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Fall 2011 - Problem 11

maximum principle

Let $\func{f}{\C}{\C}$ be a holomorphic function with $f\p{z} \neq 0$ for all $z \in \C$ . Define $U = \set{z \in \C \mid \abs{f\p{z}} < 1}$ . Show that all connected components of $U$ are unbounded.

Solution.

Since $f$ vanishes nowhere, $g = \frac{1}{f}$ is also an entire function. Thus,

U = \set{z \in \C \mid \abs{g\p{z}} > 1}.

Notice that $U$ is open since $\abs{f}$ is continuous, and so the connected components of $U$ are open as well. If $U$ has a bounded connected component, say $C$ , then we may apply the maximum principle on this domain. Since $C$ is open, $\partial{C} = \cl{C} \setminus C$ , and so $\partial{C} \subseteq U^\comp$ . Thus, $\abs{g\p{z}} \leq 1$ on $\partial{C}$ . By the maximum principle, it follows that $\abs{g\p{z}} \leq 1$ on all of $C$ , and so

\frac{1}{\abs{f\p{z}}} \leq 1 \implies \abs{f\p{z}} \geq 1

on $C \subseteq U$ . But we assumed that $\abs{f\p{z}} < 1$ on $U$ , a contradiction. Thus, no bounded connected component could have existed to begin with, which completes the proof.