Layla Velazquez

2022-07-01

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 $R\left(z\right)=\frac{{a}_{0}\left(z-{a}_{n}\right)\cdots \left(z-{a}_{1}\right)}{\left(z-{b}_{m}\right)\cdots \left(z-{b}_{1}\right)}$, 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.

### Answer & Explanation

Jake Mcpherson

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
${\psi }_{\alpha }\left(z\right)=\frac{z-\alpha }{1-\overline{\alpha }z}$
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.
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 $f\left(\alpha \right)=0$. Consider $g=f\circ {\psi }_{-\alpha }$, and note that ${\psi }_{-\alpha }\left(0\right)=\alpha$. By the preceding remarks, $g$ is a holomorphic bijection of $\mathbb{D}$ which fixes the origin. By the Schwarz lemma, one has $|{g}^{\prime }\left(0\right)|\le 1$ with equality if and only if $g$ is a rotation. The inverse map $h={g}^{-1}$ also maps $\mathbb{D}$ into itself, and satisfies $h\left(0\right)=0$, so we have $|{h}^{\prime }\left(0\right)|\le 1$ as well. But ${h}^{\prime }\left(0\right)=1/{g}^{\prime }\left(0\right)$, so we find that $|{g}^{\prime }\left(0\right)|=1$ 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 $\theta$.

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