Analysis Problem of the Day 26

Today’s problem appears as Problem 11 on the UCLA Fall 2011 Analysis Qual:

Problem 11: Let $f$ be an entire function with $f(z) \not = 0$ on $\mathbb{C}.$ Show that the connected components of $f^{-1}(\mathbb{D})$ are unbounded.

Solution: Since $f$ does not vanish, its complex logarithm exists, i.e. $f(z) = e^{g(z)}$ for some entire function $g.$ Since $|f(z)|<1$ whenever $\text{Re }(g(z)) < 0,$ it thus suffices to show that the connected components of $g^{-1}(\mathbb{H})$ are unbounded, where $\mathbb{H}$ is the left half-plane. For sake of contradiction, let $U$ be a bounded connected component of $g^{-1}(\mathbb{H}).$ Note that $U$ is open in $\mathbb{C}$ by continuity. Note that for any $p \in \partial U,$ $\text{Im} (g(p)) = 0,$ as otherwise the point would belong to the connected component $U.$ Thus, $g(\partial U)$ consists of purely imaginary numbers. But then $\inf_{z \in U} \text{Re}(g(z))<0,$ so by Heine-Borel and the open mapping theorem there exists $z_n \in U$ s.t. $z_n \to z^* \in \partial U$ with $\text{Re}(g(z^*))<0,$ which is a contradiction since $g(\partial U)$ consists of purely imaginary numbers. Thus, every connected component of $f^{-1}(\mathbb{D})$ is unbounded.

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