Fall 2009 - Problem 9

Solution.

1. Notice that the conditions imply that $f$ is double periodic, so the behavior of $f$ in the rectangle $R = \br{0, \sqrt{2}} \times \br{0, \sqrt{2}}$ determine the behavior of $f$ globally.

   Suppose first that $f$ has no poles in $R$. Since $R$ is compact, this means that $f$ is bounded on $R$, which means by double periodicity that $f$ is bounded on all of $\C$. By Lioville's theorem, this implies $f$ is constant, which is impossible. Thus, $f$ must have at least one pole $z_0 \in R$.

   By symmetry, the point furthest from every vertex of $R$ must be the center $\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}$, which has distance $1$ from every vertex. Let $v$ be a closest vertex to $z_0$ so that $\abs{z_0 - v} \leq 1$. Thus, by double periodicity, $f$ must have a pole within distance $1$ of $0$, i.e., $f$ has at least one pole in $\cl{D}$. By assumption, $f$ has exactly one pole in $\cl{D}$, as required.

2. Consider a different rectangle $R = \br{-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}} \times \br{-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}}$. Notice that $R$ has side length $\sqrt{2}$, so this is a fundamental domain for $f$, by the periodicity conditions. Moreover, $\partial{R}$ contains no poles. Indeed, $R \subseteq \cl{D}$, so by periodicity, if there is a pole on one of the sides, then there will be another pole on the opposite side, but $\cl{D}$ has exactly one pole, which is impossible.

   Let $\gamma_L, \gamma_B, \gamma_R, \gamma_T$ be the left, bottom, right, top side of $\partial{R}$ oriented counter-clockwise, respectively. By periodicity, $f$ takes on the same values on $\gamma_L$ as it does on $\gamma_R$, and $\gamma_B$ as it does on $\gamma_T$. Thus, because they are oriented oppositely,

   $$\int_{\gamma_L} f\p{z} \,\diff{z} = -\int_{\gamma_R} f\p{z} \,\diff{z} \quad\text{and}\quad \int_{\gamma_B} f\p{z} \,\diff{z} = -\int_{\gamma_T} f\p{z} \,\diff{z}.$$

   Thus, $\int_\gamma f\p{z} \,\diff{z}$ vanishes. But by the residue theorem, this means

   $$0 = \int_\gamma f\p{z} \,\diff{z} = 2\pi i \Res{f}{z_0}.$$

   Consequently, the $\p{z - z_0}^{-1}$ term in the Laurent series of $f$ vanishes, so $z_0$ cannot be a simple pole.