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I was reading Lars Ahlfors' book on Complex Analysis, 3rd edition and encountered a problem which I could not solve:

Prove that $$\left|\frac{a-b}{1-\bar{a}b}\right|=1$$ if either $|a|=1$ or $|b|=1$. what exception must be made if $|a|=|b|=1$?

At first, I substituted $|a|=1$, and after that $|b|=1$, and it worked. But I think that the problem is asking me to 'prove' it.
I noticed that $$\left|\frac{a-b}{1-\bar{a}b}\right|=\frac{|a-b|}{|1-\bar{a}b|}$$ So the equation is $$\frac{|a-b|}{|1-\bar{a}b|}=1$$ Squaring both sides gives $$\frac{|a-b|^2}{|1-\bar{a}b|^2}=1$$ Now I used the formula $$|a-b|^2=|a|^2+|b|^2-2\Re(a\bar{b})$$ which turns the equation to \begin{align} |a|^2+|b|^2-2\Re(a\bar{b})&=1+|a\bar{b}|^2+\Re(a\bar{b})\\ |a|^2+|b|^2&=1+|a\bar{b}|^2 \end{align} What can be done now?
Note: I am not too experienced in mathematics. I am studying it as a hobby, but math is not my field. It may happen that I missed something 'trivial'.

dali108
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  • You were almost there. Just note that $|a\bar{b}|^2 = |a|^2 \cdot |b|^2$, so that your last equation can be written as $(1-|a|^2)(1-|b|^2) = 0$. – Martin R Aug 13 '21 at 14:39

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