(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+dfrac (1)({z)^2-1}(z)=(z)^2+dfrac (1)({z)^2-1}(z)=(z)^2+dfrac (1)({z)^2-1}

(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

[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

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)

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

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

3.指出下列函数f(z)的解析性区域,并求出其导数:-|||-(1) ((z-1))^5 ;-|||-(2) ^3+2iz ;-|||-(3) dfrac (1

设 C: |z+1|=(1)/(2)，则 int_(C) (sin frac(pi)/(4) z)(z^2-1) dz=A. $2\pi i$B. $-\fra

[ dfrac (sin z)({z)^2},0] =-|||-A 1-|||-B .-1-|||-C dfrac (1)(2)

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