tch-prob 1

Chapter 4 Integrals Complex integral is extremely important, mathematically elegant.

30. Complex-Valued Functions w(t) First consider derivatives and definite integrals of complex-valued functions w of a real variable t.

(1) )()()( titutwLet

Real function of t

(2) )(')(')()(' titudt

tdwtw

Provided they exist.

tch-prob 2

0 0 0

0 0 0

0 0 0 0

0 0 0 0

0 0

For each complex constant

( ) ( ) ( )

( ) ( )

( ) ( )

( )(

i

x iy

x yzd dz w t u idt dt

d x u y i y xdtd dx u y i y u xdt dtx iy u

0 0

' ')

. . ( ) '( )

idi e z w t z w tdt

Various other rules for real-valued functions of t apply here.

tzeZtzedtdge 000 ..

However, not every rule carries over.

tch-prob 3

Example: Suppose w(t) is continuous on .bta

. . or ( ( ) and ( ) are continuous)i e u t t

The “mean value theorem” for derivatives no longer applies.

There is a number c in a<t<b such that

abawbwcw

)()()('

a c b

0)0()2(

1 )('

20 )(

wwbut

Zeroneveritietw

tonitetwconsider

tch-prob 4

Definite Integral of w(t) over a t b

is defined as ) ( ) ( ) (5)b b bw(t dt u t dt i t dta a a

when exists

Re ( ) Re ( ) Im Im ( ) b b b bw t dt w t dt w(t)dt w t dta a a a

Can verify that

1 2 1 2

( ) ( )

( ( ) ( )) ( ) ( )

( ) ( ) ( )

b bcw t dt c w t dta ab b bw t w t dt w t dt w t dta a ab c bw t dt w t dt w t dta a c

tch-prob 5

Anti derivative (Fundamental theorem of calculus)

( )|

( ) ( ) ( )

,

( ) ( ) ( )

( )|

( ) ( ) ( ) ( )

( )

b ba a

iV t

W t U t iV tif W'(t) w(t)then U'(t) u(t) V'(t) υ(t)b b bw t dt u t dt i t dta a a

U t

U b iV b U a iV a

W b W

( )a

tch-prob 6

00 0

00

0

0

0 0 0

0

To Prove ( )

Let

Re

Re

Re

b bw(t) dt w(t) dt a ba a

iθb b w dt r e w dt ra a

iθbr e w dta

-iθbe w dtaiθb (e w) dta

iθ iθ iθBut (e w) e w e w w

r

b w dtab bor wdt w dta a

realmust be real

Real part of real number is itself

tch-prob 7

31. Contours

Integrals of complex-valued functions of a complex variable are defined on curves in the complex plane, rather than on just intervals of the real line.

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

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

where x(t) and y(t) are continuous functions of real t.

非任意的組合

This definition establishes a continuous mapping of interval

into the xy, or z, plane; and the image points are ordered according to increasing values of t.

bta

tch-prob 8

It is convenient to describe the points of arc C by

( ) ( )

where ( ) ( ) ( )

z z t a t b

z t x t i y t

The arc C is a simple arc, or a Jordan arc, if it does not cross itself.

1 2 1 2( ) when i.e. z(t ) z t t t

When the arc C is simple except that z(b)=z(a),

we say that C is a simple closed curve, or a Jordan Curve.

tch-prob 9

0

2

0 2Simple closed curve

0 2

Ex :iθ z e ( θ π)

iθ also z z R e ( θ π)

3

0 2

Ex :iθ z e ( θ π)

10 11 2

Ex :x ix x

zx i x

0 1 2

1

y

x

2

40 2i

Ex : z e ( θ π)

tch-prob 10

arc C : z(t) x(t) i y (t)

Suppose that x’(t) and y’(t) exist and are continuous throughout bta

C is called a differentiable arc

z'( ) '( ) '( )Since t x t iy t

2 2'( ) '( ) '( ) integrable over

real valuedz t x t y t is a t b

function

The length of the arc is defined as

0

ba

z'(t) dtb

L z'(t) Δta

Δt

tch-prob 11

L is invariant under certain changes in the parametric

representation for C

To be specific, Suppose that

t (τ) (α τ β)

where is a real valued function mapping the interval

β)τ(α onto the interval .

bta We assume that

is continuous with a continuous derivative

We also assume that 0 for each'(τ) τ ( increases with )t

tch-prob 12

if arc is

the fact that

allows us to write

β L z' (τ) '(τ) dταSo., C z Z(τ) z (τ) (α τ β)

Z'(τ) z' (τ) '(τ)

β L Z'(τ) dτα

Exercise 6(b)

Exercise 10

If is a differentiable arcand 0

z z(t) (a t b) z'(t) in a t b

Then the unit tangent vector

0z'(t)( 　不代表水平，而是在此處停頓 長度不增加）

z'(t)T z'(t)

is well defined for all t in that open interval.

Such an arc is said to be smooth.

tch-prob 13

For a smooth arc z z(t) (a t b)

is continuous on (we agree)and non zero on

z'(t) a t b a t b

A contour, or piecewise smooth arc, is an arc consisting of a finite number of smooth arcs joined end to end.

If z=z(t) is a contour, z(t) is continuous , Whereas z’(t) is piecewise continuous.

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

tch-prob 14

32. Contour Integrals

Integrals of complex valued functions f of the complex variable z: Such an integral is defined in terms of the values f(z) along a given contour C, extending from a point z=z1 to a point z=z2 in the complex plane. (a line integral)

Its value depends on contour C as well as the functions f.

Written as 2

1( ) or ( ) zf z dz f z dzc z

When value of integral is independent of the choice of the contour.

Choose to define it in terms of

not as lim .w(t) dt, Σ

tch-prob 15

Suppose that z z(t) (a t b)

represents a contour C, extending from z1=z(a) to z2=z(b).

Let f(z) be piecewise continuous on C.

Or f [z(t)] is piecewise continuous on bta

The contour integral of f along C is defined a

(( ) ( 2)) ( ) bf z dz f z t z' t dt c a

Since C is a contour, z’(t) is piecewise continuous on bta t 的變化　 define contour C

Section 31

So the existence of integral (2) is ensured.

tch-prob 16

From section 30

0 0

( ) ( )

and ( ) ( ) ( ) ( )

z f z dz z f z dzc c f z g z dz f z dz g z dzc c c

0: complex constantz

Associated with contour C is the contour –C From z2 to z1

Parametric representation of -C

( )z z t (-b t -a)

z2=z(b)

z1=z(a)

tch-prob 17

order of –C must also follow increasing parameter value

order of C follows (t increasing)

bta

a b -t 大 小

atb

( ) ( )

( ) w ( )

( ) ( ) ( ) ( )( )( )

( )

z t w ta f w t ' t dtb

dw t dz t dz t d tw' t dt dt d t dt

z' t

Thus

( ) ( ) ( )af z dz f z t z' t dtc b

where z’(-t) denotes the derivative of z(t) with respect

to t, evaluated at –t. |( ) ( )or ( ) t t

dz t dz t d t dt

tch-prob 18

After a change of variable,

) ) ) )

τ ta τ b

a bf(z) dz f z(τ z'(τ dτ f z(τ z'(τ dτc b a f(z) dzc

dtd

1 2

1 2 1 0 0 2

1 2

if C C C z z z z z z

f(z) dz f(z) dz f(z) dzc c c

Definite integrals in calculus can be interpreted as areas, and they have other interpretations as well.

Except in special cases, no corresponding helpful interpretation, geometric or physical, is available for integrals in the complex plane.

z1

C1

z0z2

CC2

tch-prob 19

33. Examples

Ex. 1

By def.

Note:

on the circle

for z

y

C2

0

-2

x

when is 2 ( )2 2

i

CI zdz C z e

2

2

2 2

2 2

2 (2 ) '

2 (2 ) 4 4

i i

i i

I e e d

e ie d i d i

42, 4, or

Hence .C

z zz z zdz

iz

22

2

2 2 2

2 2

2 (2 ) '

2 (2 ) 4 0.

i i

C

i i i

I zdz e e d

e ie d i e d

tch-prob 20

x

y

A B

C1 C2

0

1+i1C1

Ex2. 1 2: contour , : segment C OAB C OB

2( ) 3 ( ).f z y x i x z x iy

1 1

0 0( ) .

2OA

if z dz yidy i ydy

1 1 12 2

0 0 0

1( ) (1 3 )1 (1 ) 3 ( ) - .

2ABf z dz x i x dx x dx i x dx i

1

1( ) .

2C

if z dz

2

1 12 2

0 0( ) 3 (1 ) 3(1 ) 1- .

Cf z dz i x i dx i x dx i

1 1 22

1( ) ( ) ( ) .

2C C C C

if z dz f z dz f z dz

:

(0 1)

OB

z x ix

x

:

0

(0 1)

OA

z iy

y

:

(0 1)

AB

z x i

x

tch-prob 21

Ex3

Want to evaluate

Note that

dep. on end points only. indep. of the arc.

Let C denote an arbitrary smooth arc z=z(t), bta

2 2 2

2 22 1

2 2

2 2

2

2 1221

( ) ( ) ( )

zz

z

zzdz

z t z b z abI

az

z

b

C aI z dz z(t) z'(t) dt

2

2z(t)d z(t) z' (t)

dt

Integral of z around a closed contour in the plane is zero

c z dz 0

tch-prob 22

Ex4. Semicircular path 3 0iθC : z e ( θ π)

起點 終點Although the branch (sec. 26) p.77. /2 0 0 2iθf(z) r e ( r , θ π)

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

12zI dzc of that branch nevertheless exists.

For the integrand is piecewise continuous on C.

C

30-3

323 3 3 sin 0

2 2right hand limit at 0 3 0.

iθz(θ) eθ θiθ /f z(θ) e cos i ( θ π)

θ is i

/ 20

3 3

Hence ( ) is continuous on 0 when its value at 0

is defined as 3 Consequently 2 3 1 .iθ iθ e ie

f z θ π θ

. , I dθ ( i)

tch-prob 23

34. Antiderivatives

－ There are certain functions whose integrals from z1 to z2 　 are independent of path.

　 The theorem below is useful in determining when 　　　　　 integration is independent of path and, moreover, when an 　 integral around a closed path has value zero.

－ Antiderivative of a continuous function f : a function F such that F’(z)=f(z) for all z in a domain D.

－ note that F is an analytic function.

tch-prob 24

Theorem: Suppose f is continuous on a domain D.

The following three statements are equivalent.

(a) f has an antiderivative F in 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.

(c) The integrals of f(z) around closed contours lying

entirely in D all have value zero.

Note: The theorem does not claim that any of these statements

is true for a given f in a given domain D.

tch-prob 25

Pf:

has am antiderivative

z

A:(a) (b) (a) f F.

d F z(t) F' z(t) z'(t) f z(t) z'(t)dt

b b f(z) dz f z(t) '(t)dt F z(t) F z(b) F z(a)ac a

2 1 F(z ) F(z )

(b)

c f(z)dz

c c c f(z)dzf(z)dzc f(z)dzf(z)dz

c f(z)dzc f(z) dz

CC C(c)B:(b)

01 1 2

02

021

21

(c)z1

C2

C1

z2

1 2 1 20 0

C :(c) (b) f(z) dz f(z) dz f(z) dzc c -c c c

(b)

tch-prob 26

0

0 0

Define ( ) ( )

( ) ( ) ( ) ( ) ( )

(choose integration path from to as

z F z f s dszz z z z z F z z F z f s ds f s ds f s dsz z z

z z ds z z z z z

1 1( )

a line segment)

( )

( ) ( ) 1( ) ( ) ( )

f zz z

z z z z f(z) ds f z dsz zz zF z z -F z f z f s - f z dszz z

( ) ( )( ) ( ) ( )

Since is continuouswhenever

1 1

when is close to so that

lim

z zz

F z z -F zf z f s f z

z

f , f(s)- f(z) ε s- z δ

ds ε z ε z z

z z z, z δ

z

( ) ( ) or0

F z z f z F'(z) f(z) z

(c) -->(a)

tch-prob 27

35. Examples2

3

131+i 2 3

00

Ex1. ( ) continuous

1 has an antiderivative ( ) throughout the plane.3

1 2 Hence (1 ) ( 1 )

3 3 3

for every contour from 0 to 1 .

i

f z z

F z z

zz dz i i

z z i

tch-prob 28

For any contour from z1 to z2 that does not pass through the origin .

20 for : 2 (- )

zdz iC z ec

Note that: zzf 1)( can not be evaluated in a similar way

though derivative of any branch F(z) of log z is , 1

z

F(z) is not differentiable, or even defined, along its branch cut. (p.77)

which

t throughoudomaina in lienot does C C. circle theintersect ray theepoint where at theexist tofails cut, branch the

from the origin is used to form raya if ,particular (InF'(z)

1'( ) .F z

z

2

22

2 1 21 1 1 2

1

z1

1 1 1

Ex2. is continuous every where except at the origin.

It has an antiderivative in the domain 0

consequently 0 0zz

zz

dz

z z z z

- .

z , (z , z )

z

tch-prob 29

.Ex3

The principal branch Log z of the logarithmic function serves as an antiderivative of the continuous function 1/z throughout D.

Hence

when the path is the arc 22 2π πiθz e (- θ )

excluded

Let : 0, Arg .D z z

222 2

Log Log(2 ) Log( 2 )

(ln2 ) (ln2 )2 2 2 2

iii i

dzz i i

z

i i i π π (- θ )

(compare with p.98)

tch-prob 30

1

12

12

12

4

Evaluate

where is the branch

2 0 0 2

Cz

Ex .

dz

z θi

z r e (r , θ π)

C1

-3

C2

3

(2)

C1 is any contour from z=-3 to z=3, that lies above the x-axis. (Except end points)

tch-prob 31

1. The integrand is piecewise continuous on C1, and the integral therefore exists.

2. The branch (2) of z 1/2 is not defined on the ray in particular at the point z=3. F(z) 不可積

3. But another branch.

is defined and continuous everywhere on C1.

4. The values of F1(z) at all points on C1 except z=3 coincide with those of our integrand (2); so the integrand can be replaced by F1(z).

)πθπ, - (riθ

er(z)F2

32

021

tch-prob 32

Since an antiderivative of F1(z) is

12

1 11

333 0 22 33 3

2 3 1

zi π

i dz f (z) dz F (z) (e e )c -

( i)

We can write

32

32 2 32 01 3 3 2 2

i θ π πF (z) z r r e (r , - θ )

(cf. p. 100, Ex4)

-3 3

12

2For z dzc

Replace the integrand by the branch

tch-prob 33

2

2

2

2 22

12

2 1

52 02 2

32 52Then 03 2 2

is an antiderivative of31 33 32 22 33 3

2 3 1

4 3

z

z

iθ π πf (z) r e (r , θ )

i θ π π F (z) r r e (r , θ )

f (z).i π

i πdz f (z) dz F (z) (e e )c

(- i)

dz -c c

1 2c c two contours can not be on the some domain

0

tch-prob 34

36. Cauchy-Goursat Theorem

We present a theorem giving other conditions on a function f ensuring that the value of the integral of f(z) around a simple closed contour is zero.

Let C denote a simple closed contour z=z(t) described in the positive sense (counter clockwise). Assume f is analytic at each point interior to and on C.

),( bta

are continuous th, , ,

( ) ( ) ( )

( ) ( ) ( ) and ( ) ( )

( ) ( ) ( )

Green's theorem: If ( , ), ( , ),x y x yP Q Q

bf z dz f z t z' t dtc aif f z u x,y iv x,y , z x t i y t .

b bf z dz ux' vy' dt i v x' uy' dt.c a a

P x y Q x y P

roughout

( )the closed region consisting of all points interior to and on , xR yQ dAP

R CPdx Qdyc

tch-prob 35

( ) ( ) )x y x yC R Rf z dz v u (u v dAdA i then

But(Cauchy-Riemann eg.) u v u v x y y x

when is analytic in and is continuous there

( ) 0

f R f'

f z dz Cauchy 's finding.c

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

Cauchy-Goursat Theorem:

If f is analytic at all points interior to and on a simple closed contour C, then

( ) 0f z dzc

tch-prob 36

37. Proof: Omit

38. Simply and Multiply Connected Domains

A simply connected domain D is a domain such that every simple closed contour within it encloses only points of D.

Multiply connected domain : not simply connected.

Can extend Cauchy-Goursat theorem to:

Thm 1: If a function f is analytic throughout a simply connected domain D, then

( ) 0f z dzc

for every closed contour C lying in D.

Not just simple closed contour as before.

n分成simpleclosed

contours

tch-prob 37

Corollary 1. A function f which is analytic throughout a simply connected domain D must have an antiderivative in D.

section 34 for continuousgiven

analytic ( ) 0co

pf : f f f z dzc f

ntinuous has antiderivative f

Extend cauchy-goursat theorem to boundary of multiply connected domain

Theorem 2. If f is analytic within C and on C except for points

interior to any Ck, ( which is interior to C, ) then

( ) ( ) 01 k

nf z dz f z dzc c

k

C: simple closed contour, counter clockwise

Ck: Simple closed contour, clockwise

C1 C2C

tch-prob 38

C2

C1Corollary 2.

Let C1 and C2 denote positively oriented simple closed contours, where C2 is interior to C1.

If f is analytic in the closed region consisting of those contours and all points between them, then

1 2

1 2

( ) ( )

from thm 2 ( ) ( ) 0

f z dz f z dzc cpf : , f z dz f z dzc -c

Principle of

deformation of paths.

tch-prob 39

Example:

C is any positively oriented simple closed contour surrounding the origin.

0

0

0

2

0

Want to show that 2 .

Construct a positively oriented circle with center at the

origin and radius so small that lies entirely inside C.

Since 2

and since 1/ is anal

C

i

iC

dzi

zC

C

dz i ed i

z e

z

ytic everywhere except at =0,

we have 2 from corollary 2.C

z

dzi

z

C0

C

0

2 22 2

0 0

2

0

?

i i i

C C

C

zdz zdz e i e d i e d

z dz

tch-prob 40

39. Cauchy Integral FormulaThm. Let f be analytic everywhere within and on a simple closed contour C, taken in the positive sense. If z0 is any point interior to C then,

00

00

)

)

( )1 (2

or( ) 2 (

f z dzf z c z- zπ i

f z dz π i f zc z z

(1) Cauchy integral formula

(Values of f interior to C are completely determined by the values of f on C)

2

2

2

9

9 ))(

. Let be 2

since ( ) is analytic within and on ,

and since is interior to .

2 (510( )(9 )

z

z

z

z

z

Ex C

f z C

-i C

z dz -i π dz π i c c z i- z z i

tch-prob 41

C

C0

z0

Pf. of theorem:

0

00

00

-

-

is continuous at

( ) ( ) when

choose a positive :

z z

z z

f z

f z f z

C

since

0

( )f zz z is analytic in the closed region consisting of C

and C0 and all points between them, from corollary 2, section 38,

00 0

00

0 00 0 0

))

( ) ( )

( ) (( ) ( dz

f z dz f z dzc cz z z z

f z f z f z dz dzf z c c cz z z z z z

tch-prob 42

0 0

0 00

00 0

0 0

0 00 0

0

0 0

) ))

) )

)

but 2

( ) ( ( ) (2 (

( ) ( ( ) (also

( ) (2

dz ic z zf z - f z f z f z

- π i f z dzc cz z z z

f z f z f z f z dz dzc cz z z z

f z f z dz πρc z z

2 π

00 0

00

)

)

( ) 2 ( 2

( ) 2 ( 0

f z dz π i f z πc z z

f z dz π i f zc z z

Non-negative constant arbitrary

tch-prob 43

40. Derivatives of Analytic FunctionsTo prove : f analytic at a point

its derivatives of all orders exist at that point and are themselves analytic there.

0 interior to

in Cauchy integral formula on

z z C

z s C

( )1( )2

f s dsf z c s zπ i

( ) ( ) ( )1 1 12

( )12 ( )( )

f z z f z f s ( ) dsc s zz s z z zπ if s ds cπ i s z z s z

S C

zd

z

when 0 ,

where is theshortest distamefrom to on .

z d

d

z s C

tch-prob 44

2

2

1 1but ( )( )( ) ( )

( )( )( )

f s dsc s z z s z s zf s ds z c s z z s z

2 2

-

Let denote the maximum value of ( ) on .

Let be the length of .Note that

( ) 0 as 0( )( ) ( )

lim ( ) ( ) (120

s z

s z z z

zz

z

M f s C

L Cd

s z z d -

MLf s ds z c s z z s z d - d

f z z - f z f s z π iz

2

2

3

( )

)

( )1'( )2 ( )

( )1 similarly for '' ( )( )

s z ds c #

f s ds f z cπ i s zf s ds f z cπ i s z

tch-prob 45

Thm1. If f is analytic at a point, then its derivatives of all orders are also analytic functions at that pint.

In particular, when

( ( ( (

( ) ( , ) ( , ) is analytic at a point z ( )

the analyticity of ' eusures the continuity of ' there '( ) ) ) ) )

are conti

f z u x y iv x y x,y

f ff z u x,y i v x,y v x,y i u x,yx x y y

u , u , v , v x y x y

( ( (

nuous at ( )

similarly ''( ) ) ) )

x,y

f z u (x,y) i v x,y v x,y i u x,yxx xx yx yx

Corollary : If ( ) ( ) ( ) is analytic at a point , then have continuous partial derivatives of all orders at that point.

f z u x,y i v x,yz x iy u, v

tch-prob 46

(0)

( )( 1)

( )0( 1)

0

( )

( )( )

Let ( ) denote ( ) and 0! 1, can use mathematicalinduction to verify that

( )!( ) ( 01 2 )2

( ) 2or!

nn

nn

s z

zz z

f z f z ,

f s dsnf z n , , ....cπ i

f z dz π i fc n

2

2

4 3 1( 0)

1: 1 ( )

( ) 2 8(0)3! 3

z

z

z

z

Ex C : f z e

f z dze π i π i dz f'''c cz

0

10 0

in , ( ) 1,

)

2:

2 0 ( 1 2 )(

Thm 2: If continuous in domain and ( ) 0 for every closed contour in ,c

n

f zEx z Cdz dz π i n , , ....c cz- z z- z

f Df z dz C D

then is analytic throughout .f D

tch-prob 47

41. Liouville’s Theorem and the Fundamental Theorem of Algebra

Let z0 be a fixed complex number.

If f is analytic within and on a circle 0z z R

( )0 1

0

( )( )

( )!then ( 1 2 ).2

nn

zz zf z dzn f n , ,.......cπi

Let MR denote the Maximum value of )(zf

( )0 1

( )0

0

( )

( )

)

! 2 ( 1 2 )2!

( 1 2 ) Cauchy's inequality.

for 1 '(

n Rn

n Rn

R

MzR

z

nf πR n , ,....π

n Mf , n , ,.....

RM

n f zR

tch-prob 48

Thm 1 (Liouville’s theorem): If f is entire and bounded in the complex plane, then f(z) is constant throughout the plane.

0

0

0

)

)

: '(

for large, '( must be zero since is arbitrary, '( ) 0 everywhere is a constant

Mpf f zR

R f zz

f zf

finite

可以 Arbitrarily large

Thm 2 (Fundamental theorem of algebra):

Any polynomial2

0 1 2

0 0)

( ) ( 0)of degree ( 1) has at least one zero.i.e., there exists at least one point s.t. ( 0

nn n

P z a a z a z ......... a z an n

z P z

Pf. by contradiction

tch-prob 49

Suppose that P(z) is not zero for any value of z.

Then is clearly entire and it is

also bounded in the complex plane.

1( )( )

f zP z

To show that it is bounded, first write

0 11 21 2

So that ( ) ( )

nn n n

nn

zz z za aa a

w ....

P z a w z

Can find a sufficiently large positive R such that

0 1 0 11 2 11 2 1

20 1 1, when ni

n i

n nn n n n n

az

z n

w z zz z z z z

a , i , , ........n- R

a a a aa a a.... .....

when2n

a z R

Generalized triangle inequality

tch-prob 50

2

( )2 2

1 2Then ( )( )

nn n

n nnn nn

nn

aa w a w

a aP z a w z z R z R

f z z Ra RP z

So, is bounded in the region exterior to .

But is continuous in is bounded in .

is bounded in the entire plane.

From Liouville's theorem, ( ) is constant.

f z R

f z R f z R

f

f z

( ) is constant.

But ( ) is not constant We have a contradiction.

P z

P z

tch-prob 51

From the (F. T. 0. A) theorem

any polynomial P(z) of degree n can be expressed as

1 1) ( )( ) ( 10 sec 42

zP z z z Q (Ex. , )

Polynomial of degree n-1

1 2 2( ) ( )but ) ( z Q z z z Q

Polynomial of degree n-2

1 2 2

1 2

) ) ( )

) ) )

( ) ( (

( ( (n

zP z z z z z Q : : c z z z z ....... z z

tch-prob 52

42. Maximum Moduli of FunctionsLemma. Suppose that f(z) is analytic throughout a neighborhood

0 0 0) of a point . If ( ) (z z ε z f z f z

for each point z in that neighborhood, then f(z) has the constant value f(z0) throughout the neighborhood .

00

2 00

200

)

( )

( )

( )1(2

121

2

iθ

iθ

iθ

f z ρe

ρe

f z ρe dθ

f z dzf z c z zπ i

iθ i ρ e dθπ i

π

0: 0 2iθCρ z z ρe ( θ π)

z1

z0

C

1 0

1 0

0

: Let be any point other than in the neighborhood.

Let be the distance between and .

: circle .

Pf z z

z z

C z z

f’s value at the center is the arithmetic mean of its values on the circle. ~ Gauss’s mean value theorem.

tch-prob 53

200 0

0 0

2 20 0 00 0

200 0

) ( )

) )

( ) ( ) )

) ( )

1( (3)2

On the other hand, since ( ( (4)

2 (

1(2

i

i

i

i

f z ρe d

f z ρe d f z

f z ρe d

f z ..........π

f z ρe f z ..........

dθ π f z

f z π

(5) ......

From (3) and (5)

200 0

20 00

0 from (4), and continuous in

) ( )

( ) ( )

1(2

or 0

i

i

f z ρe

f z f z ρe

f z dθπ

dθ

tch-prob 54

0 0

0 0

0 0

) )

)

)

( ( 0 2

( ) ( for all points in

( ) ( for all points in

From Ex 7(c), sec 22, (when the modulus of an analytic function is

if z f z ρe ( θ π)

f z f z z z- z ρ

f z f z z z- z ε

0)

constant in a domain, the function itself is constant there.)( ) ( for all points in the neighborhood.f z f z z

Thm. (maximum modulus principle)

If a function f is analytic and not constant in a given domain D, then has no maximum value in D. That is, there is no point z0 in the domain such that for all points z in it.

)(zf)0()( zfzf

tch-prob 55

d

N0 N1N2

z0 z1 z2

N n-1

Nnzn-1 zn

p

DPf:

-1's is in neighborhood

k kz z

Construct : k k

N z z d

0

is a connected open set.

There is a polygonal line lying entirely in that extends from , to any

other point in .

Let be the shortest distance from points on to the boundary of .

We then

D

L D z

P D

d L D

0 1 2

1

form a finite sequence of points , , , , (= ) along , where each point

is sufficiently close to the adjacent ones that

( 1, 2, , )

n

k k

z z z z P L

z z d k n

tch-prob 56

Assume f(z) has a max value in D at z0.

f(z) also has a max value in N0 at z0.

1 0 1 0

1 1

1 0

) ) )

)

from Lemma ) )

( ( ( in

( ) ( for all in

( ) ( (

f z f z z N

f z f z z N

f z f z f z

From Lemma, f(z) has constant value f(z0) throughout N0.

2 0 2 1

0

) ) ( in )

) )

( ( :

( ( for every in ( ) is constant a contradiction( ) has no maximum value in

n

z Nf z f z

f z f z z Df z f z D

tch-prob 57

If a function f that is analytic at each point in the interior of a closed bounded region R is also continuous throughout R, then the modulus has a maximum value somewhere in R. (sec 14) p.41

)(zf

. ..( ) for all points in .

i ef z M z R

If is a constant fumction,then ( ) for all in

ff z M z R.

not continuouson boundary

Corollary: Suppose f is continuous in a closed bounded region R.

and that it is analytic and not constant in the interior of R.

Then Maximum value of in R occurs somewhere on the boundary R and never in the interior.

)(zf

Maximum at the boundary.