Rational function which is bijective on unit disk What is the general form of a rational function which is bijective on the unit disk?
I'm stuck on this problem. If I let $R(z) = \frac{a_0(z-a_n) \cdots (z-a_1)}{(z-b_m)\cdots (z-b_1)}$, then exactly one of the $a_i$ must fall in the unit disk and none of the $b_i$ can be in the unit disk. 
Any hints on how to proceed? I see simple functions like $z$ work, but I don't know how to show that higher combinations of them are bijective.
 A: The only bijective holomorphic functions from the unit disk onto itself are of the form $e^{i\theta}\psi_\alpha$, where $\theta$ is real and
$$
\psi_\alpha(z) = \frac{z - \alpha}{1-\bar{\alpha}z}
$$
with $|\alpha| < 1$. For a proof, consider any bijective holomorphic function $f$ from the unit disk onto itself. Choose an appropriate value of $\alpha$ in the disk, and then apply the Schwarz lemma to $f\circ \psi_\alpha$ and its inverse.

Added. Here's a bit more detail, as requested by @Hope. We use the fact that a holomorphic bijection has a holomorphic inverse. First, a computation shows that $\psi_\alpha$ maps the unit disk $\mathbb D$ into itself, and since the inverse can be computed explicitly as $\psi_{-\alpha}$, it is bijective. Now assume $f : \mathbb D \to \mathbb D$ which is bijective, and choose the unique point $\alpha \in \mathbb D$ satisfying $f(\alpha) = 0$. Consider $g = f\circ \psi_{-\alpha}$, and note that $\psi_{-\alpha}(0) = \alpha$. By the preceding remarks, $g$ is a holomorphic bijection of $\mathbb D$ which fixes the origin. By the Schwarz lemma, one has $|g'(0)| \leq 1$ with equality if and only if $g$ is a rotation. The inverse map $h = g^{-1}$ also maps $\mathbb D$ into itself, and satisfies $h(0) = 0$, so we have $|h'(0)| \leq 1$ as well. But $h'(0) = 1/g'(0)$, so we find that $|g'(0)| = 1$ after all. By the case of equality in the Schwarz lemma, $g$ is a rotation, and therefore $f = e^{i\theta}\psi_{\alpha}$ for some real $\theta$.
