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

设C：|z-2|=5为正向圆周，则int dfrac (2{z)^3+3(z)^2+2z+1}(z)dz=()A、2πіB、πі; C、i;D、0;设C：|z-

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

(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

(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

z=0是函数(z)=dfrac (z)(sin {z)^2cdot ((e)^z-1)}的几级极点A 1 B 2 C 3 D 4z=0是函数的

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

设C为正向圆周|z|=2, 则下列积分值不为0的是( )A.int dfrac (z)(z-1)dxB.int dfrac (z)(z-1)dxC.

设(z)=1-overline (z), _(1)=2+3i _(2)=5-i, 则 ([ f({z)_(1)-(z)_(2))-|||-__等于设等于

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)

[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