高等數(shù)學(xué)電子教案同濟(jì)復(fù)變函數(shù)2第5章

1.

留數(shù)的定義

2.

留數(shù)定理

3.

留數(shù)的計(jì)算規(guī)則§2 留數(shù)(Residue)1.

留數(shù)的定義定義

設(shè)z0為f

(z)的孤立奇點(diǎn)，

f

(z)在z0鄰域內(nèi)的洛朗級(jí)數(shù)中負(fù)冪次項(xiàng)(z-z0)–1的系數(shù)c–1稱為f

(z)在z0的留數(shù)，記作Res[f

(z),z0]或Res

f

(z0)。+￥n=-￥nnc

(z

-

z

)0由留數(shù)定義, Res[f

(z),

z0]=

c–1

(1)設(shè)f

(z)=cdz-10-1

c

z

-

zf

(

z)dz

=

c=

2pic(

z0是f

(

z

)的弧立奇點(diǎn)

,

c包含z0在其內(nèi)部

)對(duì)上式兩邊沿簡單閉曲線c逐項(xiàng)積分得：(2)0f

(z)dzc=

1-12pi故

Res[

f

(z),

z

]

=

c2.

留數(shù)定理c

f

(z)dz

=

2pi

Re

s[

f

(z),

zk

]

(3)定理設(shè)c是一條簡單閉曲線,f

(z)在c內(nèi)有有限個(gè)弧立奇點(diǎn)z1

,z2

,,zn

,除此以外,f

(z)在c內(nèi)及c上解析nk

=1證明用互不包含,互不相交的正向簡單閉曲線ck(k

=1,2,n)，將c內(nèi)的弧立奇點(diǎn)zk圍繞,Dcznz1z3z2nncnf

(z)dz1c

f

(z)dz

=

12pi

k

=1

2pif

(z)dzc

f

(z)dz

=

cf

(z)dz

+

c21f

(z)dz

+

+

cn由復(fù)合閉路定理得：用2pi

除上式兩邊得:=

Re

s[

f

(z),

zk

]k

=1nk

=1故

c

f

(z)dz

=

2piRe

s[

f

(z),

zk

]得證！

求沿閉曲線c的積分，歸之為求在c中各孤立奇點(diǎn)的留數(shù)。3.

留數(shù)的計(jì)算規(guī)則一般求Res[f(z),z0]是采用將f(z)在z0鄰域內(nèi)展開成洛朗級(jí)數(shù)求系數(shù)c–1的方法,但如果能先知道奇點(diǎn)的類型，對(duì)求留數(shù)更為有利。以下就三類奇點(diǎn)進(jìn)行討論：(i)若z

=z0為可去奇點(diǎn)

c-1

=0

Re

s[f

(z),z0

]=0規(guī)則I(4)Re

s[

f

(

z),

z0

]

=

lim(

z

-

z0

)

f

(

z)若z0是f

(z)的一級(jí)極點(diǎn),z

fi

z0規(guī)則II

若z0是f

(z)的m級(jí)極點(diǎn)(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為極點(diǎn)時(shí)，求Re

s[f

(z),z0

]有以下幾條規(guī)則c

(z

-

z

)n

n

0-￥展開+￥(ii)若z

=

z

為本性奇點(diǎn)

f

(z)

=+c

(z

-

z

)m

+事實(shí)上，由條件-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

)

乘上式兩邊

,

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

0兩邊求m

-1階導(dǎo)數(shù)得dzmm

-1

{(z

-z0

)

f

(z)}=(m

-1)!c-1

,移項(xiàng)得(5)式.d

m

-1limz

fi

z0

當(dāng)m=1時(shí)，式(5)即為式(4).00Q'(z

)p(z0

)

(6)Q(z)p(z0

)

?

0,

Q(z0

)

=

0,

Q'(z0

)

?

0

設(shè)f

(z)

=

p(z)

p(z),

Q(z)在z

處解析,規(guī)則III1的一級(jí)極點(diǎn),Q(z)\z0為Q(z)的一級(jí)零點(diǎn),從而z0為z0是f

(z)的一級(jí)極點(diǎn),且Re

s[f

(z),z0

]=事實(shí)上，

Q(z0

)=0及Q'(z0

)?01

10Q(

z

)

z

-

zj

(z

)(j

(z

)在z0處解析且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)

得證！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則z0為f

(z)的-級(jí)極點(diǎn),由規(guī)則Iz

=2

z(z

-

1)2

5z

-

2

dz計(jì)算:例1z(z

-

1)2解

f

(z)

=

5z

-

2

在

z =

2的內(nèi)部有一個(gè)一級(jí)極點(diǎn)Re

s[

f

(

z

),0]

=

lim

zf

(

x

)

=

lim

5z

-

2

=

-2z

fi

0

(

z

-

1)

2z

=0和一個(gè)二級(jí)極點(diǎn)z

=1由規(guī)則I1

5z

-

2

}d

{(z

-1)2Re

s[

f

(z),1]

=

limzfi

1

(2

-1)!

dz

z(z

-1)2z

fi

0由規(guī)則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計(jì)算

z

dz例2解

f

(z)有4個(gè)一級(jí)極點(diǎn):–1,–i都在圓周c內(nèi)，Q'(z)

4z

3

4z

2=P(z)

=

z

1由規(guī)則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

計(jì)算解f

(z)=cos

z

有一個(gè)z

=0的三級(jí)奇點(diǎn)\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由規(guī)則I(n

?

N

)計(jì)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

為一級(jí)極點(diǎn),由法則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 由留數(shù)定理得：

(1)要靈活運(yùn)用規(guī)則及洛朗級(jí)數(shù)展開來求留數(shù)，不要死套規(guī)則。z6Q(z)f

(z)

=

P(z)

=

z

-

sin

z\z

=0是p(z)的三級(jí)零點(diǎn),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