复变函数与积分变换A卷(月)

n．nn．n年~年二学课程名称：复函数与分变换考生学号：试卷类型：A卷√卷□

专业年级：考生姓名：考试方式开□闭卷√………………一、单选题每题3分共15分）．若

z

11

，则

z

的值等于

()A．

i

B

C

D．

fz)，则．若上可导A．仅在直线

f

满足

B．在直线

上导

()C．仅在

点解析

D仅在

点可导．函数

f(z)

z(

在

z

处的泰勒展开式的收敛圆域为

()A．

|z2

B

|z

C．

|z

D．

|．z是数f(z)

1z(

的

()A．可去奇点

B．阶极点

C．二阶极点

D．阶极点．设

f(t)

是一个无穷次可微函数，

t)

为单位脉冲函数，那么

(t)f(tt

()A．

f(t)

B

(0)

C．

f

D

f(0)

(0)二、填题每题分，21分）当为的倍数时，复数1ii

n

的值为．．设

f(0)f

，limz0

f(z)z

．/

ii．设为正向圆周

|

，则

C

z

____________．．幂级数

)nz

的收敛半径为_．1．数

在其孤立奇点z处的留数为____________．设

f(t)

的傅里叶变换是

)

，则函数

f(t

的傅里叶变换是___

．．

()

s

2

1

，则

L

)

．三、计题每题分，16分）．复数

，求

的实部、虚部、模、辐角主值以及

z

的．．用留数计算反常积分

-

ix

2

dx

．四、求列分每题8分共16分）．C为向周

z

，计算积分

C

e(z2)

dz

．/

．算积分

I

C

1zsinz

，其中C为向圆周．五、解题每题分，16分）．知调和函数

y

y

，求解析函数

f

．．函数

f(z)

z(2)

分别在圆环域

z和z

内开洛级．/

六、解题每题分，16分）．知函数

f(t)

的换为

)

，求函数

()()

的Fourier换．．用变求解积分方程

f(t)at

t

sin(t

)

)d

．/

ii年~2014年二学期复变函数与积分变换A参考答案一、单选题每题3分共15分）．．D．B4C5二、填题每题分，21分）．．

．．

22

10．111．

)

．

12

2三、计题每题分，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、求列分每题8分共16分）

．

解

由

高

阶

导

数

公

式

得

C

ez(z2)

dz

=

2!

lim(z)z2

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

（6分）=

lime

z

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

（分z2解正向圆周z内函数z)

1zsinz

有唯一的奇点且阶极点·····（分）因此s[f(),0]z0

2

1zsin

z0

sincoszsinzsinzz

）故

I

C

1zsin

s[(z),0]

·········

（分五、解题每题分，16分）解方法

x2y

,

2

①,

②··········（分由①式

xdy

()

，两边对

求导得

y

，··············（分将其代入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．解在圆环域

0

内，

111z2

···············

（分所以f(z)··························································4）z(2)nn2在圆环域z内，有故zz

z

2(zn

(2(

···········

（分）所以

f(z)

(n2n(n

························································（分）六、解题每题分，16分）．解由相似性质得

1[f()]()22

··········································（）再由频域微分性质得

[()]

idi()F2d24

···················

（分．解原方程即

ft)tt)

·····················································（2）设

L

f(t))

对方程两边作拉氏变换由卷积定理可得

F(s)

a(s（分2s解