UNIT 6 CLASSIFICATION OF SINGULARITIES AND CALCULUS OF RESIDUES
Structure 6.1 Introduction Objectives 6.2 Classification of Singularities 6.2.1 Zeros of an Analytic Function 6.2.2 Singular P~ints 6.2.3 Types of Isolated Singular Points 6.2.4 Singularities at Infinity 6.3 Calculus of Residues 6.3.1 Residue at a Finite Point 6.3.2 Residue at the Point at Infmity 6.3.3 Cauchy 's Residue Theorem 6.3.4 Mittag-I,eflk-rExpansion Theorem 6.3.5 Rouche's Theorem 6.4 Summary 6.5 Answers/Solutions/Hints to Exercises
6.1 INTRODUCTION
Consider the functions
1 1 1 ;,x sin ;, exp (- k2), ---- etc. x ER x(x - l)(x - 2)'
We see that each of these fmctions is not defrned at 0. In such a case, the point x = 0 is a singular point for each of these functions in the sense that the function is defined in a deleted neighbourhood of 0. The problem of classifying singularities are not easy to solve satisfactorily for functions on R. On the other hand, the situation in C is different. Iff is analylrc at a point s = a, then there is a neighbourhood N of a inside which f is analytic. Let r be a positively oriented smooth closed curve contained in N. Then, f is analybc Inside and on r and, as seen in Unit 5, celebrated Cauchy Integral Theorem tells us that
If, however,ffails to be analytic at finitely many points interior to r, then the above argument fails; which means that there is, as we shall see in this unit, a specified number (zero or non-zero) which each of these points contribute to the value of the integral. This motivates to generalize the idea of Cauchy Integral Theorem to functions which have isolated singularities. This is reflected in the Residue Theorem. In this unit, we start with the zeros of an analyhc function. We then classify the singularities of complex-valued functions - in particular, we define isolated and non-isolated singuldies, removable singularities, poles and their order, singulsrrities at infinity and illustrate these singularities. Cm~plbxVariables The notion of residue at an isolated singularity is introduced and this notion is used to 1 prove the Residue Theorem. Using theresidue theorem, we develop and illustrate some of the basic methods employed in Complex Integration.
Objectives b 1 After studying this unit, you should be able to I I tell zeros of an analytic function and find their order, define sinnularities of an analytic function and make distinction between (i) isolated and non- isolated singularities (ii) removable and irremovable singularities, -1 determine poles, if any, of an analytic hction and their order, define singularities at infinity, define and obtain residue of an analytic function at a frnite point, define and obtain residue at the point at infinity, learn Residue Theorem and its applications in evaluating some integrals in complex plane, and learncertain applications of residue theorem.
6.2 CLASSIFICATION OF SINGULARITIES
Before we take up singularities of a complex-valued function, it is desirable to study. the zeros of an analytic function, which we take up next. 6.2.1 Zeros of an Analytic Function If a function f(z), analytic in a region R, is zero at a point zo in R, then zo is called a zero of f(z).
If f(zo) = 0 but f '(z,) # 0, then z, is called a simple zero or a zero of the first order.
If f(zo) = 0, f '(z0) = 0 ,..., f (n- ') (ZO)= 0, but f ("' (z,) + 0, then z, is called a zero of order n.
For ixample, the function z2 sin z has a zero of order three at z = 0 and simple zeros at
The function 1 - cos z has second order zeros at z = 0, + 2n, +- 4n, .. .
' If an analytic function f(z) has a zero of order n at z = z,, then its Taylor series is of the form
2 =(~-~O)~[b,+h,-l(z-~~)+b,+~(z-z~) +...I
= (z - z,)" [b, + (Z - zO)Sj, say
By taking ) z - z, ) sufficiently small, we can make
lb,dl > lz - zol . I SI Thus a neighbourhood of zo can be found where is not zero. Consequently, f (zl kasnrrcrt have another zero in this neighbourhood. ClvssiCication of Singularities and Calculus uf Residues We express this by saying that the zeros of an analytic function are isolated.
We next take singularities of complex-valued functions
6.2.2 Singular Points A complex-valued function f(z) is said to have a regular point at z, iff is analytic at z, .
The point zo is called a Singular Point or Singularity of f(z) if z, is a limit point of regular points and f(z) is not analytic at zo .
For example z = 0 is a singular point off(z) = z-', z # 0.
If the singular point z, is such. that there is no other singular point in its neighbourhood, i.e., iff (z) is analytic in some deleted neighbourhood of zo ,then the point z, is called an Isolated Singular Point or Isslated Singularity off (2).
For example, the functionflz) = lis analytic eveqwhere except at z = a.Thus f(z) z - a has an isolated singularity at z = a.
z2 + 1 T!le function has three isolated singular points, namely (z- 1) (z +4)
The function -has an infinite number of isolated singularities at sin(%) z = f 1, + V2, k V3, f l/4, ...... The origin z = 0 is also a singular point, but it is not an isolated singular point, because there are other singular points in its neighbourhood, however small we choose thf;neighbourhood.
The function &has a singularity at z = 0, which is not isolated. The other si~lgularities near the origin arise due to the discontinuity suffered by a single-valued branch of & on crossing the cut in the z-plane. Singularities of this nature always arise at 'branch points', where two branches of a multivalued function have the same value.
Other examples of such a singularity are log z'at z = 0 and sin-'z at z = f 1.
If ro is not an isolated singularity off (z), it is called a Non-isolated singular point of f(z).
Descriptions of Non-isolated singularity at z, and for isolated singularities at z,,I,, 2, are given in Figures 6.1 & 6.2, respectively. Example 6.1
"Discuss the singularities if'g(z) = '/f(z), where
Az) = U(X,y) + iv(x, y) = sin
Solution
It is easy to see that f (z) is analybc everywhere except at z = 0. Therefore the singularity of g is at z = 0 and at the points where f (z) = 0, i.e., when u (x,y) = 0 and v(x, y) = 0.
Since cos h @/I a f ) > 1, therefore u (x, y) = 0 implies that
In the case when sin(x/( z 12) = 0,we have cos( ?flzl2 ) f 0. Thus v(x,y) = 0 implies that
sinh (v/lz12)= 0 Thus (6.1) gives
and (6.2) yields y = 0. I+ I I 1 Further,y=Ogivesx=.-forn=O, + 1,-+2,...... Thusu(x,y)=Oand (n X) v(x, y)=Oholdsifandonly if x=l/(nn),n=O ,+1 ,+2 ,... .Thusthe singularities ofg(z) are at points x = l/(nn) , n = 0,+ 1 ,f 2 ,... and at their limit point z = 0. This shows that g (z) has isolated singularities at z = l/nn, n = 0, f 1, f 2, .... and has non-isolated singularity at the limit point z = 0.
Remark 1 :We note that fi)= sin the discussion of singularities of g zeros of sin ( l/z ).
Remark 2 : Any neighbourhood of a non-isolated singular point of a function f (2) contains other singularities and hence a non-isolated singular point off (2) is a limit point of the singular points of f(z). Further observe that f(z) must be discontinuous at an isolated singular point. You may now try an exercise. Exercise 1
Discuss tlte singulnritiet;PC 1/COS ( ).
We next take up the kinds of isolated singularities for an analytic function. 6.2.3 Types of Isolated Singular Points There are three kinds of isolated singularities for an analytic function i) Removable ,Singttlurity, \vhicli upon close exalniilatioii is not actually a singular point at all.
ii) Pole is a zero of the reciprocal which is an analytic iimctiun. iii) Erseitial sing?tlnrit.v, which is neither rerl~ovablenor a pole.
We next,give definition and examples of these kinds of isolated singularities.
Definition : "An isolated singularity zz,of C (2). f E N ( D \, I x-, j) , is calied rcniovable or that f (z) has a removable singularity at z, iff (z) can be defr12ed at z, so that it becomes analytic at z, ''
In other words, f (2) can be extended to z, too so that f (2) is analy~icon D
Xote : :V ( D\lx, j) is the dclctcd neighbourhood, 1.e D escluding the po~nt2,
Consider the fui~ctious.
.I;(z ) = sin z, 2 # 1
Thesc ftlncdnns 11a-c.cremovable singularitics at 1. 1 and O respcc1ive1.c.w!~icl~ call bc rcrllovcrt by letting
,/;(I) =sinl, J: (1) := 2 and I; (0):~1, ccs))ec~i\~cly
Exercise 2
" Discuss the removable singu1,uities sf the functions
sin z
L
2 (1 -cos z) and g, (zj = -- zf0 z2 and give the values of the functions ~v11ichcan remove the singularities ".
We obscrvc that if J{L)is analyl~con ol)cn scl S and z,, E S . thcn titc fi~nctionI; defined by Thus F(z) has a removable singularities at 2, , which can be removed by letting F(z0) =f '(20) . The following example illustrates this. Example 6.2
z2 + aZ "Discuss the singularity ofAz) = -7, z #- ia. z + ra Solution
Clearly Az) is analytic every where except at z = - ia Forz #- ia,we have
and hence Lt Az) = - 2ai . z + ia
ThusAz) has a removable singularity at z = - ia . Now we put
=z-ia for zit-ia g(z) = I" z + ia - 2ai for z = - ia
Then g(z) becomes analytic everywhere including at z = - ia
Remark : More generally, iffhas a removable singularity at z,, then there is a function f analytic at 2, such thatflz) = g(z)for all z in deleted neighbourhood N \ {z,} .
An important result for removable singularities is given below :-
Result I : "If f(z) has an isolated singularity at z, ,then z = z, is a removable singularity if and only if one of the following conditions holds:-
a) Az) is bounded in a deleted neighbourhood of z, ;
b) Lt Az) exists 2 + Zo
It may be observed that Lt Az) exists may give rise to the following situations:- Z + Z,
i) Az) may not be defined at to.
ii) Az) may be defined at 2, but&,) may not be equal to Lt Jz) Z -+ Zo
iii) Jz) may be defined at 2, and&,) is equal to Lt Az). 2 + Zo
In the last case (iii), f(z) is not singular at z0 . This shows that if f(z) has a removable singularity at z, ,then either (i) or (ii) holds.
We now give another result for removable singularities.
Result I1 : "Iff(z) and g(z) are analytic and if both have a zero of order n at z, , we may write Classification of Singularities and f(z) = (z - zo)"fo (z) , R(Z)= (Z- zo)" go (z) , Calculus of Residues where both f, and go are analytic and non-zero at z, . Hence
exists. This means thatf(z)/g(z) has a removable singularity at z,."
For example, each of the functions
ez- 1 z - sin z h(z) = y and h(z) = has a removable singularity at z = 0 .
We have seen that iff (2) is bounded in a deleted neighbourhood of an isolated singularity at z, ,then z, is a removable singularity off (2). Thus if z, is not a removable singularity off then f is not bounded near z, . We might then ask whether (z - z,)" Az) is bounded near z, . The answer is provided in the definition of pole given below.
Definition: "If (z - z,)"flz) is bounded near z, ,then the point z, is called apole ofAz) and the number n which is the smallest positive integer such that (z - z,)"f(z) is bounded near z, is called the order of the pole at z, . A pole of order one is called a simple pole."
For example, the functionfiz) = :/(z - 1$ has a double pole at z = J
The hnctionfiz) = 1/(1 + ei"') has simple poles at z = 1 +_ 2n ,n = 0,1, 2, .... and z = rn is the limit point of these poles. The function
z cos (m/2a) fiz) = (Z- a) (z2+ b2 )' sin5z where a and b are distinct non-zero real numbers, has poles of order 7 at + ib, 4 at z = 0: 5 at k ~r:, k = + 1, + 2 ,.... and a removable singularity at z = a. The function
thenfiz) has a pole of order 4 at z = bi and a pole of order 3 at z = - c.
The function 1 fiz) = . , af(k+1/2)n,(k=0,+1,+2,...) sin z - sin a has simple poles at z = 2 kn + a , z = (2k + 1) .rc - a.
Remark : If f(z) is analytic in a deleted neighbourhood of z,, then f has a pole of order n if and only if
i, (z - z,)"f(z) has a removable singularity at z,, i.e.
Lt (z - z,)" + ' fiz) = 0, and 2 --f z, Consider the f'ol?o\ving example
'- Disccisz thc si!~~ulari:ic~ol'
CLle poles of /. tt' an\.. nrc ck?cn~r:ned111
kietlcc /(s) 113s ,I .sllnple pull: ;~r: - 13 = i I 21 and n ~en~o\ablcs~rig~rlar~t\ iit
" ry--l{r wherc 1- is tile distance from z, to the nearest singularity off (2) other tlian z, itself., (If z, is the 01114- si!lgulari!y, than r- = u.).
Throughout thc abovc discussion: we have supposed that,f(z) is defined for all z near z,], but 1101 ~iccessarrlyat z, ~tself.Thus the classific;ition of the isolated singularity of f(%) depends only an the local hchaviour of f(z) in a neighhourhood of r,,
in (6 5). the senes with the negative powers of (E - z,)1s called Principal part nhereas the serles with non-negatlx e powers oE(z - z,) is called the Analytic part.
T'hrcz miltualll cxc1usrx.e cases arise.
Cast 1 : Xo Principal Part
111 1111s casc. (6.5) takes thc form
c-, k j(zj =: 2 (Z -- Z") . 0 < 1 z - Z" 1 < r tz 11
Thus if lve dcfine,
jl-'") = a,, = Lt flz) .
7 + Lo
then f (L) ~111be aial!?ic a1 z, This shows :, is a rernov:lhle singularity mid the Laurent scries c.;yansion of fli)must coincldc \\~lhthc Taylor serics evpalaslnrl ticarz,
Case 2 : Finite Principal Part
In this casel (6.5) takes the form
andfi~)has a pole at z =z, and if (z - it,)-"is the highest negative power in the Lauror~texpansion offlz) , the pole is said to he of order 11.
In other words:'-Jz) has a pole of ordcr n if and only if nn the cspa~~sion(6 5)of-fi~~ a-,+O.buta-k=O for k>n+ 1."
Case 3 : lntinite Principal Part.
If tlae Laurent expansion ofJz) in the neighbourhood of an isolated singular point z = z, contains infinitely many negative powers of (z - z,), i.e., if in (6.5), a- k # 0 for infinitely mmly k 2 1, thenJ(2) has an essenti:d singularity at z = z,.
In this case, LL./(zj fails to exist (including the possibility of the limit bei~iguginity). z -3 Z"
'A For example, if&) = e ,then.f is analytic on c \ (0). Further, for this function, with .r > 0 and n > 0, fixed, we have Con~plexVariables
Therefore, it follows that f is neither bounded near z = 0 nor can f have a pole of order n at z= 0. Here e' has an essential singularity at z = 0.
Similarity, it is easy to see that the function
has an essential singularity at z = b .
Also the function
1/12 ~zl=ze or e-I/' or sin (:)or cos (:]
has an essential singularity at z = 0
You may now try an exercise.
Exercise 3
Show that z = h is an essential singuliinty ot'thc ii~i~ctron
The behaviour of a function in the neighbourhood of an essential singularity is described by Casorati - Weierstran Theorem ,which states "If f(z) has an essential singularity at z, and if w, is a fixed complex number, then there exists a sequence {zn}with z,, +zo such thatf (z,) -+ wo.In other words, f(z) takes values arbitrarily close to every complex number in every neighbourhood of essential singularity." A result, deeper than above, is given by Picard according to which " in every neighbourhood of an essential singular point at z, of Az), supposed analytic in 0 < 1 z - z, ( < 6 for same 6 > 0 ,Az) takes on every value, with at most one exception, an infinite number of times. Consider the following example. Example 6.4 "Show that e '4 assunies every value infinitely many times with exception of zero, (Picard exceptional value)." Solution Consider =eIne , e;tO where e may be taken as a complex number. If e is real and non-negative, then (6.6) CIossification of Singularities and implies Calculus of Residues Thus, in particular, we have a sequence {z,),with z, + 0, such that 1/zn =e, n=1,2, ... -1 18 In general, if'e # 0 is complex number, then let z = p e , where 8 is a real number, so that (6.6) becomes p (cos 0 - i sm 0) In I e 1 + i arg e e = e => pcos 8=lnlel and p sin 8=-arge Hence, we have and pZ = (In le(f + (arg el2, where arge=arge+2&. k=0,+1,&2,.... Clearly, (6.7) and (6.8) are solvable - (6.7) is solvable since in the range 0 I8 I2n, 1 tan 8 takes on all values in ( - m, - ) and (6.8) is solvable for 0 < - < E, for each E > 0, P since arg e may be chosen as large as we please. You may try the following exercise. Exercise 4 Sllow that sin (%) assumes every 1 alue infinntely many dn~zst$ifh cxdcg.rron of zcro. So far, we have discussed the singularities of a function which are isolated and lie in the complex - plane C. We shall now be concerned with the singularity of a function at the point at ini3nity. I I 6.2.4 Singularities at Infinity We shall extend the complex plane C by adjoining one extra point at infinity which we shall denote by z = -, i.e., we consider the extended complex plane C u ( -) as a closed surface having a single point at infinity. Role of the point at infmity is understood through the function This function is defined for evcq t + 0, and maps each point z in C, except z ;t 0. into a point of ~v-plane.If we assign the point at infinity in ni-plane. to z = 0 and IV = O to the point at ~nfin~tyin z-plane, then thc above function 1s one-to-one from extended z-plane to extended w-plane. 1. Letfbc the analytic in 1 z > R , 0 _ Tllcn F is anal?tic in dcleted neighbourhood iw : 0 < 1 w I < '/R! of zero. The nature of singularity off at z = m (point at infinity ) is detined to be the same as that of F at \v = 0. For example, function f defined by has a pole of order n at z = .o, since the corresponding F defined by has a pole of order 11 at w = 0. 1 ~unction~~z)= 7 + z'" has a pole of order two at z = 0 and a pole of order m at infinity. z Similarly, function/(z) = e" has an essential singuIarity at z = m,since F defined by F(wj =/(l/',) = eL/"has an essential sillgularity at 1g = 0. Function tan z has essential singularity at z = w which is the limit point of poles of tan z and assumes every complex value with Picard's exceptional value f i. Suppose that f (z) has an isolated singularity at z = .o and R is ihe distance from origin to farthest singular point off(If the point at infinity in the only singularity-,,then R may be chosen as an arbitrary positive number). than F has an isolated singularity at w = O and so the nearest singularity of F from w = 0 is at a distance lh. Now, the function F(w) has Laurent expansion about w = O as The following cases arise: Case 1: You may recall that the function F(cv) has a pole of order k at w = 0 if and only if (6.9) takes the fonn. - I CI:~ssiticatiunof Singularities and and part a- k w- % . . .. + a_, w is the principal part of F at w = 0. Calculus of Resid~res Hence f (z)has a pole of order k at z = m if and only if its Laurent's expansion has the fonn where the principal part off(z) at z = m is Case 2 : Suppose that in (7.9)a_, = 0 for all k 2 1. Then (6.9) becomes Thus if we define F(0) = a,: then F will be analytic at w = 0. Hence F has a removable singularity at w = 0 if and only if in (6.9)a_, = 0 for dl k 2 1 , that is, if we set then f(z) defined by is analytic at z = m . In other words,f(z) has a removable singularity at z = w if and only if its Laurent expansion has the form Case 3 : The functionf (z)has an essential singularity at z = m, if its Laurent expansion has the form where an infinite number of a- k 'S are non-zero fork 2 1 We observe that "An entire function f(z) is transcedental if and only if fph) has an essential singularity at 0." Let us consider the following example. Example 6.5 c c 1 Forf(z) = - z#h, k=0,+1,*2 ,..., sinz ' discuss the singularity off at m." Con~plexVariables Solution We define and discuss the singularities of g(z) at z = 0. Here g(z) is not analytic in the neighbourhood of origin, because every neighbourhood of origin contains a singularity inside such neighbourhood. Hence g(z) has isolated essential singularities at the points z = 9i.h) ,k = 0, + 1, f 2, ...... and has a non-isolated essential singularity at z = 0 . This shows that f(z) is not analytic in {z : i z I > R} for any R > 0 and w is a non-isolated singularity for f6). You may now try the following exercises. Exercise 5 Show that w is a non-isolated singu:ar~tyfar I ~z).=(I?'- 1)- . Show that functioll g(z) = z4 sin (Qz + 1)) has a polc of' order five at the point at infinity. Exercise 7 Discuss the singularity of the function a. fi)= 7 sin (x)at inEinity. T C One of the most important and often used tool that applied scientists and engineers need from the theory of complex functions is the Calculus of Residues and we shall focus our attention on it in our next section. 6.3 CALCULUS OF RESIDUES We begin with the defintion of residue of a function at an isolated singularity. 6.3.1 Residue at a Finite Point We know that if f(z) has an isolated singularity at z,, than there is a deleted neighbourhood S = { z : 0 < ( z - z, 1 < r} in which f(z) may be expanded in a Laurent series as \ Classification of Sindarities and calcul~sof Residues ... (6.10) with where C is any circle centred at zo and lying inside S and integral along C is taken in the positive direction. Here circle C may be so small that it does not go beyond the domain of analyticity of fTz) and has no other singularity inside it. Also by Cauchy's Deformation Theorem, the integral in (6.11) will have the same value on any positively oriented curve which encloses zo but no other singularity of fb). Then, according to (6.1 l), we have The coefficient a_,of the term (z - %)-' in the Laurent expansionflz) is of great importance and is of special significance, (because of its connection with the integral of function through the formula (6.12)) and is called the residue of&) at isolated singularity to. If z, is a simple pole, then Laurent expansion (6.10) becomes 00 k+ 1 Hence (z - zo)Az)= C ak (z - 20) + a- I k= 0 The terms (z - z;lk+ ' vanish when z + 2, ; therefore This gives a simple method of calculating the residue off (2) at its simple pole 2,. If tois a pole of order m, then Laurent's expansion (6.10) off (2) becomes Differentiating (6.14) (m - B) times and t'aking the limit as z - z, , we get 1vi:ilich is the residue off!z) at z - z, , a pole of order m. We note fio~n(6.14) that Lt (z - z,)"f(z)) is finite and non-zero. This fact can bbc used L '7" Tor determining the order of a pole. Relation (6 12) provides a conven~etitwa? of evaluatuig certain integrals of a functim f(z) around an isolated smgulmt). since its value 1s s~niplyths product of 2x1 aid the coefficient of (z- zo)- ' in its Laurcnt expansion (6 10) -We observe that the most convenient way to find the residue is directly from Laurent sxpansion, af it was already available. Faemark I : Iffit) is analytic at z, , then b~ Tav1or.s Theorem, these are 110 negative process of j: - 6,)it1 thc expa~slonoff0 and has a residue offit) at zOis zero.ffz) Further, if then residue 0f.L (z) at z, is zero for n 2 2 . Remarli I1 : 1f.f (z) has a removable singularity at z,, then residue of,f(z) at z, is zero. Further, iffjr) has removable singulmties at z, , k= 1, 2, .... w. then restdue of at z, for k = 1. 2. .... r? is zero. In particular. if C is a simple closed curve containing removable singularities at z, k = 1.2 ....,n inside it, then Remark I11 : If fb) has an isolated singularity at z, and iff is even in (z - z,), i.e., Az - 4,) =f [- (z - z,)] , then residue of,f() at z, is zcro. For example, residue of even cos z - 1 function [ 1at removable singularity 0 is zero. 2 Z Similarly residue of even function -at removable singularity 0 is zero. sinZz Further, for the function el"l ,the point z = 0 is an essential singularity. Also elh2 is 2 an even function in z. Hence residue of el/\t z = 0 is zero. 2 2 Since sin (z - kn) = sin z ,thereforeAz) = -is an even function in (z - kn) and (sin z) residue off (2) at Az is 0 for k = 0, f 1, + 2 ,.... Similarly has residue at 0 as zero. sin (z2) i,:t us tske up sollie examples I Lxumljle 6.6 i I ' Dctemiine tlie poles of the l'unction and the residue at each polc." The poles of,f (2) are given by ,, I lius z = 2 is a silnple pole and z = - I is a pole of' order 2 for,f O) Now residue ofj(z) at (z.: 2) Also rcsiduc cf f iz) at double pole (I.= -- 1'1 .. "l)cter~n~~lctl~ poles of the t'un~~rtll~--'- v~dLI~s resrdrrc; ,rl <;ICJ~ !IUIG ' CllS z JZ -7K ---.--- q/2 -- I,1 - b) I-" t iaspilsl'h t-nlc , , - 5111 ; - I'l~us/@):- -I-- h,is simplc pales at t = rm t %l , /i =: cr .f I f 2. Exercise 8 1 /zn For the functionflz) = e , show that z = 0 is xi essential singular point and rcsidueoff(2) at z=Ois Orfn* 1 and 1 Ifn= 1. Exercise 9 Detennine the singularities for the function 1(z) = ,(a be~gnon-zero real ), and the residue at saeh pal$ z (a,I -a) Exercise I0 Fkd the residue of cosh z cot z at its singularities. Show that z = 0 is a simple pole for the function j$, = -----.---sin 3.2 .- 3 sin I. (sin z - z) sin z and find the residue of&) at z = 0. h reca above in eon6.2.4, a htionf(z) may have siagularity at inhity. We, naw, define the residue at the point at infinity and give method of fin* the same. 6.3.2 Residue at the ~diotat Infinity We fust consider z = '/, rmd look at the geometric aspect of above transformation. Tbis shows that as z describes the circle. I z I =Min z-plane in clackwise direction, then w describes the circle I w I = YM in w-plane in anti-clockwise direction. kthepointro=pe-'e, p~.M,outsideIzI=Moomepondstoapo~two=p-1 e i6 inaide ( w 1 = (see Figure 6.4). Let f (2) be analytic in a deleted neighbourhood of the point at infinity. Then f (2) Classif~cationof Sigularities and admits Laurent series expansion of the form Calculus of Residues We define ~={z:IzI=M>R, M issufficiently large), where C is traversed in clockwise direction (so that the point at infinity is to the left of * C as in the case of finite point zo ). 4 Let z =Me- Ie. Then, we have That is , residue off (z) at Sinity = 1 jo dz = - a- ,. 2ai C 1 In other words, residue off (2) at infmity is negative of the coefficient of - in Laurent z series expansion off (z) with centre at the point at infimity; or We observe that for z =~e-' ' with M= %', 1 residue offi) at infinity = - Az) dz 2ai ZI = - Residue of f*,,, W where C = { w : ( w I = lhI is described in the anti-clockwise direction round a small circle about the origin. f (%) Hence [residue off (2) at infinity] = - Residue of -7at zero] * = Lt [ - z Xz)].provided that this limit has a definite value 2. -i "E Important Observation It can be easily seen that the residue at a finite point is zero if the functiol~is analytic at that point hut mag not be so at infinity. e For example; consider the function for residue at infinity. The only singularity-of , z-n the function is a simple pole at z = a. The fui~ctionis analytic at infiniq*.However, by definition, the residue of --I at infinity z-a 2 n 1 I C YI~" =-- &=---I -d0 , by putting z -. a = I. el" 2ni 2- n 2~ire" Firrther, if/@) is aldytic at z = m , lt still llus a aon-zzro r~siducat infinity esccpt when 1. (4 ,. coefliercnl tmf - ln t11e Latrront's o?.pu=~s~onof/'(;) ol tlkc ncighbourhood ofWz-- 1.3. 2 IS r$ro Thus Rz J has a silnplc pdc ;it L. - - 1 Ftr!alrer, due to r'" in Ble tlhirncrdor c?f'!{/) thc;scf'ore /(2) has mr t:: 's4~lrtial. 2 singularity at 2 - (1 If wc put vr. -z !a:: liwl wc obtaln => r :m is u pcl;ir: ul'order (n - I) Stir liz) . Since z = 0 is an essential singularity ofAz), we must rely on Laurent expansion of Classification of Singularities and Calculus nf Residues Az) around zero. We have Az) = z" (1 + 2)- ' . e' 8 Collecting the terms involving '/, ,we have CO k- 1 a- , = %, which is residue ofAz) at z = 0. Ans I1 x (n + k)! To find variable ofJz) at z = w, using (6.17), we write 1 Again collecting the terms of - ,we find 'I W It (-l)n+(- l)"-l (- 11° (- 1)" - residue of F(w) at (w = 0) = - + ... 0 ! l! +-=xn! k ! Hence, residue ofAz) at (z = w) residue of* at w = b W I You may now try an exercise. # Exercise 12 "Find the residue of* Log (1 + 2) Az) = (1 +zY at each of the poles. The notion of residue of a complex valued function at the vaiious singularities will be used to prove the residue theorem, which we take up next. However, caution must be ,? ,? Cc~l~plcxVariables exercised to avoid reaching hasty conclusions based on appearances. We must identify the type of singularities and then choose a proper curve. 6.3.3 Residue Theorem or Cauchy's Residue Theorem Statement : "Iff (z) be analytic at all points inside and on a simple closed curve C, except for a finite number of singula points z,, z,, ..., zn inside C, then Az) dz = 2ni x (sum of residues at z,, z,, .. ., z,, )." L * Proof :We enclose each singular point z, z, ,..., zn by small non-intersecting circles C,, C,, .. . ,C, lying wholly inside C (see Figure 6.5). These circles C,, C,, - , Cn together with the curve form the boundary of a multiply-connected region in whichfiz) is everywhere analytic. This region can be converted'into a simply connected region by giving it C suitable cuts (along the dotted lines). The boundary l- of the singly connected region consists of curve C (traversed anti- clockwise), the circles C,, C,, . .. C,, (all traversed clockwise) and the cuts from C these (traversed both ways). Figure 6.5 : The Circles Cl, C2, ..., C,, Enclosing respectively the Su~gularPoints zl, z2,.. ., z,, Applying Cauchy Integral Theorem to within a Simple Closed Curve C. .f(z) for curve l- , we get I or, omitting the integral over the cuts since it is zero (once positive and other time negative), we get where now all integrations are taken in anticlockwise direction. ~ut1j(z) dz is 2n i times the residue ofj(z) at zk . Hence j(z> dz = 2ni x (sum of residuesof f(z) at z, . z, , ..., z.). This establishes the important residue theorem. Remark :In residue theorem, f(z) can have only a finite number of singularities, becausc otl~eenvisesingularities of f(z) would have a limit point 5 (possibly at the point at infinity ), and so 5 would not be an isolated singularity to f(z) , contrary to our assumption. Combining the residue at the point at infinity and Cauchy's residue theorem, we have "Residue Formula for the extended complex plane" as follows: "If f(z) is analytic with the exception of finitely many isolated singularities at a, in the extended complex plane9then the sum of all residues (including the residue at ' infinity) of f(z) equals zero." Equivalently (using residue theorem and relation (6.14)), we write Classilication of Sinylarities and Calculus of Residues A '4 (Residue of 7at 0 ) = (suin of residues off@) on all singularities Z a, in complex-plane). We can use Residue Formula for the extended plane, as stated above, to give another simple proof of Liouville's Theorem. .Alternate Proof of Liouville's Theorem Let f (z) be entire and bounded on the complex plane; then we must show that f (z) is constant. We pick up two distinct arbitrary points a, b in complex plane C and consider the function The function F(z) has singularities at z = a, b and possibility at the point at infinity. Then by Residue Formula for the extended complex plane, we have -Residue of F(z) at a + Residue of F(z) at b + Residue of F(z) at m = 0 . Since Lt z F(z) = 0 (because f (z) is bounded in C), we obtain Izl+m residue of F(z) at m = 0 , and therefore, we have +fo=O Residue of F(z) at a + Residue of F(z) at b = 0; i.e., -f(a) a-b b-a so that f (a) =f (b). Hence f (z) must be constant. On most of the occasions, calculation of the residues at many singular points of integrand is quite dficult. In this situation we can use Residue formula for the extended complex plane to evaluate certain integrals. zZ1 Consider~z)= . Here all the singularities off@) lie inside the circle (7' - 114 (z4- 2$ lzl= 3. Also f (z) has a simple pole at infinity with residue at z = m equal to Lt zAz) = - 4. Consequently, z/+m I /Iz) dz = - 2ni x residue of&) at (z = m) = 8ni 1z1=3 You may try the following exercises. Exercise 13 Co~i~plexVariables Exercise 14 Evaluate The application of residue theorem can be seen in the following examples. Example 6.9 "If C be the circle I z I = 3, show that Solution The integrand is analytic on I z I = 3 and at all points inside it except the poles z = - 1 (of order 2) and z = 2 (a simple pole). The residue of integrand at (z = - 1) The residue of integrand at z = 2 Hence by Cauchy's residue theorem, z+2 & = 2n.i x (sum of residues at z = - 1 and z = 2) J (z+ 1f (z-2) C Example 6.10 "Fipd'the value of where C is the circle (a) 1 z 1 = 2 (b) I z + i ( = 6.'' Solution Thusf(z) has a double pole at z = 1 and a simple pole at z = - 3/2 . a) For the circle, I z 1 = 2, both the singularity z = 1 & z = - V2 are inside the circle. Now ~esidueofflz) at 12~-7 -18-7 Residue of f(z) at (z = - 3/2) = Lt -- =-4 2 4 -3,5(~ - 1l2 - (5/2)2 :. I Az) dz = 2ni x (sum of residues of f(z) at z = +l and z = - 3/2) C : 121 = 2 b) For the circle (z + i) = 6,cartesian equation is 2 .2 2 2 x +(y+j) =3=>x +y +2iy-2=0 Here z = 1 is inside the circle, but z = - 3/2 is outside the circle. Hence I &) dz = 2ni x (Residue ofAz) at z = 1) = 2ni x 2 = 4ni Ans. 11. You may now try the following exercises. p (1: Flzld \due of %vhreC is tile circle ( z 1 = 4 described m the pcsitne SA*~Z' C (z- 1) Evaluate - clx, die integral being taken arnufid tto circle I (;;- tj2$-2) lz-~~=2. The expansion of a rational fractional function P(x)/~(x)in terms of partial fractions is a well-known procedure. Using the residue theorem, we can now establish a result, known as the Mittag - Leffler Expansion Theorem (named after the Swedish mathematician G.M. Mittag - Leffler (1 846 - 1927)), which generalizes this type of expansion to a much wider class of functions. 6.3.4 Mittag - Leffler Expansion Theorem Statement : "Iff(* is a function whose is analytic at z = O and whose only singularities are first order poles at z,, z2,z,, .. ., where the residues are respectively b,, b,, b,, .....; if { C : JzJ= R, ) is a set of circles none of which passes through any singularity of&) and where radii {R, ] become infinite as N + w; and if on each of these circles JAz)I is bounded by a constant M, which is independent of N, then Con~plexVariables Proof : Heref (z) has first-dm poles at z = z,, z,, .... . with corresponding respective residues b,, b,, ..... ; suppose that z = tis not one of these points. Then the function flz)/[(z- t) has first order poles of z = z,, z,, ...... and at z = t , At z = zk, the residue offlz)/(z- t) = Lt (z- z,) m=bk 2 + 21; z-t z,-t and at z = t , the residue is Lt (z- t)&=flt). z+t z-t Applying residue theorem to the region bounded by CN, assumed large enough to include the point z = t, we have bk 'J 'J J%z=At)+~ -, 2ni z-t 2,- t CN where summation extends order all the poles off (z) within C. Putting t= 0 in the above relation, we get Subtracting (6.19) from (6.18), we get 1 At)-A~)+xbk[L-L]=-~Az)[L-:]*+ zk- t zk 2ni z-t CN The theorem will be proved if we can show that the limit of the last integral as N + w is zero. Now for values of z on CN, we have (z-f(21z)-(t(=RN-(t( Since by hypothesis, I flz) I is bounded on 6, by a number M, which is independent of N, we find infmite. Cnl4us of Residues This coqplete the proof of the theorem. We take up an example now. Example 6.11 ! "Apply Mittag - Leffler expansion theorem to the function 1' and verify that the result is equivalent to the ordinary partial - fraction expansion." I Solution I 2z+ 1 2z+ 1 Heref(z) = - z'+6z2+ llz+6-(z+l)(z+2)(~+3)' Hence&) has simple poles at z = - 1, - 2, - 3 andfo is analytic at z = 0. 22+1 - 1 Now residue off(z) at (z = - 1) = Lt --. ,+-,(z+2)(~+3)- 2 ' 2z+ 1 - 3 residue off(z) at (z = - 2) = Lt - = 3 z+-2(z+l)(z+3)-(- l)(-l) 22+1 - - 5 5 and residue of f(z) at (z = - 3) = Lt - - z+-3(z+ 1)(z+2)-(-2)(- 1)--2 :. :. Applying Mittag - Leffler expansion theorem to f (z), we get 2z+ 1 A B C For ordinary partial fractions, let --- ++- 2+6z2+11z+6 z+1 z+2 z+3 This verifies the result. You may now try an exercise. ~onl~lexVariables Exercise 17 "Apply Mittag - Leffler expansion theorem to the function to find its expansion.' ' Before we consider some useful consequences of Residue Formula for the extended complex plane, we consider some results relating the residue of a function to the * number of zeros and poles of the function. t Result I : "IfAz) has a zero of order m at z = a, then residue of aat (z = a) = m." f(z) Result I1 : "Iff(t.) has a pole of order n at z = b, then near z = b f'(z) - --n + analytic function at z = b ." Az) 2-6 Hence at each pole z = b of f(z) ,f '(z)/'z) has a simple pole at z = b with residue equal to - n. Combining resdts I & 11, given above, we obtain a principle knqwn as Argument Principle, which states that "Letfbe meromorphic in a domain D and have only finitely many zeros and poles. If C is any simple closed curve in D such that no zeros or poles off (z) lie on C, then where Nand P denote, respectively, the number of zeros and poles off (z), inside C, each counted according to their order." We now define, A. arg f (z), that is, the change in argf (z) as z goes around C. Letfbe analytic inside and on a simple closed curve C except possibly for poles inside C andf(z) ;t: 0 on C. As z describes C once in the positive direction in the z-plane, the image point w =f(z) describes a closed curve I'=f(c) in the w-plane in a particular direction which determines the orientation of the image curve r. Sincef(z) ;t: 0 on C, r never passes through the origin in the w-plane. Let w, be an arbitrary fixed point on r - and let $ ,be a value of the argument of w,. Let arg w run continuously from i$, ,as the poitit begins at w, and traverses r once in the direction of orientation assigned to it by w =f(z). If w returns to the starting point w, ,then arg w assumes a particular value of arg w, ,which we denote by 9 , . We defme We note that the difference $ , - $ , is independent of the choice of starting point w, . We also note that the difference $ ,- $ ,is an integral multiple of 27c and integer ltion of Singulal ($ - $ ,)An is the winding numbet of around the origin in w-plane as z describes C , r Calculus of once in the positive direction. 1 Corollary I :Under the hypothesis of Argument Principle, - A c argAz) = N - P 21c Corollary I1 : Iff(z) is analytic inside and on a simple closed curve C andAz) + 0 on C, then the above formula reduces to We now take up some examples to illustrate the use of Argument Principle and its corollaries. Example 6.12 1 f'o Evaluate I = - &, C={z:Iz- 1 -.il=2} 21ci Az) C 2-2 2-2 for ' a) Az) = - b) Az) = z(z- 1) z(z - 1f Solution For our C, 0, 1,2 and i are inside C and 3, - 3, - i are outside C. Now, we have I = N - P, where Nand P are respectively the zeros and poles ofAz) inside C. Thusfor (a)I=l-2=- 1 (b)I=l-3=-2 (c)I=O- 1 =- 1 Example 6.13 Evaluate Solution LetAz) = cos z sin z ~henJtanza!z=+j -&=-If*&.asz C C C Az) SinceAz)=O=>cosz=O <=> z=(k+1/2)1c, k=O,f 1,+2,... the only zero of f(z) inside C about Y2 is z = Y2. There is no pole of f(z). Thus by Argument Principle, Note also that residue of tan z at (z = Y2) = - 1. You may now try the following exercises. The Argument Principle allows a comparison, under certain conditions, of the number of zeros of two analytic functions. This is given by Rouche's Theorem, which we take up next. 6.3.5 Rouche's Theorem Statement : "LetJ(z) and g(z) be meromorphic in a domain D and have only finitely many zeros a~~dpoles. Let C be any simple closed curve in D such that no zeros or poles off (z) and g(z) be on C. Further assume that Ig(4 I i.e., differei~cebetween the number of zeros and number of poles is the same for f and f + g, namely7 Nf- Pf= Nf+g- Pf+&." Proof :In view of * Is(z)I As z describes C by (6.21), point w = 1 + &&anverses the closed cwer lying in flz) I w - 1 I c 1 (see Figure 6.6). Figure 6.6 This implies I arg w I < Yz. This means that arg w has to return to its original value and thus Thus the result follows as a direct consequence of the above, relation (6.22) and the 1 result that -A, argflz) = N - P . 27t Sometimes Rouche's Theorem is given an equivalent formulation. Instead of assuming I g(z) I 1 flz) I on C, it is assumed that then conclusion with respect to this assumption is Remark : Iffand g have no poles in D, then from Rouche7stheorem f and f + g have the same number of zeros inside the curve C. The result of Rouche7sTheorem come handy in solving many problems easily, which we illustrate by the following example. Example 6.14 2 "Show that z4 + iz3 + 1 = 0 has all the four roots in I z I < 3/2.77 4 Solution Takeflz)=l+z4 andg(z)=iz3 .Thenon/zI=R, 4 3 IAz) 12R - 1 and (g(z)(=R Therefore on 1 z I = R, the inequality Iflz) I > 1 g(z) ( holds if R~-~>R3 . Con~plexVariables By Rouche's theorem, f and f + g have same number of zeros in I z I < R, for any R, such that R: - 1 > R: . Thus, in particular z4 + i z3 + 1 = 0 has all the four roots in I z ( < 3/2 . You may now try an exercise. Exercise 21 Show that the equation 2 + z - elZ = 0 has only one root in the entire upper half plane. The application of calculus of residues in evaluating real integral, improper integrals and Fourier integrals will be taken up in the next unit. 6.4 SUMMARY We now summarize the results of this unit. If a functionf (z), analytic L1 a region R, is zero at a point z, in R, then z, is called a zero off (2). Iff(z,) = 0, but f '(2,) # 0 , than z, is called a simple zero or zero of the first order off (2). Iff(z,) = 0 =f '(2,) = .. . =f (n - (zO)but f ("I @,) # 0 ,then z, is called a zero of order n off(z). Zeros of an analytic function are isolated. Iff(z) is not analyhc at z, ,then 2, is called a singular point off (2). If the singular point z, off(z) be such that there is no other singularity off@) in its neighbourhood, then 2, is called an Isolated Singular Point off (2). If z, is not an isolated singularity off(z), it is called a Non-isolated Point of fi4. A non-isolated singular point off (2) is a limit point of the singularities off (2). f (z) must be discontinuous at an isolated singular point. Removable singularity z, is one which on closer examination is not actually a singular point, is., iff (z) can be defined at z, so that it becomes analytic at zo . Pole is a zero of reciprocal, which is an analytic function. A singularity which is neither removable nor a pole is called an Essential , * Singularity. If (z - zo )"f(z) is bounded near z, ,then the point z, is called a pole of order n off (2). A pole of order one is called a simple pole. In Laurent series expansion of a function f(z) about an isolated singularity z, , the series with negative powers of (z - 2,) is called Principal Part and series with non-negative powers of (z - 2,) is called Analytic Part. In Laytseries, (a) if there is no principal part, then singularity is Classification of Singularities and removable (b) if there is finite principal part ,then singularity is a pole (c) Calculus of Residues and if there is infinite principal part, then singularity is an essential singularity. Behaviour of a function in the neighbourhood of an essential singularity is described by Casorati-Weierstrass Theorem, stating "Iff(z) has an essential singularity at z, and ifw, is a fixed complex number, then there exists a sequence {z,} with z, + z, such thatj(zo) - w, ." Picard's Theorem : "In every neighbourhood of an essential singularity z, of f(z), supposed analytic in 0 c lz - z, I c 6 for some 6 > O,f(z) takes on every value, with at most one exception, an infinite number of times". If w = 1/, and F (w) =f (9, then the nature of singularity off(z) at z = m is defined to be the same as that of F(w) at w = 0. The functionf(z) has a pole of order k at z = m if and only if the principal part off(z) at z = m is Functionfiz) has a removable singularity at z = w if and only if its Laurent expansion has the form Function f(z) has an essential singularity at z = w if its Laurent expansion has the form where an infinite number of a- ,'s are non-zero for k 2 1 . An entire functionf(z) is transcedental if and only iff(l1z) has an essential singularity at 0. The coefficient a- 1 of the term (z - 2,)- in the Laurent expansion off(z), namely, 00 k z)a-z) , ZE ~~:~c~z-z,~cr]with k=-c., where C is any circle with centre z, insides, is called the residue off(z) at , isolated singularity z, . Residue of simple pole zo offiz) Residue of a pole of order m at z, off(z) Con~plexVariables Iff@) is analytic at zo ,then residue off (z) at z, is zero. Iff@) has a removable singularity at 2, ,then residue off@) at z0 is zero. Iff(z) has an isolaied singularity at 2, and functionf is even in (z - z,) i.e.,fTz - z,) =A- (z - z,)), then residue off(z) at z, is zero. Residue ofJ(z) at infinity = negative of Residue of 7at zero. Z = Lt [- zf(z)], provided that this limit has a definite value. z+- Cauchy's Residue Theorem : "Iff(z) be analytic at all points inside and on a simple closed curve C, except for a finite number of singularitiesz,, z,, .. . , z, inside C, then J Az) dz = 2xi x ( sum of residues of&) at z,, z,, ..., 2"). 7, C Residue Formula for the Extended Complex Plane : "Iff(z) is analytic with the exception of finitely many isolated singularities a, in the extended complex plane, then the sum of all residues (including thk residue at infinity) off(z) equals zero." A%) Residue of --T- at zero = Sum of residues off@) on all singularities in the Z complex plane. Mittag - Leffler Expansion Theorem : "Iff(z) is a function which is analytic at z = 0 and%hose only singularities are first order poles at z,, z,, z ,,....., where residues are respectively b,, b,, b,, .... N if ( C : 1 z I = R, 1 is a set of circles none of which passes through any singularity off@) and where radii { R, } becomes infinite as N -+ = ;and if on each of these circles I fTz) I is bounded by a constant Mywhich is independent of N, then Iff@) has a zero of order m at z = a, then '(2) residue of f- at (z = a) = m Az) Iff(z) has a pole of order n at z = b, then near z = b, = --n + analytic function at (z = b) fTz) z- b i.e., at each pole z = b forAz),m has a simple pole at z = b with residue nz) equal to - n . Argument Principal : "Let fbe meromorphic in a domain D and have only finitely many zeros and poles. If C is any simple closed curve in D such that no zeros or poles off (z) lie on C, then Classification of Singularities and dz = N - P, where N and P denote, respectively, the number of Calculus of Residues 27ci fl Az) L zeros and poles off(z) inside C, each counted according to their order." Change in argument off@ as z goes around C is given by Under the hypothesis of Argument Principle, A,argf(z) =N- P. Rouche's Theorem : "Letf(z) and g(z) be meromorphic in a domain D and have only finitely many zeros and poles. Let C be any simple closed curve in D such that no zeros or poles off@) and g(z) be on C. Assume that I g(z) I < JAz)I in C. Then i.e., difference between the number of zeros and numbers of poles is the same forAz) and Az) + g(z)." Instead of I g(z) < JAz)I on C, if we assume that then A, g(z) = AJz) . ZfS(z) and g(z) have no poles in D, then from Rouche's Theorem,fand f + g have the same number of zeros inside the curve C. 6.5 ANSWERS1 SOLUTIONS1 HINTS TO EXERCISES Exercise 1 LetAz) = -. Thenf(z) has the singularities at the points where cos(%) cos (l/z) = 0 i.e., singularities of f(z) are at the point 2 z= , n=O,f l,f 2,.... (2n + 1) le Since in every neighbourhood N of z = 0, there are singular points, distinct from z = 0, within N, therefore z = 0 is a non-isolated singularity of f(z). L For the points z = , n=O,f l,f 2 ,...... , (2n+l)n there are neighbourhoods (say circles of sufficiently small radius 6 ) which contain no other singularity. This means that these singularities are isolated. Exercise 2 These functions have removable singularities at 0,O and 0 respectively, which can be removed by letting g, (0 ) = 1 , g, (0) = 1 and g, (0) = 1 respectively. Exercise 3 Hereflz) = (z - a) sin - (.' b) which shows that €, = 0 or z = b is an essential singularity of f(z). Exercise 4 Suppose for C real and C # 0 sin (I/,) = c =>z=i '.")log ic+ 1-c where k= 0, f 1, f 2, ...., we obtain a sequence {z,,)with zn + 0 such that sin (lh,,) = c, n =1,2, ... Hence sin (112) assumes evey value infinitely many times with the exception of zero. Exercise 6 Here g(z) = z4 sin (Idz+ 1)) 1 men g ($1 = 7 sin [?I= 5.sin (5) -+ 1 I which shows that w = 0 is a pole cbf order five. Hence g(z) has a pole of order five at the point at infinity. Exercise 7 Here&) = 4sin Z ($1 Then/(;) = w2 sin w, which has a removable singularity at w = 0 Hence f (z) has a removable singularity at infiity. Exercise 8 Clasqification of Singularities nr~~l Calculus of Residues 0 - nk z = 7,Jzl> 0 (n E N), using series for exponential function. From above series, z = 0 is an essential singularity offlz). Therefore, residue off (z) at (z = 0) = Coefficient of z- ' in the above expansion,, which is Lament's expansion 0 ifn7t1 1 if n=l = i Exercise 9 1 For the functionf(z) = , z (z - a') ' denominator is zero when z1 (zZ - a') = 0 =>z=O,O,+a,-a. Hence f (z) has a double pole at z = 0 and single poles at z = a and z = -(7 . We also note thatJz) =A- z) => f (z) is even in z. Hence by the result "iff(z) has isolated singularity at z,, and iff is even in' (z - z,), then residue off@) at 2, is zero," the residue off (z) at z = 0 is zero. Ans I Now residue off@) at z = a Residue off (z) at (z = - a) =Lt -=--1 I Ans. 111 + - z z a 2a3 Exercise 10 cos 2 Let cosh z cot z = cosh z -sin z Since sin z has simple zeros at z = h , k= 0, + I + 2, .... and cosh z cos z is non-zero at (z = kz) :. Residue of cosh z . cot z at (z = kz) cash Ax. cos kn cosh kn . cos hx - - = cosh h-n . Ans. - d cos h -(sinz) Iz=kn dz Con~plexVariables Exercise 11 sin3z-3 sinz Heref(z) = (sin z - z) sin z Therefore f (z) has a simple pole at z = 0 :. Residue off (z) at (z = 0) = Coefficient t' in the above expansion Exercise 12 We know that Log (1 + z) is analytic in the deleted neighboorhood C\ & w, - 1) andf(z) = Log (1 + z) has a pole of order two at z = i, - i . Hence residue of Log (1 + z) at (z = i) (z + i)" 1 + Log \5) Similarly residue of Lug(.-rj 31 I- = -. ;) = - + T)+ i - [ 8 16 (- 8 4/' Exercise 13 ClassUication of Singularities and CaIcuIus of Residues Z Letfi) = (z2 + a)" (z3 + blrn ' The denominator off(z) is zero at z,, z, = + i 1(;; and z,, z,, 2, where z' + b = 0 . Hencef (2) has a pole of order n at z,, z, each and pole of order m at z,, z,, z, each. By the conditions on a, b and R, these poles lie inside the circle I z 1 = R . Therefore by residue theorem, 2n+3m-1 I ~5= 2ni x (sum of residues at z,, z,, z,, z, and z,) c (z2 + a)" (z3 + blm By residue fonnula for extended complex plane, we have (Sum of Residues off (z) at z,, z,, z,, z, and 2,) + (Residue offiz) at ) = 0. -> sum of residue off (2) at z,, z,, z,, z, and z, = - (Residue off(z) at rn ) f (%I = (Residue of --r at z = 0) Z :. 1 z &=2nix1=2ni.Ans. c (2+ a)" (z3 + blm Exercise 14 Therefore, we have 1 zz + 1 I=IRer&=-[/lz) dz, whereJ7z) = z-a 2 z (z- a) Ct!~'=l j~!=l Hcre,f(f) has si~tlplepoles at-z = 9 and z = a and both these poles lie in I z = 1. By residue theorem. 1 I = x 2ni x (sum of residues of-f (z) at z .: 0 and z = a) -2 1 Now residue off (z) at (z = 0) = Lt zflz) = - - Z-+O a a2+ 1 and residue off (z)at (z = a) Lt (z- a)fiz)= - z-ta a Exercise 15 1 The poles of - are at points where sinh z = 0 slnh z 1 Hence z .= 0 is the simple pole of -inside 1 z / = 4 sinh z 1 and residue of -at (z = 0) s1nl1 z By residue thcorcm, Exercise 16 Hence f (I)has a double pole at z = - 1 and a simple pole at z = 2 . z-1 1 Residue 0f.f (z) at (z= 2) = Lt -1-9- - z --, 2(z + 1) d 2-1 (2-2)-(z- 1) --- 1 Residue off (z)at (z= - 1)= Lt -- -- Lt 2 - -1-2 1 (z-2) 9 For the circle lz - il = 2 , z = - 1 is the only pole inside this circle. :. By residue theorem, (Z - 1) dz = 2ni x (residue of J[z)at (z= - 1)) I (z+ 112 (z-2) !:-,;=2 - - -2xi 9 Ans. Exercise 17 Classificationof Singulrrities and Calrulus of Rcsidum The function f (2) has simple poles at z = - 1, i, - i 11 Now Residue off (2) at (z = - 1) = Lt 7= - zj-lz+l 2 1 1 - l(1 - 1) Residue off (z) at (z = i) = Lt - - jiz+l)(z+i) 2i(l+i) 2(1+1) 1 1 1 Residue off (z)at (z = - i) = Lt --- - + (z+l)z-i) 1-i -21 By Mittag - Leffler expansion, 1 1 11-z --- -+-.- Ans. 2 z+l 2 z2+1 Exercise 18 LetJTz) = sin xz and C = { z : lzl = nl. 1 .f'(4 (,z Let I = j mtzdz = - j X A=) (iJ=x C SinceAz) = sin x z = 0 <=> z = k, k = 0, f 1, + 2, ...., No\\; inside the circle C, there are seven zeros ofJ(z), namely, at the points 0.f 1,+2,+3. Tilerefore by Argument Principle, Complex VariaMcs Exercise 19 qinh z ~et~=[tmllzdz=J-dz=J f !@I u, cosh z f(z) z-1(=2 C C eZ+e-z Sincef(z) = cosh z = ,- --0 <=> e2"+1=0=>eZ =fn= +j Thus there are two zeros off@) inside the circle C Therefore, by Argument Principle, I = 25ci x 2 = 45ci Ans. Exercise 20 Consider and if z = - x, with x >O, then Thus (6.24) and (6.25) imply that f (z)has no real positive roots and no real negative roots respectively. Further, ~fz = iy with y-real, thenf(z) =j: + 1 - i y" and thercforef has no purely imaginary roots. iRb ~ei0 ~ntheotherhandif~=[z:z=Re10 . 050~5cb] R.X i.e., C is taken round the part of the first quadrant Figure6.7: C=Z:lZ!=R,OSargZ bounded by I z I = R for sufficiently large R (Figure 6.7) then IfR is sufficiently large, the second term is practically 1 and so. On the axis of y, Here y ranges from to O along the positive imaginary axis, the initial and final values of arg J(z) are zero. Thus tlle total change when R is sufficiently large is given by which means that there is a root lying in the first quadrant. Exercise 21 Classification of Singularities and Calculus of Residues 2 Take Jz) = 2+ z and g(z) = e' For z = x, JX) =2 +x-' > 1 elx/=l=lg(x)l, that is IJz) 1 > 1 g(z) 1 for z = x , x-real. Ontheotherhand,ifz=Re 19, OIOIsc,then, 2 2i8 . (f(~e'~)(= (2+R e 12~~-2 and /g(~e")I=(e'~'I=e-Rsin e Therefore on the upper semi-circle, we have -R sin e IJz) I > I g(z) I if R~-2> e If we choose R > 6,then z - Rsin e R -2>3-2=12e for 0lOllr. By Rouche's Theorem, the number of roots off+ g in ~,=(z:lzlsR and R~z~o?, for any R > 6 is equal to the number of zeros of the equation 2 .,flz) = 2 + 2 = 0 In Q,. Thus, in the entire upper left-plane, the equation 2+z2 -eIz=0 has only ont root.