Question

For complex numbers z and `omega`, prove that `|z|^2omega -|omega|^2z = z - omega` if and only if `z = omega` and `z. baromega = 1`?

Answer 1

See below.

Explanation:

From

`|z|^2omega -|omega|^2z = z - omega` and rearranging

`z/(1+absz^2) = omega/(1+absomega^2) rArr z = omega`

and now

`|z|^2omega -|omega|^2z = z - omega` then

`z bar z omega - omega bar omega z = z - omega` or

`omega(z bar z-bar omega z) = z-omega`

now multiplying both sides for `bar z`

`bar z omega(abs z^2-bar omega z) = abs z^2-barz omega`

now as `bar z omega = omega bar z` we have

`bar z omega = (abs z^2-barz omega)/(abs z^2-bar omega z) = 1`