- Suppose that \(a _ { 1 } , \ldots , a _ { r }\) and \(b _ { 1 } , \ldots , b _ { r }\) are the zeros and poles, respectively, in the fundamental parallelogram of an elliptic function \(f .\) Show that
\[ a _ {1} + \dots + a _ {r} - b _ {1} - \dots - b _ {r} = n \omega_ {1} + m \omega_ {2} \]
for some integers n and \(m .\)
[Hint: If the boundary of the parallelogram contains no zeros or poles, simply integrate \(z f ^ { \prime } ( z ) / f ( z )\) over that boundary, and observe that the integral of \(f ^ { \prime } ( z ) / f ( z )\) over a side is an integer multiple of \(2 \pi i\) . If there are zeros or poles on the side of the parallelogram, translate it by a small amount to reduce the problem to the first case.]
Solution
Translate the fundamental parallelogram slightly, if necessary, so that its boundary contains no zero or pole of \(f\). Let its vertices be \[ z_0,\quad z_0+\omega_1,\quad z_0+\omega_1+\omega_2,\quad z_0+\omega_2, \] and put \[ g(z)=\frac{f'(z)}{f(z)}. \] The function \(g\) is elliptic with the same periods as \(f\).
By the residue theorem, \[ \frac1{2\pi i}\int_{\partial P} z g(z)\,dz =\sum_j a_j-\sum_j b_j, \tag{1} \] where zeros and poles are repeated according to multiplicity.
Pair the bottom side with the oppositely oriented top side. Periodicity under \(\omega_2\) gives \[ \int_{\text{bottom}}zg(z)\,dz+ \int_{\text{top}}zg(z)\,dz =-\omega_2\int_{\text{bottom}}g(z)\,dz. \] Similarly, pairing the left and right sides gives \[ \int_{\text{right}}zg(z)\,dz+ \int_{\text{left}}zg(z)\,dz =\omega_1\int_{\text{left}}g(z)\,dz. \] Therefore \[ \frac1{2\pi i}\int_{\partial P}zg(z)\,dz =m\omega_1+n\omega_2 \tag{2} \] for some integers \(m,n\), because on either side joining two points differing by a period, \[ \frac1{2\pi i}\int \frac{f'}f\,dz \] is the winding number of the closed curve traced by \(f\) and hence is an integer.
Combining (1) and (2), \[ \sum_j a_j-\sum_j b_j=m\omega_1+n\omega_2. \] If the original fundamental parallelogram had zeros or poles on its boundary, a sufficiently small translation avoids them without changing the divisor modulo the period lattice, so the same conclusion holds.