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Ch7.2 Calculus of Residues講者： 許永昌 老師

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ContentsResidue TheoremEvaluation of Definite IntegralsCauchy Principle values

Some poles on the integral path.Pole expansion of Meromorphic FunctionProduct Expansion of Entire function

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Residue Theorem( 請預讀 P378~P379)

Laurent expansion

Closed contour

integration

Cauchy’s integral theorem

Residue Theorem

0

n

nn

f z a z z

1

1

2f z dz a

i

10

2for analytic function

except for isolated poles.

f z dzi

STOP TO THINK: How about multivalent function?

1,

1Each around ,

2 ii

i i zCC z f z dz a

i

1,

1

2 izf z dz a

i

Besides, {residue at z} -

S{residues in the finite z-plane}

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ResidueResidue:

In some books, e.g. J. Bak and D.J. Newman, Complex Analysis, it denotes by Res(f ; zi).

Exercise: Hint: Homework 7.1.1

a-n.

-a-n. (Q: 為何不是 -a-1*0=0? 所以要搞清楚 Laurent expansion.) 用 ordinary series expansion 比較快找到值。

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Res ;- n

az a z

1

Res ;0- nz a z

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Evaluation of Definite Integrals ( 請預讀 P379~P384)

4 types we will discussed here:

Hint: (1) |z|=1, (2) cosq=(z+z-1)/2, (3)sinq=(z-z-1)/(2i)

Related to Jordan’s Lemma.

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1 0sin ,cos ,I f d

2 ,

(1) no pole on the path (2) lim 0 for upper (lower) half planez

I f z dz

zf z

3 ,

(1) no pole on the path (2) 0 (3) lim 0 for upper half plane

iax

z

I f z e dz

a f z

4 0

keyhole contour

, Use a multivalent function ln to solve it.

ln , ln 0 when 0 & .2

I f z dz z

iz f z dz z z f z z z

0lim ln 0.zz z

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Exercises ( 請預讀 P379~P384)

Step 1: find the singular points.Step 2: find a suitable contour.Step 3:

For branch point, we must consider the branch cut. For poles, find a-1 on each pole.

Step 4: Residue theorem.

2

1 0

2 2

4 30

, 11 cos

,1

,1

dI

dxI

xdx

Ix

Code: quadgk(@(z)(1./(1+z.^2)),-inf,inf)quadgk(@(z)(1./(1+z.^3)),0,inf)

小心， (1) 確定 q 的範圍 (2) 一整圈為 2p 。

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Upper half circle whose radius is

iazR

R

I f z e dz

Jordan’s Lemma( 請預讀 P383)

If (1) a>0, aR, (2) lim|z| f (z)=0, 0 arg(z) p,We get limR|IR|=0,

Proof:

cos sin

0

sin

,max 0

22

,max 0

,max

2

1

.

i iaR aR iR

aR

C

aR

C

aR

C

I f R e e e iR e d

R f e d

R f e d

f ea

a

0 pi/2 pi0

0.5

1

1.5

2

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Cauchy Principle Value( 請預讀 P384)

Situation:Some singular points are directly on the

contour of integration.We define

When f (x0) is finite, the principle value limit is unnecessary.

For a simple pole, Therefore,

0

00lim

x

xf x dx f x dx P f x dx

semicircle

1

2 circlef z dz f z dz

.C C Semicircle

P f x dx f z dz f z dz f z dz

C

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Pole Expansion of Meromorphic Functions ( 請預讀 P390~P391)

Mittag-Leffler theorem: A meromorphic function can be written as

If all the poles of this function are simple poles, we get

,

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1make the sum converge entire functionpk

k nn

n k

k kk

a

z z

f z S z z E z

STOP TO THINK: 請問與 Taylor and Laurent expansion 有何不同 ?

1

1

1 10 if and ,

where is a radius circle includes ,..., but no other poles.

kk kC

k k k

k k k

f z f b f z R k Nz z z

C R z z

1

11

1

00 ' 0 ...

!

if and .k

pp pk

pk k n

pkC

z f b zf z f zf

p z z z

f z R k N

We need to prove that their remainder converges to zero.

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Pole Expansion of Meromorphic Functions (continue)Proof for |f(z)|<eRk case:

The remainder for |f(z)|<eRkp+1 case:

1

1

if is analytic at 0 and z 1. Laurent expansion at ,2. Cauchy's integral formula,3. Simple pole at

1

2

Res ;0 Res ; Res ;

0 1 1 1

k

n

k C

F

k

nn

k

nn n n

f z

fI d

i z

F F z F z

f f zb

z z z z z z

.

.

nz

max on

1 1 12

2 2kkk

k

k kCR zk Cf R

f fI d R

z R z

1

1.

2 kk pC

fI d

i z

記得，是 F(w) 的 residue 。

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ExercisesTest the pole expansion for:

Test them with the remainder to understand the meaning of |f(z)|<eRk

p+1.

11 ,

1f z

z

11 ,

1f z z

z

1

1.

2 kk pC

fI d

i z

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Example ( 請預讀 P391)

Pole expansion of cotangent: p cotpz=

Method I: Its pole is located at z=n, nZ. We will find that Choose Rk=k+0.5, we get (I did not test it) |f(z)|

<eRk. Therefore, based on Mittag-Leffler theorem, we get

Method II: The product expansion of

cos ln sin.

sin

z d z

z dz

1

coslim 1, they are simple poles whose 1 for all .

sinz n

zz n a n

z

0

1 1 1cos

nn

zz z n n

2

2 21

sin 1 .n

zz z

n

13

1

,kn

q

kk

f z g z z z

Product Expansion of Entire Functions ( 請預讀 P392~P394)

An entire function with zeros at z1,…, zn can be written as where g(z) is an entire function with no zero.Questions:

How to find the number of zeros in a region? How to do this product expansion for an entire

function?Key concept:

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entire functionSimple poles

ln ' '.

nk

k k

d f z f z g zq

dz f z z z g z

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Product Expansion of Entire Functions (continue)

How to find the number of zero points?

How to do the expansion? From the pole expansion, if |F(z)/Rk|<e, we get

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entire functionSimple poles

ln ' '.

nk

k k

d f z f z g zq

dz f z z z g z

1

'1

2

1arg . arg ument

2

n

kCk

C

f zdz q

i f z

f z f z

代表繞一圈後 的 的變化。

1

' ' 0 1 1.

0 kk k k

f z fq

f z f z z z

01

' 0

0

1

' ' ' 0' ln ln 1

' 0 0

0 1 .k k

k

zk

kk k k

qf q zz

f z

k k

f z f z f q zzdz z q

f z f f z z

zf z f e e

z

小心：此處要求 '

.kg z

Rg z

'

ln ln argf z

dz f z f z i f zf z

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Example

Zeros: mp, m0. f(0)=1 f ’(0)=0

STOP TO THINK: cos(z) ?

sin:z

z

' 0

0

1

0 1 .k k

k

qf q zz

f z

k k

zf z f e e

z

,

0

' 1 1 1cot . 1.k

nn

f zz q

f z z z n n

2

, 10

sin1 1

z

n

n nn

z z zf z e

z n n

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Rouché’s theoremIf f(z) and g(z) are analytic inside and on a

closed contour C, and |g(z)|<|f(z)| on C, then f(z) and f(z)+g(z) have the same number of zeros inside C.Proof:

DC arg(1+g(z)/f(z))=0 DC arg f(z)=DC arg [f(z)+g(z)]Z-plane w-plane

1g z

wf z

1

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Homework1, 2, 3, 4, 6, 9, 14, 16, 21, 22

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NounsResidue of f(z) at z=z0: a-1, the coefficient

of (z-z0)-1 in the Laurent expansion: Res(f ; z0).

Jordan’s Lemma: P383Cauchy Principle Value: P384Pole Expansion: P390Product Expansion: P392