Proof that if $z = 1$, then $|z-w| = |1- \overline{w}z|$, $\forall w \in$ $\mathbb{C}$
My attempt below:
$(z-w)\overline{(z-w)} = |z-w|^2$ $(z-w)\overline{(z-w)} = (z-w)(\overline{z}-\overline{w}) = |z|^2 -z\overline{w} -\overline{z\overline{w}} +|w|^2 = |z|^2 - 2Re(z\overline{w}) + |\overline{w}|^2*1 = 1 - 2Re(z\overline{w}) + |\overline{w}|^2*|z|^2 = 1 - 2Re(z\overline{w}) + |\overline{w}z|^2 = 1 - 2Re(\overline{w}z) + |\overline{w}z|^2$
$|z-w|^2 = 1 - 2Re(\overline{w}z) + |\overline{w}z|^2$
I was trying identfiy a remarkable product, but I didn't have success. Someone can help me with this question?