Complex Variables I

7/29/2019 Complex Variables I

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Complex Variables 2010/2011

EEM2046 Engineering Mathematics IV 1

Complex Variables

Learning Objective:

Throughout this chapter, we will study

the basic notion of complex functions of a complex variable (in

Section 2) and limits and continuity (in Section 3);

analyticity, the Cauchy-Riemann Equations, harmonicity and

harmonic conjugacy (in Section 4).

Integration in the complex plane, the fundamental theorem of calculusof contour integral, the principle of contour independence, Cauchy’s

theorem, and the Cauchy’s integral formula (in Section 5);

Sequences of complex number, series of complex numbers, test of

convergence, power series, Taylor series, and Laurent series (in

Section 6);Residue theory, Cauchy residue thorem, and discuss some applications

of the residue calculus in evaluation of improper integrals (in Section

7).

Teaching Outcome.

A set of tutorial questions and exercises will be provided to extend the

understanding of the basic notion presented in this chapter. This will provide training to acquire and apply fundamental principles of

mathematical sciences.

A mid-term test will be given for testing the students understanding of

the mathematical concept that have been covered. The test will trainthe students to identify, formulate and apply mathematical principles

in problem solving.



Complex Variables - Trimeser 3

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Complex Variables

1. Regions of the Complex Plane

1.1 Open Sets, Closed Sets and Connected Sets

Let r be a positive real number and 0 z a point in the complex plane . The r-

neighborhood of 0 z is the collection of all the points inside a circle of radius ,r

centered at 0 . z These are the points satisfying 0 . z z r

iy

0 z

x

A r-neighborhood of 0 z from which we have deleted the center 0 z is called a

deleted r-neighborhood of 0 z which consists of the points inside a circle centered at

0 z but excludes the point 0 z itself. These points satisfy 00 z z r .

iy

0 z

y

Let S be a subset of the complex plane . A point z S is called an interior

point of S if we can find a neighborhood of z that is wholly contained inside .S A

subset of is called an open set if every point in is an interior point of .

Here are some common and useful examples of open sets to keep in mind.

For any ,r and any fixed ,w r -neighborhood of w and deleted r -

neighborhood of w are open sets.

For any ,r and any fixed ,w the set : z z w r is open.

For any fixed ,w and any 1 2,r r with 2 1,r r the set

1 2( : , ) A w r r 1 2: z r z w r

is open because 1 2( : , ) A w r r is the intersection of two open sets

1: z z w r and 2: . z z w r

The set 1 2( : , ) A w r r is called open annulus centered at .w





A subset S of is said to be closed if cS is open. For example, the empty set

and the complex plane are open and closed. For any ,r and any fixed

,w the set : A z z w r is closed because its complement is open.

We next introduce the notion of connectedness. Generally, we say that a subset

S of is connected if it is “all in one piece”, that is, if it is not the union of two

nonempty subsets that do not touch each other. More precisely, we said that a subset

S of is said to be disconnected if it is union of two nonempty subsets 1S and 2S ,

neither of which intersects the clouse of the other one. The set S is connected if it is

not disconnected.

There is another notion of connectedness that is important in many situations. A

subset S of is called pathwise connected if any two points in S can be joined by a

polygonal line, which is contiunous curve, consisting of a finite number of straight

line segments jointed end to end, and that lies entirely in .S

pathwise connected not pathwise connected

In fact, if a subset S of is open, then S is pathwise connected if and only if S is connecteed. A nonempty subset S of is called a region if the set S is open and

connected. Consequently, a region is an open set which is pathwise connected.

Here are some useful examples of regions.

The complex plane is a region.

For any ,r and any fixed ,w r -neighborhood of w and deleted r -

neighborhood of w are regions.

For any ,r and any fixed ,w the set : z z w r is a region.

For any fixed ,w and any 1 2,r r with 2 1,r r the open annulus

centered at ,w 1 2( : , ) A w r r is a region.

2. Complex Functions

2.1 Complex Functions of a Complex Variable

Let S and B be nonempty subsets of . A complex function f from S to , B

denoted by : , f S B is a rule which assigns to each element z S and only one

element ,w B we write ( )w f z and call w the image of z under f . The set S is

the domain of , f and the set B is the codomain of . f The set of all images




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( ) ( ) : f S f z z S

is called the range of . f It must be emphasized that both a domain and a rule are

needed in order for a function to be well defined . When the domain is not mentioned,we agree that the largest possible set is to be taken.

Suppose that ivuw is the value of a complex function : , f S B at the

point , z x iy S so that ( ) ( ) . f x iy w u iv f S B Here the complex

function f maps from one complex plane, the z -plane, into another complex plane,

the w -plane.

Each of the real numbers u and v depends on the real variables x and , y it

follows that )( z f can be expressed in terms of a pair of two variables real functions

of the real variables x and y as follows:

( ) ( , ) ( , ), f z u x y iv x y . z x iy (2.1.1)

By (2.1.1), it is clear that ( , ) Re ( )u x y f z and ( , ) Im ( )v x y f z are respectively

called the real and imaginary parts of the complex function ( ). f z For instance, if we

consider ( ) z f z e with , z x iy , , x y then ( ) cos ( sin ). x x f z e y i e y Here

( , ) cos xu x y e y and ( , ) sin . xv x y e y

Often an expression for a complex function ( ) ( , ) ( , ), f z u x y iv x y given in

terms of x and y , can be rewritten rather simply in terms of z and z -notation. In

any case, the identities

2

z z x and

2

z z y

i

are useful.

Example Find the real and imaginary parts of the complex functions 2( ) f z z and

( ) Re . g z z z

Solution (i) Let , z x iy , . x y Then

2 2 2 2 2 2( ) ( ) 2 ( ) (2 ) f z z x iy x xy i iy x y i xy ,

and so, 2 2Re( ( )) f z x y and Im( ( )) 2 . f z xy Also,2( ) Re( ) ( )( ) ( ). g z z z x iy x x i xy

Hence, 2Re( ( )) g z x and Im( ( )) . g z xy

For more examples, see Chapter 7: Complex Variables, Section 7.2, Engineering

Mathematics Volume 2 (second edition), Pearson Prentice Hall, Malaysia, 2006.





The elementary algebra of complex functions of a complex variable does not

differ radically from that of real-valued functions of a real variable. Let A and B be

nonempty subsets of . Presented with a pair of complex functions : f A and

: g B , we multiply f by a complex number c to obtain the function ,cf add f

and g to form , g f multiply these functions to produce g f , and take the quotient

. g f Here, we have their definitions:

)())(( z cf z cf

for all , z A

)()())((

),()())((

z g z f z g f

z g z f z g f

for all , z A B and

( / )( ) ( ) ( ) f g z f z g z

with the domain-set : ( ) 0 . z A B g z

Example Let ( ) , f z z ( ) z g z e and ( ) sinhh z z for all . z Determine the

formula of the following function. (i) 4 , f and (ii) . f g h

Solution (i) 2 2( ) ( ) ( ) , f z f z f z z z z and 3 2( ) ( ) ( ) ( ) ( ) ( ) f z f z f z f z f z f z 2 3 , z z z and 4 3 4( ) ( ) ( ) . f z f z f z z

(ii) ( )( ) f g h z ( ) ( ) ( ) sinh z f z g z h z z e z for all . z

A more interesting way of combining functions to manufacture a new function is

the composite function.

Definition 2.1.1 Let A and B be nonempty subsets of . Let : f A and

: g B be complex functions, if the range of f is contained in the domain of ; g

i.e., ( ) , f A B then the composite function of g with , f : g f A is well

defined , and is defined by

))(())(( z f g z f g for all . z A

Example Let : \ 0 f and : g be complex functions defined as

( ) 1 f z z for all \ 0 , z and 2( ) g z z for all . z It is clear that

( \ 0 ) \ 0 , f and since 2 2( ) 2 g z x y i xy for all , z x iy we have

( ) . g Hence ( \ 0 ) , f and g f is well defined and ( )( ) g f x ( ( )) g f x

21 z for all \ 0 . z Find and explain your answer for . f g

2.2 Transformations and Mappings

Properties of real functions of a real variable are often exhibited and visualized by thegraph of the function in 2-dimensitonal Cartesian plane. When w z f )( , where z

and w are complex numbers, no such convenient graphical representation of the




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complex function f is available because each of the numbers z and w is located in the

complex plane , rather than on the real line . However one can display some

information about the function by indicating pairs of corresponding points ),( y x z

in the domain of , f and ( , )w u v in the range of . f To do this, it is generally

simpler to draw the z and w planes separately. When a function f is thought of in this

way, it is often referred to as a transformation or mapping . Let us see some examples.

Example

To illustrate some of these ideas, consider ,)( 2 z z f defined on the first quadrant of

the z- plane: ,0 x .0 y Then ( ) ( , ) ( , ) f z u x y iv x y2 2 2 ; x y i xy

i.e.,22),( y x y xu and .2),( xy y xv

Since x0 and ,0 y it follows from the form of u and v that

u and .0 v Since we cannot draw a graph of ,)( 2 z z f and we

do not wish to plot the surface ,),( 22 y x y xu .2),( xy y xv A more appealing

and commonly used device is to display the images of representative curve. For

instance, the image of y x 0 ,1 is given parametrically by ,1 2 yu

yv 2 or eliminating the parameter y, by the parabola .0 ,41 2 vvu

Similarly, the image of x y 0 ,1 is given by the parabola ,122

vu

v0 , and their images are shown.

3. Limits and Continuity

3.1 Limits of Complex Functions

In calculus of one real variable, the reader learned the notion of the limit of a function

as well as the definition of continuity as applied to real variables. These concepts

apply with some modification to complex functions of a complex variable. Let

0 , z and let be an open set containing 0 . z Let : f be a complex

function defined on , except possibly at 0 z itself. We say, roughly speaking, thatthe number 0w is the limit of the function )( z f as z approaches 0 z if )( z f stays





arbitrarily close to 0w whenever z is sufficiently near to 0 z . In precise terminology,

the definition of limit may be reformulated as follows.

Definition 3.1.1 Let 0 , z and let be an open set containing 0 . z Let

: f be a complex function defined on , except possibly at 0 z itself. We say

that )( z f has a limit 0w as z approaches 0 , z and we write

0)( w z f as 0 z z ,

or equivalently,

00lim ( ) ,

z z f z w

if for every 0, there exists a positive number 0 such that for every complex

number z satisfying 00 z z we have 0( ) . f z w We note that if a limit

of a complex function )( z f exists at a point 0 z z then the limit is unique.

Example Show that 2

lim 2 4 z

z and2

2

4lim 4

2 z

z

z

Solution Let ( ) 2. f z z For any given 0, we choose . Thus 0, and

if 0 2 z , then ( ) 4 2 4 2 f z z z . Hence, we conclude

that2

lim 2 4 z

z .

Method I: Let ( ) g z 2 4

.

2

z

z

We first note ( ) f z is not defined at 2. z Also,

we have2 4 ( 2)( 2)

22 2

z z z z

z z when 2. z Consequently, when 2, z we

see that ( ) 4 2 4 2 g z z z . Now given any 0, if we select ,

Then 0 2 z implies ( ) 4 2 g z z . This concludes that

2

2

4lim 4

2 z

z

z . Method II: We note that 2 z implies 2, z so

2

2

4lim

2 z

z

z

2 2

( 2)( 2)lim lim 2

2 z z

z z z

z . Hence,

2lim 2 4 z

z .

In real variable calculus, when we investigated the limit as 0 x x of the real

function )( x f we were actually concerned the two-sided limits with values of x

approaches 0 x from the right and left. Nevertheless, in the complex plane , the

concept of limit is more complicated because there are infinitely many paths, not just

two directions, along which we can approach sufficiently near to 0 z .

Therefore, if 0

lim ( ) z z f z exists, )( z f must tend toward the same complex value

no matter which of the infinite number (arbitrary manner) of paths of approach to 0 z

is selected. However, if one can construct two difference paths of approach to 0 z with

different limits, then this is sufficient to conclude that0

lim ( ) z z f z does not exist.




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Example Let z z z f )( for any \ 0 . z Show that 0lim ( ) z f z does not exist .

Solution If z z z f )( , then. For, if it did exist, it could be found by letting the

point ),( y x z approaches the origin in any manner. But when )0,( x z is a

nonzero point on the real axis,

;11lim0

0

lim)(lim )0,0()0,()0,0()0,(0 x x z i x

i x

z f

and when ),0( y z is a nonzero point on the imaginary axis,

.1)(

lim0

0lim)(lim

)0,0(),0()0,0(),0(0 iy

iy

iy

iy z f

y y z

Thus, by letting z approaches the origin along the real axis, we would find that the

desired limit is 1. An approach along the imaginary axis would, on the other hand,

yields limit – 1. Since the limit is unique, we must conclude that limit does not exist.

3.2 Theorems on Limits

Theorem 3.2.1 Let be an open set containing .000 iy x z Let )( z f be a

function defined on , except possibly at 0 z itself . Suppose ),(),()( y xiv y xu z f

for every 0\ z x iy z and .000 ivuw Then0 0lim ( ) z z f z w if and only if

0),(),(

),(lim00

u y xu y x y x

and .),(lim 0),(),( 00

v y xv y x y x

Example Let 0 . z Show that 0

2 2

0lim ( ) , z z z z and hence, 0 0lim ( )n n

z z z z

for all .n Solution Let . z x iy Then 2 2 2 2 2 2( ) 2 ( ) ( ) (2 ) z x iy x xyi iy x y i xy .

Then 2 2( , )u x y x y and ( , ) 2v x y xy which are two polynomial functions of two

real variables x and . y Thus, u and v are continuous functions of two real variables

x and , y and so

0 0 0 0 0

2 2

0 0 0 0( , ) ( , )

lim ( , ) lim ( , ) ( , ) , z z x iy x y x y

u x y u x y u x y x y

0 0 0 0 00 0

( , ) ( , )lim ( , ) lim ( , ) 2 .

z z x iy x y x yv x y v x y x y

Consequently,





0 0 0 0 0

2

0 0 0 0( , ) ( , ) ( , ) ( , )

22 2 2

0 0 0 0 0 0 0

lim lim ( , ) lim ( , ) ( , ) ( , )

) ( 2 ) .

z z x y x y x y x y z u x y i v x y u x y iv x y

x y i x y x iy z

By induction mathematics, we can show that0 0lim ( )n n

z z z z for all .n

Theorem 3.2.2 Let be an open set containing 0 . z Let )( z f and ( ) g z be functions

defined on , except possibly at 0 z itself . If 0

lim ( ) z z f z A and 0

lim ( ) , z z g z B

then

(i) 0 0 0

lim ( ) ( ) lim ( ) lim ( ) . z z z z z z

f z g z f z g z A B

(ii) 0 0 0


f z g z f z g z A B

(iii) 0 0 0


f z g z f z g z AB

(iv) 0

0

0

lim ( )( )

lim( ) lim ( )

z z

z z

z z

f z f z A

g z g z B, provided .0 B

Theorem 3.2.3 Let be an open set containing 0 , z and let be an open set

containing 0 .w Let )( z f and ( ) g z be functions defined on and except possibly

at 0 z and 0 ,w respectively, such that 0\ . f z If 0 0lim ( ) , z z f z w then

0 0

lim( )( ) lim ( ). z z z w

g f z g z

An immediate consequence of previous theorems, we obtain the following

interesting properties.

Corollary 3.2.4 Let be an open set containing 0 . z Let c be any complex constant ,

and let )( z f be function defined on , except possibly at 0 z itself . If

0lim ( ) , z z f z A then:

(i) 0

lim ; z z

c c

(ii)

0 0lim ( ) lim ( ) ; z z z z cf z c f z cA

(iii) 0 0

lim Re ( ) Re lim ( ) Re ; z z z z

f z f z A

(iv) 0 0

lim Im ( ) Im lim ( ) Im ; z z z z

f z f z A

(v) 0 0

lim ( ) lim ( ) ; z z z z

f z f z A

(vi) 0 0

lim ( ) lim ( ) ; z z z z

f z f z A .

(vii) 0 0

lim ( ) lim ( )n

n n

z z z z

f z f z A for any positive integer ;n




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(viii) 0

0

1 1 1lim ,

( ) lim ( ) z z

z z f z f z A

provided .0 A

Example Compute 02lim , z z z 2

2lim 2 z z and 0lim .i z

z ye

Solution Since0 0lim , z z z z it follows from Theorem 3.2.2 that

0

2lim z z z

0 0

2

0lim lim .

z z z z z z z Also, since 2 2

2lim 2 4 z z and 2lim 2 2, z it

follows that2

2lim 2 4 2 6. z

z Clearly, 1i z

e for all nonzero complex

numbers z and 0lim 0. z y Then we conclude that0lim 0.

i z

z ye

For more examples, see Chapter 7: Complex Variables, Section 7.3.2, Engineering


3.3 Continuity

Definition 3.3.1 Let 0 z be a complex number, and let be an open set containing

0 . z Let )( z f be a function defined in . The function ( ) f z is said to be continuous

at 0 z if for any ,0 there exists a positive number 0 such that for any complex

number z satisfying0 z z we have

0( ) ( ) , f z f z or equivalently,

00lim ( ) ( ). z z f z f z

Moreover, the function : f is said to be continuous on , i.e.,

: f is continuous if ( ) f z is continuos at every point . z Also, f is said

to be continuous if it is continuos on its domain set.

Example For any 0 , z show that the complex polynomial 2 1

0 1 2 1( ) ,n n

n n p z a a z a z a z a z 0 1, , , ,na a a

is continuous at 0 . z

Solution Since0 0lim n n

z z z z for all ,n and by Theorem 3.2.4 (ii), we have

0

2 1 2 1

0 1 2 1 0 1 0 2 0 1 0 0lim( ) .n n n n

n n n n z z

a a z a z a z a z a a z a z a z a z

So, we have0 0lim ( ) ( ), z z p z p z and thus, the complex polynomial ( ) p z is

continuous at 0 . z Since 0 z is an arbitrary point, it follows that complex

polynomial functions are continuous.

Theorem 3.3.1 Let be an open set containing .000 iy x z Let )( z f be a

function defined on . Then ),(),()( y xiv y xu z f is continuous at 0 z if and only

if both functions ),( y xu and ),( y xv are continuous at ).,( 00 y x In other words, Re f

and f Im are both continuous at 0 0( , ) x y if and only if f is continuous at 0 . z





Example For any 0 , z show that ( ) sin f z z is continuous at 0 . z

Solution Let , z x iy , , x y and let 0 0 0 . z x iy Then ( ) sin( ) f z x iy

sin cosh (cos sinh ). x y i x y Clearly, Re sin cosh f x y and Im cos sinh f x y are

continuous functions at 0 0( , ). x y Therefore, ( ) sin f z z is continuous at 0 . z Similarly, we can show that cos z is continuous at any 0 . z

Theorem 3.3.2 Let 1 and 2 be open subsets of containing ,0 z and let .c

If both functions 1: f and 2: g are continuous at ,0 z then

,cf , g f , g f , g f ,Re f ,Im f , f and , f

are continuous at 0 z provided that 0( ) 0 g z in the case of . g f

Example Show that 2 3 3( ) sin 5 z z z z is continuous for all . z

Solution Clearly 2( ) f z z , ( ) sin g z z and 3( )h z z are continuous for all . z

Thus, 2( )( ( )) 5 ( ) f z g z h z 2 3 3sin 5 z z z is continous for all . z

Theorem 3.3.3 Let 1 be an open set containing ,0 z and let 2 be a set . If

1: f is continuous at 0 z , and if 2: g is continuous at )( 00 z f w with

1 2( ) , f then ( )( ) g f z is continuous at 0 ; z i.e., the composition of two

continuous functions is continuous function.

Example Show that 3( ) sin(5 2) z z is continuous for all . z

Solution Clearly, 3( ) 5 2 f z z and ( ) sin g z z are continuous for all z , and

( ) . g f D Thus, 3( ) ( )( ) sin(5 2) z g f z z is continous for all . z

As a side remark, if a function that is not continuous at a given point 0 z is said to

be discontinuous at 0 , z that means, either ( ) f z is not well defined at the point 0 z z ,

or limit does not exist at 0 , z or the limit exists at 0 z but is different from 0( ) f z , or the

limit exists at 0 z but ( ) f z is not well defined at the point 0 . z z For the third and

final case, it is called a removable discontinuity, because if we simply redefine thefunction at 0 z to equal its limit then we will have a continuous function there. If the

discontinuity at a point cannot be removed, then it is called a nonremovable

discontinuity.

Example

Consider the complex function 2 1( ) , z i

z f z \ , . z i i It is easy to verify that

2

1( )( ) 21

lim ( ) lim lim lim . z i z i i z i z i z i z z i z i z i z i

f z




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However, we see that ( ) f z is discontinuous at .i That is, the limit lim ( ) z i

f z exists at

the point of discontinuity . z i We can remove the discontinuity of ( ) f z and make

it continuous at z i by redefining

2 1

2

\ , ,( )

.

z i

z

i

z i i f z

z i

However, we observe that 1lim ( ) lim . z i z i z i

f z Clearly, the point of discontinuity

z i cannot not be removed. Hence, the function ( ) f z has a removable

discontinuity point at , z i and a nonremovable discontinuity point at . z i

For more examples, see Chapter 7: Complex Variables, Section 7.3.4, Engineering


4. Analytic Functions

4.1 Differentiability and Complex Derivatives

Definition 4.1.1 Let be an open subset of containing .0 z We say that a

complex function : f is differentiable at point 0 z if the limit

0

0

0

( ) ( )lim

z z

f z f z

z z (4.1.1)

exists. The value of the limit, denoted by ),(' 0 z f 0

( )df z

z z dz or 0( )df z

dz is called the

derivative of f at 0 , z i.e., 0

0 0

( ) ( )

0'( ) lim . f z f z

z z z z f z

Moreover, a complex function : f is said to be differentiable on

if '( ) f z exists throughout . z If the limit (4.1.1) fails to exist, then ( ) f z is not

differentiable at 0 z and 0'( ) f z is undefined, and so, ( ) f z has no derivative at 0 . z

For the derivative to exist, we are requiring that the limit (in (4.1.1) exists no matter

how we approach 0 z . Let us see some examples.

Example For any , z show that the functions ( ) g z z and 2( ) f z z are

differentiable at , z and find their derivatives.

Solution For any fixed point 0 , z we see that

0 0 0

0 0

0 0

( ) ( )lim lim lim1 1,

z z z z z z

g z g z z z

z z z z

and





0 0 0 0

2 2

0 0 0 00 0

0 0 0

( ) ( ) ( )( )lim lim lim lim 2 .

z z z z z z z z

f z f z z z z z z z z z z

z z z z z z

Thus both limits exist, and since the point 0 z is arbitrary chosen, so we conclude that

'( ) 1 g z and '( ) 2 f z z for all . z

Example Show that the function ( ) f z z is not differentiable at 0. z

Solution Firstly, we consider z approaches the origin along the real axis:

0 ( ,0) (0,0) 0

( ) (0) 0lim lim lim 1.

0 0 z x x

f z f x i x

z x i x

We next consider an approach toward origin along the imaginary axis:

0 (0, ) (0,0) 0

( ) (0) 0

lim lim lim 1.0 0 z y y

f z f iy y

z iy y

Since the limit is unique, we must conclude that limit does not exist. So, ( ) f z z is

not differentiable at 0. z

Theorem 4.1.2 Let 1 and 2 be open subsets of containing a point ,0 z and let

.c Let 1: f and 2: g be complex functions. Then we have the

following properties:

(i) If f is differentiable at 0 z then it is continuous at .0 z

(ii) If f and g are differentiable at 0 z , then their ,cf sum , g f product , g f and quotient , g f where ,0)( 0 z g are also differentiable at ,0 z and

0 0

0 0 0

0 0 0 0 0

0 0 0 0

0 02

0

( ) '( ) '( ),

( ) '( ) '( ) '( ),

( ) '( ) '( ) ( ) ( ) '( ),

'( ) ( ) ( ) '( )( ) , ( ) 0.

( ( ))

cf z cf z

f g z f z g z

f g z f z g z f z g z

f z g z f z g z f z provided g z

g g z

Furthermore, we obtain

1 0 0 0

0

0 02

0

( ) ( ) '( ) 2,

'( )1( ) ( ) 0.

( )

n n f z n f z f z for any positive integer n

f z z provided f z

f f z

Example Show that for any ,n and ,c 1n ndcz dz

cnz for all . z

Solution We give a proof by induction. The case 1n was already shown in

previous example Suppose as an induction hypothesis holds true for 1,n k and

we will prove that it holds for .n k Consider ( ) .k h z z Then we can write it as1( ) k h z z z . Applying the induction hypothesis and Theorem 4.1.2, we have




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1 1 2 1 1'( ) ( ) ' ' ( 1) ,k k k k k h z z z z z k z z z k z

as desired. Again, in view of Theorem 4.1.2, we conclude that 1n nd dz

cz cnz for any

n and any , .c z

Theorem 4.1.3 (The Chain Rule or Composite Rule) Let 1 and 2 be open sets.

Let 1: f and 2: g be complex functions for which 1 2( ) . f If f

is differentiable at 0 1 , z and g is differentiable at ),( 0 z f then the composition

function f g is differentiable at ,0 z and

0 0 0( ) '( ) ' ( ) '( ), g f z g f z f z

or

( ) (( )( )) ( ).

( )

d g f d g f z df z

dz df z dz

Furthermore, a composition of two differentiable functions is again a differentiable

function.

Example Find the derivative 2( ) f z z for any , z and

2( 1)( )

1 3( )

z z i

z i g z for any

\ 1 3 . z i Hence, or otherwise, find the derivative of 2 2

( 1)( )

1 3

z z i

z ifor all

\ 1 3 . z i

Solution We see that '( ) 2 f z z for any , z and by Theorem 4.1.2, we have

2 2

2

( 1 3 )(( ) ( 1)2( )) ( 1)( )'( )

( 1 3 )

z i z i z z i z z i g z

z i

3 2

2

2 (4 7 ) (14 2 ) (6 5 )

( 1 3 )

z i z i z i

z i

for all \ 1 3 . z i Let ( )h z 2 2

( 1)( )

1 3

z z i

z i for all \ 1 3 . z i It is easy to

verify that ( ) ( )( )h z f g z for all \ 1 3 . z i By Thereom 4.1.3, we have

2 3 2

2

( 1)( ) 2 (4 7 ) (14 2 ) (6 5 )

1 3 ( 1 3 )'( ) ( ) '( ) ' ( ) g '( ) 2

z z i z i z i z i

z i z ih z f g z f g z z for all \ 1 3 . z i

Theorem 4.1.4 ( Rule pital o H L ˆ' ) Let ( ) f z and ( ) g z be complex functions. If )( z f

and )( z g are differentiable at 0 z with 0 0( ) ( ) 0, f z g z but ,0)(' 0 z g then

0

0

0

'( )( )lim .

( ) '( ) z z

f z f z

g z g z





Example Find the limit 2 7

6

( 1)

1lim .

z

z i z

Solution It is clear that 2 7( ) ( 1) f z z and 6( ) 1 g z z are differentiable at , z i

and ( ) ( ) 0. f i g i Then, by using the ˆ' L Hopital rule, we see that

2 7 2 6

6 5

( 1) 7( 1) 2

1 6lim lim 0. z z z

z i z i z z

For more details and examples about ˆ' L Hopital rule, see Chapter 7: Complex

Variables, Section 7.4.1, Engineering Mathematics Volume 2 (second edition),

Pearson Prentice Hall, Malaysia, 2006.

4.2 Analytic Functions

Definition 4.2.1 Let be an open subset of containing .0 z We say that a

complex function : f is said to be analytic on if f is differentiable for all

z (i.e., '( ) f z exists for all z ), and '( ) f z is continuous for all . z We

say that f is analytic at 0 z if f is analytic on some open set containing the

point 0 . z z A complex function that is analytic throughout the entire complex plane

is called an entire function.

It is important to note that while differentiability is defined at a specific point,

analyticity is defined on an open set. Even when we say “analytic at a point”, we

actually mean analytic on some open set containing this point. Therefore, if a

complex function f is not differentiable at any point on an open set , or is only

differentiable at a set of finite points on , then f is nowhere analytic (not analytic

at any point) on .

Example Show that the function ( ) f z z is nowhere analytic.

Solution Let 0 0 0 z x iy be an arbitrary fixed complex point. We first approach 0 z

by the direction 0 0 0 0( , ) ( , ) x y h x y as 0.h Then

0 0 0 0 0

0 0 0 0 0

( , ) ( , ) 00 0 0 0 0

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

( ) ( ) z z x y h x y h

f z f z x i y h x iy ih

z z x i y h x iy ih

Next, we approach 0 z by the direction 0 0 0 0( , ) ( , ) x h y x y as 0.h Thus

0 0 0 0 0

0 0 0 0 0

( , ) ( , ) 00 0 0 0 0

( ) ( ) ( )lim lim lim 1.

( ) z z x h y x y h

f z f z x h iy x iy h

z z x h iy x iy h

Since the limit is unique, we must conclude that limit does not exist. So, ( ) f z z is

not differentiable at the arbitrary point 0 . z z Hence, ( ) f z z is nowhere analytic.

Theorem 4.2.3 Finite sums, finite products and finite linear combinations of analytic

functions on an open subset of , are all analytic on . If g f and are analytic

on , then the quotient g f , is analytic on except for those points in for

which g vanishes.




201116

Theorem 4.2.4 (Chain Rule or Composite Rule). Let 1 and 2 be open sets. Let

1: f and 2: g be complex functions for which 1 2( ) . f If f is

analytic at 0 1 , z and g is analytic at ),( 0 z f then the composition function f g

is analytic at ,0 z and 0 0 0( ) '( ) ' ( ) '( ). g f z g f z f z Furthermore, a composition of two analytic functions is again an analytic function.

If a complex function ( ) f z is not analytic at 0 z but is analytic for at least one

point in every neighborhood of ,0 z then 0 z is called a singularity (or a singular point )

of ( ) f z .

For more discussion and examples on analyticity, see Chapter 7: Complex



4.3 The Cauchy-Riemann Equations Theorem 4.3.1 (Cauchy-Riemann Equations) Let ),(),()( y xiv y xu z f be a complex

function defined throughout some open set containing a point 000 iy x z .

Suppose that u and v and the first-order partial derivatives of the functions u and v

with respect to x and ; y i.e., xu , yu , xv and yv exist and are continuous throughout

some neighborhood of 0 z that is wholly contained in . Then ( ) f z is differentiable

at 0 z if and only if those partial derivatives satisfy the Caushy-Riemann equations

yv

xu and

xv

yu

at 0 . z Moreover , the derivative )(' 0 z f exists, and 0'( ) , x x y y f z u i v v i u where

these partial derivatives are to be evaluated at ).,( 00 y x Corollary 4.3.2 Let be an open subset of . Suppose that : , f u iv

and the first-order partial derivatives of u and v with respective to x and y exist and

are continuous on . Then f is analytic on if and only if f satisfies the Cauchy-

Riemann equations on .

Example Show that ( ) z f z e is entire and '( ) z f z e for all . z Hence, or

otherwise, concldue that ( ) ( )'( ) f z f z d dz

e f z e whenever ( ) f z is analytic.

Solution Firstly, we have ( ) cos ( sin ) z x x f z e e y i e y , . z x iy Thus, we

have ( , ) cos xu x y e y and ( , ) sin . xv x y e y Differentiating u and v with respect to

x and y , and we have cos , x

xu e y sin , x

yu e y sin x

xv e y and cos . x

yv e y

Comparing these derivatives, we see clearly that x yu v and . y xu v Hence the

Cauchy-Riemann equations are satisfied at all points. Also, it is easy to see that the

continuity of the partial derivatives holds valid, and thus, ( ) z f z e is analytic at all

points. Also, we have '( ) cos sin x x z f z e y ie y e for all . z By combining





the chain rule with the obtained result, we have ( ) ( )'( ) f z f z d dz

e f z e whenever ( ) f z is

analytic.

For more interesting examples on the Cauchy-Riemann equations, see Chapter 7:

Complex Variables, Section 7.4.3, Engineering Mathematics Volume 2 (secondedition), Pearson Prentice Hall, Malaysia, 2006.

4.4 Differentiation Formulas of Elementary Functions

(a) ,0cdz

d where c is a complex constant.

(b) 1.d

z dz

(c) ,1nn nz z dz d .n

(d) ,1nn nz z dz

d ,0 z and n is a negative integer.

(e) . z z eedz

d

(f) .cossin z z dz

d

(g) .sincos z z dz

d

(h) .sectan 2 z z dz

d

(i) .csccot 2 z z dz

d

(j) .tansecsec z z z dz

d

(k) .cotcsccsc z z z dz

d

(l) .coshsinh z z

dz

d

(m) .sinhcosh z z dz

d

(n) .sechtanh 2 z z

dz

d

(o) .cschcoth 2 z z

dz

d

(p) .tanhsechsech z z z dz

d

(q) .cothcschcsch z z z dz

d




201118

The logarithm z log , and complex exponent c z and z c are multi-valued. When

principal branches are used, these relations become single-valued complex functions

and analytic.

(a) ,

1

Log z z dz

d

where \ 0 , z arg . z

(b) ,1cc cz z dz

d where \ 0 , z arg . z

(c) ,log cccdz

d z z where c and a value clog is specified.

The inverse complex trigonometric and the inverse complex hyperbolic are

multiple-valued. When principal branches of the square root and logarithmic are

used, all these multiple-valued relations becomes single-valued complex functions

and analytic because they are the composition of analytic functions.

(a) 1

2

1

2

1Sin .(1 )

d z dz z

(b) 1

2

1

2

1Cos .

(1 )

d z

dz z

(c) 1

2

1Tan .

1

d z

dz z

(d) 1

2

1

2

1Sinh .

(1 )

d z

dz z

(e) 1

2

1

2

1Cosh .

( 1)

d z dz z

(f) 1

2

1Tanh .

1

d z

dz z

4.5 Harmonic Functions and Laplace’s Equation

Let be an open subset of . A complex function : f is said to be

harmonic in if, throughout this open set, it has continuous partial derivatives of thefirst and second order and satisfies the partial differential equation

02

2

2

2

y

f

x

f (4.5.1)

on . The equation (4.5.1) is known as Laplace’s equation. The operator

2

2

2

2

y x

is called the Laplace operator , or Laplacian, and sometimes it is denoted by the

symbol 2 . We write





2

2

2

22

y x. (4.5.2)

We now consider an analytic function ).,(),()( y xiv y xu z f Then, by Cauchy-

Riemann equations, we obtain

,2

2

y

v

x x

u

x x

uand .

2

2

x

v

y y

u

y y

u

Since the continuity of the partial derivatives of u and v ensures that

x

v

y y

v

x.

It follows that

.02

2

2

2

2

y

u

x

uu

Likewise,

.02

2

2

22

y

v

x

vv

Hence we have the following important theorem.

Theorem 4.5.1 If a function ),(),()( y xiv y xu z f is analytic in an open subset

of , then its component function ),( y xu and ),( y xv are harmonic in .

Example Show that the following is a harmonic function in the stated region.2 2( , ) 2( )u x y x y on .

Solution It is easy to verify that u is harmonic by direct verification of Laplace’s

equation, that is,

2 2 2 2 2 2 2 2

2 2 2 2

(2( )) (2( )) (4 ) ( 4 )4 4 0

u u x y x y x y

x y x y x y

holds true for all . x iy However, we recognize that 2 2( , ) 2( )u x y x y is the

real part of the entire function function 2 2 2( ) 2 2( ) 4 . f z z x y i xy Thus, by

Theorem 4.5.1, we can also conclude that u is harmonic on .

4.6 Harmonic Conjugates

If two given functions u and v are harmonic in a region and their first order

partial derivatives are continuous and satisfy the Cauchy-Riemann equations

throughout the domain , then v is said to be a harmonic conjugate of .u In other

words, given a harmonic function in a region , a harmonic conjugate of ,u let us

call ,v is a harmonic function in such that f u iv is analytic throughout .

A region is said to be simply connected if is connected and any simpleclosed curve (see Section 5.2) has its interior lying entirely in .




201120

Theorem 4.6.1 A complex function ),(),()( y xiv y xu z f is analytic in a simply

connected region if and only if v is a harmonic conjugate of u on .

Example Show that 2 2( , )u x y x y x is harmonic in the entire plane, and find a

harmoinc conjugate.

Solution Evidently, that u is harmomic in the entire plane follows from 2 xxu and

2. yyu To find a harmonic conjugate ,v we make use of Cauchy-Riemann

equations as follows. Since we want f u iv to be analytic throughout the entire

plane, it follows that v must satisfy the Cauchy-Riemann equations

x yu v and . y xu v

Since 2 1 xu x , it follows that 2 1. yv x To obtain v we integrate both sides of

this equation with respect to . y However, because v is a function of x and y , the

constant if integration may a function of , x thus we yield

2 1 yv dy x dy implies ( , ) (2 1) ( ),v x y x y c x

where ( )c x is a function of . x Plugging this into the equation , y xu v we have

2 2 ( ) ,d y x dx

y u v y c x

whence

( ) 0d dx

c x implies ( )c x ,C

where C is a real constant. Thus ( , ) 2 .v x y xy y C By choosing 0,C ( , )v x y 2 xy y is a harmonic conjugate of ( , ) ,u x y and ( ) ( , ) ( , ) f z u x y iv x y

2 2 (2 ) x y x i xy y C is an entire function.

For more numerical examples on harmonic conjuage, see Chapter 7: Complex



5. Integration in the Complex Plane

5.1 Riemann Integral of Complex Functions of a Real Variable

Given an open subset of , a complex-valued function of a real variable on ,

: , f is written as

( ) ( ) ( ) f t u t iv t for all ,t

where ( )u t and ( )v t are both real-valued functions of a real variable t on . Let

,a b with ,a b and let : , f a b be a complex-valued piecewise





continuous function of a real variable on , .a b The Riemann integral or definite

integral of ( ) ( ) ( ) f t u t iv t over ,a b is defined as

( ) ( ) ( ) .

b b b

a a a

f t dt u t dt i v t dt (5.1.1)

In view of (5.1.1), it is clear that we can treat the definite integral as a complex linear

combination of definite integrals of real functions. An immediately consequence

from the previous definition, we obtain the following simple and useful properties.

Theorem 5.1.1 Suppose that , : , f g a b are complex-valued continuous

functions on , ,a b and . Then we have the following properties.

(i)

( ) ( ) .

b b

a a f t dt f t dt

(ii) ( ) ( ) ( ) ( ) .b b b

a a a

f t g t dt f t dt g t dt

(iii) Re ( ) Re ( ) .b b

a a

f t dt f t dt

(iv) Im ( ) Im ( ) .

b b

a a

f t dt f t dt

(v) ( ) ( ) ( ) ,b c b

a a c

f t dt f t dt f t dt where .a c b

(vi) ( ) ( ) .

b a

a b

f t dt f t dt

(vii) ( ) ( ) .

b b

a a

f t dt f t dt

Example Evaluate2

0( ) , f x dx where

2

(1 ) 0 1,( )

(2 ) 1 2.

i x x f x

i x x

Solution 2 1 2 1 2

2

0 0 1 0 1

1 2

2

0 1

( ) ( ) ( ) (1 ) (2 )

1 14 1 31( ) 2 .

2 3 2 6

f x dx f x dx f x dx i x dx i x dx

i ii i x dx i x dx i

5.2 Curves and Contours in the Complex Plane

In calculus, we may recall that the concept of a curve in the 2-dimensioanl Cartesian

coordinate as being the graph of a function ( ). y f x More generally, we may

represent a curve by a parametric form, which illustrates a curve by expressing x and




201122

y as functions of third real variable .t Generally, a parametric form of a curve

( ), y f x is a representation of the curve by a pair of equations ( ) x x t and

( ), y y t where t ranges over a set of real numbers, usually a parametric closed

interval [ , ]a b . Each value [ , ] ,t a b determines a point ( ) ( ( ), ( )),t x t y t which

traces the curve as t varies. When t varies from a to b the point ( )t traces the

curve in a specific direction, starting from ( )a , the initial point of , and

terminating at ( ),b the terminal point of . As usual, for continuous curve, this

direction is denoted by an arrow on the curve . In the complex plane , as

, z x iy it makes sense to adopt the notation ( ) ( ) ( ).t x t iy t Therefore, it is

natural for us to adopt the parametric form of a curve , which lies in the complex

plane, using the complex notation as

( ) ( ) ( ),t x t iy t [ , ] .t a b

Consequently, we can think of the curve ( ) ( ) ( ),t x t iy t as a complex-valued

function of a real variable .t ( ) y t

( )b

t ( ) x t

a b ( )a

A curve ( ) ( ) ( )t x t iy t is smooth on [ , ]a b if ( )t and '( )t continuous

throughout the entire interval [ , ]a b ; i.e., ( )t and '( )t continuous on ( , )a b and both

limits lim '( )t a

t and lim '( )t b

t exist. By a contour ( )t we mean a continuous

piecewise smooth curve ( )t over a parametric closed interval [ , ] .a b More precisely,

a contour is the curve consisting of a finite number of smooth curve joined end to

end; i.e., there exists a finite subdivision, 0 1 1i i na t t t t t b of the

parametric closed interval [ , ]a b such that ( )t is smooth on 1,i it t for all1,2, , .i n A contour is called simple if )()( 21 t t whenever 1 2t t . The

contour on [ , ]a b is closed if the initial point is joined to the terminal point; i.e.,

( ) ( ).a b Furthermore, if the contour is simple except for the initial and

terminal points of , we say that is a simple closed contour .





Given a curve on , ,a b for the sake of simplicity, we shall denote by R the

range of ; i.e., ( ) [ , ] R t t a b . If is a simple closed contour, we use Int

and Ext to denote the set of all interior points contained inside the closed contour

, and the set of all exterior points contained outside , respectively.

Ext

Int

Example

(i) Consider the contour 1 ( ) ,it t e 0 2 .t Then 1 is the positively oriented

unit circle, and hence, it is a simple closed contour.

(ii) Consider the contour 2 0( ) ,it t z re .a t b Then 2 is the positively

oriented contour with its initial point is 2 ( )a and its terminal point 2 ( ).b

(iii) Let 1 z and 2 z be two arbitrary complex numbers. A directed line segment from

1 z to 2 z on the complex plane is the contour 3 over [0,1] defined by

3 1 2( ) (1 )t t z tz [0,1].t

Clearly, its initial point is 3 1(0) z and its terminal point is 3 2(1) . z

For more detailed explanation on coutours, see Chapter 7: Complex Variables,

Section 7.5.2, Engineering Mathematics Volume 2 (second edition), Pearson Prentice

Hall, Malaysia, 2006.

5.3 Contour Integrals

We study now the kind of integral encountered most often with complex functions of

a complex variable; i.e., the contour integral . Such an integral is defined in terms of

the values ( ) f z along a given contour , extending from a point 1 z z to a point

2 z z in the complex plane. Thus, its value depends, in general, on the contour as

well as on the complex function . f It is written

( ) f z dz or 2

1

( )

z

z

f z dz

Let be a contour on a parametric closed interval [ , ] ;a b i.e., ( ) ( ) ( ),t x t iy t

is a continuous piecewise smooth complex function of the real variable t on [ , ] .a b




201124

As t advances from t a to t b in the interval [ , ]a b , the locus of )(t in the

complex plane is the piecewise smooth curve connecting a point ( ) ( ( ), ( ))a x a y a

to a point ( ) ( ( ), ( ))b x b y b in the complex plane. If we differentiate this equation

with respect to t , then we get

'( ) '( ) '( ).t x t iy t

Let ( ) f z be a piecewise continuous complex function define on the graph of (or on

). R The line integral or contour integral of f along the contour is defined by

( ) ( ( )) '( )b

a

f z dz f t t dt . (5.3.1)

Note that since is a contour, '( )t is also piecewise continuous on the

parametric closed interval [ , ] ,a b it follows that the existence of integral (5.3.1) is

ensured. In fact, the contour integral becomes a Riemann integral of a piecewisecontinuous complex-valued function of a real variable over [ , ] .a b

Example Evaluate the contour following integral. , z e dz where is the polygonal

line from 3 z i to 0 z , and then from 0 z to 2. z

Solution

Let 1 and 2 be the directed line segments from 3 z i to 0 z , and from 0 z to

2 z respectively. Then we have 1( ) (1 )( 3 ) 0 3( 1) ,t t i t t i 0 1,t and

2 ( ) (2 )0 ( 1)2 2 2,t t t t 1 2.t Then

1 2

3( 1) 2 2

0 1

1 1 2

2 2

0 0 1

2

(3 ) (2)

3 cos(3 3) sin(3 3) 2

cos3 sin3.

z t i t

t

e dz e i dt e dt

i t dt i t dt e dt

e i

Example Let C be the positively oriented unit circle with center at 0 . z Show that

0

1 2 . z z C

dz i

Solution Parametrize C by 0( ) ,it t z e 0 2 .t Then

0 0 0

2 2 2

1 1 1

0 0 0

( ) ( ) 2 .it it

it it

z z z e z eC dz ie dt ie dt i dt i

Theorem 5.3.1 Let be such that ),(t [ , ] ,t a b is a contour and let f and g be

piecewise continuous complex functions on an open subset of containing [ , ]a b

and let and be complex constants. Then we have the following properties.

(i) ( ) ( ) ( ) ( ) . f z g z dz f z dz g z dz





(ii) If :[ , ]c d C is a contour such that ),()( cb then

( ) ( ) ( ) , f z dz f z dz f z dz

where is the sum of the contours and . (iii) Define a function

1 : ,a b such as )1()(1 t t for all .bt a We

get 1 is a contour , with )()(1 t t and is called the reversed contour of

. Then

1

( ) ( ) ( ) . f z dz f z dz f z dz

(iv) Re ( ) Re ( ) , f z dz f z dz

(v) Im ( ) Im ( ) , f z dz f z dz

(vi) ( ) ( ) . f z dz f z dz

Example Let C be the positively unit circle. Compute

(i) ,C z dz

(ii)C z dz and

(iii) .C x dz

Solution Parametrize C by ( ) ,it t e 0 2 .t By (i), we see that

2 2

0 0

( ) cos 2 sin 2 0.it it

C z dz e ie dt i t i t dt

Then, by (ii), we have 2 2 2

0 0 0

( ) ( ) 2 .it it it it

C z dz e ie dt i e ie dt i dt i

We next solve the part (iii). Firstly, we see that 12( ), x z z then

1 1 1 1 12 2 2 2 2(0) (2 ) .C C C C x dz z z dz z dz z dz i i

We say that 1 ( ),t ,a t b and 2 ( ),t ,c t d are equivalent

parametrizations if there exists an increasing continuously differentiable function

( )t from [ , ]c d onto [ , ]a b such that ( )c a and ( )d b and 2 1( ) ( )( )t t

for all [ , ].t c d

Theorem 5.3.2 Let 1 ( ),t ,a t b and 2 ( ),t ,c t d are equivalent

parametrizations of same path and let ( ) f z be a piecewise continuous complex

function on . Then2 1

( ) ( ) . f z dz f z dz




201126

5.4 Antiderivatives and The Principle of Path Independence

In general, the value of a contour integral of a complex function ( ) f z from a fixed

point 1 z to a fixed point 2 z depends on the path or contour that is taken. However, in

this section, we will study one of the most important results in the theory of complex

functions, which extends the Fundamental Theorem of Calculus to contour integrals.

It implies that in certain situations the contour integral of a complex function is

independent of the particular path joining the initial and terminal points; in fact, it also

completely characterizes the conditions under which this property holds. Definitely,

with existence of such elegant result, it can simplify the evaluation of many contour

integrals.

In developing the notion, we need the concept of the antiderivative of a complex

function ; f Indeed, the idea is completely analogues to that for a real function. A

complex function ( ) F z is said to be an antiderivative of a continuous complex

function ( ) f z in a region if

( )( )

dF z f z

dz for all z .

Note also that an antiderivative is, of necessity, an analytic function, and an

antiderivative of a given function f is unique except for an additive complex constant.

Example

(i) An antiderivative of 3( ) 7 16 f z z z is24 7

4 2( ) 16 , z z F z z and the identity

( ) ( )d dz

F z f z holds true for all . z

(ii) The function ( ) sin f z z is continuous for all . z An antiderivative is

( ) cos , F z z and the identity ( ) ( )d dz

F z f z holds true for all . z

A contour integral of f along the contour is said to be independent of path in

a region if for any pair of complex numbers 1 z and 2 z in , the value of the

contour integral is the same for all contours in joining from the two end points

1 z to 2 z . We now have the first result.

Theorem 5.4.1 The contour integral of a continuous complex function f is

independent of path in a region if and only if f has an antiderivative in (i.e.,

)()(' z f z F for some analytic function on ).

2 z

1 z





Theorem 5.4.2 ( Fundamental Theorem of Calculus of Contour Integral ). Suppose

that the complex function )( z f is continuous in a region , and it has an

antiderivative )( z F throughout (i.e., )()(' z f z F for some analytic function on

). Then the contour integral of for any contour lying in with initial point 1 z

and terminal point 2 z is

).()()( 12

z F z F dz z f (5.4.1)

Example Evaluate z e dz where is the semi- circle ( ) ,it t e 0 .t

Solution The function ( ) z f z e is continuous in the entire complex plane , with

an antiderivative . z e The initial point of is (0) 1 z and its terminal point is

( ) 1. z By Theorems 5.4.1 and 5.4.2, we conclude that

1 1 11 2sinh(1). z z z z e dz e e e

Another important conclusion that can be drawn from Theorems 5.4.1 and 5.4.2

is that when a complex function )( z f has an antiderivative, its contour integral

depends only on the endpoints; i.e., the contour integral is independent of the contour

joining these two endpoints. This property is called the Principle of Path

Independence. Next, we give an interesting result on closed contour integrals.

Theorem 5.4.3 Let )( z f be continuous in a region . Then the following are

equivalent .

(i) )( z f has an antiderivative in .

(ii) The contour integrals of f are independent of path in ; i.e., if 1 and 2 are

any two contours in sharing the same initial and terminal points, then

.)()(21

dz z f dz z f

(iii) Every closed contour integral of f in vanishes; i.e.,

0)(

dz z f for any closed contour in .

For more examples, see Chapter 7: Complex Variables, Section 7.5.2,Engineering Mathematics Volume 2 (second edition), Pearson Prentice Hall,

Malaysia, 2006.

5.5 Continuous Deformation of Contours and Simply

Connected Regions

In this section we will briefly introduce some homotopically equivalent relation of

contours in a region of the complex plane. We start by describing three basic

geometric illustrations to describe the fundamental concept of deformation of contours.




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Suppose that 0 z and 1 z are two fixed points in a region , and that 1 and 2

are contours joining 0 z to 1 z . We say that 1 is continuously deformable into 2

relative to if we can continuously move 1 over 2 while keeping the two end

points fixed at 0 z to 1 z without leaving .

Suppose that 1 and 2 are two closed contours in a region . We say that 1 is

continuously deformable into 2 relative to if we can continuously move 1 without leaving , in such a way that 1 overlaps with 2 in direction and position.

If is a closed contour in a region , we say that is continuously deformable

into a point 0 z relative to if we can continuously shrink into the point 0 z in

without leaving . Indeed, this is the same situation as the continuously deformable

of the closed contour 1 into another closed contour 2 , where, in this case, 2 is

degenerated into a point; i.e., 2 0( )t z for all .t

Two contours 1 and 2 are said to be mutually deformable in a region if 1 is

continuously deformable into 2 relative to , and also, 2 is continuously

deformable into 1 relative to .

Definition 5.5.1 A region is called simply connected if all contours joining any

two fixed points 0 z and 1 z in are mutually continuously deformable relative to .

If a region is not simply connected, then it is called multiply connected .





5.6 Cauchy’s Theorem and The Principle of Deformation of

Paths

Theorem 5.6.1 (Cauchy’s Theorem) Let be a simple closed contour . If a complex

function ( ) f z is analytic at all points in and on the contour (i.e., Int z R ),

then .0)(

dz z f

R

Int

Theorem 5.6.2 If a complex function f is analytic throughout a simply connected

region , then ( ) 0 f z dz for all closed contours lying inside .

Example Evaluate 2 9

z e

z dz where is the positive oriented circle centered at origin

with radius 2.

Solution Singularities of 2 9

z e

z occur at 3 z and 3. z The integrand is therefore

analytic inside and on . By Cauchy-Goursat ’s theorem, we conclude that the value

of the contour integral is zero.

Corollary 5.6.3 A complex function f which is analytic throughout a simply

connected region must have an antiderivative in .

Since the complex plane is a simply connected region, it follows from

Corollary 5.6.3 that all entire complex functions always possess antiderivatives.

Theorem 5.6.4 ( Principle of Deformation of Paths) Let 1 and 2 denote positively

(counterclockwise) oriented simple closed contours with 2 is interior to 1 . If a

complex function )( z f is analytic in the closed region consisting of those contours

and all points between them (except for points interior any2

), then

.)()(21

dz z f dz z f

1 2




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Theorem 5.6.5 Let and i , ,,,2,1 ni be the simple closed contours such that

(i) is a simple closed contour, described in the counterclockwise direction;

(ii) i , 1, , ,i n are simple closed contours, all described in the counterclockwise

directions, that are interior to and whose interiors have no points in common. If a function ( ) f z is analytic throughout the closed region consisting of all

points within and on except for points interior any i , then

1

( ) ( )i

n

i

f z dz f z dz .

Example Let be the positively oriented simple closed contour , and let 0 z be a

point not on . Show that

0

00

0 ,1

2 .

if z is in the exterior of dz

i if z is in the interior of z z

Solution If 0 z is the exterior of , then the function 0( ) 1 ( ) f z z z is analytic on

and inside , and hence, by Cauchy’s theorem, its integral along is zero. For the

second case, where 0 z is the interior of , then we cannot claim that the function

0( ) 1 ( ) f z z z is analytic on and inside . So, we cannot apply Cauchy’s theorem,

however by previous examples, we conclude that

0

12 .dz i

z z

5.7 Cauchy’s Integral Formula and Derivatives of Analytic

Functions

Theorem 5.7.1 (Cauchy’s Integral Formula) Let ( ) f z be analytic throughout within

and on a simple closed positively oriented contour . If 0 z is any point interior to ,

then

.)(

2

1)(

0

0 dz z z

z f

i z f (5.7.1)

Note that formula (5.7.1) tells us that if a complex function ( ) f z is to be analytic

within and on a simple closed contour , then its values everywhere interior to a




simple closed contour are completely determined by the values of ( ) f z on .

When the Cauchy’s integral formula is written

0 0

( )2 ( ),

f z dz if z

z z

(5.7.2)

it can be used to evaluate certain integrals along simple closed contours.

Example Let be the positively oriented circle 3 z . Evaluate 2

1

4.

z dz

Solution We have 1 and 2 are positively oriented contours defined as1

1 4( ) 2 it t e and 1

2 4( ) 2

it t e , 0 2 .t Then it is clear that the function

2

1

4( )

z f z is analytic throughout all points on , 1 and 2 , and their interiors.

Then we have

1 2 1 2

1 12 2

2 2 2

( ) ( )1 1 10.

4 4 4 ( 2) 2

z z dz dz dz dz dz z z z z z

Theorem 5.7.2 (Generalized Cauchy’s Integral Formula). If a function )( z f is

analytic at a point (or , within a region ), then it possesses derivatives of all orders

at that point (or , in that region ). These derivatives are themselves analytic

functions at that point (or , in that region ). If )( z f is analytic on and in the

interior of a positively oriented simple closed contour and it 0 z is inside , then

,)()(

2!)(

1

0

0

)( dz z z

z f i

n z f n

n 0,1,2, .n (5.7.3)

Moreover , this also provides the useful formula

( )

0

1 0

2 ( )( ),

!( )

n

n

i f z f z dz

n z z 0,1,2, .n (5.7.4)

Example Compute the following integral 3

,( )

iz edz

z where is the positively

oriented circle 4 z .

Solution By (5.7.4), we have

2

3 2

2( ) .

2!( )

iz iz i

z

e i d dz e ie i

z dz

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