《复变函数与积分变换（刘建亚）》作业答案(10)

11z z ' ??=- ??? ,且 1

0011111i (i(1i i (ii i i i 1i n

n n n n n z z z z z ∞∞

+==--??==?=-=- ?-+-??+∑∑, 所以12111(i(1i n n n n n z z -∞ +=-=--∑,从而2121 11(i(1(ii n n n n n z z z -∞ -+=-=--∑.

当1i z <-<∞时,由于 i 11 z <-,所以 1

0011111i i (1i i (ii i i (i1i n

n n n n n z z z z z z z ∞∞

+==??==?=?-=- ?+-----??+-∑∑, 且2 11z z ' ??=- ??? ,从而221

1(1i (1(in n n n n z z ∞ +=+=--∑,所以 2 3

11(1i (1(i(in n n n n z z z ∞ +=+=---∑. (7 由于 21z <且3 1z <,所以 1

0000111111(2(332131213213232n n n n n n n n n n n n z z z z z z z z z z z z z ∞∞∞∞ +====?? =-=?- ?

------??

??--????=-==?? ? ?????????∑∑∑∑ 习题5:

1、求下列函数的孤立奇点并确定它们的类别,若是极点,指出它们的级. (1 22 1 (1z z +; (3

3sin z z ; (4 ln(1z z +; (7 21 (1 z

z e -; (11 1sin 1z -. 解:(1 易见 0z =,i z =±是 22 1 ((1f z z z = +的孤立奇点.由于 22 1

lim

(1z z z →=∞+, 22i 1 lim

(1z z z →±=∞+,所以0z =,i z =±是极点. 0z =,一级极点,i z =±,二级极点. (3 3 0sin lim z z

z →=∞,所以0z =是极点.0z =,二级极点. (4 易见0z =是ln(1(z f z z +=的孤立奇点,且0ln(1

lim 1z z z

→+=,所以0z =是可去奇点; (7 0z =,三级极点,2i 1,2,z k k π==±± (,一级极点; (11 1z =,本性奇点.

5、求下列各函数在有限奇点处的留数. (2 (21 1z z -; (3 ( 2221z z +; (6

21 sin z z . 解:(2 记 ( 2 1 (1f z z z =

-,则易见0,1±是(f z 的孤立奇点,且他们都是一级极点.由规则Ⅰ, (2 01

Res[(,0]lim 0(lim 11z z f z z f z z →→=-==-, (1 1 11

Res[(,1]lim 1(lim (12

z z f z z f z z z →→-=-==-+,