complex numbers equality question Let $a$ be given a complex number. Show that 
$$\left|\frac{z-a}{1-a^*z}\right|=1$$
for $z$ with $|z|=1$ and $a^*z\neq 1$. 
If $|z|=1$, that means $z$ can be equal to $i$, $-i$, $1$ or $-1$ right?
Also I tried multiplying top and bottom by the conjugate of the bottom but it didn't really help.

It is enough to show 
$$\left|\frac{z-a}{1-a^*z}\right|^2=1$$
But 
$$\begin{align}\left|\frac{z-a}{1-a^*z}\right|^2&=\frac{z-a}{1-a^*z}\frac{z^*-a^*}{1-az^*}\\&=\frac{zz^*+aa^*-az^*-a^*z}{1+aa^*zz^*-az^*-a^*z}\\&=\frac{zz^*+aa^*-az^*-a^*z}{zz^*+aa^*-az^*-a^*z}\ \ \ \ \text{using that }zz^*=|z|^2=1\\&=1\end{align}$$
