Let $z,w$ complex numbers such that $|z| = |w|$

Question

Prove that $$|\frac{z-w}{1-\bar{z}w}| = 1$$.

Suggestion: Remember that $|z|^{2} = z\bar{z}$

I don't know where to start. I've been thinking this for a few hours. Any idea?

Answer 1

If we only have $|z|=|w|$ then a counterexample is $z=2$ and $w=2i$ since $|z-w| = 2\sqrt{2}$ while $|1 - \bar{z} w| = \sqrt{17}$. The claim is true if you have the stronger assumption $|z|=|w|=1$ (edit: actually $|z|=1$ is enough).

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It is equivalent to show $$|z-w| = |1-\bar{z}w|.$$ By the hint, this is equivalent to $$(z-w)(\overline{z-w}) = (1-\bar{z}w)(\overline{1-\bar{z} w}).$$ Expand both sides, simplify, and use the assumption $|z|=1$.

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Correction: to be clear, the necessary and sufficient condition is $|z|^2 + |w|^2 = 1 + |z|^2 |w|^2$. This makes $|z|=1$ sufficient, as Rezha Adrian Tanuharja noted in the other answer.

Answer 2

$|z|=1$ or equivalently $z\bar{z}=1$ is sufficient.

$$ \begin{aligned} \frac{|z-w|}{|1-\bar{z}w|}&=\frac{|z||z-w|}{|z-z\bar{z}w|}\\ &=\frac{|z-w|}{|z-w|}\\ &=1 \end{aligned} $$

Answer 3

Aliter. Taking $z=e^{i\theta}$ and $w=e^{i\phi}$ we get $|\frac{z-w}{1-\overline{z}w}|=|\frac{e^{i\theta}- e^{i\phi}}{1-e^{-i\theta}e^{i\phi}}|=1 ,$ and this ca be obtained just by multiplying and dividing the last expression by $e^{i\theta}$.