How to show that for $a,b\in \mathbb{C}: |a-b|=|1-a\overline{b}| \Leftrightarrow |a|=1 \, \lor \, |b|=1$

Question

How to show that for $a,b\in \mathbb{C}: |a-b|=|1-a\overline{b}| \Leftrightarrow |a|=1 \, \lor \, |b|=1$

I defined $a:=x+iy$ and $b:=u+iv$ which resulted in quite an algebra mess. Is there a simpler method to solve such statements?

Answer 1

$$|a-b|=|1-a\overline{b}|$$ $$\Longleftrightarrow(a-b)(\overline{a}-\overline{b})=(1-a\overline{b})(1-\overline{a}b)$$ $$\Longleftrightarrow (a\overline{a}-1)(b\overline{b}-1)=0$$