高等数学电子教案同济复变函数2第5章

1.

留数的定义

2.

留数定理

3.

留数的计算规则§2 留数(Residue)1.

留数的定义定义

设z0为f

(z)的孤立奇点，

f

(z)在z0邻域内的洛朗级数中负幂次项(z-z0)–1的系数c–1称为f

(z)在z0的留数，记作Res[f

(z),z0]或Res

f

(z0)。+¥n=-¥nnc

(z

-

z

)0由留数定义, Res[f

(z),

z0]=

c–1

(1)设f

(z)=cdz-10-1

c

z

-

zf

(

z)dz

=

c=

2pic(

z0是f

(

z

)的弧立奇点

,

c包含z0在其内部

)对上式两边沿简单闭曲线c逐项积分得：(2)0f

(z)dzc=

1-12pi故

Res[

f

(z),

z

]

=

c2.

留数定理c

f

(z)dz

=

2pi

Re

s[

f

(z),

zk

]

(3)定理设c是一条简单闭曲线,f

(z)在c内有有限个弧立奇点z1

,z2

,,zn

,除此以外,f

(z)在c内及c上解析nk

=1证明用互不包含,互不相交的正向简单闭曲线ck(k

=1,2,n)，将c内的弧立奇点zk围绕,Dcznz1z3z2nncnf

(z)dz1c

f

(z)dz

=

12pi

k

=1

2pif

(z)dzc

f

(z)dz

=

cf

(z)dz

+

c21f

(z)dz

+

+

cn由复合闭路定理得：用2pi

除上式两边得:=

Re

s[

f

(z),

zk

]k

=1nk

=1故

c

f

(z)dz

=

2piRe

s[

f

(z),

zk

]得证！

求沿闭曲线c的积分，归之为求在c中各孤立奇点的留数。3.

留数的计算规则一般求Res[f(z),z0]是采用将f(z)在z0邻域内展开成洛朗级数求系数c–1的方法,但如果能先知道奇点的类型，对求留数更为有利。以下就三类奇点进行讨论：(i)若z

=z0为可去奇点

c-1

=0

Re

s[f

(z),z0

]=0规则I(4)Re

s[

f

(

z),

z0

]

=

lim(

z

-

z0

)

f

(

z)若z0是f

(z)的一级极点,z

fi

z0规则II

若z0是f

(z)的m级极点(5)100lim

{(z

-

z

)m

f

(z)}Re

s[

f

(z),

z

]

=dzm

-1d

m

-1(m

-

1)!

zfi

z00

Re

s[

f

(z),

z0

]

=

c-1(iii

)若z

=z0为极点时，求Re

s[f

(z),z0

]有以下几条规则c

(z

-

z

)n

n

0-¥展开+¥(ii)若z

=

z

为本性奇点

f

(z)

=+c

(z

-

z

)m

+事实上，由条件-m

-2

-1f

(z)

=

c-m

(z

-

z0

)

+

+

c-2

(z

-

z0

)

+

c-1

(z

-

z0

)+c0

+

c1

(z

-

z0

)

+,(c-m

„

0)m以(

z

-

z0

)

乘上式两边

,

得(z

-

z

)m

f

(z)

=

c

+

c

(z

-

z

)

++

c

(z

-

z

)m-10

-m

-m+1

0

-1

0f

(z)}

=

(m

-

1)!c-1

+

m!(z

-

z0

)

+mdz

m

-1

{(z

-

z0

)d

m

-10

0两边求m

-1阶导数得dzmm

-1

{(z

-z0

)

f

(z)}=(m

-1)!c-1

,移项得(5)式.d

m

-1limz

fi

z0

当m=1时，式(5)即为式(4).00Q'(z

)p(z0

)

(6)Q(z)p(z0

)

„

0,

Q(z0

)

=

0,

Q'(z0

)

„

0

设f

(z)

=

p(z)

p(z),

Q(z)在z

处解析,规则III1的一级极点,Q(z)\z0为Q(z)的一级零点,从而z0为z0是f

(z)的一级极点,且Re

s[f

(z),z0

]=事实上，

Q(z0

)=0及Q'(z0

)„01

10Q(

z

)

z

-

zj

(z

)(j

(z

)在z0处解析且j

(z0

)„0)因此,

=10z

-

z且g(z0)„0),故f

(z

)=g(

z

)(

g(

z

)

=

j

(

z

)

p(

z

)在z

解析,000=(Q'(z

)„0)

得证！Q'(z

)p(z

)

0

z

-

z0Q(

z)

-

Q(z

)p(z)Re

s[

f

(z),

z0

]

=

lim(z

-

z0

)

f

(z)=

limz

fi

z0z

fi

z00则z0为f

(z)的-级极点,由规则Iz

=2

z(z

-

1)2

5z

-

2

dz计算:例1z(z

-

1)2解

f

(z)

=

5z

-

2

在

z =

2的内部有一个一级极点Re

s[

f

(

z

),0]

=

lim

zf

(

x

)

=

lim

5z

-

2

=

-2z

fi

0

(

z

-

1)

2z

=0和一个二级极点z

=1由规则I1

5z

-

2

}d

{(z

-1)2Re

s[

f

(z),1]

=

limzfi

1

(2

-1)!

dz

z(z

-1)2z

fi

0由规则IIz=

lim(

5z

-

2

)'

=

lim

2

=

2zfi

1

z

2zfi

1=\

z

2

f

(z)dz

=

2pi

Re

s[

f

(z),0]

+

2pi

Re

s[

f

(z),1]

=

0c

z

4

-

1c

:

正向

z

=

2计算

z

dz例2解

f

(z)有4个一级极点:–1,–i都在圆周c内，Q'(z)

4z

3

4z

2=P(z)

=

z

1由规则I-

1]

=

01

14 4

-

4

4dzzc

z4

-1=

2pi[1

+=

2pi{Res[

f

(z),-1]+

Res[

f

(z),1]+

Res[

f

(z),i]+

Res[

f

(z),-i]}故z

1=z

3cos

z

dz例3

计算解f

(z)=cos

z

有一个z

=0的三级奇点\z2cos

z

dz

=

2pi

Re

s[

f

(z),0]

=

2pi(-

1

)

=

-pi=1z

32

213=

1

lim

(cos

z

)'

'

=

-

1d

2Re

s[

f

(

z

),0]

=

lim

2

[

z f

(

z

)]z

fi

0z

fi

0

(

3

-

1)!

dzz3由规则I(n

˛

N

)计算z n

tan

pzdz=例4解22即,

z

=

k

+

1

(k

=

0,–1,–2,)解得pz

=kp

+ptanpz

=sinpz

令cospz

=0cospz22„

0z

=k

+

1z

=k

+

1(cot

pz)'

=

-p

csc2pz2\z

=k

+1

为一级极点,由法则III得2=

-

1

(k

=

0,–1,)p2

(cospz)'

z=k

+

1sinpzRe

s[tanpz,

k

+

1]

=2k

+

1

<nz

=nptanpzdz

=

2pi

Re

s(tanpz)

=

2pi(-

2n

)

=

-4ni故 由留数定理得：

(1)要灵活运用规则及洛朗级数展开来求留数，不要死套规则。z6Q(z)f

(z)

=

P(z)

=

z

-

sin

z\z

=0是p(z)的三级零点,由于p(0)

=

0

p'(0)

=

(1

-

cos

z)

z=0

=

0p"(0)

=

sin

z

z=0

=

0

p'"(0)

=

cos

z

z=0

=

1

„

0如