|z|lt dfrac (1)(4)-|||-(3) (z)=dfrac ({z)^-1-a}(1-a{z)^-1} , |z|gt a-|||-(4) (z)=dfrac (1-dfrac {1)(4)(z)^-1}(1-dfrac {8)(15)(z)^-1+dfrac (1)(15)(z)^-2} ..dfrac (1)(5)lt |z|lt dfrac (1)(3)

2.14分别用长除法、留数法、部分分式法求下列Z反变换:-|||-(1) (z)=dfrac (1-2{z)^-1}(1-dfrac {1)(4)(z)^-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

(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

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

直线-1=dfrac (y+3)(-2)=z-4与直线-1=dfrac (y+3)(-2)=z-4的夹角为( )-1=dfrac (y+3)(-2)=z-4-1

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

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