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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.

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By bijective on the unit disk, I suppose you mean a bijection between the unit disk and itself? And $z^{2n+1}$ doesn't work, for $n\ge1$. –  Harald Hanche-Olsen Mar 27 '12 at 17:07
    
Yes, this was my intended meaning. –  Stuart Mar 27 '12 at 17:16
    
Note that $R$ extends to a bijection on the boundary, so to the closed disc. By the reflection principle, it extends to a bijection on the projective line ($R$ is a bijection on the complement of the closed disc in $\mathbb{P}^1$). That rather restricts the possible degree of $R$. –  WimC Mar 27 '12 at 17:43
    
It hasn't been covered, but I will think about it and see if I get something. –  Stuart Mar 27 '12 at 20:27
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1 Answer 1

up vote 4 down vote accepted

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 $\theta$.

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I have omitted some details because you tagged the question as homework, but if you would like me to elaborate, please don't hesitate to say so. –  Nick Strehlke Mar 27 '12 at 20:32
    
Thanks, I was thinking more about the properties of rational functions and I hadn't seen the Schwarz lemma before. –  Stuart Mar 27 '12 at 23:03
    
@Nick Strehlke Could you elaborate more on your methodology? This problem is interesting and I would like to see your approach. –  Hope Mar 28 '12 at 6:25
    
I added some detail. –  Nick Strehlke Mar 28 '12 at 6:42
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I found this answer sensual and arousing. Mathematics has never made me feel this way. This experience, while confusing, is not unpleasant. Thank you. –  Daniel Montealegre Apr 1 '12 at 8:00
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