complex number question involving modulus 
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!
 A: 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, $|1-z^*w|=|w(z^*-w^*)|=|w||z^*-w^*|$ 
$=|z^*-w^*|=|(z-w)^*|=|z-w|$
A: Hint: Replace $1$, as appropriate, by $z\bar{z}$ or $w\bar{w}$. 
A: 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 
$$|w^*-z^*| = |-(w^*-z^*)|=|z^*-w^*| = |(z^*-w^*)^*| = |z-w| \, . $$ 
A: After squaring, the initial equation can be rewritten as $$(z-w)(z^*-w^*)=(1-wz^*)(1-w^*z).$$
Then expanding and simplifying,
$$ww^*-wz^*-w^*z+zz^*=1-w^*z-w^*z+ww^*zz^*,$$
$$ww^*+zz^*=1+ww^*zz^*,$$
$$(1-ww^*)(1-zz^*)=0.$$
