例8 已知 (z)=dfrac (1)(2pi )(|)_(|i|=1)dfrac (cos xi )({(xi -z))^3}ds, 证明:当 |z|neq 1 时,f(z)解析,-|||-并求f`(z).

5.证明 (z)=cos (z+dfrac (1)(z)) 用z的幂表出的洛朗展开式中的系数为-|||-_(n)=dfrac (1)(2pi )(int )_(

(z)=(z)^2+dfrac (1)({z)^2-1}，则其解析区域为（）(z)=(z)^2+dfrac (1)({z)^2-1}(z)=(z)^2+dfra

4.设 (z)=dfrac (1)(z)-zsin dfrac (1)({z)^2}, 则 [ f(z),0] =-|||-(A)1; (B)2; (C)0;

(z)=(z)^2+dfrac (1)({z)^2+1}，则其解析区域为( )(z)=(z)^2+dfrac (1)({z)^2+1}(z)=(z)^2+dfr

计算(int )_(c)_(c)dfrac (cos z)((z-dfrac {1)(2))(z-1)}dz，其中(int )_(c)_(c)dfrac (co

求下列函数的奇点:(1)dfrac (z+1)(z({z)^2+1)}; (2)dfrac (z+1)(z({z)^2+1)}求下列函数的奇点:(1)

[单选题](z)=(z)^2+dfrac (1)({z)^2-1}f(z)=( )[单选题]A.B.C.D.

[题目]如果复数z1,z2,z3满足等式 dfrac (({z)_(2)-(z)_(1))}(({z)_(3)-(z)_(1))}=dfrac (({z)_(1

|z|lt dfrac (1)(4)-|||-(3) (z)=dfrac ({z)^-1-a}(1-a{z)^-1} , |z|gt a-|||-(4) (z)

(3.)求下列极限:-|||-(1) lim _(xarrow infty )dfrac (1)(1+{z)^2} ;-|||-(3) lim _(zarrow