Difficult Complex Number Proof. Given $|w| =1$ or $|v|=1$

Question

Let $z, w$ be distinct complex numbers. Show that if $|z| = 1$ or $|w| = 1$, then

$$\left|\frac{w-z}{1-\overline{w}z}\right| = 1$$

Hint: Note that $|a|^2 = a\overline a$

I have been stuck on this problem for a while and cant seem to find somewhere to start.

Edit: The question linked is a different question

Edit: I have been able to prove the |w|=1 case, but when proving the z case I take these steps and cant seem to figure out where to go next. $|1-\overline{w}z| = |z\overline{z} - \overline{w}z| = |z||\overline{z} - \overline{w}| = |\overline{z} - \overline{w}|$

Answer 1

We can suppose that $|w| = 1$ (the problem is equivalent if $|z| = 1$). $$|1 - \overline{w}z| = \left|1 - \frac zw\right| = \frac1{|w|} |w - z| = |w - z| $$

Answer 2

Ok so we know that $z$ and $w$ are distinct and so $$\left|\frac{w-z}{1-\bar{w}z}\right|\neq 0,$$ right? So try proof by contradiction and remember how norms work with division: $$\left|\frac ab\right|=|a||b|$$. Note that the only positive real number that squares to $1$ is $1$.