Let \(f(z)\) be a nonconstant meromorphic function on \(\mathbb{C}\). A number \(\omega\in\mathbb{C}\) is a period for \(f(z)\) if \(f(z+\omega)=f(z)\) for all \(z\). Show that any bounded set of periods of \(f\) must be finite.
Solution
Suppose a bounded set of periods were infinite. It would contain a sequence of distinct periods \((\omega_n)\) having a convergent subsequence. Passing to that subsequence, the differences \[ \tau_n=\omega_{n+1}-\omega_n \] are nonzero periods and satisfy \(\tau_n\to0\).
Choose \(z_0\) which is not a pole of \(f\). For all sufficiently large \(n\), \(z_0+\tau_n\) is also not a pole, and periodicity gives \[ f(z_0+\tau_n)=f(z_0). \] Hence \[ f'(z_0) =\lim_{n\to\infty}\frac{f(z_0+\tau_n)-f(z_0)}{\tau_n}=0. \] Thus \(f'\) vanishes at every point where \(f\) is holomorphic. On each component of the complement of the poles, \(f\) is constant; by analytic continuation the meromorphic function is constant on \(\mathbb C\), contradicting the hypothesis.
Therefore \(0\) is isolated in the period group. A bounded subset of a discrete subset of \(\mathbb C\) is finite, proving the claim.