Question

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

Answer 1

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