lecture 31 33

BITS Pilani Pilani Campus

MATH F112 (Mathematics-II)

Complex Analysis

BITS Pilani Pilani Campus

Lecture 31-33

Integrals

Dr Trilok Mathur,

Assistant Professor,

Department of Mathematics

BITS Pilani, Pilani Campus

.

1

1&

04

01)(

3

C

dzzf

.xyi z

i zC

yy,

y, zf

evaluate Then

curve the along to

fromarc the is

LetEx.3


C AO OB

dzzfdzzfdzzf

haveWe

A

O

B 1+i

-1-i

01,

:

3 xixxiyxz

AOAlong


i

xix

dxxidzzfAO

1

3

3

)31(1)(

0

1

3

0

1

2


i

dxixx

dxxiydzzf

x,xixzOB

OB

21

314

314)(

10

21

0

3

21

0

3

, Along

iiidzzfC

32211


b.t; atzC:z

zf

)(

)(

contour a on defined function

continuous piecewise a be Let

constant negative-non some

forsatisfies

, contour the on that Suppose

M.

Mzf zf

C

)()(


b

aC

dttztzfdzzf ')(Then

dttztzfb

a'))((

b.taC

dttzL

MLdttzM

b

a

b

a

in of

length the iswhere

'

,'


.24

.1

)140.(

4

Cz

dz

zi z

C p

thatshow origin, the to closest the is

midpoint the segment, line that on points the

all of that, observingBy to from

segment line the denote Let.2 Q


/4

/4

/4

A

B

D

2

1

4sin

4sinsin

OBOD

OB

OD

ABD of point mid the is

O 1

i


2

1z

AB,z

then

line the on pointany is If

41

4z


24

2

11

4

22

MLz

dz

ABL

C

of length


.60

430

.

C

z dzze

i, ,

Cp

:thatShow

direction. ckwisecounterclo withoriented

, vertices withtriangle the

ofboundary the be Let:140). ( Q.3,


C

z zezfzdzfI ,Let

0

3i

-4

C

MLdzzfIC

Then

triangle the of perimeter

curve the of length the

on where

C L

CM zf

&

)(


12453

443 22

3

triangle the of PerimeterL

zezezf

zezf

zz

z haveWe


zezf z)(

5

541

M

60MLdzzfC


.31

222

2

C

thatShow

quadrant. first the in lies

that to from circle

the ofarc the be Let :.140)(Q1,

z

dz

i zzz

Cp


Let f (z) be continuous function in a

domain D.

If there exists a function F(z) such

that

, in all for DzzfzF )()(


then F(z) is called an antiderivative of

f (z) in D.

Remark 1: 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: f (z) has an antiderivative F(z) in D;

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;


The integral of f(z) around closed contours

lying entirely in D all have value zero.

D.z zfzF

D.zfzFD

zf

in all for

in of tiveantideriva an is and

domain a in continuous is Let :Corollary

)()(

)()(

)(


).()()(

,

12

21

21

zFzFdzzf

D

z&zC

Dzz

C

Then . inENTIRELY lying

and joining contourany is and

in points twoany be and Let


.

2/i

i

zdzeevaluate to tiveantideriva an UseEx.1:

.)(

)(

z

z

ezF

ezf

tiveantideriva

an has that Note:Soln

ieeiFiFdze ii

i

i

z 11)()2/( 2/2/


Advised:

See W.O.E. 3, p. 143

See W.O.E. 4, p. 145

See Q. No. 5, p. 149


If a function f is analytic at all points

interior to and on a simple closed

contour C, then

0)(C

dzzfAugustin Cauchy

(1789-1857)

Edourd Goursat

(1858-1936)

lecture 31 33

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