# $|1-z \bar w|^2 - |z-w|^2 = (1-|z|^2)(1-|w|^2)$ [closed]

Given $z,w \in \mathbb C$, show that $$|1-z \bar w|^2 - |z-w|^2 = (1-|z|^2)(1-|w|^2)$$

I think I need to use the equation $|z|^2 = z \bar z$

Thanks for any help.

$$|1-z \bar w|^2 - |z-w|^2 = (1-z\bar w)(1-\bar z w) -(z-w)(\bar z-\bar w)= \\ = 1-z\bar w -\bar z w +|z|^2|w|^2 -|z|^2+w\bar z + z \bar w -|w|^2 =\\ = 1 -|z|^2 -|w|^2 +|z|^2|w|^2 = (1-|z|^2)(1-|w|^2)$$