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ChapterV

Residue

§1

isolatedsingularity

Functiondoesnotresolvethepointof

singularity.

Ifthefunction

f(z)

althoughnoresolution

in

z

0,

z

0

inacertainneighborhoodtothecenter

0

<|

z

z

0|

withintheanalytic

everywhere,

then

z

0

as

f(z)

ofthe

isolated

singularity.

Thefunction

f(z)

initsisolatedsingularity

z

0

totheheartoftheneighborhood

0

<|

z

z

0|

bestartedintoaLaurent

series.

Accordingtothedifferentexpansionconditionsforclassificationofisolated

singularities.

Removablesingularity

freeiftheLaurentseriesin

z

z

0

ofthenegativepowerkey,

thenisolated

Legislationknownasthesingularpoint

z

0

f(z)

oftheremovablesingularity.

Then,

f

(z)

=

c

0

+

c

1

(z

z

0)

+...+

c

n

(z

z

0)

n

+....

0

<|

z

z

0

|

Inthecirculardomain

|

z

z

0

|

inthere

f

(z)

=

c

0

+

c

1

(z

z

0

)+...+

c

n

(z

z

0)

n

+...,

Thusthefunction

f(z)

at

z

0

onaresolutionof

the.

So

z

0

iscalledremovable

singularity.

2.

Pole

intheLaurentseries,ifonlyalimitednumberof

z

z

0

ofthenegativepoweritems

Andoneonthe

(z

z

0)

1

themaximumpowerforthe

(z

z

0)

m,

thatis

f

(z)

=

c

m

(z

z

0)

m

+...+

c

2

(z

z

0)

2

+

c

1

(z

z

0)

1

+

c

0

+

c

1

(z

z

0

)+...

(M

1,

c

m

0),

thenisolatedsingularpoint

z

0

iscalledthefunction

f(z)

the

m-classpole.

Thiscanalsobewrittenas

Where

g(z)

=

c

-

m

+

c

-

m

+1

(z

-

z

0)

+

c

-

m

+2

(z

-

z

0)2

+...,

In

|

z

-

z

0

|

<d

areanalyticfunctions

within,

and

g(z

0)

0.

Conversely,

whenafunction

f(z)

canbeexpressedas

(*)form,

and

g(z

0)

0,

then

z

0

is

f(z)

the

m-class

pole.

If

z

0

is

f(z)

the

poles,

by

(*),

we

have

3.

Thenatureofsingularity

IftheLaurentseriescontainsaninfinitenumberof

z-

z

0

ofthenegativepower

key,

then

z

0

iscalledanisolatedsingularity

f(z)

thenatureof

singularity.

In

summary:

Wecanusethislimittodistinguishthedifferentsituationsofthetypeisolatedsingularity.

4.A

function

of

therelationshipbetweenthezeroandpole

Isnotidenticallyequaltozeroanalyticfunction

f(z)

canbeexpressedasif

f

(z)

=

(z

z

0)

m

(z),

where

(z)

analyticin

z

0

and

(z

0)

0,

m

isapositive

integer,

then

z

0

as

f(z)

of

m-level

zero.

Forexample,when

f

(z)

=

z(z

1)

3

時(shí),

z

=0

and

z

=1

isitsoneandthree

zeros.

Accordingtothis

definition,

wecangetthefollowing

conclusions:

If

f(z)

analytic

in

z

0,

then

z

0

is

f(z)

of

m-level

isnecessaryandsufficientconditionzero

f

(n)(z

0)

=

0,

(n

=0,1,2,...,

m

1),

f

(m)(z

0)

0.

Thisis

because,

if

f(z)

at

z

0

resolution,

youwillbeabletostartaneighborhoodof

z

0

fortheTaylor

series:

f(z)

=

c

0

+

c

1

(z

z

0

)+...+

c

m

(z

z

0)

m

+...,

EasyCard

z

0

is

f(z)

of

m-level

necessaryandsufficientconditionzerocoefficientoftheformer

m

c

0

=

c

1

=...=

c

m

1

=

0,

c

m

0,

whichisequivalentto

f

(n)(z

0)

=

0,

(n

=0,1,2,...,

m

1),

f

(m)(z

0)

0.

Forexample,

z

=1

is

f(z)

=

z

3

-1

zero,

because

f

'(1)=

3

z

2

|

z

=1=3

0,

so

z

=1

isknown

f(z)

ofazero

.

Since

f(z)

=

(z

-

z

0)

m

j(z)

of

j(z)

analytic

in

z

0,

and

j(z

0)

0,

soit'sintheneighborhoodof

z0

isnot

zero.

Thisbecause

(z)

analyticin

z

0,

z

0

willbein

continuous,

sogiven

So

f

(z)

=

(z

z

0)

m

(z)z

0

intheneighborhoodtotheheartisnot

zero,

thatisnotidentically

Zerosofanalyticfunctionsofzerois

isolated.

Theorem

If

z

0

is

f(z)

the

m-class

pole,

then

z

0

isthe

m-level

zero,

Inturnset

up.

Determinethefunctionofthistheoremprovidesasimplepole

method.

Example

2

Example

3

To

Discussthefunction

In

DepartmentofState.

5.

FunctionInfinitybehavior

ifthefunction

f(z)

atinfinity

z

Totheheartoftheneighborhood

R

<|

z

|

withinthe

resolution,

saidthepoint

as

f(z)

oftheisolated

singularity.

Totransform

Theexpansionofthe

z

planetotheheart

neighborhood

R

<|

z

|<+

mappedtoexpandthe

w

planetotheoriginofheartNeighborhood:

Another.

Inthis

way,

wecantakeheartintheneighborhoodto

R

<|

z

|

on

f(z)researchintothe

Within

(w)

for

research.

Clearly

(w)

in

Withintheresolution,

so

w

=0

isanisolated

singularity.

f(z)

atinfinity

z

Thetypeofsingularpoint

Equivalentto

(w)

at

w=0

thesingularitytype.

That

z

is

f(z)

ofremovable

singularities,

poles,orthenatureof

singularity,

totallylimit

Theexistenceof

(finitevalue),

forinfinityortheinfinitedoesnotexistarenot

determined.

Example

1

Example

2

Example

3

§2

residue

Definitionoftheresidueandtheresidue

theorem,

ifthefunction

f(z)

z

0

in

aneighborhoodof

D,

Resolution,

thentheCauchyintegraltheorem

However,

if

z

0

is

f(z)

ofanisolatedsingularity,

Thenalongthecenter

z

0

ofaneighborhoodto

0

<|

z

z

0|

<R

z

0

containsanyofabeingasimpleclosedcurve

C

oftheintegral

Generallynotequalto

zero.

Therefore

f

(z)

=...+

c

n

(z

z

0)

n

+...+

c

1

(z

z

0)

1

+

C

0

+

c

1

(z

z

0

)+...+

c

n

(z

z

0)

n

+...

0

<|

z

z

0

|

<R

Bothendsofthe

pointsonebyone

alongthe

C:

That

C

1

as

f(z)

z

0

inthe

residue,

denotedbyRes

[f(z),z

0],

ie

Theorem1

(ResidueTheorem)

Setfunction

f(z)

inregion

D,

exceptafinitenumberofisolatedsingularpoint

z

1,

z

2,

...,

z

n

analyticeverywhere

outside.

C

is

D,

allsurroundedbyapositivesingularsimpleclosed

curve,

then

D

z

1

z

2

z

3

z

n

C

1

C

2

C

3

C

n

C

[Evidence]

tothe

C

intheisolatedsingularpoint

z

k(k

=1,2,...,

n)

areincludedwitheachother

Tothesimpleclosedcurve

C

k

around

up

thereundercompositeclosed-circuittheorems

Notethattheconditionsofthetheoremtobesatisfied.

Suchas

Cannotapplytheresiduetheorem.

Demandfunctionintheisolatedsingularity

z

0

OfficeoftheresiduethatisseekingitsLaurentseriesin

(Z

z

0)

1

termcoefficient

c

1

can

be.

Butifyouknowthetypeofsingularpoint

of

Seekingtostayafewmaybemore

favorable.

If

z

0

is

f(z)

oftheremovablesingularity,

Then

Res

[f(z),z

0]

=

0.

If

z

0

isthenatureof

singularity,

thenhadtobestartedbyLaurent

series.

If

z

0

is

Pole,

therearesomerequirements

c

1

useful

rule.

2.

Residuecalculationrules

Rule

1

if

z

0

is

f

(z)

ofapole,

Then

Rule

2

If

z

0

is

f

(z)

the

m-class

pole,

then

In

fact,

f

(z)

=

c

m

(z

z

0)

m

+...+

c

2

(z

z

0)

2

+

c

1

(z

z

0)

1

+

c

0

+

c

1

(z

z

0

)+...,

(Z

z

0)

m

f(z)

=

c

m

+

c

m

+1

(z

z

0

)+...+

c

1

(z

z

0)

m

1

+

c

0

(z

z

0)

m

+...,

Sothatbothendsofthe

z

z

0,

therightendofthelimitis

(m

1)!

C

1,

bothendsofthedividedby

(m

1)!

That

Res

[f(z),z

0],

whichmayrule

2,

when

m

=

1,

istherule

1.

Thatwas

Rule

3.

Bytherule

1,

may

Wecanalsousetherulestoseektoremainthenumber

3:

Thisis

simpler

than

someof

therules

1.

Example

5

Solution:

So

Originalstyle

=

Example

4

Solution:

z=0

isapole.

3.

Infinityresidue

setfunction

f(z)

intheringdomain

R

z

|<

Analytic,

C

fortheringareaaroundtheoriginofanyofthesimpleclosed

curve,

thentheintegral

Thevalue

hasnothingtodo

with

C,

called

f(z)

pointsin

residue,

denoted

f(z)

intheringdomain

R

z

|<

Analytic:

Interpretedasaringaroundthearea

Asimpleclosedcurveofany.

That

is,

f(z)

tostay

in

pointisequaltoitsnumberofpointstothe