5.(单选题)f(z)=(1)/(z(z-1)^2)在0A. $(A)\sum_{n=0}^{\infty}(-1)^{n}z^{n}$B. $(B)\frac
[单选题](z)=(z)^2+dfrac (1)({z)^2-1}f(z)=( )[单选题]A.B.C.D.
计算(int )_(c)_(c)dfrac (cos z)((z-dfrac {1)(2))(z-1)}dz,其中(int )_(c)_(c)dfrac (co
-2-|||-bigcirc b.0-|||-bigcirc C.2-|||-○ d.不存在
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
[ dfrac (sin z)({z)^2},0] =-|||-A 1-|||-B .-1-|||-C dfrac (1)(2)
5.[单选题]设C为正向圆周|z|=2,则oint_(c)(cos z)/((z-1)^2)dz等于()A. (A)-sin1;B. (B)0;C. (C)co
4.设 (z)=dfrac (1)(z)-zsin dfrac (1)({z)^2}, 则 [ f(z),0] =-|||-(A)1; (B)2; (C)0;
1.将下列各函数在指定圆环内展为洛朗级数.-|||-(1) dfrac (ln (2-z))(z(z-1)) lt |z-1|lt 1;-|||-(2) dfr
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.