《复变函数与积分变换（刘建亚）》作业答案(5)

3z C

e dz z -? ,:31C z -=; (2 22 C dz z a -? ,: C z a a +=; (3 i 2 1 z C

e dz z +? ,4:2i 3C z -=; (4 3 C zdz z +? ,:2C z =; (5

23(1(1C dz z z +-? ,:1C z r =<; (6 3cos C z zdz ?

,C 为包围0z =的闭曲线; (7 22(1(4C dz z z +-? ,3:2 C z =; (8 sin C z dz z ? ,:3C z =; (9 2 cos 2C z dz z π? ? - ??

?? ,:3C z =; (10

5z C e dz z ? ,:1C z =. 解:(1 33 2i 2i 3 z z

z C e dz e e z ππ==?=-? ; (2 22 12i i C z a dz z a z a a π π=-=?=- --? ; (3

i i 2i 2i 1i z z C z e e dz z z e π

π==?=++? ; (4 03 C zdz z =+? ; (5 230(1(1C dz z z =+-? ; (6 3 cos 0C z zdz =? ; (7

222222i i 1(1(42i (i(4(i(4111 02i 44C C C z z dz dz dz z z z z z z z z =-=??-=-?? +-+---??

??-=-= ?--????? ;

(8

0sin 2i sin 0z C z dz z z π==?=? ; (9 (2 2

cos 2sin 21!2C z z i

dz z i z ππππ= =?-=-? ? - ?? ?? ; (10 50 2(51!12 z z C z e i i dz e z ππ==?=