Let \(f\) be an elliptic function and \(P\) be its fundamental parallelogram. Suppose that \(f\) is nonconstant. Show that \(f\) has at least two poles in \(P\), counted with multiplicity.
Solution
Write the period lattice as \(\Lambda=\omega_1\ZZ+\omega_2\ZZ\), and translate the fundamental parallelogram if necessary so that \(f\) has no poles on \(\bd P\).
If \(f\) had no poles, periodicity would make it an entire bounded function: it is bounded on the compact closure of \(P\), hence bounded on all of \(\CC\) by translation through \(\Lambda\). Liouville’s theorem would then force \(f\) to be constant. Thus \(f\) has at least one pole.
Suppose, toward a contradiction, that \(f\) has only one pole counted with multiplicity. Then it is a single simple pole \(z_0\in P\), so \[ \Res_{z=z_0}f\neq0. \]
Write \(\bd P = \sum_{1\leq k \leq 4} \gamma_k\) where the \(\gamma_k\) are the edges traversed counterclockwise. By periodicity, opposite edges contribute equal integrals with opposite orientations, so \[ \int_{\bd P}f(z)\,dz=0. \] But the residue theorem gives \[ 0=\int_{\bd P}f(z)\,dz =2\pi i\Res_{z=z_0}f(z), \] contradicting the simplicity of the pole.
More generally, the same boundary cancellation always shows \[ \sum_{z_k\in P}\Res_{z=z_k}f(z)=0 \] for every elliptic function whose fundamental parallelogram has no poles on its boundary.