! 1 Res f ,0 coefficient of 1 z 6 Residue Theorem or Cauchy Residue Theorem
Let f be analytic in a domain G except for the isolated singular points z1, z 2 ,...... zm .If x is a simple closed curve which does not pass through an of the points zk then
m f z d z2 i Re s ( f ; zk ) x k 1 2i ×[sum of residue of f at its pole inside x ], where x is traversed in anticlockwise direction. (2008) 256
Proof: Given that, z1, z 2 , z 3 ...... zm are isolated singular points in G. Assume these points lie inside x Choose positive numbers r1, r 2 ,...... rm so small that no two circles xk: z z k r k intersect k 1,2...... m and every circle xk k1,2,...... m is inside x . x 1 x2 x3 x4
r r r r z1 1 z2 2 z3 3 z4 4
x m
zm rm
Fig 13.1
the function f is analytic in a domain which is bounded by non- intersecting closed curves x1, x 2,...... xm and on the curves. By using Cauchy Deformation Theorem fzdz fzdz fzdz .... fzdz x x1 x 2 xm m f z dz f z dz ………….. 1 k 1 x xk f has an isolated singular point at z zk f has Laurent expansion about z zk n f z an z z k 0 z zk r k n Any point on the circle = centre + radiusei Consider, n an z z k n n z zk dz xk f z dz x k xk n
Any point on the circle xk is, i z zk r k e 0,2 i dz irk e d 257
2 2 i n1 fzdz a() rei n ired i a. irn1 e d n k k n k n xk 0 n 0
2 Note: i( n 1) e d 2 , n 1 0 0 ,n 1
1 1 f z dz a1 i rk 2 2i a1 2i . Re s f i zk by xk definition of residue. Put this value in equation (1) to get. m m f z dz 2 i Re s f ; zk 2i Re s f , zk x k 1 k 1
Example 1: Use Residue Theorem to evaluate z2 1 where x is the circle z 3 2 x z z 2 z 4
Solution:- By using Residue Theorem f z dz 2 i sum of residues of f at its poles inside x ]……..(1) x z2 1 Here f z 2 z z 2 z 4 f has a simple pole at z 0 and z 2 i and z 4 is a pole of order 2 .
But, simple poles z 0, z 2 lies inside x
-4 -3 0 2 3
Fig 13.2 258
For simple pole at z=0 Res f ; lim z f z z z2 1 Res f ;0 lim z 0 z0 z z 2 z 42 2 2 2 1 z2 1 0 1 lim lim = 2 2 2 z0 2 z 2 z 4 z0 z 2 z 4 0 2 0 4 1 32
Similarly, ez e z 0 ez e z 0 e z e z 2 e2z 1 e (2 n 1) i 2 z (2 n 1) i z (2 n 1) i z 3 2 z x1 5 Res f ;2 72 from equation (1) z2 1 2i Re s f ;0 Re s f ;2 2 x z z 2 z 4 1 5 1 5 9 20 11 2i 2 i 2i 2i 32 72 8 4 8 9 72 4 72 4 2
z2 1 11 i 2 144 x z z 2 z 4
(2) Use Residue Theorem to evaluate sin z dz where z is a closed square bounded by x 2, and 3 x z 4 y 2 .
Solution : By using residue Theorem, f z dz 2 i [sum of residue of f at its poles inside x ]…….(1) x 259
sin z Here f z 3 z 4 f has a pole of order 3 at z 4 But, the pole z lies inside x 4 For pole of order m 3, m1 1 d m Res ( f ; lim z f z 3 1 ! z dz3 1
2
1 z=11 f (z) 4 -2 2 -1 -1 1
-2
Fig 13.3
3 1 d3 1 sin 2 Res f ; lim z 3 1 4 3 1 !z dz 4 3 z 4 1 d 2 1 lim sin z lim sin z 2 2 2 z dz z 4 4 1 1 1 sin 2 4 2 2 2 1 Res f ; 4 2 2 sin z 1 1 f z dz dz 2i i 3 2 2 2 x x z 4
ez (3) dz where x is circle z 3 cosh z 260
Solution :- by using residue Theorem f z dz 2 i [sum of residue of f at its poles x ez Here f() z cosh z Let coshz=0 ez e z 0 ez e z 0 e z e z 2 e2z 1 e (2 n 1) i 2 z (2 n 1) i z (2 n 1) i 2 2n 1 f has simple poles at z i 2 But , the simple poles z i lies inside z 3 2 For simple pole Res f ; lim z f z z 2 e 0 Res f ; i lim z i form 2 2 cosh z 0 z i 2 z 2 z i e e 1 2 cos i sin 0 i .1 lim 2 2 1 sinh z i.1 z i i sin 2 2 e3 Res f ; i lim z i 2 2 cosh z z i 2 2 2 z i e e 1 i 2 0 e 2 cos i sin lim 2 2 1 sinh z isin z i sinh i 2 2 2 ez 2i 1 1 4 i cosh z x z2 1 (4) where x is the circle z 4 2 x z1 z 2 2 5 261
13.3 EVALUATION OF STANDARD TYPES OF INTEGRALS BY THE RESIDUE CALCULUS METHOD :
* Application of Residue Theorem to evaluate real Integrals.
Type-I 2 2 Integral of the type Fcos,sin d, Where F cos ,sin is 0 0 rational function of cos and sin 2 Consider, Fcos ,sin d 0 dz dz Put z ei, 0 , 2 , dz i ei d d i. ei iz ei e i z z1z1 z 2 1 cos z 2 2 2 2z ei e i z z1z1 z 2 1 sin z 2i 2i 2 i 2 iz 2 z21 z 2 1 dz Fcos , sin d F , ,Where 2z 2 z iz 0 x z21 z 2 1 i f z F , 2z 2 z iz 2i [sum of residue of f at its pole inside x ] (by residue Theorem) b b2 4 ac ax2 bx c is polynomial than the root are x 2a
2 d (1) Use Residue Theorem to evaluate 2 cos 0 2 d Solution :- the given integral is 2 cos 0 Put z ei, 0 , 2 , dz i ei d dz dz d d i. ei iz ei e i z z1 z 2 1 cos 2 2 2z 262
2 d 1 dz where x is the unit circle z 1 2 cos z2 1 iz 0 x 2 2z 1 1 1 2 1 dz dz i z 4z z2 1 i z2 4 z 1 x x 2 z ………(1) 1 Put f z z2 4 z 1 4 16 4 2 2 3 f has simple pole at z i.e. 2 2 z 2 3 but, the simple pole 2 3 2 1.73 lies inside x For simple pole, Res f ; lim z f z z 1 0 Res f ; 2 3 lim z 2 3 form z2 3 z2 4 z 1 0 1 1 1 lim z2 3 2z 4 2 2 3 4 2 3 by using Residue Theorem f z dz 2 i [sum of residue of f at its poles inside x ] x dz 1 i 2i 2 2 3 3 x z 4 z 1 Put this value in equation (1) 2 d 2 i 2 cos i 3 0 2 d 2 2 cos 3 0 263
2 d (2) use residue Theorem to evaluate a b a bsin 0 2 d Given , a bsin 0 d2 dz Put z ei0, 2 dz iei d d iei iz ei z z1 z 2 1 sin 2i 2 i 2 iz 2 d dz iz where x is the unit circle z 1 a bcos z2 1 x x a b 2iz dz 1 dz 2 ………..(1) iz 2 2 x 2aiz bz b x bz2 aiz b 2a iz 1 Put f z bz2 2 aiz b 2 4a2 i 4 b 2 f has simple pole at z 2 ai 2b 2ai 2 b2 a 2 2b
b2 a 2 a a2 b 2 i.e. f has simple pole at z ai i b b
a a2 b 2 a a2 b 2 Let 1 i and i b b 1 1 1 1...... b a But the pole lies inside x For simple pole, Res f ; lim z f z z 1 0 Res f ; lim z . form 1 1 2 z1 bz 2 aiz b 0 264
1 lim z1 2bz 2 ia 1 1 a a2 b 2 2ai 2i a2 b 2 2 ai 2 b i 2ia b 1 Res f ;1 2i a2 b 2 By using Residue Theorem, f z dz 2 i [sum of residue of f at its poles inside x ] x 1 2i 2i a2 b 2 f z dz 2 2 x a b Put this value in equation (1) we get 2 d 2 2 a bcos 2 2 2 2 0 a b a b d 3. Use residue theorem to evaluate a>b! a bcos 0 2 d Hint: First evaluate a bcos 0 2 d d Then use the property a bcos a b cos 0 1 2cos 4. Evaluate d & 5 3cos 0 2 cos3 5. Evaluate d & 5 4cos 0 d 6. Evaluate a 1 a cos 0 265
Type-II Integral of the type f() x dx P() x Improper Integral where f( x ) , where p ( x ), Q ( x ) are Q() x polynomial in x . M Example: If f() z for z Rei where K>1 and M are RK constants.
Then lim f ( z ) dz 0 where, xR is the semi circle are of radius R xR R as shown in figure.
M Proof: Given f() z for z Rei RK
x x R
-R O R
Fig 13.4
M f()() z dz f z dz dz K x x R x RR R Put z Rei dz Rie i d M M f() z dz Riei d R RK RK xR
M f( z ) dz 0 as R RK 1 xR lim f ( z ) dz 0 R x R 266
Example: Evaluate f() x dx , where P() x f(),()() x P x and Q x are polynomials in x . Q() x Solution: Consider f() z dz where x is a closed curve consisting of x large semicircle x of radius R and the real axis from –R to R R traversed in the anticlockwise direction.
qÎé0,pù x ëê ûú R
-R O R
£- plane
Fig 13.5
We choose only those poles of f which lie in the upper half of the complex plane. by residue theorem, f( z ) dz 2 i [sum of residues of f at its pole inside x ] x R f( z ) dz f ( x ) dx 2 i [sum of residues of f at its pole inside x ] x R on real axis z x R lim f () z dz lim f () x dz 2 i [sum of residues of f at its RR x R R poles inside x ] O f( x ) dx 2 i [ sum of residues of f at its poles inside x ] ……….. (by previous example lim f ( z ) dz R x0 267
P() z Note: If f() z where, P(z) and Q(z) are polynomial in z such Q() z that, (i) Q(z) = 0 has no real roots. (ii) The degree of Q(z) is greater than that of P(z) by atleast 2, then, f( z ) dz 2 i [Sum of residue of f at its poles inside x ] x
Example: x2 3 (1) Evaluate dx 2 2 x1 x 4
x x R
-R O R
Fig 13.6
Where x is a closed curve consisting of large semicircle x of R radius R and real axis form –R to R traversed in the anticlockwise direction. degQ(z) degP(z)+2 P( z ) z2 3 & Q( z ) z2 1 z 2 4 this general method is applicable Consider, z2 3 z21 z 2 4 z3 3 Here, f() z z21 z 2 4 Here f has simple poles at z i and z 2 i but the poles z i and z 2 i lies inside x . For simple pole 268
Res f; lim z f ( z ) z Res z2 3 f; i lim z i z i z21 z 2 4 z i z2 3 z 2 3 lim lim z i z i z i z2 4 z i z i z 2 4 i2 31 3 3 1 2 i i i2 4 2i 1 4 2 i 4 1 2 i (3) 1 Res f ; i 3i
z2 3 Res f;2 i lim z 2 i z2 i z21 z 2 4 2 z 2 i z 3 z2 3 lim lim z2 i2 z 2 i 2 z1 zr 2 i z r 2 i z 1 z 2 i (2i )2 3 4 i 2 3 4 3 2 2 2i 1 2 i 2 i 4i 1 4 i 4i 4 1 1 Res f ;2 i P2 i By using Residue Theorem z2 3 dz2 i [Sum of Residue of f at its pole lies inside 2 2 x z1 z 4 x ] 1 1 4 1 5 2i 2 i 3i 2 12 i 6 R z2 3 x 2 3 5 dz dx 2 2 2 2 6 xz1 z 4 R x 1 x 4 R R z2 3 x 2 3 5 lim dz lim dx ----(1) RR 2 2 2 2 6 xz1 z 4 R x 1 x 4 R 269
z2 3 T.P.T. lim 0 R 2 2 x z1 z 4 R
2 2i i z2 3 R e 3 Re d dz 2 2 2 2i 2 2 i x z1 z 4 0 R e1 R e 4 R i 2 put z Re , [0, 2 ] R 3 R d i RR21 2 4 dz iRe d 1 1 z z z z 2 e 2i 1 R 2 e 2 i 1 1 2 1 2 2 R2 e 2i 1 R 1 3 R 1 3 z3 3 2 dz R d 2 2 2 x z1 z 4 R 1 1 1 4 0 R RR2 2 1 3 1 2 R 0as R R 1 1 1 4 RR2 2 z2 3 lim dz 0 R z21 z 2 4 xR Put this value in equation (1) x2 3 5 0 dx 2 2 6 x1 x 4
x2 3 5 dx 2 2 6 x1 x 4
(2) Use Residue theorem to evaluate x2 dx 2 0 1 x2 x2 Hint: First calculate dx and then use the property 2 1 x2 270
x2 x 2 dx 2 dx 2 2 1 x2 0 1 x 2 z2 Solution: Consider dz where x is a closed curve 2 x 1 z2 consisting of large semicircle xR of Radius R and real axis from –R z2 to R traversed in the anticlockwise direction. Here f(), z 2 1 z2 Here, f has pole of order z at z i but, the pole z i lies inside x . For pole of order 2 1 d 2 Res f; i lim z i f ( z ) 2 i ! z i dz 2 2 2 2 d2 z z i z d z lim z i lim lim z idz 2 z i 2 z i dz 2 1 z2 z i z i z i x2 x 2 (3) dx 4 2 x10 x 9
Type-III: Integral of the type sim mx f() x dx or cos(mx ) f ( x ) dx P() x m > 0 where f x Q() x
x x R
-R O R
Fig 13.7 Consider eimz f() z dz where x is a closed curve consisting of large x semicircle xR of radius R and the real axis from –R to R traversed in the anticlockwise direction. 271
by residue theorem eimz f( z ) dz 2 i [Sum of residue of f at its pole lies inside x ] x R eimz fzdz( ) e imz fxdx ( ) 2 i [ Sum of residue of f at its pole RR lies inside x ] R lim eimz f ( z ) dz lim e imx f ( x ) dx 2 i [Sum of residue of f RR x R at its pole inside x ] ------(1) lim eimz f ( z ) dz 0 ------(by next example) R xR from equation (1), 0 Cosmx i sin mxfxdx ( ) 2 i [Sum of residue of its poles inside x ] cos3x Example: 1) Use Residue Theorem, to evaluate 2 dx . x 4 e3iz Solution: Consider dz 2 x z 4
x x R z q=p q=0 -R O R
Fig 13.8
Where x is a closed curve consisting of large semicircle xR of radius R and the real axis from –R to R traversed in the anticlockwise direction. e3iz Put f() z z2 4 Here, f is a simple poles at z 2 i but, the pole z zi lies inside x . For simple pole, 272
Res f; lim z f ( z ) z2 i 3iz e3iz z 2 i e Res f;2 i lim z 2 i lim z2 iz2 4 z 2 i z 2 i z 2 i 2 e3iz e 3 i (2 i ) e 6 i lim z2 i z 2 i 2 i 2 i 4 i
e6 Res f ;2 i 4! by residue theorem, f( z ) dz 2 i [Sum of residue of f at its all poles inside x ] x e3iz e 6 dz 2 i 2 4i x z 4
e3iz . e 6 dz 2 2 x z 4 R e3iz e 3 iz dz dx e6 z2 4 x 2 4 2 xR R R e3iz e 3 ix lim dz lim dx e6 ------(1) RR z2 4 x 2 4 2 xR R e3iz T.P.T. lim dz 0 R z2 4 xR Put z Rei 0,2
dz i.. R ei d 1 1 R2 e 2i 4 R 2 4 R2 e 2i 4 R2 4
3i .Rei i e iRe d e3iz dz z2 4 R2 e 2i 4 xR 2 R R e3iR Cos isin d e3R Sin 2 2 R 4 0 R 4 0 273
3iR Cos … e 1 20 Sin , 0, 2
y=Sinq
1 y=Sinq 2q y = p
0 p p x=q 2
Fig 13.9 3iz 6R e R 2 dz 2 e d ……. Sin z2 4 R2 4 xR 2 6R Sin 3R Sin 6R 2 6RR 6 2R e 2R 0 e 2 e R2 4 6R 2 6R R 4 0 e3R 1 1 e3R 0as R 0 2 3R 4 3R2 1 4 R2 e3iz lim dz 0 R z2 4 xR Put this value in equation (1) Cos3 x i Sin 3 x 0 e6 2 4 x 4 Equating Real part from both sides Cos3 x dx e6 2 2 x 4 274
Sinmx (2) dx m 0 a R 2 2 0 x x a Cosmx (3) dx [ z i is a pole of order 2 ] 2 0 1 x2 Sinx (4) dx 2 x 4 x 5
Type-IV: Poles on the Real axis In this case, we cannot use residue theorem because the pole z = 0 lie on real axis and hence on x .
sin x Example: Prove that dx x 2 0 eiz Solution: Consider dz z 0 Y
x x R
X -R -r o r R
Fig 13.10
Where x is a closed curve consisting of (i) a real axis from r to R
(ii) a large semicircle xR of radius R. (iii) a real axis from –R to –r
(iv) a small semicircle xr of radius r. the pole z = 0 lie outside x , by Cauchy- Goursat theorem, eiz dz 0 z x 275
R r eix e iz e ix e iz dx dz dx dz 0 x z x z r xR R x r In 3rd integral x is negative dx will be dx , using this negative sign make limits r to R. RR eix e ix e iz e iz dx dx dz dz 0 x x 2 z r r xR x r R eix e ix e iz e iz 2i dx dz dz 0 2i x z z r xR x r R Sinx eiz e iz eiz e iz 2i dx dz dz 0 Sinz x z z 2 r xR x r Taking limit R and r 0 R Sinx eiz e iz lim 2i dx lim dz lim dz 0 ------(1) R x R z r z r0 r xR eiz T.P.T. lim dz 0 (Proof as similar as Type III) R z xR eiz Consider dz z xR Put z r ei dz ir e i d iz i r ei e e i dz ire d ------ f f z r ei xr 0 x x Taking limit as r 0 eiz lim dz i e0 d i x0 z xr 0 from equation (1), Sinx 2i dx 0 i 0 x 0 Sinx dx 0 i 0 x 276
Sinx i Sinx x2 i x 2 0
Meromorphic Function (M.F): If f is defined and analytic in an open set G except for poles. Then f is a Meromorphic function on G. z2 e..() g f z z z 5
Theorem (1) If f has a zero of order m at z , then Res f() z ; m f() z
Proof: Given that, f has a zero of order m at z an analytic function g:; B R m s.t. f() z z g z where g( ) 0 Diff. w.r.t. z m1 m f()()() z m z g z z g z m1 f z z m g()() z g z m g z g z m f z z g z z g z f z Laurent expansion for about z f z f z m g z , 0 z R f z m g z g z is analytic in BR, and hence it has Taylor series g z expansion about z .
By the definition of residue, f z Res ; m = Coefficient of z in the Laurent series f z expansion.
Theorem (2) If f has a pole of order n at z then Res f z ; n f z 277
Proof: Given that, f h as a pole of order n at z an analytic function g:; B R g() z s.t. f() z where g 0 z n Diff. w.r.t. z zn g z g z n z n1 f z 2 z n n1 n f z z z g z ng z z n n f z z z g z z g z n g z g z n z g z g z z f z n g z 0 z R f z z g z g z is analytic in ;R and hence it has Taylor series g z expansion about z by definition f z Res ; n f z 13.4 THE ARGUMENT PRINCIPLE :
Let f be a Meromorphic function in a domain G and have only finitely many zeros and poles. If x is a simple closed curve in G s.t. no zeroes and poles of f lie, on x, then f z 2i dz Z P Where, ZP, , denote respectively the f z f f f f x number of zeros and poles of f inside x, each counted according to their order or multiplicity. (2004, 2007)
Proof: Given that, f is a Meromorphic function in domain G. f z Put F z f z the singular points of F inside x are the zeros and poles of f. 278
by Residue Theorem, f z F z dz dz 2 i [Sum of residues of F at its singular f z x x points inside x . ------(1) f z If is a zero of f of order m , then Res ; m j j j j f z f z If B is a pole of f of order n , then Res ; n k k k f z From equation (1) f z f z f z dz 2 i Re s ; Re s ; B f z f zj f z K x 1 f z dz m n 2i f z j K x i K 1 f z dz Z P 2i f z f f x
Example: (1) Use Argument Principal to evaluate f z z 2 dz where f z and x is the circle z 3 . f z z z1 z 4 x Solution: by Argument Principal 1 f z dz Z P where, Z and Z are the no. of zeros and 2i f z f f f f x poles of f inside x each --(1) Given function, z 2 f z z z1 z 4 Herefhassimplezerosatz=2andz=0,Z=1andz=4aresimple poles
x
0 1 2 3 4
Fig 13.11 279
Given equation of circle z 3 But, simple zero z = 2 and simple poles z = 0, 1, lies inside x . from equation (1) 1 f z f z dz1 2 1 2i 2i f z f z x x
f z (2) Use the argument principal to evaluate dz where f z x z 1 f z and x is the circle z z2 z 2 z 3
Solution: By argument principle 1 f z dz Z P 2i f z f f x z 1 Given function f z z2 z 2 z 3 Here f has simple zero at z1 and z 0 is a poles of order 2 and z = 2, 3 are simple poles. Given equation at circle, z But simple zero z = 1 and simple poles z = 2,3 and z = 0 is pole of order 2 lie inside x . From equation (1), f z 2i 1 4 2 i 3 6i . f z x (3) Evaluate cot z by using argument principle z cos z 1 .cos z Solution: cot z dz dz z z Sin z z Sin z by argument principal 1 cotzdz 2 i Z P 2i Z P …………. (1) p f f f z Here f() z Sin z has simple zeros at z n n z n n 280
z 0, 1, ,2, 3,..... are simple zeros of f and f has no poles but zeros z 0,1, 2, lies inside x i.e. z
cot(z ) 2 i 7 0 14i z (4) Evaluate tanz dz where x is the circle z x Take as an Exercise.
13.5 ROUCHE’S THEOREM :
Suppose f and g are Meromorphic functions in a nbd. of BR; with no zeros and poles on the circle x z; z R. If zf,, z g P f P g are the no. of zeros (poles) of f and g counted according to their order and if f z g z g z2 x then, ZPZP (2005, 2008,2009) f f g g .
Proof: Given that, f and g are Meromorphic functions in a nbd. of BR; f z Put F z g z F is a Meromorphic function in a neighbourhood of BR; Let Z x t be any point on the circle x . Then, z x t Reit t 0,2
F s=Fox x s (t) R a o 1 2p
z- plane B(i ;1) w- plane Fig 13.12 x is a simple closed curve in Z-plane and F is analytic. F. x is also a closed curve in w-plane for any t 0,2 281
f x t t 1 F x t 1 1 ------(1) g x t Given that f z g z g z z x f() z 1 1 g() z Put z x t f x t 1 1 g x t Put the above value in equation (1), we get t 1 1 B((1,0);1) 0 belongs to the unbounded component of w \ by definition of winding no; ; 0 0 ( 0 lies outside the curve ) 1 d 0 2i 0 Put t d t dt t 0,2 2 1 t dt 0 ------(2) 2i t 0 f x t Now, t g x t x t gxtfxt fxtgxt xt t 2 g x t t gxt gxtfxt gxt x t 2 t f x t g x t 2 g x t f x t g x t f x t g x t x t 2 2 gxt gxt fxt gxt 282
t f x t g x t x t t f x t g x t 2 1 f x t g x t 0 x t dt ------(2) 2i f x t g x t 0 Put z x t dz x t dt 1 f z g z 0 dz 2i f z g z x 1f z 1 g z dz dz 2i f z 2 i g z x x by the argument principle ZPZPf f g g
Example: (1) Use Rouche’s theorem to prove that all zeros of the polynomial z7 5 z 3 12 0 lie between the circles z 1 and z 2 . Solution:
Consider the circle x1 : z 1 Let f z z7 5 z 3 12 and g z 12
g has no zeros inside x1
For any point z x1 7 3 f z g z z7 5 z 3 12 12 z7 5 z 3 z 5 z 3 17 5 1 1 5 6 12 9 z f()()() z g z g z v z x) Fox
f z g z g z z x1 by Rouche’s theorem,
ZZf g (Here, there are no poles)
f has no zeros inside x1
Consider the Circle x2 : z 2 Let f z z7 5 z 3 12 and g z z 7
g has 7 zeros, counting order, inside x2 283
For any point z x2 f z g z z7 5 z 3 12 z 7 5z3 12
3 5z3 12 5 z 12 5 8 12 40 12 52 z7 g z
f z g z g z z x2
Hence all zeros of the polynomial z7 5 z 3 12 0 lie between the circles z1 and z 2
(2) Use Rouche’s theorem, to prove that ez az n a e has n zeros (roots) inside the circle z 1
Solution: Consider the circle x: z 1 x1 Let f z azn e z and g z azn F g has n zeros, counting order, inside x. For any point z x, f z g z azn e z az n Fig 13.13 ez e z e1 a g z f z g z g z z x
by Rouche’s theorem ZZf g f has n-zeros inside the circle z 1
(3) Use Rouche’s theorem to prove that every polynomial of degree n has n zeros.
13.6. SUMMARY:
1) Residues: Let f has an isolated singularity at z and n Let f() z an z be its Laurent expansion about z Res n 1 f ; = Coefficient of z in Laurent series = a1. 284
2) Calculation of Residues: 1) If f has a pole of order mat z then, 1 d m1 Res f; lim ( z )m f ( z ) Where m = order m1 m 1 ! z dz
2) If f has a simple pole at z , then, Res f; lim z f z z Let f be analytic in a domain G except for the isolated singular points z1, z 2 ,...... zm :If x is a simple closed curve which does not pass through an of the points zk than.
3) Cauchy – Residue Theorem: m f z d z2 i Re s ( f ; zk ) x k 1 2i [sum of residue of f at its pole inside x ] Where, x is traversed in anticlockwise direction
4) Meromorphic Function (M.F): If fis defined and analytic inanopen setG except for poles. Then f is a Meromorphic function on G.
5) The Argument Principle: Let f be a Meromorphic function in a domain G and have only finitely many zeros and poles. If x is a simple closed curve in G s.t. no zeroes and poles of f f z lie, on x, then 2i dz Z P where, ZP, , denote f z f f f f x respectively the number of zeros and poles of f inside x, each counted according to their order or multiplicity.
6) Rouche’s Theorem: Suppose f and g are Meromorphic functions in a nbd. of BR; with no zeros and poles on the circle x z; z R. If zf,, z g P f P g are the no. of zeros (poles) of f and g counted according to their order and if f z g z g z2 x then,
ZPZPf f g g 285
7) Schwarz’s Lemma Let D z : z 1 be the unit disk and suppose f is analytic in D with, (i) f (0) = 0 and (ii) f z 1 for z D Then, f z z z D and f 0 1
8) A function f: D D is said to be an Analytic automorphism or Automorphism of the Unit disc D, if f is bijective and if both f, f 1 are analytic in D.
13.7. UNIT END EXCERCISES:
1) Show that the Polynomial P( z ) 2 z10 4 z 2 1 has exactly two zeroes in z 1. Solution: Let f( z ) 4 z2 , g ( z ) 2 z 10 1 f()() z g z for every number on the Unit circle. By Rouche’s theorem, the number of zeroes of f g inside the curve Z 1= the number of zeroes of inside the curve Z 1 . 2z10 4 z 2 1 has exactly two zeroes in the curve z 1. (Here the number of zeroes of inside the curve Z 1. 1f 1 8 z dz 2. 2iz1 f 2 i z 1 4z2
2) Suppose that f is entire and f() z is real if and only if z is real .Use the Argument Principle to show that f can have atmost one zero. ( Hint: Consider the image of the circle z R . Here f maps the entire upper semicircle z R, y 0 into either the upper half plane or the lower half plane . Similarly, f maps the entire lower semicircle z R, y 0 into either the upper or lower half plane , because Arg() w is atmost π in any upper/ lower half plane Agf( z ) 2 as z traverses through the circle z R . The Number of zeroes of f() z in z R 1 1 Argf( z ) 2 1. 2 2 286
1 3) Find the number of zeroes of f() z ez z in z 1. ( Hint : 3 1 Let f(),() z z g z ez f( z ) g ( z ) z , z 1) 3
4) Find the number of zeroes of f( z ) z6 5 z 4 3 z 2 1 in z 1( Hint : Take f( z ) 5 z4 , g ( z ) z 6 3 z 2 1 on z1, f ( z ) 5 z4 5 z 6 3 z 2 1
Also f()() z g z only at i . There are 4 zeroes of f in z 1.
5) Show that for each R 0 if n is large enough then z2 zn P( z ) 1 z ... has no zeroes in z R . n 2!n ! z (Hint: Pn () z e as n .)
6) If f is Meromorphic on G and f : G is defined by f() z if z is pole of f f() z otherwise. Show that f is continuous on G.
dx 7) Find . x4 1 i 3i Solution: Here z e 4 and z e 1 2 4 1 Represent the poles of in the upper half plane. Since z4 1 each of these is a simple pole 1 The residues are given by the values of f() z at these 4z3 poles. i 1 1 Res ;e4 2 i 2 and z4 1 8 3i 1 1 Res ;e 4 2 i 2 z4 1 8 i dx 1 13i 2 2i Re s ; e4 Re s ; e x41 z 4 1 z 4 1 2 287
sin(x ) 8) Evaluate dx . x eix Solution: We know that the function has pole at x 0 . x We modify the integral as follows: ix sin(x ) e 1 dx Im dx . x x ix iz iz M e1 1 e e dx dz i dz M xIM z IM z iz ix e e 1 lim dz 0 dx i M IM z x ix e1 sin( x ) Im dx . dx . x x
dx 9) Evaluate 0 1 x3 i 1 3 Solution: Let f(). z Then f has poles at z1 e and 1 z3 3i i 3 z2 e 1, z 3 e i log(z )3 i 1 i 3 Res ; z1 e 1 z3 9 2 2 i log(z ) 3 i Res ; z2 e 1 z3 3 si log(z )3 Si 1 i 3 Res ; z3 e 1 z3 9 2 2 log(z ) 2 Res ; zk 3 k 1 z3 9 dx log( z ) 2 Res ; zk 3 0 1 x2k 1 z 3 9
dx 10) Evaluate . 0 x(1 x ) a1 x 1 ( Hint: The integral has the form dx with 0 a 1 0 p() x 2 xa1 z a 1 Use the formula 1 e2i ( a 1) dx Re s ; z the 0 k k p()() x p x sum on R.H.S. is taken over the zeroes of the function p() z . 288
2 d 11) Evaluate 0 2 s ( ) 2 d 2 dz Solution: Consider 0 2 cos( ) i z 1 z2 4 z 1 1 2 4 Res ; 3 2 3 z2 4 z 1 3
x2 12) Evaluate the integral dx . 2 2 x 1 13) Use Cauchy-Residue theorem to evaluate
sin z dz where is the circle z i . 2 z 1
cos5x 14) Use Cauchy-Residue theorem to evaluate dx . 2 0 x 4
129 7
MOBIUS TRANSFORMATION
Unit structure: 7.0 Objective 7.1 Introduction 7.2 Conformal Mapping 7.3 Some standard transformation 7.4 Mobius Transformation Or Bilinear Transformation Or Linear Fractional Transformation 7.5 Summary 7.6 Unit End Exercise
7.0. OBJECTIVE
After going through this unit you shall come to know about
Special type of functions called transformation from The combination of special function to give rise to a transformation called Mobius Transformation Special properties of Mobius Transformation like fixed point and cross ratio. Method to find the bilinear transformation using various method.
7.1. INTRODUCTION
There are certain transformation that can be readily described in terms of geometry. In this chapter, we are mainly concerned with certain geometric interpretations of functions and finding the image of a given figure under a given bilinear function .
7.2. CONFORMAL MAPPING
2 A differentiable map f : is said to be conformal map if
detDfz 0 z & Df z , Df z ,, {0} Thus, conformal map is preserves the angle between two intersecting curves in 130
Proposition: Let be a domain in and f : be a map. Then f is any analytic function with f z 0 z if and only if f is conformal map with det(Dfz) > 0 z Proof: Let f be analytic, Dfz f z . Then Re f z f z Re f z, f z , , /{0} f z f z Thus f is conformal map. Let f(z)=u(z)+iv(z), z . By Cauchy Riemann equation, the u u u v Jacobian of f =( u(x,y),v(x,y)) is x y x x . v v x y vx u x 2 2 Hence det(Dfx)=ux + uy 0 . Now f z ux iv x & f z 0 .
2 2 2 Hence det(Dfx)=ux + uy = f z >0. Conversely, Fix z . Since f is conformal map,
Dfz1,0 , Df z 0,1 1, i 0 2 So, Dfz1,0 Df z 0,1. Let Dfz 1,0 a , b . Then
Dfz 1,0 b , a .
Since detDfz 0, Df z 0,1 b , a .Let a ib. Then
Dfz () (Verify) and f is complex differentiable.
7.3. SOME STANDARD TRANSFORMATION
(i) Translation: w z c where c is a complex constant. The transformation w z c is simply a translation of the axes and as such as preserves the shape if the region in z-plane.
Note: in particular translation maps circle in z-plane onto circles in the w-plane.
(ii) Rotation and Reflection: w cz where c is a complex constant. Let w Rei , z re i , c e i Now, Rei re i e i r e i() R& r Thus the transformation maps a point P(r, ) in the z-plane onto a point P(,) r in the w-plane. i.e. the image is magnified (or diminished) by r and rotated by . 131
Note: in particular w cz maps circle in z-plane onto circles in the w-plane. 1 (iii) Inversion: w z Let w Rei , z re i 1 Rei rei 1 R& r 1 i.e. transformatiom w maps P(r, ) in the z-plane onto a point z P(,) r in the w-plane.
1 Note: in particular w maps circle in z-plane onto circles in the z w-plane.
7.4. MOBIUS TRANSFORMATION OR BILINEAR TRANSFORMATION OR LINEAR FRACTIONAL TRANSFORMATION
az b Definition: A transformation s z w , where a,,, b c d are cz d complex constant and ad bc 0 is called Mobius Transformation or Bilinear Transformation or Linear fractional transformation.
b d Note : If ad bc 0 ad bc . a c b az b a z a w a constant. cz d c z d c a Thus, ad bc 0 is a necessary condition for the Mobius az b Transformation : s z w cz d 1) If S is a Mobius transformation, then S 1 is the inverse 1 1 mapping of S i.e. S S z z S S z. az b Let S z where a,,, b c d are complex constant and cz d ad bc 0 . dz b S1 z cz a 132
1 1 dz b S S z S S z S cz a dz b 1 a b cz a dz b c d cz a adz ab bcz ab cz a adz bcz cz a cdz bc cdz ad bc ad ad bc z z ad bc 1 Similarly, S S z z . a z () b ()az b Consider s() z ( 0) ()()()c z d cz d The efficients a, b, c, d are not unique.
a1 a 2 b 1 c2 aadd 1212 bbcc 1212 bcd 211 d2
a1 a 2 b 1 c2 abcd 2112 abcd 1221 bcd 211 d2 =
aadd1212 abcd 2221 bbcc 1212 abcd 2112
= adad1122 bc 22 bcbc 1122 ad 22 = adad1122 bc 22 bcad 1122 bc 22
= a1 d 1 b 1 c 1. a 2 d 2 b 2 c 2 0 . i.e. the coefficients a,,, b c d are not unique.
3) Let S be a Mobius transformation on . az b S z where a,,, b c d are complex constant and ad bc 0 . cz d z a b b z a S z z z c d c d z z a S when c 0 c when c 0 az b az b Again, S z cz d c z d c d S when c 0 c 133
4) If S and T are Mobius Transformations, then (TS ), (composition of T andS)isalsoa Mobius Transformation. a z b Let S z 1 1 1 where a,,, b c d are complex constant and c1 z d 1 a2 b 2 a1 d 1 b 1 c 1 0 and T where a2,,, b 2 c 2 d 2 are c2 d 2 complex constant and a2 d 2 b 2 c 2 0 .
a1 z b 1 T S z T S z T c1 z d 1
a1 z b 1 1 T S z a2 b 2 c1 z d 1 a1 z b 1 c2 d 2 c1 z d 1
a a z a b b c z b d c1 z d 1 1 2 2 1 2 1 2 1 c1 z d 1 a1 c 2 z b 1 c 2 c 1 d 2 z d 1 d 2 a a z a b b c z b d a a b c z a b b d z 1 2 2 1 2 1 2 1 1 2 2 1 2 1 2 1 a1 c 2 z b 1 c 2 c 1 d 2 z d 1 d 2 a1 c 2 c 1 d 2 z b 1 c 2 d 1 d 2 xz where, a1 a 2 b 2 c 1, a 2 b 1 b 2 d 1, x a1 c 2 c 1 d 2, b1 c 2 d 1 d 2
Now, x aa12 bc 21 bc 12 dd 12 ab 21 bd 21 ac 12 cd 12
aabc1212 aadd 1212 bbcc 1212 bcdd 2112 aabc 1212
abcd2112 abcd 1221 bcdd 2112 0 Hence, composition of S and T is also a Mobius transformation.
Proposition : If S is a M.T. then, S is a composition of translation rotation, inversion and Magnification.
az b Proof : Consider a Mobius Transformation, S z where, cz d a,,, b c d are complex constant and ad bc 0
Case I : When c 0 az b a b S z z d d d a b If S z z and S z z 1 d 2 d 134
a a b then, S2 S 1 S 2 S 1 z S 2 z z S z d d d SSS 2 1 In this case, Mobius transformation is a composition of translation, rotation and Magnification.
Case II : When c 0
c az b az bc ad ad bc ad a cz d S z c cz d ccz d c cz d c cz d bc ad 1 a . 2 z d c c c d 1 bc ad a If S1 z z ,, S 2 z S 3 z z and S4 z z c z c2 c d S S S S z S S S S z S S S z 4 3 2 1 4 3 2 1 4 3 2 c 1 SS4 3 z d c c bc ad c bc ad SS S S 4 3 4 2 4 cz d c c2 d c cz d bc ad a c cz d c a bc ad S z c cz d SSSSS 4 3 2 1 In this case, M.T. is a composition of translation, rotation, inversion and magnification.
Fixed Points : Definition : Let G be a subset of and f: G . Then point z0 G is said to be a fixed point of f if f z0 z 0 . e.g. i) Let f z z2 .Here, f has fixed points 0, 1 and . ii) Let f z 1 . Here, f has fixed point 1 and – 1. z iii) Let f z z 3 i .Here, f has fixed point .
Example 1 : What are the fixed points of Mobius transformation? 135
az b Solution : Consider a M.T., S z where, a,,, b c d are cz d complex constant and ad bc 0 . For fixed points, Put az b S z z z az b cz2 dz cz d cz2 d a z b 0 d a d a2 4. b . c z 2c a d d a2 4 bc z 2c A. M.T. can have atmost 2 fixed points unless S z z z iz 2 2) S z z 1
Solution : For fixed points iz 2 Put S z z z iz 2 z2 z z2 1 i z 2 0 z 1 1i 1 i2 8 z 2
Definition : For any three distinct points z2,, z 3 z 4 in C , the cross ratio of four points z1,,, z 2 z 3 z 4 is defined to be z1 z 3 z 2 z 4 z1,,, z 2 z 3 z 4 . z1 z 4 z 2 z 3 z2 z 3 z 2 z 4 For z1 z 2 , z2, z 2 , z 3 , z 4 1 z2 z 4 z 2 z 3 z3 z 3 z 2 z 4 For z1 z 3 , z3, z 2 , z 3 , z 4 0 z3 z 4 z 2 z 3 z4 z 3 z 2 z 4 For z1 z 4 , z4,,, z 2 z 3 z 4 z4 z 4 z 2 z 3
Definition : If z1 then the cross ratio z1,,, z 2 z 3 z 4 is the image of z1 under the unique. Mobius transformation which takes z2 to 1, z3 to 0 and z4 to . i.e. S z1 z 1,,, z 2 z 3 z 4 . 136
Note : If M is any Mobius transformation and w2,, w 3 w 4 are complex number s.t. M w2 1, M w 1 0 and M w4 then M z z,,, w2 w 3 w 4 .
Proposition : If z2,, z 3 z 4 are distinct points in and T is any Mobius transformation then, zz1,,,,,, 2 z 3 z 4 TzTzTzTz 1 2 3 4 for any fix z1.
Proof : Let S z z1,,, z 2 z 3 z 4 S is a M.T. (1) and S z2 1, S z 3 0 and S z4 Given that, T is any M.T. Put MST 1 1 M Tz2 S T Tz 2 S z 2 1 1 M Tz3 S T Tz 3 S z 3 0 1 M Tz4 S T Tz 4 S z 4
M z z,,, Tz2 Tz 3 Tz 4 Put z Tz1 MTz 1 Tz 1,,, Tz 2 Tz 3 Tz 4 1 1 ST Tz1 TzTz 1,,, 2 TzTz 3 4 MST S z1 Tz 1,,, Tz 2 Tz 3 Tz 4 zz1,,,,,, 2 z 3 z 4 TzTzTzTz 1 2 3 4 S is M.T. i.e. The cross ratio is invariant under Mobius Transition.
Proposition : If z2,, z 3 z 4 are distinct points in and w2,, w 3 w 4 are also distinct points in then, there is a unique M.T. S s.t. S z2 w 2, S z3 w 3 and S z4 w 4 .
S z w 3 w 3 z2 2 z4 w S is a M.T. 4
z p la n e w plane
Fig 7.1 137
Proof : Let T z z,,, z2 z 3 z 4 and M z z,,, w2 w 3 w 4 .
T and M are Mobius Transformations and T z2 1, T z3 0 and T z4 , M w2 1, M w 3 0 and M w4 . Put SMT 1 1 1 S z2 M Tz 2 M1 w 2 1 1 S z3 M Tz 3 M0 w 3 1 1 S z4 M Tz 4 M w 4 Let R be another M.T. s.t.
R zj w j for j 2, 3, 4 1 1 1 R S z2 R S z 2 R w 2 z 2 1 1 1 R S z3 R S z 3 R w 3 z 3 1 1 1 R S z4 R S z 4 R w 4 z 4 1 1 Here, RS composite map of S and R has 3 fixed points. 1 RSI (Identify map) SR Hence, S is the unique transformation.
Proposition : Let z1,,, z 2 z 3 z 4 be distinct points in then the cross ratio z1,,, z 2 z 3 z 4 is a real number all four points lies on a circle (or a straight line). (2009) Proof : Let S : be defined by S z z,,, z2 z 3 z 4 = Real number.
S Mobius transformation S ¡¥
Fig 7.2 1 Then, S z: z , z2 , z 3 , z 4 is real i.e. image of under the Mobius Transformation is a circle. 138
We will prove that the image of under the Mobius transformation is a circle. az b Let S z where a,,, b c d are complex constant and cz d ad bc 0 . 1 If z x and w S x , then S w x . S w is purely real number. S w s w aw b a. w b cw d c. w d awb cwd awb cwd 2 2 acw adw bcw bd acw awd bcw bd ac ac u 2 ad bcw bc adw bd bd 0 (1)
Case I : If ac is not real, then ac ac 0. If ac is real then ac ac ac ac ac 0 From equation (1), adbc adbc bdbd w 2 w w 0 acac acac acac ad bc bd bd Put x and ac ac ac ac w 2 xw xw 0
w 2 xw xw x 2 x 2 w x R w () x R which is the equation of the circle with centre at (-x, 0) and radius equal to R.
Where, R x 2
2 ad bc bd bd ad bc 0 a. c ac ac ac ac ac 139
Case II : If ac is real then ac ac 0 From equation (1) ad bcw bc adw bd bd 0 Put ad bc & i bd bd w w i0. 2 Im ( )i 0. Im ( ) 0.
w or Im 0 (2) 1
Note : Consider a straight line L in the Complex Plane . If a is any point inL and b is its direction vector then,
L
b
a
Fig 7.3
z a L z : Im 0 z a b L z : Im 0 b OR z a bt: t , i.e. t The point w lies on a line determined by equation (2) for fixed and . Theorem : A Mobius transformation takes circles onto circles. 140
Proof : Let be any circle in (z-plane). Let S be any Mobius Transformation. Suppose z2,, z 3 z 4 are distinct points on the circle . G G¢ w 3 z 3 S w 4 z is a 4 M.T. w z2 2
w z æ ö S çT ÷=G¢ è ø
Fig 7.4
Put Szj w j for j 2, 3, 4 .
w2,, w 3 w 4 are distinct points in (w-plane).
These three distinct points w2,, w 3 w 4 determine a circle in w- plane. T.P.T. S
Since, z2,, z 3 z 4 are distinct points in and S in a M.T.
z,,,,,, z2 z 3 z 4 Sz Sz 2 Sz 3 Sz 4 for any point z.
Sz, w2 , w 3 , w 4 ... (1)
If z then the cross ratio z,,, z2 z 3 z 4 is a real number. z,,, z2 z 3 z 4 all lie on a circle … Sz,,, w2 w 3 w 4 is a real number. (by equation (1))
These four points Sz,,, w2 w 3 w 4 lie on a circle . Put Sz w As z moves on the circle , then the corresponding point w moves on the circle under a M.T. ‘S ’. S . Hence, a Mobius Transformation takes circles onto circles. 7.6. UNIT END EXERCISE
Example : Find a M.T. which maps points z 1, 0,1 onto the points w 1, i ,1. Also find the image of unit circle z 1 in the -plane under this M.T. 141
Solution : Given points z z, z2 1, z 3 0, z 4 1 and w w , w2 1 w3 i , w4 1. Y v i
-1 1 y = 0-1 1 X u
-i w=1
z- plane w- plane
Fig 7.5
M.T. is given by z,,,,,, z2 z 3 z 4 w w 2 w 3 w 4 (Proposition on page no.81(2)) z z z z w w w w 3 2 4 3 2 4 z z4 z 2 z 3 w w 4 w 2 w 3 z 0 1 1 w i 1 1 z1 1 0 w 1 1 i z iz w1 1 z w i zw z izw iz w i zw iz z izw w i z i S z w i (1) z i which is a required bilinear transformation. z i From equation (1), w i z i wz iw iz1 wz iz 1 iw z w i 1 iw 1 iw z w i Given equation of unit circle is z 1.
1 iw 2 2 z 1 1 iw w i 1 iw w i w i 1 iw 1 iw w i w i 1 iw 1 iw w i w i 142
w i 1 w i z z
z z2 z1 , 0 & 0 z1 z 1
w w2 w3 , 0 & 0 w3 w 3 1 iw 1 iw w i w i 1 iw iw w 2 w 2 iw iw 1
2iw 2 iw 0 w w 0 w w Put w u iv , w w u iv u iv u iv u iv 2iv 0 v 0 which is the equation of real axis.
1) Find a Mobius transformation , which send 1, i, 1 onto 1,i ,1 respectively.
a b ai b a b Solution : f(1) 1, f ( i ) i , f ( 1) 1.. c d ci d c d a b c d, ai b di c a d and b c , b a d c c d d c d 0 a 0 b c 1 f z , since c 0 . cz cz z 1 f z is the required bilinear transformation. z
z 2) Find the fixed points of the mapping w . z 1
Solution: Let z0 be the fixed point of the mapping
z z0 2 w f z z0 z 0 0, let z0 x 0 iy 0 x 0 0 z1 z0 1 and y0 0 . z z 0 is the fixed point of the mapping w . 0 z 1
3) Find the Mobius Transformation bilinear mapping sending 1 i, i ,2 i onto ,0, respectively. 3 143
az b Solution: Let f z ad bc 0 be the required bilinear cz d d d mapping . We know that f maps onto i, c c az b d ic f z cz ic ai b ai b Now f i 0 f i 0 b ai ci ic2 ci az ai a z i 1a 1 f z . f2 i f 2 i cz ic c z i 3 3c 3 a c . a z i z i f z , here a 0 c z i z i
(Alternate method for problems involving infinity)
4) Find the bilinear transformation which maps the points z , i ,0 onto the points z 0, i , (w w )( w w ) ( z z )( z z ) Solution: We have transformation 1 2 3 1 2 3 (w1 w 2 )( w 3 w ) ( z 1 z 2 )( z 3 z )
Since, z1 and w3 , we define N and D of LHS and RHS by w3 and z1 respectively.
z z2 w w2 z1 , 0 & 0 and w3 , 0 & 0 z1 z 1 w3 w 3 (w w )( 1) ( 1)(z z ) 1 2 3 ()()w1 w 2 z 3 z Put z2=i, z3=0 and w1=0, w2=i w i 1 w i z z
5) Find a Mobius transformation , which send 1, i, 1 onto 1,i ,1 respectively.
a b ai b a b Solution : f(1) 1, f ( i ) i , f ( 1) 1.. c d ci d c d a b c d, ai b di c a d and b c , b a d c c d d c d 0 a 0 b c 1 f z , since c 0 . cz cz z 1 f z is the required bilinear transformation. z 144
z 6) Find the fixed points of the mapping w . z 1
Solution: Let z0 be the fixed point of the mapping
z z0 2 w f z z0 z 0 0, let z0 x 0 iy 0 x 0 0 z1 z0 1 and y0 0 . z z 0 is the fixed point of the mapping w . 0 z 1
7) Find the Mobius Transformation/ bilinear mapping sending 1 i, i ,2 i onto ,0, respectively. 3
az b Solution: Let f z ad bc 0 be the required bilinear cz d d d mapping . We know that f maps onto i, c c az b d ic f z cz ic ai b ai b Now f i 0 f i 0 b ai ci ic2 ci az ai a z i 1a 1 f z . f2 i f 2 i cz ic c z i 3 3c 3 a c . a z i z i f z , here a 0 c z i z i
8) Let z1,,, z 2 z 3 z 4 be four distinct points in . Then show that z1,,, z 2 z 3 z 4 is a real number iff all four points lie on a circle.
(Hint: Define s : by s z z,,, z2 z 3 z 4 Show that 1 s is a circle. Here s =the set of all z such that
z,,, z2 z 3 z 4 is real.
z 1 9) Prove that all the points z C satisfying 2 lie in a circle. z 4 Find its radius and centre (2009) 145
10) Find the image of the circle x2+ y2+2x=0 in the complex plane 1 under the transformation w= (2008) z
11) Find the Mobius transform which maps the points z=1, I ,-1 onto the points Mobius transformation (2008)
12) Let H={z C / Im(z) 0 }and let D={z C / | z | 1}. Find the Mobius transformation g s.t .g(H)=D and g(i)=0. Justify your claims (2007)
13) Show that Mobius transformation has 0 and as its only fixed points if and only if it is dilation (magnification) (2007)
14) Show that Mobius transformation has as its only fixed points if and only if it is a translation (2007)
15) Find the Mobius transform which maps the real axis R onto the circle (2006) z 1. az b a 16) Fix a,b,c,dC with c 0.show that as z cz d c
1i z 17) Verify that the Mobius transformation w maps the i z exterior of the circle z 1 in the z-plane into the upper half plane Imw 0 in the w-plane.
18) Find the image of the circle z3 i 3 in the complex plane 1 under the transformation w . Illustrate the results graphically. z
1 1 19) Find the image of an infinite strip y in the complex plane 4 2 1 under the transformation w z