Show that $\big|\frac{a-b}{1-\overline{a}b}\big|=1$, when either $|a|=1$ or $|b|=1$.

Question

The question is from Ahlfors Complex Analysis text.

Show that $\big|\frac{a-b}{1-\overline{a}b}\big|=1$, when either $|a|=1$ or $|b|=1$.

A solution I'm reading lets $|a|=1$ and $|b|\neq 1$. then they show that $|a-b|=|1-\overline{a}b|$, by first squaring $|a-b|$ which gives $a\overline{a}+b\overline{b}-a\overline{b}-b\overline{a}=a\overline{a}+b\overline{b}-2Re(\overline{a}b)$.

My Question

I'm unsure of why $a\overline{b}+b\overline{a}=2Re(\overline{a}b)$?

They did write that this is because $b\overline{a}=\overline{\overline{a}b}$ but this didn't clear things up for me. Thanks

Answer 1

Recall that for any $w$

$$\Re(w)=\frac{w+\bar w}{2}$$

and let $w=\bar ab \iff \bar w=a\bar b$.

Answer 2

Alt. hint:   $\,\bar a b \ne 1\,$ for the equality to make sense, then using that $\,|w|^2 = w \bar w\,$:

$$\require{cancel} \begin{align} \left|\frac{a-b}{1-\overline{a}b}\right|=1 \;\;&\iff\;\;|a-b|=|1-\bar a b| \\ &\iff\;\; |a-b|^2=|1-\bar a b|^2 \\ &\iff\;\; (a-b)(\bar a - \bar b) = (1 - \bar a b)(1 - a \bar b) \\ &\iff\;\; |a|^2+|b|^2 - \cancel{a \bar b} - \bcancel{\bar a b} = 1 + |a|^2|b|^2 - \cancel{a \bar b} - \bcancel{\bar a b} \\ &\iff\;\; |a|^2|b|^2 - |a|^2 - |b|^2 +1 = 0 \\ &\iff\;\; \ldots \end{align} $$