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Fall 2012 - Problem 11

periodic functions

Find all holomorphic functions $\func{f}{\C}{\C}$ satisfying $f\p{z + 1} = f\p{z}$ and $f\p{z + i} = e^{2\pi}f\p{z}$ for all $z \in \C$ .

Solution.

Observe that $g\p{z} = e^{-2\pi iz}$ is such a function. Given any other $f\p{z}$ satisfying the same identities, set $h\p{z} = \frac{f\p{z}}{g\p{z}}$ . Then

\begin{gathered} h\p{z + 1} = \frac{f\p{z + 1}}{g\p{z + 1}} = \frac{f\p{z}}{g\p{z}} = h\p{z} \\ h\p{z + i} = \frac{f\p{z + i}}{g\p{z + i}} = \frac{e^{2\pi}f\p{z}}{e^{2\pi}g\p{z}} = h\p{z}. \end{gathered}

In other words, $h$ is doubly periodic. Notice that $g\p{z}$ never vanishes, so $h$ is also entire. In particular, $h$ is continuous on $\br{0, 1} \times \br{0, 1}$ , so by double periodicity, $h$ is bounded everywhere. By Liouville's theorem, $h$ is constant, so $f\p{z} = Ce^{-2\pi iz}$ for some $C \in \C$ , which completes the proof.