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.int dfrac (z)(z-1)dxD.int dfrac (z)(z-1)dx

曲线C为正向圆周|z-1|=3, (int )_(c)^3dfrac (3)(2)dz=|z-1|=3, (int )_(c)^3dfrac (3)(2)dz=

曲线C为正向圆周|z|=2, (int )_(c)dfrac (cos z)({(z-1))^3}dz=曲线C为正向圆周A.0B.C.D.

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

曲线 C 为正向圆周 |z-1|=3，int_(C) (1)/(z^3(z-2)^2) , dz=A. $\frac{3}{8}\pi i$B. $\frac{

5.利用留数计算下列积分.-|||-(3) (int )_(|z|=2)dfrac ({e)^2z}((z+1){(z-1))^2}dz

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

求直线dfrac (x-1)(1)=dfrac (y)(1)=dfrac (z-1)(-1)-|||-__ __在平面dfrac (x-1)(1)=dfrac

1.求下列函数的留数:-|||-(1) (z)=dfrac ({e)^z-1}({z)^5} 在 z=0 处;-|||-(2) (z)=(e)^dfrac (1

直线dfrac (x-1)(1)=dfrac (y-1)(0)=dfrac (z-1)(-1)-|||-__与直线dfrac (x-1)(1)=dfrac (y

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-