|COMPLEXE NUMBERS| Show that?... thank you!

Show that:
Is z ∈ ℤ, than is z^-1 = z^- / |z|^2
The z in the numerator is supposed to be the complex conjugate of z.
Thank you!

1 Answer
Jan 14, 2018

Yes

Explanation:

take z= x + iy
then z^-1=1/(x + iy)
|z|^2=x^2 + y^2= (x+iy)(x-iy) = zz^_

hence we get
RHS

z^_ / (zz^_) = 1/z = z^-1 = LHS
hence proved
Hope u find it helpful :)