FACULTY of MATHEMATICAL STUDIES MATHEMATICS for PART I ENGINEERING Lectures MODULE 22 COMPLEX NUMBERS II 1. Revision 2. Relation

FACULTY OF MATHEMATICAL STUDIES

MATHEMATICS FOR PART I ENGINEERING

Lectures

MODULE 22 COMPLEX NUMBERS II 1. Revision 2. Relationships between trigonometric and hyperbolic functions 3. Logarithm of a complex number 4. De Moivre’s theorem and complex roots 5. Loci in the complex plane

1. Revision Let us ﬁrst recall the main points of module 5 on complex numbers. The key deﬁnition is √ j = −1, → j2 = −1 .

A complex number is then written z = x + jy ,whereboth x and y are real. x is the real part of z , written Re z ; y is the imaginary part of z , written Im z . [Note that the imaginary part does not involve j and therefore is a real number.] Examples of complex numbers are 2 + j3, −1 − j, a + bj . The product j3isthesameas3jand so both forms are used in writing complex numbers. It should also be pointed out that in the third example the real and imaginary parts are algebraic – this is ﬁne but letters represent real quantities, from the deﬁnition of a complex number. Complex numbers are displayed using an Argand diagram, in which the the real and imaginary parts of a complex number form the x-and y-coordinates with respect to the usual xy Cartesian axes. y 3 2+3j =(2,3) 2 1 -1 -2 1 2 3 x -1 -1+j =(-1,1) -2 ( polar form A complex number can also be expressed in modulus-argument form exponential form These alternative forms essentially arise from expressing a point (x, y) in the Cartesian plane in polar coordinates (r, θ) , i.e. r = x + jy = r(cos θ + j sin θ) , where r denotes the modulus of z (written |z| ,ormodz); θ is the argument of z (usually written arg z ). p Hence r denotes the distance from the point (x, y) to the origin with r = x2 + y2 . Recall that θ is the angle between the positive x-axis and the line joining the point to the origin. In the usual way, angles measured in an anti-clockwise direction are positive, whereas those measured in the opposite clockwise direction are negative. The principal value of the argument lies in the range (−π, +π] , but clearly

1 integral multiples of 2π can be added to the argument without altering the position of the complex number in the Argand diagram. To illustrate the ideas consider z = 2+2j. Draw the Argand diagram, which is extremely useful in deciding the quadrant in which the complex number lies, and hence the magnitude of the argument. y 2 r 1 θ 1 2 x -1

In this case it is clear from the diagram√ that √ r2 =22+22 = 4 + 4 = 8 hence r = 8=2 2, 2 1 π cos θ = √ = √ , → θ = (= 45o). 2 2 2 4

Next consider z =1−j and again draw the corresponding Argand diagram. y 2 1 1 2 α r x -1

From the triangle including α it is easy to deduce that √ 1 π r2 =12+12 =1+1=2, so r= 2, and cos α = √ , → α = . 2 4 π In this example the angle is being measured in a clockwise direction so arg (1 − j)=− . 4 [The determination of the correct angle is usually most easily calculated with the aid of a diagram, as above. Note that in the second example the angle was found from a triangle, using the positive lengths of the sides. The argument of the given complex number could easily be found from the triangle using other trigonometric 1 π functions, e.g. from tan α = = 1, which again gives α = .] 1 4 Before writing the complex number in the alternative forms recall that in module 5 we stated Euler’s formula ejθ =cosθ+jsin θ . Using this, and the usual formulae for polar coordinates, we can write: √ π π √ 2+2j=2 2 cos + j sin =2 2ejπ/4, 4 4 √ π π √ 1 − j = 2 cos − + j sin − = 2 e−jπ/4. 4 4

2. Relationships between trigonometric and hyperbolic functions Euler’s formula states ejθ =cosθ+jsin θ . Replacing θ by −θ then implies

e−jθ =cos(−θ)+j sin(−θ)=cosθ−jsin θ,

(since cos(−θ)=cosθ, sin(−θ)=−sin θ). Adding and subtracting the above expressions leads to

ejθ + e−jθ ejθ − e−jθ cos θ = , sin θ = . 2 2j

2 From the earlier work on hyperbolic functions you know that

ex + e−x ex − e−x cosh x = , sinh x = , 2 2 (these definitions can also be found on the Formula sheet and in the Engineering Data book) and, by comparing the above two sets of results, relationships between trigonometric and hyperbolic functions can be established: ejx + e−jx ejx − e−jx cosh(jx)= =cosx, sinh(jx)= = j sin x. 2 2 Note that the definitions of the hyperbolic functions were initially given for real variables but in writing the above formulae their extension to complex variables has been introduced, e.g. it has been assumed that ez + e−z cosh z = ,where z is complex. 2 From the definitions of cos, sin, cosh and sinh it also follows that

2 2 ej(jx) + e−j(jx) ej x + e−j x e−x + ex cos(jx)= = = =coshx, 2 2 2 ej(jx) − e−j(jx) ej2x − e−j2x j e−x − e+x ex − e−x sin(jx)= = = = j = j sinh x. 2j 2j j 2j −2(−1)

Two comments should be made about the above relationships: (i) you are not expected to remember the above identities, but you should be able to derive them if you are asked to do so; (ii) the identities hold for all variables x , even when the latter is complex. (For example, replacing x by jx in cos(jx)=coshx leads to the earlier result cos x =cosh(jx). nπ o Ex 1. Find the value of sin (1 + j) . 4 Using the Formula Sheet for the expansion of sin(A + B) , and the above identities nπ o nπ πo π π π π sin (1 + j) =sin +j =sin cos j +cos sin j 4 4 4 4 4 4 4 1 π 1 π 1 π π = √ cosh + √ j sinh = √ cosh + j sinh 2 4 2 4 2 4 4 (=0.9366 + j 0.6142, to four decimal places)

Ex 2. Find all values of z such that cos z =2. Put z = x + jy where x and y are real, then

cos z =cos(x+jy)=cosxcos(jy) − sin x sin(jy) =cosxcosh y − sin x (j sinh y), =2(given).

Equating real and imaginary parts in the latter expression yields

cos x cosh y =2 (1) sin x sinh y =0 (2) sin x =0, i.e. x = nπ, where n is an integer (2) → or sinh y =0, i.e. y =0. If y = 0 then cosh y = 1 and hence (1) → cos x = 2 . This is not possible for real x so the possibility y = 0 leads to an unacceptable answer.

3 Hence must have x = nπ .Nowcos(nπ)=(−1)n and substituting this result into equation (1) implies

2 cosh y = . (−1)n

−1 But cosh y ≥ 1, andso n must be an even integer, giving cosh y = 2 which leads to y = cosh 2. The ﬁnal solution therefore is

−1 z = mπ j cosh 2,m=0,2,4, ...

which is more usually written as

−1 z =2nπ j cosh 2,n=0,1,2, ...

Hence cos z = 2 has inﬁnitely many solutions.

3. Logarithm of a complex number The easiest way to obtain the answer is to express z in exponential form, take logs and use the standard results for exponentials and logarithms:

z = rejθ → ln z =ln rejθ =lnr+ln ejθ =lnr+jθ.

Integral multiples of 2π can always be added to θ without changing z, so the above definition can be extended z = rej(θ+2nπ),n=0,1,2, ... → ln z =lnr+ln ej(θ+2nπ) i.e. ln z =ln|z|+j(θ+2nπ)=ln|z|+j(arg z +2nπ),n=0,1,2, ... Note that for a particular value of z there are an infinite number of values of ln z, one for each value of n, which implies ln z is not a standard function). When the argument of z takes its principal value, lying in the range (−π, +π], then define the principal value of ln z by Ln z =ln|z|+jarg z. Observe that a capital L is used to denote that the log is calculated using the principal value of the argument.

Ex 3. Obtain all values of ln(1 − j). In section 1 it was shown that √ √ j − π j − π +2nπ 1 − j = 2 e ( 4 ) = 2 e ( 4 ),n=0,1,2, ... hence √ n o j −π+2nπ ln(1 − j)=ln 2+ln e ( 4 ) ,n=0,1,2, ... √ π =ln 2+j − +2nπ ,n=0,1,2, ... 4

4. De Moivre’s theorem Since z = rejθ using the properties of exponentials it follows that n zn = rejθ = rnejnθ.

4 Using the modulus argument form for z on the LHS and expanding the exponential on the RHS using Euler’s theorem for angle nθ, we obtain the usual form of de Moivre’s theorem n [r(cos θ + j sin θ)] = rn [cos(nθ)+jsin(nθ)] . The theorem is usually proved ﬁrst for positive integer n , by multiplying out and using trigonometric identities for small n , and by induction for the general case. The above exponential approach requires proof of the exponential results for complex exponents, if you want to prove the result rigorously.

Ex 4. Express 2 + 2j in polar form, and hence evaluate (2 + 2j)12. In section 1 it was shown that √ √ n π π o 2+2j=2 2ejπ/4 =2 2 cos + j sin , 4 4 hence n o √ π π 12 √ 12π 12π (2 + 2j)12 =(2 2)12 cos + j sin =(2 2)12 cos + j sin 4 4 4 4 =(23/2)12 {cos(3π)+j sin(3π)} =218 {cos π + j sin π} =218 {−1+j(0)} = −218 = −262144

De Moivre’s theorem is most commonly used to ﬁnd the roots of complex numbers. Suppose z = r(cos θ + j sin θ), then using de Moivre’s theorem with n = p/q we obtain p p pθ p2nπ pθ p2nπ z q r q j ,n , , , ..., q − . = cos q + q + sin q + q =0 1 2 ( 1) The parameter n is needed to obtain all the qth roots of zp. The angles appearing in the general result above may not all lie in the range −π to +π so further amend- ments are often necessary, see Ex 5 below. 1 1 Ex 5. Given z = − + j evaluate z1/2. 2 2 For z 1/2 , a square root, two solutions are expected. y

1/2 r φ θ -1/2 x

From the diagram 2 2 1 1 1 1 1 1 r2 = + = + = , → r = √ 2 2 4 4 2 2 1/2 π π 3π tan φ = =1, → φ= , → θ=π− = . 1/2 4 4 4 Therefore 1 1 1 3π 3π z = − + j = √ cos + j sin 2 2 2 4 4 3π 3π =2−1/2 cos +2nπ + j sin +2nπ 4 4 1/2 3π 2nπ 3π 2nπ → z1/2 = 2−1/2 cos + + j sin + 8 2 8 2 3π 3π =2−1/4 cos + nπ + j sin + nπ ,n=0,1. 8 8

5 Hence the two roots are 3π 3π 11π 11π 2−1/4 cos + j sin , 2−1/4 cos + j sin . 8 8 8 8

The second argument lies outside the required range so must be modiﬁed by adding or subtracting integral multiples of 2π . In this case it is necessary to subtract 2π making the second solution 5π 5π 2−1/4 cos − + j sin − . 8 8

Note that when n =2 3π 3π 3π 3π z1/2 =2−1/4 cos +2π +j sin +2π =2−1/4 cos + j sin , 8 8 8 8 the same as the ﬁrst solution. In a similar way all higher values of n lead to one or other of the two stated solutions. N.B. The two solutions are equally spaced around the circumference of a circle of radius 2−1/4 .

3π/8 2-1/4 5π/8

Ex 6. Evaluate (−1)1/3. Here with a cube root three answers are required. The point −1 lies on the negative x-axis so in an obvious way r = 1 and the argument equals π. Hence

−1=1{cos π + j sin π} =1{cos(π +2nπ)+j sin(π +2nπ)} ,n=0,1,2, ... π 2nπ π 2nπ (−1)1/3 =(1)1/3 cos + + j sin + ,n=0,1,2, 3 3 3 3

with other values of n leading to one of the above three answers. Since (1)1/3 = 1 the roots can be written √ π π 1 3 (n =0) cos +jsin = + j 3 3 2 2 (n =1) cosπ+jsin π = −1+j0=−1 √ 5π 5π π π 1 3 (n=2) cos +jsin =cos − +jsin − = − j 3 3 3 3 2 2

x 2π/3 π/3 x π/3 1 2π/3

6 5. Loci in the complex plane. A locus is a set of points with a specified property – e.g. the locus of points in two-dimensions that are a fixed distance from a given point is a circle. Note that loci is the plural of locus. |z| =1 → all points whose distance from the origin equals 1, hence the locus is a circle, centre the origin, radius 1 (see figure 5a); |z − 1| =2 → all points whose distance from the point (1, 0) is 2, hence the locus is a circle, centre (1, 0), radius 2 (see figure 5b); |z + j| =1 or |z−(−j)|=1 → all points whose distance from the point −j is 1, therefore locus is a circle, centre −j (i.e. at (0, −1)), radius 1 (see figure 5c).

-1 1 3 1 0 -1

figure 5a figure 5b figure 5c

z+1 Ex 7. Find the loci in the complex plane given by (a) Re(z)=2, (b) =2. z−1 (a) Put z = x + jy,thenRez=Re(x+jy)=x=2.

0 1 2

z +1 |z+1| (b) Now = = 2 implies |z +1|=2|z−1|. z−1 |z−1| Put z = x + jy then z +1=x+1+jy and z − 1=x−1+jy , and the above equation gives |x +1+jy| =2|x−1+jy| 1/2 1/2 → (x +1)2 +y2 =2 (x−1)2 + y2 then squaring → (x +1)2 +y2 =4 (x−1)2 + y2 x2 +2x+1+y2 =4(x2−2x+1+y2) 3x2 +3y2 −10x +3=0 10 x2 +y2 − x +1=0. 3 Completing the square leads to 2 2 5 5 x − − + y2 +1=0 3 3 2 2 2 5 5 16 4 x− +y2 = −1= = 3 3 9 3 5 4 which is a circle, centre , 0 , radius . 3 3