Fall 2014 - Problem 9

Solution.

Suppose $f$ were not injective, so there exist $z_1 \neq z_2$ in $\Omega$ such that $w = f\p{z_1} = f\p{z_2}$. Since $\Omega$ is connected, there exists a path $\gamma$ such that $\gamma\p{0} = z_1$ and $\gamma\p{1} = z_2$.

Let $\delta = \frac{1}{2}d\p{\gamma, \Omega^\comp} > 0$ and cover $\C$ with squares with side length $\delta$. Let $\Omega_\gamma$ be the union of the squares which intersect $\gamma$ non-trivially, and by our choice of $\delta$, we see that $\Omega_\gamma \subseteq \Omega$. Let $\Gamma = \partial\Omega_\gamma$, which is a finite union of segments of length $\delta$ since $\gamma$ is compact. Thus, $\Gamma$ is a closed, simple curve containing $z_1$ and $z_2$.

On the interior of $\Gamma$, $f\p{z} - w$ has only finitely many zeroes since its zeroes are isolated ($f$ is non-constant) and $\Gamma$ is compact. Hence, by shrinking $\delta$ if necessary, we may assume that $\Gamma$ contains no zeroes of $f\p{z} - w$. In particular, $\epsilon = \inf_{z \in \Gamma} \abs{f\p{z} - w} > 0$. Since $\Gamma$ is compact, (local) uniform convergence gives $N \in \N$ such that $\abs{f_N\p{z} - f\p{z}} < \epsilon$ on $\Gamma$. Thus,

so by Rouché's theorem, $f\p{z} - w$ and $\p{f\p{z} - w} + \p{f_N\p{z} - f\p{z}} = f_N\p{z} - w$ have the same number of zeroes $M$ in $\Gamma$. Since $f\p{z_1} = f\p{z_2} = w$, $M \geq 2$, but because $f_N$ is injective, we have $M \leq 1$, a contradiction. Hence, $f$ must have been injective to begin with, which completes the proof.