復(fù)變函數(shù)與積分變換第一章復(fù)數(shù)

第六節(jié)復(fù)變函數(shù)的極限和連續(xù)性一、函數(shù)的極限

二、函數(shù)的連續(xù)性三、小結(jié)與思考一、函數(shù)的極限1.函數(shù)極限的定義:設(shè)函數(shù)w

=f

(z)定義在z0

的去心鄰域0

<z

-z0

<r

內(nèi),如果有一確定的數(shù)A

存在,對于任意給定的e

>0,相應(yīng)地必有一正數(shù)d(e)那末稱A

為f

(z)當(dāng)z

趨向于z0

時的極限.<d(0

<d

￡

r)時,有f

(z)-A

<e使得當(dāng)0

<z

-z0zfi

z0

fi

A)記作lim

f

(z)=A.(或f

(z)zfi

z0注意:定義中z

fiz0

的方式是任意的.lim

u(

x,

y)

=

u0

, lim

v(

x,

y)

=

v0

.xfi

x0yfi

y0xfi

x0yfi

y0zfi

z02.

極限計算的定理定理一設(shè)

f

(

z)

=

u(

x,

y)

+

iv(

x,

y),

A

=

u0

+

iv0

,z0

=

x0

+

iy0

,

那末

lim

f

(z)

=

A的充要條件是證

(1)

必要性.如果lim

f

(z)=A,zfi

z0根據(jù)極限的定義當(dāng)0

<(x

+iy)-(x0

+iy0

)<d

時,(u

+iv)-(u0

+iv0

)<e,2

2(x

-x0

)

+(y

-y0

)

<d

時,或當(dāng)0

<(u

-

u0

)

+

i(v

-

v0

)

<

e,

u

-

u0

<

e,v

-

v0

<

e,xfi

x0yfi

y0xfi

x0yfi

y0故

lim

u(

x,

y)

=

u0

, lim

v(

x,

y)

=

v0

.lim

v(

x,

y)

=

v0

,xfi

x0yfi

y0xfi

x0yfi

y0(2)

充分性.

若

lim

u(

x,

y)

=

u0

,2

2(x

-x0

)

+(y

-y0

)

<d

時,那么當(dāng)0

<2

20<

e

,

v

-

v

<

e

,0有u

-uf

(z)

-

A

=

(u

-

u0

)

+

i(v

-

v0

)￡

u

-

u0

+

v

-

v0故當(dāng)

0

<

z

-

z0

<

d

時,f

(z)

-

A

<

e,所以lim

f

(z)=A.zfi

z0說明[證畢]該定理將求復(fù)變函數(shù)f

(z)=u(x,y)+iv(x,y)的極限問題,轉(zhuǎn)化為求兩個二元實(shí)變函數(shù)u(x,y)和v(x,y)的極限問題.定理二g(z)

B(3)

lim

f

(z)

=

A

(B

?

0).設(shè)lim

f

(z)

=

A, lim

g(z)

=

B,

那末zfi

z0

zfi

z0lim[

f

(z)

–

g(z)]

=

A

–

B;zfi

z0lim[

f

(z)g(z)]

=

AB;zfi

z0zfi

z0與實(shí)變函數(shù)的極限運(yùn)算法則類似.證(一)不存在.z例1

證明函數(shù)f

(z)=Re(z)當(dāng)z

fi

0

時的極限令z

=x

+iy,,則f

(z)=x2

+

y2xxu(

x,

y)

=

,

v(

x,

y)

=

0,x2

+

y2當(dāng)z

沿直線y

=kx

趨于零時,x2

+

y2lim

u(

x,

y)

=

limxfi

0y=kxxfi

0y=kxxx2

+

(kx)2=

limxfi

0x=

limxfi

0xx2

(1

+

k

2

) 1

+

k

2,1=

–隨k

值的變化而變化,所以lim

u(x,y)不存在,xfi

x0yfi

y0lim

v(

x,

y)

=

0,xfi

x0yfi

y0根據(jù)定理一可知,

lim

f

(z)

不存在.zfi

0令z

=r(cosq

+i

sinq

),證(二)r則f

(z)=r

cosq

=cosq,當(dāng)z

沿不同的射線arg

z

=q

趨于零時,f

(z)趨于不同的值.例如

z

沿正實(shí)軸

arg

z

=

0

趨于零時,

f

(z)

fi

1,2沿arg

z

=π

趨于零時,f

(z)

fi

0,故lim

f

(z)不存在.zfi

0證限不存在.z例2

證明函數(shù)f

(z)=z

(z

?0)當(dāng)z

fi

0

時的極f

(z)

=

u

+

iv,令z

=x

+iy,x2

-

y2則

u(

x,

y)

=

,x2

+

y2,2

xyv(

x,

y)=

x2

+

y2當(dāng)z

沿直線y

=kx

趨于零時,2

xylim

v(

x,

y)

=

limy=kxxfi

0

x2

+

y2xfi

0y=kx,1

+

k

22k=隨k

值的變化而變化,所以lim

v(x,y)不存在,xfi

x0yfi

y0根據(jù)定理一可知,

lim

f

(z)

不存在.zfi

0二、函數(shù)的連續(xù)性1.

連續(xù)的定義:如果lim

f

(z)=f

(z0

),那末我們就說f

(z)zfi

z0在z0

處連續(xù).如果f

(z)在區(qū)域D內(nèi)處處連續(xù),我們說f

(z)在D內(nèi)連續(xù).函數(shù)f

(z)在曲線C

上z0

處連續(xù)的意義是lim

f

(z)

=

f

(z0

),

z

?

C

.zfi

z0定理三函數(shù)f

(z)=u(x,y)+iv(x,y)在z0

=x0

+iy0連續(xù)的充要條件是:u(x,y)和v(x,y)在(x0

,y0

)處連續(xù).例如,f

(z)

=

ln(

x2

+

y2

)

+

i(

x2

-

y2

),u(x,y)=ln(

x2

+y2

)在復(fù)平面內(nèi)除原點(diǎn)外處處連續(xù),

v(

x,

y)

=

x2

-

y2

在復(fù)平面內(nèi)處處連續(xù),故f

(x,y)在復(fù)平面內(nèi)除原點(diǎn)外處處連續(xù).定理四在z0

連續(xù)的兩個函數(shù)f

(z)和g(z)的和、差、積、商(分母在z0

不為零)在z0處仍連續(xù).如果函數(shù)h

=g(z)在z0

連續(xù),函數(shù)w

=f

(h)在h0

=g(z0

)連續(xù),那末復(fù)合函數(shù)w

=f

[g(z)]在z0

處連續(xù).特殊的:有理整函數(shù)(多項式)w

=

P(z)

=

a

+

a

z

+

a

z2

+

+

a

zn

,0

1

2

n對復(fù)平面內(nèi)的所有點(diǎn)z

都是連續(xù)的;有理分式函數(shù)w

=

P(z),Q(z)其中P(z)和Q(z)都是多項式,在復(fù)平面內(nèi)使分母不為零的點(diǎn)也是連續(xù)的.例3也連續(xù).證明:如果f

(z)在z0

連續(xù),那末f

(z)在z0證設(shè)f

(z)=u(x,y)+iv(x,y),則f

(z)=u(x,y)-iv(x,y),由f

(z)在z0

連續(xù),知u(x,y)和v(x,y)在(x0

,y0

)處都連續(xù),于是u(x,y)和-v(x,y)也在(x0

,y0

)處連續(xù),故f

(z)在z0

連續(xù).三、小結(jié)與思考通