15 cauchy gourst thrm & cauchy integral formula

ANTIDERIVATIVES Let f(z) be continuous function in a domain D. If there exists a function F(z) such that

then F(z) is called an

antiderivative of f(z) in D.

D,in z allfor )()( zfzF

Remark1: An antiderivative of a given function f is an analytic function. Remark 2: An antiderivative of a given function f is unique except for an additive complex constant.

Theorem: Suppose that a function f(z) is continuous on a domain D. If any one of the following statement is true, then so are the others:

i. f(z) has an antiderivative F(z) in D;

ii. 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 same value;

iii.the integral of f(z) around

closed contours lying entirely in D all have value zero.

Corollary:

D.in z allfor )()(

D.in f(z) of tiveantideriva

an is F(z) and D,domain

ain continuous is f(z)Let

zfzF

).()()(

Then D.in ENTIRELY

lying and ,z and z joining

contour any is C and D,in points

any two be z and zLet

12

21

21

zFzFdzzfC

Example:

.)( tiveantideriva

an has )( that Note

.

evaluate totiveantiderivaan Use2/

z

z

i

i

z

ezF

ezf

dze

i

ee

iFiFdze

ii

i

i

z

1

1

)()2/(

2/

2/

Cauchy - Goursat Theorem:

If a function f is analytic at all points

interior to and on a simple closed

contour C, then

0)( C

dzzf

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

0)exp( 3 C

dzz

because the function )exp()( 3zzf is analytic

everywhere.

Defn: A simply connected

domain D is a domain

such that every simple

closed contour within it

encloses only points of D.

The set of points interior to a

simply closed contour is an

example.

A domain that is not

simply connected is said

to be multiply connected

for example, the annular

domain between two

concentric circles.

The Cauchy – Goursat

theorem for a simply

connected domain D is

as follows:

Theorem: If a function f is analytic throughout a simply connected domain D, then

0)(

C

dzzf

for every closed contour C lying in D.

Result: Let C1 and C2 denote

positively oriented simple

closed contours, where C2 is

interior to C1 .

If a function f is analytic in

the closed region consisting

of those contours and all

points between them, then

21

)()(CC

dzzfdzzf

Ex.1 Evaluate

dzzf

C

when zzezf , C: |z|=1.

Ans: 0 (Why??)

Ex.2 Evaluate

dzzf

C

when

.2:,4

sin2

zCz

zzzf

Ans: 0 (Why??)

Qs 3/154. Let C0 denote the circle Rzz 0 , taken counter clockwise

using the parametric representation

izz Re0 for C0 to derive the following integrations:

no. realany is 0 where

,sin2

)(

,...2,1,0)( (b)

2 )(

0

10

0

10

0 0

a

aa

iRdzzzc

ndzzz

izz

dza

a

C

a

C

n

C

Sol. We have Rzz 0

iddz

zzi

i

.Re

Re0

a)

ii

id

zz

dzI

i

i

C

2

Re

.Re

0 0

b)

tion)simplifica(after 0

Re11

1

00

dieR

dzzzI

inin

C

n

c)

dieR

dzzzI

iaia

C

a

Re11

1

00

aSina

Ri a2

Exercise:

• Does Cauchy – Goursat Theorem hold separately for the real or imaginary part of an analytic function f(z) ? Justify your answer.

Cauchy Integral Formula

Czz

dzzf

izf .

)(

2

1)(

then sence,

positive in the taken C,contour

closed simple aon and inside

everywhere analytic be fLet

00

Derivative Formula

then C,

ointerior tpoint any is If sence.

positive in the taken C,contour closed

simple aon and inside everywhere

analytic is ffunction a that Suppose

0z

C

C

zz

dzzf

izfb

zz

dzzf

izfa

,)(

)(

2

)!2()()

,)(

)(

2

1)()

30

0

20

0

Cn

n

zz

dzzf

i

nzf

c

.)(

)(

2

)!()(

)

10

0)(

Theorem:

If f(z) is analytic at z0, then its

derivatives of all orders exist at

z0 and are themselves analytic

at z0.

Qs.1(a)/163: Let C denote the positively oriented boundary of the square whose sides lie along the

lines 2x and 2y . Evaluate

the following integral .

)8(

cos2

Czz

dzz

Ans : i/4.

Qs. 2(b)/163: Find the value of the integral of g(z) around

the circle 2 iz in the positive sense when

22 4

1

zzg

.

16

2

12

224:

22

2222

iz

CC

izdz

di

iziz

dz

z

dzSol

Qs.4/163: Let C be any simple closed contour, described the positive sense in the z- plane and write

dz

wz

zzwg

C

3

3 2

Show that

iwwg 6

when w is inside C and that

0wg

when w is outside C.

wfi

dzwz

zfwg

zzzf

C

2

2

,

Then .2)(Let

C. inside is Let w :I Case

3

3

zzzf 23

wwf

zzf

zzzf

6

6

3 2

iwwgI 6

Case 2. When w is outside C,

then by Cauchy Goursat

Theorem 0wg .

Qs. 5/163: Show that if f is analytic within and on a simple closed contour C and z0 is not on C, then

dzzz

zfdz

zz

zf

CC

200

Sol. Let

dzzz

zfI

dzzz

zfI

C

C

20

2

01 and

Case I: Let z0 is inside C, then

0

01

2

20

zfi

zfidzzz

zfI

zzC

and

0

20

2

2 zfi

dzzz

zfI

C

21 II .

Case II: Let z0 is outside C

Then I1 = I2 = 0.

(WHY ???)

Morera’s Theorem:

D.in analytic is f(z) then D,in

lying Ccontour closedevery for

,0f(z)dz

if and Ddomain ain throughout

continuous is f(z)function a If

C

LIOUVILLE’S THEOREM If f is entire and bounded in the complex plane, then f(z) is constant throughout the plane.

Fundamental Theorem of Algebra

.0)P(z

such that zpoint on least at exist

thereisThat zero. oneleast at

has )1( degree of )0(

...)(

polynomialAny

0

0

2210

nna

zazazaazP

n

nn

Theorem: Suppose that

(i) C is a simple closed contour, described in the counter-clockwise direction,

(ii) Ck (k = 1, 2, …., n) are finite no. of simple closed contours, all described in the clockwise direction, which are interior to C and whose interiors are disjoint.

.0f(z)dzf(z)dz

then,C ointerior t points for the

except Con and within points

all of consistingregion closed

t the throughouanalytic is f(z) If

1

n

k kCC

k

Ex. Evaluate C zz

dz

)1( 2 for all

possible choices of the contour C that does not pass through any of the points 0,

i .

Solution: Case 1. Let C does not enclose 0,

i . Then

.012 TheoremCGby

z

dzI

C

Case 2a. Let C encloses only 0. Then

ifi

zzf

z

dzzfzz

dzI

C

C

2)0(2

)1(

1)(,

0

)()1(

2

2

Exercise: Case 2b. Let C encloses only i. Ans: I = -i Case 2c. Let C encloses only -i. Ans: I = -i

Case 3 a). Let C encloses only 0, -i. then

iCCizizz

dz

izizz

dzI

))(())((0

where C0 and C-i are sufficiently small circles around 0 and –i resp.

iC

C

dzizizz

dzz

iziz

)()(

1

))((1

0

iii

ii

i

)2(

1)2(

1)2( 2

Case 3 b). Let C encloses only 0, i. then

iCCizizz

dz

izizz

dzI

))(())((0

where C0 and Ci are sufficiently small circles around 0 and i resp.

i

iiii

dzizizz

dzz

izizI

iC

C

2.

122

)()(

1

))((1

0

Case 3 c). Let C encloses only -i, +i. Then

i

iii

iii

dzizizz

dzizizz

I

i

i

C

C

22.

12

2.

12

)()(

1

)(1

Case 3 d). Let C encloses all of the points 0, -i, +i. Then

0

2

)()(

1

)(1

1

1

0

2

iii

dzizizz

dzizizz

dzz

zI

iC

iCC

0

2

)()(

1

)(1

1

1

0

2

iii

dzizizz

dzizizz

dzz

zI

iC

iCC