✔ 最佳答案
Let P(z)=a+bi and Q(w)=x+yi
Then
(a+bi)(1-x-yi)=(x+yi)
a+bi
=(x+yi)/(1-x-yi)
=(x+yi)(1-x+yi)/(1-x-yi)(1-x+yi)
=[(x-x^2-y^2)+yi]/[(1-x)^2+y^2]
Compare
a=(x-x^2-y^2)/[(1-x)^2+y^2]
Since 2a + 1= 0
We have
2(x-x^2-y^2)=-[(1-x)^2+y^2]
2x-2x^2-2y^2=-1+2x-x^2-y^2
x^2+y^2=1
Q describes a circle whose centre is at the origin.