Complex Functions Analysis $f(z)=\frac{z-a}{1-\bar{a}z}$ Here's a question I'm working on:
let $a,z \in \mathbb{C}$, $f(z)=\frac{z-a}{1-\bar{a}z}$ 
If $|z| = 1$, find $|z-a|^2 - |1-\bar{a}z|^2$
Here's my progress: 
let $z = x+iy$  $a = b+ic$ 
$\Rightarrow$ $|z-a|^2 - |1-\bar{a}z|^2 = (x-b)^2 - (y-c)^2 - (1-bx-cy)^2 - (cx - by)^2$
But I don't know how to use the fact that $|z|=1$ or if I could use the function given somehow, and I'm not sure exactly what the question is asking. Am I missing something here?
Anyway, thanks.
 A: Splitting into real and imaginary parts is not generally helpful. 
Use
$$|z-a|^2 = (z-a)(\overline{z-a}) = |z|^2 - z\bar a - \bar z a + |a|^2$$
and
$$|1 - \bar a z|^2 = (1 - \bar a z)(\overline{1 - \bar a z}) = 1 - z\bar a - \bar z a + |z|^2|a|^2.$$
What happens if $|z| = 1$?
A: A much better idea is to use the polar form: if $\lvert z \rvert=1$, then there is a real $\theta$ so $z=e^{i\theta}$. Then
$$ \lvert z-a \rvert = \lvert e^{i\theta}-a \rvert = \lvert e^{i\theta}(1-ae^{-i\theta}) \rvert = \lvert 1-ae^{-i\theta} \rvert = \lvert 1-\bar{a}e^{i\theta} \rvert, $$
using $\lvert z w \rvert = \lvert z \rvert \lvert w \rvert $ and $\lvert z \rvert = \lvert \bar{z} \rvert$.
A: Following the lead of @Chappers, we let $z=e^{i\theta}$. In addition, let $a=\hat ae^{i\theta}$, where $\hat a =|a|$. Then
$$
|z-a|=|e^{i\theta}-\hat ae^{i\theta}|=|1-\hat a|\\
|1-\bar az|=|1-\hat ae^{-i\theta}e^{i\theta}|=|1-\hat a|\\
|z-a|^2 - |1-\bar{a}z|^2=0
$$
A: Given $|z|^2=1 \iff z \bar z = 1 \iff \bar z = \cfrac{1}{z}\,$:
$$|z-a|^2 - |1-\bar{a}z|^2=|z-a|^2-|z|^2\cdot\left|\frac{1}{z}-\bar a\right|^2=|z-a|^2 - |\bar z - \bar a|^2=|z-a|^2 - \left|\overline{z - a}\right|^2=0$$
