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

$$\frac{|z − w|}{|1 − z^*w|} = 1$$

[Hint: Note that $|a|^2 = aa^*$.]

Hey guys, couldn't get my thinking cap on for this question. Any helpful input? Will appreciate it!

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up vote 2 down vote accepted

If $|z|=1$, $$1-z^*w=z^* z-z^*w=z^*(z-w)$$

Taking modulus, $|1-z^*w|=|z^*(z-w)|=|z^*||z-w|=|z-w|$

If $|w|=1$, $$1-z^*w=w^*w-z^*w=w(z^*-w^*)$$

Taking modulus, $|1-z^*w|=|w(z^*-w^*)|=|w||z^*-w^*|$


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Hint: Replace $1$, as appropriate, by $z\bar{z}$ or $w\bar{w}$.

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+1: nice hint ! – enzotib Sep 18 '12 at 16:50

First, assume that $|z|=1$. This is true if and only if $z^*z = 1$. It follows that $$\frac{|z-w|}{|1-z^*w|} = \frac{|z-w|}{|z^*z-z^*w|} = \frac{|z-w|}{|z^*||z-w|} = \frac{1}{|z^*|} = 1$$ since $|z^*| = |z| = 1.$ Next, assume that $|w| = 1$. This is true if and only if $w^*w = 1$. It follows that $$\frac{|z-w|}{|1-z^*w|} = \frac{|z-w|}{|w^*w-z^*w|} = \frac{|z-w|}{|w^*-z^*||w|} = \frac{1}{|w|} = 1$$ since $|-v| = |v|$ and $|v^*| = |v|$ for all $v \in \mathbb{C}$, and hence $$|w^*-z^*| = |-(w^*-z^*)|=|z^*-w^*| = |(z^*-w^*)^*| = |z-w| \, . $$

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After squaring, the initial equation can be rewritten as $$(z-w)(z^*-w^*)=(1-wz^*)(1-w^*z).$$

Then expanding and simplifying, $$ww^*-wz^*-w^*z+zz^*=1-w^*z-w^*z+ww^*zz^*,$$ $$ww^*+zz^*=1+ww^*zz^*,$$ $$(1-ww^*)(1-zz^*)=0.$$

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