復(fù)變函數(shù)與積分變換A卷(月)

n．nn．n年~年二學(xué)課程名稱：復(fù)函數(shù)與分變換考生學(xué)號(hào)：試卷類型：A卷√卷□

專業(yè)年級(jí)：考生姓名：考試方式開□閉卷√………………一、單選題每題3分共15分）．若

z

11

，則

z

的值等于

()A．

i

B

C

D．

fz)，則．若上可導(dǎo)A．僅在直線

f

滿足

B．在直線

上導(dǎo)

()C．僅在

點(diǎn)解析

D僅在

點(diǎn)可導(dǎo)．函數(shù)

f(z)

z(

在

z

處的泰勒展開式的收斂圓域?yàn)?/p>

()A．

|z2

B

|z

C．

|z

D．

|．z是數(shù)f(z)

1z(

的

()A．可去奇點(diǎn)

B．階極點(diǎn)

C．二階極點(diǎn)

D．階極點(diǎn)．設(shè)

f(t)

是一個(gè)無窮次可微函數(shù)，

t)

為單位脈沖函數(shù)，那么

(t)f(tt

()A．

f(t)

B

(0)

C．

f

D

f(0)

(0)二、填題每題分，21分）當(dāng)為的倍數(shù)時(shí)，復(fù)數(shù)1ii

n

的值為．．設(shè)

f(0)f

，limz0

f(z)z

．/

ii．設(shè)為正向圓周

|

，則

C

z

____________．．冪級(jí)數(shù)

)nz

的收斂半徑為_．1．?dāng)?shù)

在其孤立奇點(diǎn)z處的留數(shù)為____________．設(shè)

f(t)

的傅里葉變換是

)

，則函數(shù)

f(t

的傅里葉變換是___

．．

()

s

2

1

，則

L

)

．三、計(jì)題每題分，16分）．復(fù)數(shù)

，求

的實(shí)部、虛部、模、輻角主值以及

z

的．．用留數(shù)計(jì)算反常積分

-

ix

2

dx

．四、求列分每題8分共16分）．C為向周

z

，計(jì)算積分

C

e(z2)

dz

．/

．算積分

I

C

1zsinz

，其中C為向圓周．五、解題每題分，16分）．知調(diào)和函數(shù)

y

y

，求解析函數(shù)

f

．．函數(shù)

f(z)

z(2)

分別在圓環(huán)域

z和z

內(nèi)開洛級(jí)．/

六、解題每題分，16分）．知函數(shù)

f(t)

的換為

)

，求函數(shù)

()()

的Fourier換．．用變求解積分方程

f(t)at

t

sin(t

)

)d

．/

ii年~2014年二學(xué)期復(fù)變函數(shù)與積分變換A參考答案一、單選題每題3分共15分）．．D．B4C5二、填題每題分，21分）．．

．．

22

10．111．

)

．

12

2三、計(jì)題每題分，16分）．解

z)Im(z)

，······················

（分z

,

z

2

,

2

)i

（分）解

ix

2

dx

i[i]2

································（4）

i()i

z

e

····························································

（分四、求列分每題8分共16分）

．

解

由

高

階

導(dǎo)

數(shù)

公

式

得

C

ez(z2)

dz

=

2!

lim(z)z2

···············································

（6分）=

lime

z

······················································································

（分z2解正向圓周z內(nèi)函數(shù)z)

1zsinz

有唯一的奇點(diǎn)且階極點(diǎn)·····（分）因此s[f(),0]z0

2

1zsin

z0

sincoszsinzsinzz

）故

I

C

1zsin

s[(z),0]

·········

（分五、解題每題分，16分）解方法

x2y

,

2

①,

②··········（分由①式

xdy

()

，兩邊對(duì)

求導(dǎo)得

y

，··············（分將其代入②式得

)x

，所以

vxyx

.

··················

（分/

n即

f(zx2(2xyx

.······················································

（8分方法2x，x，，·······························（分xyfxizi···········································（分）f

所

以.·······························································分．解在圓環(huán)域

0

內(nèi)，

111z2

···············

（分所以f(z)··························································4）z(2)nn2在圓環(huán)域z內(nèi)，有故zz

z

2(zn

(2(

···········

（分）所以

f(z)

(n2n(n

························································（分）六、解題每題分，16分）．解由相似性質(zhì)得

1[f()]()22

··········································（）再由頻域微分性質(zhì)得

[()]

idi()F2d24

···················

（分．解原方程即

ft)tt)

·····················································（2）設(shè)

L

f(t))

對(duì)方程兩邊作拉氏變換由卷積定理可得

F(s)

a(s（分2s解