Analysis Problem of the Day 82

Today’s problem appeared as Problem 11 on the UCLA Spring 2009 Analysis Qual:

Problem 11. Let $f: \mathbb{D} \to \mathbb{C}$ be a holomorphic function that is injective on some annulus $\{z \in \mathbb{C}: r<|z|<1\}.$ Show that $f$ is injective on $\mathbb{D}.$

Solution: Since $f(z)-c$ is injective for any constant $c$ if and only if $f$ is injective, it suffices to show that $f$ has at most one zero on the unit disc. By the argument principle, it further suffices to show that $\frac{1}{2\pi i}\int_\gamma \frac{f'(z)}{f(z)}dz \leq 1$ for $\gamma$ being a parametrization of a sufficiently large circle $|z|=R$ with $r<R<1.$ But since $f$ is injective, by the change of variables formula for integration, the above integral can be expressed as

$\frac{1}{2\pi i}\int_\gamma \frac{f'(z)}{f(z)}dz = \frac{1}{2\pi i}\int_{f(\gamma)} \frac{1}{z} dz \leq 1,$

since $f(\gamma)$ is a simple closed curve by the Jordan curve theorem (and the value of the integral is either $0$ or $2\pi i,$ depending on whether $0 \in \text{int}(f(\gamma))$ or not). Thus, $f$ is injective on $\mathbb{D}.$

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