$f(z) = \frac{z-a}{1 - z\bar{a}}$ interpretation?

Question

I recently started a complex-analysis course and our teacher proposed this exercise to us:

Given $a, z \in \mathbb{C}$, consider the complex function $$ f(z) = \frac{z-a}{1 - z\bar{a}}.$$

I've proved already that either $\vert{z}\vert = 1$ or $\vert{a}\vert = 1$ implies $$\left\vert{\frac{ z-a}{1 - z\bar{a}}}\right\vert = 1.$$

So, assume $\vert{a}\vert = 1$. My teacher asked "What exception needs to be made if $\vert{z}\vert = \vert{a}\vert = 1$?"

I can tell that if e.g., $ z = \frac{1}{\bar{a}}$, the function goes all the way to infinity, and that there is a $\frac{0}{0}$ type indetermination if $z = a$, but:

1. Is there any way to solve/understand the indetermination?
2. Or a way to interpret the function geometrically? I've already tried plotting it on davidbau.com, but I can't really understand what's going on. Or is it just because I'm not used to complex functions?

Answer 1

If $|a|=1$ then $\bar{a} = \frac 1a$, so that for $z \ne \frac{1}{\bar{a}}$ $$ f(z) = \frac{z-a}{1 - z\bar{a}} = \frac{z-a}{1 - z/a} = -a $$ So $f$ has a removable singularity at $z=\frac{1}{\bar{a}}$, and can be continuously extended to the constant function $z \mapsto -a$.