If $\mid z_{0} \mid = 1, z_{0} \in \mathbb{C} $ prove that then $\forall z\in \mathbb{C} , z \neq z_{0}$
$$\left| \frac{z-z_{0}}{1- \bar{z}z_{0}} \right|= 1$$
P.S. if $z = x+y i$, $\bar{z} = x - y i$.
I tried by multiplying given fraction with $ \frac{\bar{z}_{0}}{\bar{z}_{0}}$ but I got nowhere.
(Since I posted a comment) Hint: $1=\overline{z_0} \cdot z_0 \Rightarrow 1- \overline{z}z_{0}=\overline{z_0} \cdot z_0 - \bar{z}z_{0}=z_0 (\overline{z_0-z})$ and $|\overline{z_0-z}|=|z-z_0|$.
$|a|=1$ iff $|a|^2=a\overline{a}=1$ hence for all $z\neq z_0$ $$ \left|\frac{z-z_0}{1-\overline{z}z_0}\right|^2=\frac{(z-z_0)(\overline{z}-\overline{z_0})}{(1-\overline{z}z_0)(1-z\overline{z_0})}=\frac{|z|^2-2\mathrm{Re}(z\overline{z_0})+1}{1+|z|^2-2\mathrm{Re}(z\overline{z_0})}=1. $$
