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.
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 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 $\lvert z \rvert = \lvert \bar{z} \rvert$.
Following the lead of @Chappers, we let $z=e^{i\theta}$. In addition, let $a=\hat ae^{i\theta}$, where $\hat a =|a|$. Then
$$ |z-a|=|e^{i\theta}-\hat ae^{i\theta}|=|1-\hat a|\\ |1-\bar az|=|1-\hat ae^{-i\theta}e^{i\theta}|=|1-\hat a|\\ |z-a|^2 - |1-\bar{a}z|^2=0 $$
Given $|z|^2=1 \iff z \bar z = 1 \iff \bar z = \cfrac{1}{z}\,$:
$$|z-a|^2 - |1-\bar{a}z|^2=|z-a|^2-|z|^2\cdot\left|\frac{1}{z}-\bar a\right|^2=|z-a|^2 - |\bar z - \bar a|^2=|z-a|^2 - \left|\overline{z - a}\right|^2=0$$
