z^2-4iz-(4-9i)=0的根为_____。

1.解方程z^2-4iz-(4-9i)=0.1.解方程$z^{2}-4iz-(4-9i)=0$.

5.4已知复势为:(1) W(z)=(1+i)z ;(2) (z)=(1+i)ln (dfrac (z+1)(z-4)); (3) W(z)=-6iz+-|||

z=1+i，那么z=1+iz=1+iz=1+i，那么

z=2i 为函数 f(z)=(mathrm(e)^z)/(z^2)(z^(2+4)^2) 的（A. 可去奇点B. 本性奇点C. 极点D. 解析点

在点 (0,1,(pi)/(2)) 处的法平面方程为 A 4x-z+(pi)/(2)=0 B 4y-z+(pi)/(2)=0 C 4x-z-(

1.3 解方程组 ) 2(z)_(1)-(z)_(2)=i (1+i)(z)_(1)+i(z)_(2)=4-3i .

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

_(0)=2;-|||-(3) dfrac (1)({2)^2} , _(0)=-1;-|||-(4) dfrac (1)(4-3z), _(0)=1+i;-|

‘1系统稳定时:i?i2 k’?i??i2?4i?4?i1?4i1?4k??(k (i1?4)(i0?4)i?4?4,?k?0?0.811616i0?16k??

球面^2+(y)^2+(z)^2+4x+6y+2z+10=0的球心坐标为A.^2+(y)^2+(z)^2+4x+6y+2z+10=0B.^2+(y)^2+(z)