- Suppose that a meromorphic function \(f\) has two periods \(\omega _ { 1 }\) and \(\omega _ { 2 }\) , with \(\omega _ { 2 } / \omega _ { 1 } \in \mathbb { R }\)
Suppose \(\omega _ { 2 } / \omega _ { 1 }\) is rational, say equal to \(p / q ,\) , where \(p\) and \(q\) are relatively prime integers. Prove that as a result the periodicity assumption is equivalent to the assumption that \(f\) is periodic with the simple period \(\begin{array} { r } { \omega _ { 0 } = \frac { 1 } { q } \omega _ { 1 } } \end{array}\) [Hint: Since p and q are relatively prime, there exist integers m and n such that \(m q + n p = 1\) (Corollary 1.3, Chapter 8, Book I).]
If \(\omega _ { 2 } / \omega _ { 1 }\) is irrational, then \(f\) is constant. To prove this, use the fact that \(\{ m - n \tau \}\) is dense in R whenever \(\tau\) is irrational and m, n range over the integers.
Solution
Write \(\tau=\omega_2/\omega_1\in\mathbb R\).
For (a), suppose \(\tau=p/q\) with \((p,q)=1\). If \(f\) has period \(\omega_0=\omega_1/q\), then both \[ \omega_1=q\omega_0,\qquad \omega_2=p\omega_0 \] are periods. Conversely, if \(\omega_1\) and \(\omega_2\) are periods, choose integers \(m,n\) with \[ mq+np=1. \] Then \[ m\omega_1+n\omega_2 =\left(m+\frac{np}{q}\right)\omega_1 =\frac{mq+np}{q}\omega_1 =\omega_0. \] Integer linear combinations of periods are periods, so \(\omega_0\) is a period. Thus the two periodicity conditions are equivalent.
For (b), suppose \(\tau\) is irrational. The additive subgroup \[ G=\{m-n\tau:m,n\in\mathbb Z\} \] is dense in \(\mathbb R\). Hence there is a sequence of nonzero periods \[ h_j=(m_j-n_j\tau)\omega_1\longrightarrow0. \] A meromorphic function cannot have a pole: if \(p\) were a pole, then every \(p+h_j\) would also be a pole, producing distinct poles accumulating at \(p\), contrary to isolatedness of poles. Thus \(f\) is entire.
For every \(z\in\mathbb C\), \[ f'(z)=\lim_{j\to\infty}\frac{f(z+h_j)-f(z)}{h_j}=0, \] because each \(h_j\) is a period. Therefore \(f'\equiv0\), so \(f\) is constant.