Maximum Modulus Principle:

If f is analytic and not constant in a given domain D, then |f(z)| has no maximum value in D.

That is, there is no z0 in the domain such that |f(z)||f(z0)| for all points z in D.

Proof: Assume that |f(z)| does have a maximum value in D.

2

0

00 )(2

1)( drezfzf i

2

0

00 )(2

1)( drezfzf i

z0

R

CR

Alternatively

Theorem: If f is analytic, continuous and not constant in a closed bounded region D, then the maximum value of |f(z)| is achieved only on the boundary of D.

Some other aspects of the maximum modulus theorem:

Assume that f(z) is not 0 in a region R. Then if f(z) is analytic in R, then so is 1/f(z).Result: minimum of |f(z)| also occurs on the boundary.

. then ),(),()( If )( uivuivuzg eeeeeyxivyxuzg

Since the max and min of |f(z)| are on the boundary, so is the max and min of u(x,y).

Same applies to v(x,y).

Indented Contour

• The complex functions f(z) = P(z)/Q(z) of the improper integrals (2) and (3) did not have poles on the real axis. When f(z) has a pole at z = c, where c is a real number, we must use the indented contour as in Fig 19.13.

Fig 19.13

Suppose f has a simple pole z = c on the real axis. If

Cr is the contour defined by

THEOREM 19.17Behavior of Integral as r →

,irecz then,0

rCr

czfidzzf ),)((Res)(lim0

THEOREM 19.17 proofProofSince f has a simple pole at z = c, its Laurent series is f(z) = a-1/(z – c) + g(z) where a-1 = Res(f(z), c) and g is analytic at c. Using the Laurent series and the parameterization of Cr, we have

(12)

21

001 )(

)(

II

derecgirdre

irea

dzzf

iii

i

Cr

THEOREM 19.17 proofFirst we see

Next, g is analytic at c and so it is continuous at c and is bounded in a neighborhood of the point; that is, there exists an M > 0 for which |g(c + rei)| M.

Hence

It follows that limr0|I2| = 0 and limr0I2 = 0.We complete the proof.

) ),((Res9

1

01011

czfiia

idadre

ireaI i

i

rMMdrderecgirI ii 002 )(

Example 5Evaluate the Cauchy principal value of

Solution Since the integral is of form (3), we consider the contour integral

dx

xxx

x

)22(

sin2

)22(

1)( ,

)22( 22

zzzzf

zzz

dzeC

iz

Fig 19.14

f(z) has simple poles at z = 0 and z = 1 + i in the upper half-plane. See Fig 19.14.

Example 5 (2)

• Now we have

(13)

Taking the limits in (13) as R and r 0, from Theorem 19.16 and 19.17, we have

)1,)((iRes2

)0,)((Res)22(

P.V. 2

iezf

ezfidxxxx

e

iz

izix

C C

r

R C

R

r

iz

R riezfi )1,)((Res2

Example 5 (3)

Now

Therefore,

)1(4

)1 ,)((Res

21

)0 ,)((Res

1

ie

iezf

ezf

iiz

iz

)1(4

i221

)22(P.V.

1

2 ie

idxxxx

e iix

Example 5 (4)

Using e-1+i = e-1(cos 1 + i sin 1), then

)]1cos1(sin1[2)22(

sinP.V.

)1cos1(sin2)22(

cosP.V.

12

12

edxxxx

x

edxxxx

x

Indented Paths

iBdzzf

RxzC

B

Rxz

xzf(z)

C

00

20

0

20

0

)(lim Then,

taken.isdirection clockwise the

and where,|| circle a of halfupper thedenotes (ii)

; residue with and

96) (Fig. ||0disk punctured ain tion representa seriesLaurent a

with axis, real on the point aat pole simple a has function a (i)

thatSuppose :Theorem

x0

C

2

sinlim

0

20

dx

x

xI

dzz

edx

x

x iz

Imsin

C

CR

I1I2

021 CCII R

applies) Lemma s(Jordan' 011

Rx

2

0

2

0

sin2

0

sincos2

0

)sin(cos

2

0

)sin(cos2

0

)sin(cos

zero? togoes over integral theHope

dededeediedzz

e

diedeie

edz

z

e

C

iii

C

iz

iiii

ii

C

iz

21

sin

)(

)sin(sinIdu

u

udu

u

udx

x

xI

C

CR

I1I2

021 CCII R

origin. at the pole simple a has z

eiz

1

!3!2!11

z

1

exists.tion representa seriesLaurent

0

32

B

iziziz

z

eiz

iiBdzz

e

C

iz

0

1

)(

find ely toAlternativ

00

0

eB

ez

Biz

dzz

edx

x

x

idzz

e

dzz

edz

z

edz

z

e

iz

iz

C

iz

C

iziz

R

00

limImsin

0

Contour Integration Example

The graphical interpretation

)2( .2/

)sin(

2

1)0( Find

dp

x

xxf

)sin()(

>> x=[-10*pi:0.1:10*pi];>> plot(x,sin(x)./x)>> grid on>> axis([-10*pi 10*pi -0.4 1])

axis Real on the is )sin()sin(

)( iyxzz

z

x

xzf

>> x=[-10*pi:0.1:10*pi];>> plot3(x,zeros(size(x)),sin(x)./x)>> grid on>> axis([-10*pi 10*pi -1 1 -0.4 1])

z

ezg

iyxzz

ezf

iz

iz

)(

axis Real on the is Im)(

z

zzg

)cos()(Re

z

zzg

)sin()(Im

>> mesh(x,y,sin(z)./z)

axis Real on the is )sin()sin(

)( iyxzz

z

x

xzf

>> x=[-10*pi:0.1:10*pi];>> y = [-3:0.1:3].';>> z=ones(size(y))*x+i.*(y*ones(size(x)));>> mesh(x,y,cos(z)./z)

z

zzg

)cos()(Re

z

zzg

)sin()(Im

Cauchy’s Inequality: If f is analytic inside and on CR and M is the maximum value of f on CR, then

z0

R

CR

),2,1(!

)( 0)( n

R

Mnzf

nn

Proof:

2

0)1(1

01

0

00

)( )(

2

!

)(

)(

2

!)( dRie

eR

eRzf

i

ndz

zz

eRzf

i

nzf i

nin

i

Cn

in

R

2

0)1(1

02

0)1(1

0)(

2

!)(

2

!dRie

eR

eRzfndRie

eR

eRzf

i

n i

nin

i

inin

i

nn

i

nin

i

R

MndR

R

MndRie

eR

eRzfn !

2

!)(

2

! 2

01

2

0)1(1

0

iR eRzzC 0 :as written becan on point Any :Note

),2,1(!

)( 0)( n

R

Mnzf

nn

R

Mzf )(' 0

As R goes to infinity, then f’(z) must go to zero, everywhere. Then f(z) must be constant.

Liouville’s Theorem: If f is entire and bounded in the complex plane, then f(z) is constant throughout the plane.

Proof:

2

0

0

2

0

0

00 )(

2

1)(

2

1)(

2

1)( drezfdire

re

rezf

idz

zz

zf

izf ii

i

i

CR

Gauss’s Mean Value Theorem: If f is analytic within and on a given circle, its value at the center is the arithmetic mean of its values on the circle.

0

11

plane halfleft On the

22

3

dzzz

ei

i

z

C1

dzz

zdy

iy

iydy

iy

iydx

x

x i

i

22222222 1

3cosh

1

3cosh

1

3cosh

1

3cos

dz

zz

eedz

zz

zdz

z

z i

i

zzi

i

i

i

22

33

2222 112

1

11

3cosh

1

3cosh

The integral does not go to zero on the circle, the integral can’t be solved this way.

RRiRz

RRiRz

RiRz

eee

eee

ee

3sin3cos33

3sin3cos33

sin3cos33

,For

,For

dxx

exdx

x

exdx

x

axx iaxiax

4Im

4Im

4

sin4

3

4

3

4

3

4/34/4/4/3

334

4

2,2,2,2

4,4,4,4

04

iiii

iiii

eeeex

eeeex

x

044 4

3

4

3

R

R

R

x

x

4Res2

4Res2

44 4

3

)4/3exp(24

3

)4/exp(24

3

4

3

z

zi

z

zi

z

ezdz

z

eziziz

C

iaziaz

RC

0R

Rx

y

;|| circle a oexterior t

are that 0 plane halfupper in the points allat analytic is )(function a (i)

that Suppose

0Rz

yzzf

; where)0( semicircle a denotes (ii) 0RReRzC iR

0lim where

,|)(|such that constant positive a is there,on points allfor (iii)

RR

RRR

M

MzfMCz

0)(lim ,constant positiveevery for Then, RC

iaz

Rdzezfa

Jordan’s Lemma

dzz

edx

x

x zi

22

3

22 1Re

1

3cos

dz

iziz

edz

z

e zizi

22

3

22

3

1

01

1

1

12222

RRM

Rz

C1

C2

dz

iziz

edz

z

e zizi

22

3

22

3

1 01

1

1

12222

RRM

Rz

C2

4323

2

3

23)('

)(

iz

eizizeiz

iz

ez

iz

zizi

zi

333

22

3

16

412Res

ie

ieei

iziz

e zi

iz

33

22

3 2)(2

1 eieidz

z

e zi