chapter 4. integrals weiqi luo ( 骆伟祺 ) school of software sun yat-sen university email ：...

Chapter 4. Integrals

Weiqi Luo (骆伟祺 )School of Software

Sun Yat-Sen UniversityEmail ： [email protected] Office ： # A313

mailto:[email protected]

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Derivatives of Functions w(t) Definite Integrals of Functions w(t) Contours; Contour Integrals; Some Examples; Example with Branch Cuts Upper Bounds for Moduli of Contour Integrals Anti derivatives; Proof of the Theorem Cauchy-Goursat Theorem; Proof of the Theorem Simply Connected Domains; Multiple Connected Domains; Cauchy Integral Formula; An Extension of the Cauchy Integral Formula; Some

Consequences of the Extension Liouville’s Theorem and the Fundamental Theorem Maximum Modulus Principle

2

Chapter 4: Integrals

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Consider derivatives of complex-valued functions w of real variable t

where the function u and v are real-valued functions of t. The derivative

of the function w(t) at a point t is defined as

37. Derivatives of Functions w(t)

3

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Properties For any complex constant z0=x0+iy0,


4

u(t) v(t)

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Properties


5

where z0=x0+iy0. We write

0 0 0 0 0( )0 0cos sinz t x iy t x t x te e e y t ie y t

u(t) v(t)

0 0 00 0( cos ) ' ( sin ) 'z t x t x td

e e y t i e y tdt

Similar rules from calculus and some simple algebra then lead us to the expression

0 0 0 0( )0 0 0( )z t x iy t z td

e x iy e z edt

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Example Suppose that the real function f(t) is continuous on an

interval a≤ t ≤b, if f’(t) exists when a<t<b, the mean value theorem for derivatives tells us that there is a number ζ in the interval a<ζ<b such that


6

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Example (Cont’)

The mean value theorem no longer applies for some complex functions. For instance, the function

on the interval 0 ≤ t ≤ 2π .

Please note that

And this means that the derivative w’(t) is never zero, while


7

(2 ) (0) 0w w (2 ) (0)

'( ) 0, 0 22 0

w ww

（，）

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When w(t) is a complex-valued function of a real variable t and is written

where u and v are real-valued, the definite integral of w(t) over an interval a ≤ t ≤ b is defined as

38. Definite Integrals of Functions w(t)

8

( ) ( ) ( )b b b

a a a

w t dt u t dt i v t dt Provided the individual integrals on the right exist.

Re ( ) Re ( ) & Im ( ) Im ( )b b b b

a a a a

w t dt w t dt w t dt w t dt

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Example 1


9

1 1 12 2

0 0 0

2(1 ) (1 ) 2

3it dt t dt i tdt i

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Properties The existence of the integrals of u and v is ensured if

those functions are piecewise continuous on the interval a ≤ t ≤ b. For instance,


10

( ) ( ) ( )b c b

a a c

w t dt w t dt w t dt

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Integral vs. Anti-derivative Suppose that

are continuous on the interval a ≤ t ≤ b.

If W’(t)=w(t) when a ≤ t ≤ b, then U’(t)=u(t) and V’(t)=v(t). Hence, in view of definition of the integrals of function


11

( ) ( ) ( ), ( ) ( ) ( )w t u t iv t W t U t iV t

( ) ( ) ( )b b b

a a a

w t dt u t dt i v t dt ( ) ( ) [ ( ) ( )] [ ( ) ( )]b b

a aU t iV t U b iV b U a iV a

( ) ( ) ( ) baW b W a W t

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Example 2

Since

one can see that


12

1 1( )

itit it itd e d

e ie edt i i dt i

440

0

itit e

e dti

4 1

ie

i i

1 1(1 )

2 2i

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Example 3 Let w(t) be a continuous complex-valued function of t defined on an

interval a ≤ t ≤ b. In order to show that it is not necessarily true that there is a number c in the interval a <t< b such that

We write a=0, b=2π and use the same function w(t)=eit (0 ≤ t ≤ 2π) as the Example in the previous Section (pp.118). We then have that

However, for any number c such that 0 < c < 2π

And this means that w(c)(b-a) is not zero.


13

( ) ( )( )b

a

w t dt w c b a

2 220

0 0

( ) 0it

it ew t dt e dt

i

| ( )( ) | | | 2 2icw c b a e

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pp. 121

Ex. 1, Ex. 2, Ex. 4

38. Homework

14

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Arc A set of points z=(x, y) in the complex plane is said to be an

arc if

where x(t) and y(t) are continuous functions of the real parameter t. This definition establishes a continuous mapping of the interval a ≤ t ≤ b in to the xy, or z, plane.

And the image points are ordered according to increasing values of t. It is convenient to describe the points of C by means of the equation

39. Contours

15

( ) & ( ),x x t y y t a t b

( ) ( ) ( )z t x t iy t

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Simple arc (Jordan arc)

The arc C: z(t)=x(t)+iy(t) is a simple arc, if it does not cross itself; that is, C is simple if z(t1)≠z(t2) when t1≠t2

Simple closed curve (Jordan curve)

When the arc C is simple except for the fact that z(b)=z(a), we say that C is simple closed curve.

Define that such a curve is positively oriented when it is in the counterclockwise direction.

39. Contours

16

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Example 1 The polygonal line defined by means of the equations

and consisting of a line segment from 0 to 1+i followed by one from 1+i to 2+i is a simple arc

39. Contours

17

, 0 1

, 1 2

x ix if xz

x i if x

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Example 2~4 The unit circle about the origin is a simple closed curve, oriented in the

counterclockwise direction.

So is the circle

centered at the point z0 and with radius R.

The set of points This unit circle is traveled in the clockwise direction. The set of point

This unit circle is traversed twice in the counterclockwise direction.

39. Contours

18

, (0 2 )iz e

0 , (0 2 )iz z e

, (0 2 )iz e

2 , (0 2 )iz e

Note: the same set of points can make up different arcs.

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The parametric representation used for any given arc C is not unique

To be specific, suppose that

where Φ is a real-valued function mapping an interval α ≤ τ ≤ β onto a ≤ t ≤ b.

39. Contours

19

( ),t

Here we assume Φ is a continuous functions with a continuous derivative, and Φ’(τ)>0 for each τ (why?)

: ( ) ( ( )),C Z z

: ( ),C z t a t b

The same arc C

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Differentiable arc Suppose the arc function is z(t)=x(t)+iy(t), and the components

x’(t) and y’(t) of the derivative z(t) are continuous on the entire interval a ≤ t ≤ b.

Then the arc is called a differentiable arc, and the real-valued function

is integrable over the interval a ≤ t ≤ b.

In fact, according to the definition of a length in calculus, the length of C is the number

39. Contours

20

2 2| '( ) | [ '( )] [ '( )]z t x t y t

| '( ) |b

a C

L z t dt ds Note: The value L is invariant under certain changes in the representation for C.

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Smooth arc A smooth arc z=z(t) (a ≤ t ≤ b), then it means that the derivative z’(t)

is continuous on the closed interval a ≤ t ≤ b and nonzero throughout the open interval a < t < b.

A Piecewise smooth arc (Contour) Contour is an arc consisting of a finite number of smooth arcs joined

end to end. (e.g. Fig. 36)

Simple closed contour When only the initial and final values of z(t) are the same, a contour

C is called a simple closed contour. (e.g. the unit circle in Ex. 5 and 6)

39. Contours

21

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Jordan Curve Theorem

39. Contours

22

Refer to: http://en.wikipedia.org/wiki/Jordan_curve_theorem

Jordan Curve Theorem asserts that every Jordan curve divides the plane into an "interior" region bounded by the curve and an "exterior" region containing all of the nearby and far away exterior points.

Interior of C (bounded)

Jordan curve

Exterior of C (unbounded)

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pp. 125-126

Ex. 1, Ex. 3, Ex.4

39. Homework

23

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Consider the integrals of complex-valued function f of the complex variable z on a given contour C, extending from a point z=z1 to a point z=z2 in the complex plane.

40. Contour Integrals

24

( )C

f z dzor 2

1

( )z

z

f z dz

When the value of the integral is independent of the choice of the contour taken between two fixed end points.

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Contour Integrals Suppose that the equation z=z(t) (a ≤ t ≤b) represents a

contour C, extending from a point z1=z(a) to a point z2=z(b). We assume that f(z(t)) is piecewise continuous on the interval a ≤ t ≤b, then define the contour integral of f along C in terms of the parameter t as follows


25

( ) ( ( )) '( )b

C a

f z dz f z t z t dt Contour integral

Note the value of a contour integral is invariant under a change in the representation of its contour C.

On the integral [a b] as defined previously

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Properties


26

0 0( ) ( )C C

z f z dz z f z dz

[ ( ) ( )] ( ) g( )C C C

f z g z dz f z dz z dz

-

( ) ( )C C

f z dz f z dz

Note that the value of the contour integralsdepends on the directions of the contour

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Properties


27

1 2

( ) ( ) f ( )C C C

f z dz f z dz z dz

The contour C is called the sum of its legs C1 and C2 and is denoted by C1+C2

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Example 1 Let us find the value of the integral

when C is the right-hand half

41. Some Examples

28

C

I zdz

2 , ( )2 2

iz e

/2 /2

/2 /2

( ( )) '( ) 2 (2 ) 'i iI f z z d e e d

4 i

C

4zdz iC

4zz

dz iz

C

44dz i

z

C

dzi

z

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Example 2 C1 denotes the polygnal line OAB, calculate the integral

41. Some Examples

29

1

1 ( )C

I f z dz ( ) ( )OA AB

f z dz f z dz

Where2( ) 3 , ( )f z y x i x z x iy

The leg OA may be represented parametrically as z=0+iy, 0≤y ≤1

1

0

( )2OA

if z dz yidy

In this case, f(z)=yi, then we have

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Example 2 (Cont’)

41. Some Examples

30

1

1 ( )C

I f z dz ( ) ( )OA AB

f z dz f z dz

Similarly, the leg AB may be represented parametrically as z=x+i, 0≤x ≤1

In this case, f(z)=1-x-i3x2, then we have

12

0

1( ) (1 3 ) 1

2AB

f z dz x i x dx i

Therefore, we get

1 1( )

2 2 2

i ii

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Example 2 (Cont’)

C2 denotes the polygonal line OB of the line y=x, with parametric representation z=x+ix (0≤ x ≤1)

41. Some Examples

31

2

12

2

0

( ) 3 (1 ) 1C

I f z dz i x i dx i

1 2

1 2

OABO -C

-1+i( ) ( ) =I -I =

2C

f z dz f z dz

A nonzero value

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Example 3 We begin here by letting C denote an arbitrary smooth arc

from a fixed point z1 to a fixed point z2. In order to calculate the integral

41. Some Examples

32

( ), ( )z z t a t b

( ) '( )b

C a

zdz z t z t dt

Please note that

2[ ( )]( ) '( )

2

d z tz t z t

dt

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Example 3 (Cont’)

41. Some Examples

33

2 22 2 22 1[ ( )] ( ) ( )

( ) '( )2 2 2

bba

C a

z zz t z b z azdz z t z t dt

The value of the integral is only dependent on the two end points z1 and z2

2

1

2 22 1

2

z

C z

z zzdz zdz

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Example 3 (Cont’) When C is a contour that is not necessarily smooth since a

contour consists of a finite number of smooth arcs Ck (k=1,2,…n) jointed end to end. More precisely, suppose that each Ck extend from wk to wk+1, then

41. Some Examples

34

1k

n

kC C

zdz zdz

1 2 2

1

1 1 2

k

k

wn nk k

k kw

w wzdz

2 2 2 21 1 2 1

2 2nw w z z

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Example 1 Let C denote the semicircular path

from the point z=3 to the point z = -3.

Although the branch

of the multiple-valued function z1/2 is not defined at the initial point z=3 of the contour C, the integral

42. Examples with Branch Cuts

35

3 (0 )iz e

1/2 1( ) exp( log ), (| | 0,0 arg 2 )

2f z z z z z

1/2

C

I z dz nevertheless exists.

Why?

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Example 1 (Cont’) Note that


36

( ) 3 iz e

/2 3 3[ ( )] '( ) ( 3 )(3 ) 3 3 sin 3 3 cos

2 2i if z z e ie i 0

At θ=0, the real and imaginary component are 0 and 3 3

Thus f[z(θ)]z’(θ) is continuous on the closed interval 0≤ θ ≤ π when its value at θ=0is defined as 3 3i

1/2 3 /2

0

3 3 i

C

I z dz i e d

3 /2

0

22 3(1 )

3ie i

i

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Example 2 Suppose that C is the positively oriented circle


37

Re ( )iz

-R

Let a denote any nonzero real number. Using the principal branch

1( ) exp[( 1) ]af z z a Logz (| | 0, )z Argz

of the power function za-1, let us evaluate the integral

1a

C

I z dz

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Example 2 (Cont’) when z(θ)=Reiθ, it is easy to see that

where the positive value of Ra is to be taken.

Thus, this function is piecewise continuous on -π ≤ θ ≤ π, the integral exists.

If a is a nonzero integer n, the integral becomes 0 If a is zero, the integral becomes 2πi.


38

[ ( )] '( ) a iaf z z iR e

1 2[ ] sin

ia aa a ia a

C

e RI z dz iR e d iR i a

ia a

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pp. 135-136

Ex. 2, Ex. 5, Ex. 7, Ex. 8, Ex. 10

42. Homework

39

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Lemma If w(t) is a piecewise continuous complex-valued function

defined on an interval a ≤ t ≤b

43. Upper Bounds for Moduli of Contour Integrals

40

| ( ) | | ( ) |b b

a a

w t dt w t dt Proof:

00 ( )

bi

a

r e w t dt

| ( ) | | ( ) |b b

a a

w t dt w t dt holds( ) 0b

a

w t dt Case #1:

00( ) 0

bi

a

w t dt r e Case #2:

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Lemma (Cont’)


41

00 ( )

bi

a

r e w t dtNote that the values in both sizes of this equation are real numbers.

0 00 Re[ ( ) ] Re[ ( )]

b bi i

a a

r e w t dt e w t dt

0 0 0Re[ ( )] | ( ) | | || ( ) | | ( ) |i i ie w t e w t e w t w t

0 | ( ) | | ( ) |b b

a a

r w t dt w t dt Why?

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Theorem Let C denote a contour of length L, and suppose that a

function f(z) is piecewise continuous on C. If M is a nonnegative constant such that

For all point z on C at which f(z) is defined, then


42

| ( ) |f z M

| ( ) |C

f z dz ML

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Theorem (Cont’) Proof: We let z=z(t) (a ≤ t ≤ b) be a parametric

representation of C. According to the lemma, we have


43

| ( ) | | [ ( )] '( ) | | [ ( )] '( ) |b b

C a a

f z dz f z t z t dt f z t z t dt

| [ ( )] '( ) | | '( ) |b b

a a

f z t z t dt M z t dt | ( ) |f z M

| '( ) |b

a

M z t dt ML

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Example 1

Let C be the arc of the circle |z|=2 from z=2 to z=2i that lies in the first quadrant. Show that


44

3

4 6| |

1 7C

zdz

z

| 4 | | | 4 6z z 3 3| 1| || | 1| 7z z

Based on the triangle inequality,

Then, we have

3

4 6| |

1 7

z

z

And since the length of C is L=π, based on the theorem

3

4 6| |

1 7C

zdz

z

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Example 2 Here CR is the semicircular path

and z1/2 denotes the branch (r>0, -π/2<θ<3π/2)

Without calculating the integral, show that


45

Re ,0iz

1/2 /21exp( log )

2iz z re

1/2

2lim 0

1RCR

zdz

z

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Example 2 (Cont’)


46

1/2

2lim 0

1RCR

zdz

z

Note that when |z|=R>1

1/2 /2| | | |iz Re R 2 2 2| 1| || | 1| 1z z R

1/2

2 2| |

1 1R

z RM

z R

Based on the theorem1/2

2 2| | ( )

1 1R

CR

z Rdz M L R

z R

2lim ( ) 0

1R

RR

R

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pp. 140-141

Ex. 3, Ex. 4, Ex. 5

43. Homework

47

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Theorem Suppose that a function f (z) is continuous on a domain D. If any

one of the following statements is true, then so are the others

a) f (z) has an antiderivative F(z) throughout D;

b) the integrals of f (z) along contours lying entirely in D and extending from any fixed point z1 to any fixed point z2 all have the same value, namely

where F(z) is the antiderivative in statement (a);

a) the integrals of f (z) around closed contours lying entirely in D all have value zero.

44. Antiderivatives

48

2

2

1

1

2 1( ) ( ) ( ) ( )z

zz

z

f z dz F z F z F z

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Example 1 The continuous function f (z) = z2 has an antiderivative

F(z) = z3/3 throughout the plane. Hence

For every contour from z=0 to z=1+i

44. Antiderivatives

49

1 32 1

0

0

2( 1 )

3 3

iiz

z dz i

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Example 2

The function f (z) = 1/z2, which is continuous everywhere except at the origin, has an antiderivative F(z) = −1/z in the domain |z| > 0, consisting of the entire plane with the origin deleted. Consequently,

Where C is the positively oriented circle

44. Antiderivatives

50

20

C

dz

z

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Example 3 Let C denote the circle as previously, calculate the integral

It is known that

44. Antiderivatives

51

1

C

I dzz

1(log ) ' , ( 0)z z

z

10

C

I dzz

?

For any given branch

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Example 3 (Cont’) Let C1 denote

The principal branch

44. Antiderivatives

52

2 , ( )2 2

iz e

ln , ( 0, )Logz r i r

22

2

1 2

1 1(2 ) ( 2 )

ii

i

C i

dz dz Logz Log i Log i iz z

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Example 3 (Cont’) Let C2 denote

Consider the function

44. Antiderivatives

53

32 , ( )

2 2iz e

ln , ( 0,0 2 )logz r i r

22

2

2 2

1 1(2 ) ( 2 )

ii

i

C i

dz dz logz log i log i iz z

Why not Logz?

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Example 3 (Cont’) The value of the integral of 1/z around the entire circle

C=C1+C2 is thus obtained:

44. Antiderivatives

54

1 2

2C C C

dz dz dzi i i

z z z 0

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Example 4 Let us use an antiderivative to evaluate the integral

where the integrand is the branch

44. Antiderivatives

55

1/2

1C

z dz

1/2 /21( ) exp( log ) , ( 0,0 2 )

2if z z z re r

Let C1 is any contour from z=-3 to 3 that, except for its end points, lies above the X axis.

Let C2 is any contour from z=-3 to 3 that, except for its end points, lies below the X axis.

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Example 4 (Cont’)

44. Antiderivatives

56

/21

3, ( 0, )

2 2if re r

1/2 /21( ) exp( log ) , ( 0,0 2 )

2if z z z re r

f1 is defined and continuous everywhere on C1

3 /21

2 3( ) , ( 0, )

3 2 2iF z r re r

1/2 31 3

1

( ) 2 3(1 )C

z dz F z i

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Example 4 (Cont’)

44. Antiderivatives

57

/22

5, ( 0, )

2 2if re r

1/2 /21( ) exp( log ) , ( 0,0 2 )

2if z z z re r

f2 is defined and continuous everywhere on C2

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Basic Idea: (a) (b) (c) (a)

1.(a) (b) Suppose that (a) is true, i.e. f(z) has an antiderivative F(z) on the domain D being considered.

If a contour C from z1 to z2 is a smooth are lying in D, with parametric representation z=z(t) (a≤ t≤b), since

45. Proof of the Theorem

58

( ( )) '[ ( )] '( ) ( ( )) '( ), ( )d

F z t F z t z t f z t z t a t bdt

2 1( ) ( ( )) '( ) ( ( )) ( ( )) ( ( )) ( ) ( )b

ba

C a

f z dz f z t z t dt F z t F z b F z a F z F z

Note: C is not necessarily a smooth one, e.g. it may contain finite number of smooth arcs.

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b) (b) (c)

Suppose the integration is independent of paths, we try to show that the value of any integral around a closed contour C in D is zero.


59

1 2

( ) ( )C C

f z dz f z dz

1 2

( ) ( ) ( )C C C

f z dz f z dz f z dz

C=C1-C2 denote any integral around a closed contour C in D

1 2

( ) ( ) 0C C

f z dz f z dz

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c) (c) (a) Suppose that the integrals of f (z) around closed

contours lying entirely in D all have value zero. Then, we can get the integration is independent of path in D (why?)


60

0

( ) ( )z

z

F z f s dsWe create the following function

and try to show that F’(z)=f(z) in D

i.e. (a) holds

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61

0

( ) ( )z

z

F z f s ds

0 0

( ) ( ) ( ) ( ) ( )z z z z z

z z z

F z z F z f s ds f s ds f s ds

z z

z

ds z

Since the integration is independent of path in D, we consider the path of integration in a line segment in the following. Since

1( ) ( )

z z

z

f z f z dsz

( ) ( ) 1( ) [ ( ) ( )]

z z

z

F z z F zf z f s f z ds

z z

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62

Please note that f is continuous at the point z, thus, for each positive number ε, a positive number δ exists such that

| ( ) ( ) |f s f z When | |s z

( ) ( ) 1 1| ( ) | | [ ( ) ( )] | | |

| |

z z

z

F z z F zf z f s f z ds z

z z z

Consequently, if the point z+Δz is close to z so that | Δ z| <δ, then

0

( ) ( )'( ) lim ( )

z

F z z F zF z f z

z

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pp. 149

Ex. 2, Ex. 3, Ex. 4

45. Homework

63

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Cauchy-Goursat TheoremGive other conditions on a function f which ensure that

the value of the integral of f(z) around a simple closed contour is zero.

The theorem is central to the theory of functions of a complex variable, some modification of it, involving certain special types of domains, will be given in Sections 48 and 49.

46. Cauchy-Goursat Theorem

64

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65

Let C be a simple closed contour z=z(t) (a≤t ≤b) in the positive sense, and f is analytic at each point. Based on the definition of the contour integrals

( ) ( ( )) '( )b

C a

f z dz f z t z t dt

And if f(z)=u(x,y)+iv(x,y) and z(t)=x(t) + iy(t)

( ) ( ( )) '( ) ( ' ') ( ' ')b b b

C a a a

f z dz f z t z t dt ux vy dt i vx uy dt

C C

udx vdy i vdx udy

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66

Based on Green’s Theorem, if the two real-valued functions P(x,y) and Q(x,y), togetherwith their first-order partial derivatives, are continuous throughout the closed region R consisting of all points interior to and on the simple closed contour C, then

( )x y

C R

Pdx Qdy Q P dA

( )C C C

f z dz udx vdy i vdx udy

( ) , ( , )x y

R

v u dA P u Q v ( ) , ( , )x y

R

u v dA P v Q u If f(z) is analytic in R and C, then the Cauchy-Riemann equations shows that

,y x x yu v u v Both become zeros

( ) 0C

f z dz

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Example If C is any simple closed contour, in either direction,

then

This is because the composite function f(z)=exp(z3) is analytic everywhere and its derivate f’(z)=3z2exp(z3) is continuous everywhere.


67

3exp( ) 0C

z dz

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Two Requirements described previously The function f is analytic at all points interior to and on

a simple closed contour C, then The derivative f’ is continuous there

Goursat was the first to prove that the condition of continuity on f’ can be omitted.


68

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Cauchy-Goursat Theorem If a function f is analytic at all points interior to and on a

simple closed contour C, then


69

Interior of C (bounded) ( ) 0C

f z dz

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Simple Connected domain A simple connected domain D is a domain such that

every simple closed contour within it encloses only points of D. For instance,

48. Simply Connected Domains

70

The set of points interior to a simple closed contour

Not a simple connected domain

A simple connected domain

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Theorem 1 If a function f is analytic throughout a simply connected

domain D, then

for every closed contour C lying in D.


71

( ) 0C

f z dz

Basic idea: Divide it into finite simple closed contours. For this example,

4

1

( ) ( ) 0k

kC C

f z dz f z dz

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Example If C denotes any closed contour lying in the open disk |

z|<2, then


72

2 50

( 9)

z

C

zedz

z

This is because the disk is a simply connected domain and the two singularities z = ±3i of the integrand are exterior to the disk.

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Corollary A function f that is analytic throughout a simply

connected domain D must have an antiderivative everywhere in D.

Refer to the theorem in Section 44 (pp.142), (c)(a)


73

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Multiply Connected Domain A domain that is not simple connected is said to be

multiply connected. For instance,

The following theorem is an adaptation of the Cauchy-Goursat theorem to multiply connected domains.

49. Multiply Connected Domains

74

Multiply connected domain Multiply connected domain

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Theorem Suppose that (a) C is a simple closed contour, described in the counterclockwise direction;

(b) Ck (k = 1, 2, . . . , n) are simple closed contours interior to C, all described in the clockwise direction, that are disjoint and whose interiors have no points in common.

If a function f is analytic on all of these contours and throughout the multiply connected domain consisting of the points inside C and exterior to each Ck, then


75

1

( ) ( ) 0k

n

kC C

f z dz f z dz

Main Idea: Multiple Finite Simple connected domains

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CorollaryLet C1 and C2 denote positively oriented simple closed

contours, where C1 is interior to C2. If a function f is analytic in the closed region consisting of those contours and all points between them, then


76

2 1

( ) ( )C C

f z dz f z dz

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Example

When C is any positively oriented simple closed contour surrounding the origin, the corollary can be used to show that


77

2C

dzi

z

0

2C

dzi

z

For a positively oriented circle C0 with center at the original

0

( ) ( ) 2C C

f z dz f z dz i

pp. 136 Ex. 10

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pp. 160-163

Ex. 1, Ex. 2, Ex. 3, Ex. 7

49. Homework

78

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Theorem Let f be analytic everywhere inside and on a simple closed contour

C, taken in the positive sense. If z0 is any point interior to C, then

which tells us that if a function f is to be analytic within and on a simple closed contour C, then the values of f interior to C are completely determined by the values of f on C.

50. Cauchy Integral Formula

79

00

1 ( )( )

2 C

f zf z dz

i z z

Cauchy Integral Formula

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80

00

1 ( )( )

2 C

f zf z dz

i z z

Proof:Let Cρ denote a positively oriented circle |z-z0|=ρ, where ρ is small enough that Cρ is interior to C , since the quotient f(z)/(z-z0) is analytic between and on the contours Cρ and C, it follows from the principle of deformation of paths

0 0

( ) ( )

C C

f z dz f z dz

z z z z

This enables us to write

0 00 0 0 0

( ) ( )( ) ( )

C C C C

f z dz dz f z dz dzf z f z

z z z z z z z z

2πi

00

0 0

[ ( ) ( )]( )2 ( )

C C

f z f z dzf z dzif z

z z z z

00

( )2 ( )

C

f zdz if z

z z

pp. 136 Ex. 10

School of Software


81

00

0 0

( ) ( )( )2 ( )

C C


z z z z

Now the fact that f is analytic, and therefore continuous, at z0 ensures that corresponding to each positive number ε, however small, there is a positive number δ such that when |z-z0|< δ

0| ( ) ( ) |f z f z

00

0 0

( ) ( )( )| 2 ( ) | | | ( )(2 ) 2

C C


z z z z

00

( )2 ( )

C

f z dzif z

z z

The theorem is proved.

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82

00

( )2 ( )

C

f zdz if z

z z

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

00

1 ( )( )

2 C

f zf z dz

i z z

ExampleLet C be the positively oriented circle |z|=2, since the function

2( )

9

zf z

z

is analytic within and on C and since the point z0=-i is interior to C, the above formula tells us that

2

2

/(9 )2 ( )

(9 )( ) -(-i) 10 5C C

z z z idz dz i

z z i z

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51. An Extension of the Cauchy Integral Formula

83

( )01

0

( ) 2( )

( ) !n

nC

f z idz f z

z z n

0,1,2,...n

The Cauchy Integral formula can be extended so as to provide an integral representation for derivatives of f at z0

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84

Example 1 If C is the positively oriented unit circle |z|=1 and

then

( ) exp(2 )f z z

4 3 1

exp(2 ) ( ) 2 8'''(0)

( 0) 3! 3C C

z f z i idz dz f

z z

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Example 2 Let z0 be any point interior to a positively oriented

simple closed contour C. When f(z)=1, then

And


85

0

2C

dzi

z z

10

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

C

dzn

z z

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Theorem 1 If a function f is analytic at a given point, then its derivatives of

all orders are analytic there too.

CorollaryIf a function f (z) = u(x, y) + iv(x, y) is analytic at a point z = (x, y),

then the component functions u and v have continuous partial derivatives of all orders at that point.

52. Some Consequences of the Extension

86

( )0 1

0

! ( )( )

2 ( )n

nC

n f zf z dz

i z z 0,1,2,...n

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Theorem 2

Let f be continuous on a domain D. If

For every closed contour C in D. then f is analytic throughout D


87

( ) 0C

f z dz

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Theorem 3

Suppose that a function f is analytic inside and on a positively oriented circle CR, centered at z0 and with radius R. If MR denotes the maximum value of |f (z)| on CR, then


88

( )0

!| ( ) | ( 1, 2,...)n R

n

n Mf z n

R

The Cauchy’s Inequality

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pp.170-172

Ex. 2, Ex. 4, Ex. 5, Ex. 7

52. Homework

89

chapter 4. integrals weiqi luo ( 骆伟祺 ) school of software sun yat-sen university email ：...

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