复数W=（A—i)/(1—i),且A大于0,Z=W(W+i)的虚部减去它的实部所得的差为3/2,求复数Z

W=（A—i)/(1—i)
=(A-i)(1+i)/[(1-i)(1+i)]
=[A+Ai-i+1]/2
=(A+1)/2+(A-1)i/2
W^2=(A+1)^2/4-(A^2-1)i/2-(A-1)^2/4
Wi=-(A-1)/2+(A+1)i/2

Z=W(W+i)
=(A+1)^2/4-(A^2-1)i/2-(A-1)^2/4-(A-1)/2+(A+1)i/2
=(A+1)/2+(A+1)(2-A)i/2

依题意有:
(A+1)(2-A)/2-(A+1)/2=3/2
(A+1)/2*(1-A)=3/2
A=根号2i
所以
W=(A+1)/2+(A-1)i/2
=(根号2i+1)/2+(根号2i-1)i/2
=1/2-根号2/2+(根号2i-1/2)i

z=(1+i)*(a-i)/2
w=z*(z+1)=...=(a+1)/2+a(a+1)i/2
中间计算略去
a(a+1)/2-(a+1)/2=3/2
a>0得a=2
w=3/2+3i