Question

复数z=(1+i)/(1-i)+(1-i)^2的虚部等于

Answer 1

z=(1+i)/(1-i)+(1-i)^2

=(1+i)^2/2+(1-i)^2

=i+(-2)i

=-i

复数z=(1+i)/(1-i)+(1-i)^2的虚部等于-i

你能明白