Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 2 / 23F (Riemann Surfaces)

Put $b=\mathrm{Im}\tau >0$. On $|\mathrm{Im}z|\le R$, the modulus of the $n$th summand is at mostThe resulting Gaussian majorant is summable, independently of the unbounded real part of $z$. The Weierstrass test proves uniform convergence on every such strip. Since the summands are entire, locally uniform convergence implies $\psi$ is entire. It is not identically zero: integrating $\psi (x)e^{-\pi ix}$ over $0\le x\le 1$ extracts its nonzero $n=0$ Fourier coefficient.

Logarithmic differentiation of the two supplied transformation laws givesThese hold as meromorphic identities, with poles at zeros of $\psi$. For a translated fundamental parallelogram with no boundary zeros, opposite edges parallel to $\tau$ cancel by the first identity. The other edge pair contributes $2\pi i$ by the second, so the argument principle counts exactly one zero, including multiplicity, in that parallelogram.

Changing the summation index $n\mapsto -n-1$ shows $\psi (-z)=-\psi (z)$, hence $\psi (0)=0$. Quasiperiodicity makes every lattice point a zero. Every generic fundamental parallelogram contains exactly one lattice point, which already accounts for its one zero; varying its position rules out any additional zeros and proves the lattice zeros are simple. ThereforeIn particular the centred parallelogram in the question has no lattice points on its boundary and its unique zero is zero. Using generic translated contours first avoids assuming this boundary fact before determining the zeros.

For the proposed sum, the first period always holds, while $f(z+\tau )-f(z)=-2\pi i{\sum }_j{\lambda }_j$. Thus it is $\Lambda$-periodic exactly when ${\sum }_j{\lambda }_j=0$. Meromorphicity follows from the meromorphic logarithmic derivatives. Finally differentiation removes the additive shift in $\ell$, so $\frac{d}{dz}[{\psi }^{\prime }(z-a)/\psi (z-a)]$ has both periods for every $a$.