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7/31/2019 Documents and Settings Vc1 My Documents HIT HIT223 Eng Math

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Chapter 1

Complex Variables

1.1 Introduction

Some basic denitions underlying Complex Analysis.

Denition 1.1.1 Set of points - any collection of points in the complex plane iscalled an aggregate or a set of points. each point is called a member or element of theset.

Denition 1.1.2 Neighbourhood of a point - Let |z zo| < , where zo is a given point and is a small positive number. Then the neighbourhood of zo consists of all points lying inside but not on a circle of radius with center at zo .

Denition 1.1.3 Limit point - a point zo is said to be the limit point of a set of points S in the Argand plane if every neighbourhood of zo contains points of set S other than zo . Thus, each point on the circumference of the circle |z| = r is a limit point of the set |z| < r and these limit points do not belong to the set. Thus, limit points of a set are not necessarily the points of the set. Accordingly, there are two types of limit points:

(a) Interior points - A limit point zo of the set S is an interior point if in the neigh-bourhood of zo there exists entirely the points or the set S .

(b) Boundary points - A limit point zo which is not an interior point is said to be a boundary point.

Denition 1.1.4 Bounded and unbounded sets - a set of points is said to bebounded if there exists a positive number k such that |z| k is satised for all pointsz of the set. If there does not exist such a number k , then the set is unbounded .

Denition 1.1.5 Open and Closed sets - a set containing entirely the inner pointsis called an open set. E.g., the set dened by |z| < r is an open set since it containsonly the inner points. But |z| r is a closed set.

Denition 1.1.6 Compact set - a closed and bounded set is called compact.

Denition 1.1.7 Convex or Connected set - an open set is said to be connected if every pair of its points can be joined by a nite number of straight line segments all of whose points belong to the set. For example, the set |z| < 1 is a connected set.

Denition 1.1.8 Domain - A connected open set is called an open region or domain.

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Denition 1.1.9 Closed domain - The set obtained by adding to an open connected set its boundary points is called a closed domain or region.

Denition 1.1.10 Circular disk - the inequality |z zo| < represents the interior of a circle of radius with center zo . This region is called an open circular disk.The region given by |z zo| is called the closed circular disk.

1.2 Functions of a Complex VariableA complex variable is an ordered pair of real variables, that is,

z = ( x, y )= x + iy

Let z = x + iy and w = u + iv be two complex variables. If for each value of z in S (set of complex numbers) there corresponds one or more values of w in a well denedway, the w is said to be a function of z ; thus

w = f (z)

Examples : w = 1z

, w = z2, w = ez , etc.

If to each value of z , there corresponds only one value of w, then the function w(z) iscalled a single valued function.

r s r r s u

z1z2z3

---

w1w2w3

one-to-one mapping

If w(z) takes more than one value corresponding to a value of z , then w(z) is a multi-valued function of z . For example,

w = zIn this course we will concern ourselves with single valued functions.Suppose w = u + iv is the value of f at z = x = iy where x,y,u and v are real,

u + iv = f (x + iy ) f (z) = u (x, y ) + iv (x, y )

Example 1.2.1 If f (z) = z2,

u + iv = ( x + iy )2

= x 2 y2 + 2 ixy

u (x, y ) = x2 y2

and v (x, y ) = 2 xy



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Exercise 1.2.1 Let z = x + iy . Find u (x, y ) and v(x, y ) if f (z) = ez .

Solution 1.2.1

ez = ex + iy

= ex (cos y + i sin y)

u (x, y ) = ex cos y and v (x, y ) = ex sin y

Now try the following exercise problems.

Practice Questions

Express the following in the form u(x, y ) + iv (x, y )

1. sin z 2. log z 3. cos z 4. z3 5. ez 6. z2 + 2 z

1.3 LimitsDenition 1.3.1 Let w = f (z) be a function of z dened at all points in some neigh-bourhood of zo . Then, the function f (z) is said to have a limit as z approaches zo if for every positive real number there exists a positive real number such that

|f (z) | < , z, z o , z = zoin the domain 0 < |z zo| < .

The limit of f (z) as z approaches zo is a number and is given by

limz z o

f (z) =

The evaluation of limits can be done using the following theorems on limits.

Theorem 1.3.1 If limz z o

f (z) = and limz z o

(z) = m , then

(i) limz z o

[f (z) + (z)] = + m

(ii) limz z o

[f (z)(z)] = m

(iii) limz z o

f (z)(z)

= m

, m = 0 .

Example 1.3.1 Find limz 1

z3 1z 1

.

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Solution 1.3.1 since z = 1 , we have:

z3 1z 1

= z2 + z + 1

limz 1

z3 1z 1

= limz 1

(z2 + z + 1)

= 3

Theorem 1.3.2 Let f (z) = u (x, y ) + iv (x, y ), zo = x o + iy o and wo = u o + iv o. Then,the necessary and sufficient conditions that lim

z 0f (z) = wo are that

(i) lim(x,y ) (x o ,y o )

u (x, y ) = u o

(ii) lim(x,y ) (x o ,y o )

v(x, y ) = vo

From Theorem 1.3.2 we can evaluate Example 1.3.1 as follows:

limz 1z3

1

z 1 , as z 1, x 1 and y 0. Thus,z2 + z + 1 = ( x + iy )2 + ( x + iy ) + 1

= x 2 + x y2 + i(y + 2 xy ) + 1= ( x 2 + x y2 + 1) + i(y + 2 xy )

which tends to 3 as x 1 and y 0. Hence, limz 1z3 1z 1

= 3.

1.4 Continuity

Let f (z) be dened at all points of the complex plane and let z1 be a point of the plane.

Denition 1.4.1 The function f (z) is said to be continuous at z = z1 if for every positive number there exists another number such that

|f (z) f (z1)| < whenever |z z1| < If the point z1 is a limit point of the complex plane, then continuity of f (z) at z = z1means that

limz z 1

f (z) = f (z1)

A necessary and sufficient condition that f (z) = u (x, y ) + iv (x, y ) be continuous is thatthe real functions u (x, y ) and v(x, y ) should be continuous.

Denition 1.4.2 A function f (x, y ) of two independent real variables x and y is said to be continuous at point (x o , yo) if



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(i) f (x o , yo) is nite

(ii) limx x oy y o

= f (x o , yo) in whatever way x x o and y yo.

Example 1.4.1 Given that

f (x, y ) =

x 3

y3

x 3 + y3 , x, y = 0

0 , x = y = 0

Discuss the continuity of f (x, y ) at (0,0).

Solution 1.4.1

lim(x,y ) (0 ,0)

x 3 y3x 3 + y3

= limx 0y 0

x 3 y3x 3 + y3

= limy 0

y3

y3= 1

and limx 0y 0

x 3 y3x 3 + y3

= limx 0

x 3

x 3= 1

Further, when y = mx

limx 0y 0

x 3 y3x 3 + y3

= limx 0

x 3 m 3x 3x 3 + m 3x 3

= 1 m 31 + m 3Since the limit is not unique, it follows that the function is not continuous at (0,0).

Another example!!

Example 1.4.2 Discuss the continuity of f (z) at z = 0 .

f (z) =

x 3

y3

x 2 + y2 +ix 3 + y3

x 2 + y2, z

= 0

0 , z = 0

We proceed as follows in investigating the continuity of f (z) at (0,0).



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Solution 1.4.2 From f (z) we deduce that

u (x, y ) =x 3 y3x 2 + y2

v(x, y ) =x 3 + y3

x 2 + y2Now,

limx 0y 0

u (x, y ) = limy 0(y) = 0limy 0x 0

u (x, y ) = limx 0

(x ) = 0

Similarly,

lim(x,y ) 0

v(x, y ) = 0 in whatever way x and y 0.

Hence u (x, y ) and v(x, y ) are continuous at (0,0). It follows that f (z) is continuous at z = 0 .

1.5 Differentiability

As in the case of limits and continuity, the concept of differentiation of a function of acomplex variable is dened in the same way as the real variable case.

Denition 1.5.1 The derivative of the function f (z) at zo is f (z) and is given by

f (z) = limz z o

f (z) f (zo)z zo

= limh 0 f (zo + h ) f (zo)h , where h = z zo

Example 1.5.1 Find the derivative of z2 at z = zo .

Solution 1.5.1 We proceed as follows:

Let f (z) = z2

f (zo) = limh 0

(zo + h )2 z2oh

= limh 0

2hz o + h2

h= 2 zo

Example 1.5.2 Show that the derivative of f (z) = z does not exist anywhere.



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Solution 1.5.2

f (z) = z = x iyf (z + z) = x + x i(y + y)

hencef (z + z) f (z)

z=

x i y x + i y

Now, lim x 0 y 0

x i y x + i y = lim y 0 i yi y = 1

and lim y 0 x 0

x i y x + i y

= lim x 0

x x

= 1

Since the limit is not unique, f (z) does not exist anywhere.

Theorem 1.5.1 If f (z) is differentiable at zo, then f (z) is continuous at z = zo .Let us show a little proof of the above theorem.

We have,

limz 0

[f (z) f (zo)] = limz 0f (z) f (zo)

z zo(z zo)

= limz 0

f (z) f (zo)z zo

limz 0

(z zo)= 0

Hence, limz 0

f (z) = f (zo), which shows that f (z) is continuous at z = zo .

1.5.1 Rules for differentiation

(i)d

dz[c1f (z) + c2(z)] = c1f

(z) + c2 (z), where c1 and c2 are complex constants.

(ii)d

dz[f (z)(z)] = f (z)(z) + f (z) (z).

(iii)d

dzf (z)(z)

=f (z)(z) f (z)

[(z)]2 , (z) = 0.

(iv)d

dzf ((z)) = f ((z))

(z)

1.6 Cauchy-Riemann equations

Let f (z) = u (x, y ) + iv (x, y ) be differentiable at zo = x o + iy o . Then the rst orderpartial derivatives of u and v with respect to x and y must exist at ( xo , yo) and must



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satisfy the following equations.

ux

=vy

uy

= vx

Cauchy-Riemann Equations

Since the Cauchy-Riemann equations are only a necessary condition, they can be used

to locate points at which f

(z) does not exist.

Example 1.6.1 Show that the function dened by f (z) = |z| is not differentiable at the origin but is continuous everywhere.

Solution 1.6.1 Let f (z) = u + iv = |z| = x 2 + y2Thus, u = x 2 + y2, and v = 0Now,

ux =

x

x 2 + y2 ,uy =

y

x 2 + y2vx

= 0 ,vy

= 0

When x = 0 , and y = 0 we have

ux

= 0 =vy

and vx

= 0 , uy

= 0 .

Similarly, when x = 0 , and y = 0 we have

ux

= 0 ,vy

= 0

and uy

=vx

= 0

Finally, when x = y = 0 ,

x

u (0, 0) = limh 0

u (h, 0) u (0, 0)h = |h |h = 1which shows that the limit is not unique. Since the Cauchy-Riemann equations are alsonot satised at the origin, it follows that f (z) = |z| is not differentiable at the origin though its continuous everywhere.



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Example 1.6.2 Determine whether or not the C-R equations are satised by the func-tion f (z0 = z2 3z .Solution 1.6.2 We rst express f (z) in terms of u and v. That is,

f (z) = 3x + x 2 y2 + i(3y + 2 xy )Thus, u (x, y ) = 3x + x 2 y2 and v(x, y ) = 3xy + 2 xy

Now you can verify that u

x =v

y , and u

y = v

x

Practice Questions

Verify whether the Cauchy-Rieman equations are satised in each of the following.

1. f (z) = 3 z2 3z + 1 2. f (z) = 1z 3. f (z) = |z| 4. f (z) = |z|2

5. f (z) =z 1z + 1

1.7 Analytic FunctionsA function f (z) is said to be analytic in the domain Dif it is differentiable at every pointin the domain D. Such a function is also called regular, holomorphic or monogenic.If a function f (z is analytic at each point in the entire nite plane, then f (z) is anentire function.If f(z) is not analytic at z = zo but is analytic at some point in the neighbourhood of zo , the z o is called a singular point or singualrity off .

(a) If two functions are analytic in a domain D, their sum and their product are bothanalytic inD(b) The quotient of two analytic functions is also analytic provided the function in

the denominator does not vanish at any point in D.

1.7.1 Harmonic Functions

Any function f which possesses continuous partial derivatives of the rst order andsecond order and satises the Laplace equation is called a harmonic function .

2f x 2

+ 2f y 2

= 0 Laplaces equation

If f (z) = u + iv is an analytic function, then the component functions u and v are bothharmonic functions.Since f is analytic, the rst order partial derivatives of u and v satisfy the C-R equa-tions.

ux

=vy

anduy

= vx



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If we differentiate both sides of the equations with respect to x we obtain

2ux 2

= 2v

xyand

2uxy

= 2vx 2

..................... (1)

Differentiating with respect to y gives

2uyx

= 2v 2y

and 2uy 2

= 2v

yx..................... (2)

Since partial derivatives are continuous,

2uxy

= 2u

yxand

2vxy

= 2v

yx

Equating (1) and (2) gives us

2ux 2

+ 2uy 2

= 0 and 2vx 2

+ 2vy 2

= 0

which shows that u and v are harmonic.If two given functions are harmonic in the domain Dand their rst order partialderivatives satisfy C-R equations at every point of D, then v is a harmonic conjugateof u .

1.8 Complex Integration

Let Cbe a smooth curve in the complex z-plane having end points zo and zn . Letz1 , z2, z3, . . . , z n 1 be any arbitrary number of intermediate points which subdivide thecurve Cinto n arcs zoz1, z1z2, z2z3 , . . . , z n 1zn as shown in Figure 1.1.

Let 1, 2 , . . . , r , . . . , n be the points chosen on the curve Csuch that 1 lies on arczoz1, 2 lies on z1z2 , . . . , r lies on zr 1zr and so on.Let

S n =n

r =1

f (r ) zr , where zr = zr zr 1 .If the curve is divided into smaller and smaller parts so that n , | r | 0and is the summation tends to a unique limit , that is, independent of the choice of the intermediate points and the manner in which the subdivision is performed, the theunique limit S n is called the line integral (or denite integral ) of the complex functionf (z) taken along the curve Cbetween the points z o and zn and is denoted by

limn

S n = z nz o f (z) dzHere are some basic denitions you must be familiar with in dealing with complexintegrals:



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-

6

p pq p p q

p q q

zo

z1

z2z3 . . . pzr

zn

zn 2

q p p1

2

3r

nzn 1

zr 1

p p p p q q rq

q p p p q p p p

p p p

p p p p

Figure 1.1

Continuous arc : a set of points z = ( x, y ) satisfyingz = x (t ) + iy (t ) t

where x (t ) and y(t ) are continuous functions of the real variable t .

Multiple points : a point z1 is a multiple point of the arc if the equationz1 = x (t ) + iy (t )

is satised by two or more values of t in the given range.

Jordan arc : a continuous arc having no multiple points.

Jordan Curve or Simple closed curve:

Given the continuous arc z = x (t ) + iy (t ) t , If x (t ) = ( ) andy( ) = y( ) and if there is no other multiple point on the continuous arc denedby z , then this continuous arc is a Jordan curve or a simple closed curve.The circle x = a cos t, y = a sin t (0 t 2 ) is a simple closed curve.

Contour : a curve composed of a chain of nite number of regular Jordan arcs.

Simply Connected and multiply Connected domains

A domain is simply connected if the interior of every closed contour in the domainconsists only the points of the domain.In general, a domain is said to be m -fold connected if its boundaries consist of mdistinct parts without common points.



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1.8.1 Contour Integration

Let f (z) be dened from z = a to z = b. The the integral of f (z) from z = a to z = bis dened in terms of the values of f (z) at points along the contour Cextended fromz = a to z = b. Such an integral is called a line integral , its value depending uponthe choice of Cas well as upon f, a and b.

C f (z) dz = b

af (z) dz

dsds

Example 1.8.1 Find the integral of z2 along the contour Cwhere Cis given by z(s ) =s + is 2 , 0 s 1.Solution 1.8.1 From the denition of contour integration we have,

C z2 dz = C f (z) dzds dsNow, z(s ) = s + is 2 and

dzds

= 1 + 2 is

C z2 dz = 1

0(s + is 2)2(1 + 2 is )ds

= 10 (s 2 s 4 + 2 is 3)(1 + 2 is )ds= 10 (s 2 5s 4)ds + i

1

0(4s 3 2s 5)ds

=s 3

3 s5

1

0+ i s 4

s 6

3

1

0

=

2

3+ i

2

3Exercise 1.8.1 (a) Integrate C z2 dz where Cis the contour given by

z(s ) = 2s , 0 s 12 + i(s 1) , 1 s 2



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(b) Evaluate C z2 dz where Cis the straight line joining the origin O to the point P (2, 1) on the complex plane.

1.8.2 Cauchys Integral Theorem

Theorem 1.8.1 If a function f (z) is analytic at all points inside and on a closed contour C, then

C f (z) dz = 0We need to prove Theorem 1.8.1 inorder to appreciate its usefulness in evaluatingcertain types of complex integrals. the proof is as follows.Proof: Let Rbe a closed region consisting of all points inside and on a closed contourCand let Cbe described in the counter clockwise sense.From Greens theorem, if P (x, y ), Q (x, y and their partial derivatives

P y

andQx

are

continuous single-valued functions over a closed region Rbounded by the curve C, then

C (P dx + Qdy ) = R Qx P y dxdyNow, let f (z) = u (x, y ) be analytic throughout Rin the z plane. Then dz = dx + idy .Thus,

R

f (z) dz = C

(u + iv )(dx + idy )

= C

(udx vdy ) + i C

(vdx + udy ) (1.1)

Since f (z) is analytic, f (z) exists in Rand then u and v and their rst partial deriva-tives are continuous in R.Applying Greens Theorem on the right-side of equation 1.1,

R

f (z) dz = R

vx

uy

dxdy + i R

ux

vy

dxdy

Since f (z) is analytic, the C-R equations must be satised.

ux

=vy

anduy

= vx

C f (z) dz = 0Example 1.8.2 Verify Cauchys theorem for the function f (z) = z where Cis thecircle |z| = 2 .HIT223 - 2009 vchikwasha Harare Institute of Technology


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Solution 1.8.2 We know that we can express z in polar form as z = re i . So, for

|z| = 2 we can write this function as z = 2 ei . Thus, if z = 2 ei , then dz = 2 ie i d .

C f (z) dz = 2

02ei 2ie i d

= 4 i 20 2ei d= 4 i

e2i

2i

2

0

= 2( e4i 1)= 2(cos 4 + i sin4 1)= 0

1.8.3 Cauchy Integral Formula

Consider the integral

I = C

1z a

dz (a = constant)

f (z) =1

z ais not analytic at z = a . This function is not dened at this point and

thus cannot posses a derivative.Let Cinvolved in this denition be a simple closed curve. If the point z = a is outsidethe curve, then Cauchys theorem holds and I = 0. That is,

C

1z a

dz = 0

If z = a is inside C, I = 0.Now, if Cis a circle of radius R centered at z = a , we evaluate the integral by settingz = a + Re i . Thus, dz = iRe i d .

I =2

0

1a + Re i a

iRe i d

=2

0

i d

I = 2 i

This is true for any simple closed path which encircles z = a .If f (z) is analytic inside and on a closed contour Cand zo is any point interior to C,then,

f (zo) =1

2i C

f (z)z zo

dz



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Exercise 1.8.2 (i) Evaluate C

z2

z 2dz , Cbeing the circle |z| = 3 .

(ii) Show that C

z dz9 + 9 z z2 z3

= 14

i , Cbeing the circle |z| = 2 .

1.9 Derivatives of an Analytic Function

If f (z) is holomorphic in a region Dbounded by a simple closed curve C, then itsderivatives of all orders exist and are analytic at each point of the domain D.Theorem 1.9.1 If f (z) is analytic inside and on a closed contour Cand zo is a point inside C, then

f (n )(zo) =n !

2i C

f (z)(z zo)(n +1)

dz

Example 1.9.1 Evaluate C2z2

z

2

(z 2)4 dz , where Cis the circle |z| = 3 .

Solution 1.9.1 We begin by letting f (z) = 2 z2 z 2. Comparing Theorem 1.9.1with our integral, zo = 2 and n = 3 . Thus we can write the integral asf (3) (2) =

62i

C

f (z) dz(z 2)4

Thus,

f (z) = 4 z

1

f

(z) = 4f

(z) = 0

C

2z2 z 2(z 2)4

dz =i3

f (3) (2)

= 0

Exercise 1.9.1 Evaluate

(i) C2z2

z

2

z2 4z + 4 dz, C: |z| = 3 (ii) C2z2

z

2

(z 2)3 dz, C: |z| = 3

(iii) C

dzz2(z 3)

, C: |z| = 2



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1.10 Taylors series

Theorem 1.10.1 Taylors Theorem Let f (z) be analytic inside a circle Cwith center at zo and radius r o. Then at each point inside C, f (z) can be expanded in ascending powers of (z zo and may be written as

f (z) = f (zo) +(z zo)

1!f (zo) +

(z zo)22!

f

(zo) + . . . +(z zo)n

n !f (n )(zo) + . . .

=

n =0

(z zo)nn ! f (n ) (zo)

This is Taylors series for f (z) with center at zo . The series converges to f (z).In particular case when zo = 0 , we obtain the Maclaurin series for f (z).

Example 1.10.1 Find the Taylor series expansion of 1

1 zabout the point z = 0 .

Solution 1.10.1 f (z) =1

1 z, zo = 0 .

f (z) = f (zo) +(z

zo)

1! f

(zo) +(z

zo)2

2! f

(zo) + . . .Now,

f (z) =1

z 1 f (zo) = f (0) = 1

f (z) =1

(1 z)2 f (zo) = 1!

f (z) =2

(1 z)3 f (zo) = 2!

f

(z) =6

(1 z)4 f

(zo) = 3!

f iv (z) =24

(1 z)5 f (zo) = 4!

......

f (n )(z) =n !

(1 z)n +1 f (n ) (zo) = n !

Finally, we can write the series as

f (z) = f (zo) + (z zo)1! f (zo) + (z zo)2

2! f

(zo) + . . .

= 1 + z +z2

2! 2! +z3

3! 3! + . . .

1

1 z= 1 + z + z2 + z3 + . . . .



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Exercise 1.10.1 Expand by Taylors series:

(i)1z

about the point z = 1 .

(ii)1

1 + z2about z = 0 .

You can also verify the following important power series:

ez = 1 +z1!

+z2

2!+

z3

3!+ . . . +

zn

n !+ . . . , |z| <

sin z =z1!

z3

3!+

z5

5!+ . . . + ( 1)n +1

z2n 1

(2n 1)!+ . . . , |z| <

cos z = 1 z2

2!+

z4

4! . . . (1)n z

2n

(2n )!+ . . . , |z| <

11 + z

= 1 z + z2 z3 + . . . + ( 1)nz n + . . . , |z| < 1ln(1 + z) = z

z2

2+

z3

3 . . . (1)n +1 z

n

n+ . . . , |z| < 1

1.10.1 Laurents Series

In expanding a function f (z) by means of Taylors series at a point zo, f (z) must beanalytic at zo. Laurents series gives an expansion of f (z) at a point zo even if f (z) isnot analytic at zo .

Theorem 1.10.2 If f (z) is analytic between and on two circles C 1 and C 2 having common radii r 1 , r 2(r 1 > r 2), then at each point z between them, f (z) can be expanded by a convergent series of positive and negative powers of (z zo), namely,

f (z) = a o+ a 1(zzo)+ a 2(zzo)2+ . . . + a n (zzo)

n

+b1

z zo +b2

(z zo)2 + . . . +bn

(z zo)n + . . .where

a n =1

2i c1

f (z )(z zo)n +1

dz , n = 0 , 1, 2, 3, . . .

and

bn =1

2i c2

f (z )(z zo)n +1

dz , n = 1 , 2, 3, . . .

with the integrals taken in the counter clockwise sense.

Example 1.10.2 Expand f (z) =z

(z 1)(2 z)in Laurent series valid for



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(i) |z| < 1 (ii) 1 < |z| < 2 (iii) 0 < |z 2| < 1Solution 1.10.2 f (z) =

z(z 1)(2 z)

can be expressed in terms of its component

fractions as f (z) =1

z 1+

22 z

.

Now, for

(a) |z| < 1 we have1

z 1 = 1

1 z = (1z) 1

= (1+ z+ z2

+ z3

+ . . . ) ............(i)

and 2

2 z=

2

2 1 z2

=1

1 z2

= 1 z2

1= 1+

z2

+z2

4+

z3

8+ . . . ... ....... (ii )

Combining (i) and (ii) we have

f (z) = (1 + z + z2 + z3 + . . . ) + (1 +z2

+z2

4+

z3

8+ . . . )

= z2

+34

z2 +78

z3 + . . .

(b) 0 1 we have1

z 1=

1

z 1 1z

=1z

1 1z

1

=1z

1 +1z

+1z2

+ . . . ........... (iii )

and when |z| < 2 ,2

2 z= 1

z2

1= 1+

z2

+z2

4+

z3

8+ . . . .............. (iv )

Thus, f (z) =1z

1 +1z

+1z2

+ . . . + 1 +z2

+z2

4+

z3

8+ . . .

(c) 0 < |z 2| < 1First let u = z 2 0 < |u | < 1 f (z) =

z(z 1)(2 z)

= u + 2

(u + 1) u=

2u

+1

u + 1

when |u | < 1 (1 + u) 1 = 1 u + u 2 + . . .

Thus, f (z) = 2

z 2+ 1 (z 2) + ( z 2)2 . . .

1.11 Zeros of a Function

Let f (z) be an analytic function. If f (z) = 0, the z0 is a zero of f (z).Since f (z) is analytic at z0, there exists a neighborhood of z0 at which f (z) can beexpanded in Taylors series

f (z) = a 0 + a 1(z z0) + a 2(z z0)2 + . . . , + a n (z z0)n + . . . , |z z0| < r 0.vchikwasha Harare Institute of Technology


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where a 0 = f (z0) and a n =f (n ) (z0)

n !.

If z0 is a zero of f (z), then f (z0) = 0 and a 0 = 0.If a 0 = 0 and a 1 = 0, then z0 is a simple zero.If a 0 = a 1 = a 2 = . . . = a m 1 = 0 but a m = 0, the z0 is a zero of order m .Thus, we can dene a zero of order m by the condition

f (z0) = f

(z0) = . . . = f (m 1)

(z0) = 0 and f (m )

= 0 .So we can write f (z) = ( z z0)m g(z) where g(z) is analytic and g(z0) = a m = 0.

Theorem 1.11.1 Let z0 be a zero of an analytic function f (z). There is a neighbour-hood of f 0 throughout which f (z) has no other zeros, unless f is identically zero. Thus,if an analytic function is not identically zero, then its zeros are isolated.

Example 1.11.1 1. f (z) = ( z + 1) 4 has a zero of order 4 at z = 1.2. f (z) = cos z has simple zeros at z =

2,

3

2, . . . .

3. Find the zeros of f (z) =z 1z2 + 1

2

Solution 1.11.1 f (z) = 0 when (z 1)2 = 0 . Thus, z = 1 is a zero of order 2.

1.12 Singularities of a Function

If a function f (z) is analytic at every point in the neighbourhood of a point z0 exceptat z0 itself, z0 is called a singularity or singular point of the function.

Example 1.12.1 Locate and name all singularities of

f (z) =z6 + 2 z3 + 3

(z + 1) 2(z2 + 1)

Solution 1.12.1 (z + 1) 2(z2 + 1) = 0z = 1, z = i

1.12.1 Isolated Singularity

For f (z) =1z

, z = 0 is an isolated singularity since f (z) is analytic everywhere exceptat z = 0.



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1.12.2 Removable Singularity

The singularity point z = z0 is called removable if

limz z 0

f (z) exists

f (z) =z2 a 2z a

, z = a is a removable singular point.

Theorem 1.12.1 Let f (z) be analytic in the domain Dexcept at z = z0 . Then z0 isan isolated singularity of f (z) and we can draw two concentric circles with center at z0such that f (z) can be expanded in a Laurents series in the annular region between thetwo circles. Thus,

f (z) =

n =0

a n (z z0)n

regular part of f(z)+

n =1

bn(z z0)n

principal part in the(1.2)

neighbourhood ofz 0

z0 is a removable singularity if bn = 0 ,n , or if limz z 0 f (z) exists.

1.12.3 Essential Singularity

If bn in (1.2) contains an innite number of terms, then z = z0 is an isolated essential singularity of f (z).Thus, there exists no nite n such that

limz z 0

(z z0)f (z) = a nite constant .

1.13 Poles of a Function

Consider againf (z) =

n =0

a n (z z0)n +

n =1

bn(z z0)n

principal partIf the principal part terminates, at n = m (say), then f (z) becomes

f (z) = a 0 + a 1(z z0) + . . . +b1

z z0+ . . . +

bm(z z0)m

The isolated singular point z = z0 is called a pole of order m .A pole of order 1 is called a simple pole.

A pole of order 2 is called a double pole.

Example 1.13.1 Determine the order of the pole of the function f (z0) =1

z3(z + 4).

Solution 1.13.1 f (z0) has a pole of order 3 at z = 0 and a simple pole at z = 4.



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1.14 Residues

Let z0 be an isolated singularity of f (z). Then there exists a positive number r , suchthat f (z) is analytic at each point of the region 0 |z z0| < r , and in which it can berepresented by the Laurents series

f (z) =

n =0

a n (z z0)n +

n =1

bn(z z0)n

wherea n =

12i C f (z

)(z z0)n +1

dz , n = 0 , 1, . . . .

andbn =

12i C f (z

)(z z0)n +1

dz , n = 1 , 2, . . . .

b1 =1

2i C f (z )dz = 12i C f (z)dz (1.3)where Cis any closed contour around z0(counterclockwise) such that f (z) is analyticon

Cand interior to

C, except at z = z0.

b1 is the coefficient of 1

z z0and is called the residue of f (z) at the isolated singular

point z0.Thus, Res z = z 0 f (z) = b1From (1.3), Res z = z 0 f (z) =

12i C f (z)dz , where f (z) is analytic everywhere on and

within Cexcept at z0.If f (z) has a simple pole at z = z0, the residue of f (z) is given by

Res z= z 0 f (z) = limz z 0 (z z0) f (z)

Example 1.14.1 Find the residue of f (z) = 1z(z 4)

at each of its poles.

Solution 1.14.1 f (z) has simple poles at z = 0 and z = 4 .Thus,

when z = 0

Res z=0 f (z) = limz 0

z 1

z(z 4)=

14

and when z = 4

Res z=4 f (z) = limz 4

(z 4) 1

z(z 4)=

14



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Suppose f (z) has a pole of order m ,

f (z) =

n =0

a n (z z0)n +b1

z z0+

b2(z z0)2

+ . . . +bm

(z z0)m

(z z0)m f (z) =

n =0

a n (z z0)m + n + b1(z z0)m 1 + . . . + bm (1.4)

Differentiating (1.4) ( m 1) times,dm 1

dz m 1[(z z0)m f (z)] =

n =0

a n (m + n )(m + n + 1) . . . (n + z)(z z0)n +1 + b1(m 1)!

thus, b 1 =1

(m 1)!lim

z z 0

dm 1

dz m 1[(z z0)m f (z)]

Res z = z 0 f (z) =1

(m 1)!lim

z z 0

dm 1

dz m 1[(z z0)m f (z)]

Example 1.14.2 Calculate the residue of z2 e z

(z a )3 .

Solution 1.14.2 z = a is a pole of order 3,

Res z = a f (z) =12!

limz a

d2

dz 2(z a )3

z2ez

(z a )3=

12

limz a

d2

dz 2z2ez

=12

limz a

(z2ez + 4 ze z + 2 ez )

=12

[ea (a 2 + 4 a + 2)]

=ea

2(a 2 + 4 a + 2)

Exercise 1.14.1 Find the residues of

1.z2 + 5

z(z + 3) 2

2.z

(z 1)(z 2)2at the poles.

1.14.1 Cauchys Residue Theorem

If Cis a closed contour within and on which a function f (z) is analytic except for anite number of singular points z1, z2, . . . , z n interior to C, then

C f (z) dz = 2 i (R 1 + R 2, . . . + R n )where R j is the residue of f (z) at z = z j .



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Example 1.14.3 Find C zz + 1 dz , where Cis the circle |z| = 2 .Solution 1.14.3 z = 1 is a simple pole,

Res z = 1f (z) = limz 1

(z + 1)z

z + 1= 1

C zz + 1 dz = 2 i (1)= 2i

Exercise 1.14.2 Evaluate

1. C sin zz6 dz where Cis a unit circle at the origin.2. C dzz3(z 1) , C: |z| = 23. C z + 1z2 2z dz , C: |z| = 34. C e

z

(z 1)3dz , C: |z| = 3

5. C ez

z2 1dz

(a) given that Cis the circle |z 1| = 1(b) given that Cis the circle |z i| = 1
