This problem is exercise 6.1 from Silverman's Arithmetic of Elliptic Curves.
6.1 Let $\Lambda=\mathbb{Z}\omega_1$+$\mathbb{Z}\omega_2$ be a lattice.Suppose $\theta(z)$ is an entire function,with the property that there are constants $a_1,a_2\in \mathbb{C}$ such that,$\theta(z+\omega_1)=a_1\theta(z)$ and $\theta(z+\omega_2)=a_2\theta(z)$.Show that there are constants $b,c\in\mathbb{C}$ such that, $$\theta(z)=be^{cz}$$
I've seen the answer in this Proving a function to be exponential given that it satisfies certain equations.. But I do not think the answer there is right because the constant $B$ defined does not satisfy $e^{Bz+B\omega_1}=a_1e^{Bz}$ Can someone give me a hint on how to go about proving it?
Assume that $\theta(z)$ is not identically $0$.Consider the function $g(z)=\theta'(z)/\theta(z)$.Clearly $g(z)$ is an elliptic function with periods $\omega_1$ and $\omega_2$ since $\theta'(z)$ is quasi-periodic with same quasi-periods.Let $P_0$ be the fundamental parallelogram of the lattice.Then, $$\frac{1}{2\pi i}\int_{\partial P_0} g(z)=0$$ since the sum cancels out on the opposite sides of the parallelogram.But this is same as saying, $$\frac{1}{2\pi i}\int_{\partial{P_0}}\frac{\theta'(z)}{\theta(z)}=0$$ which by the Argument principle is the number of zeros-number of poles of $\theta(z)$ in $P_0$.But $\theta(z)$ is an holomorphic function on $\mathbb{C}$ meaning it has no poles in $P_0$.Hence,by the above equation it has no zeros as well.Hence $\theta(z)$ is a non-vanishing entire function.Thus,we have $g(z)$ is holomorphic on $\mathbb{C}$ and since it is an elliptic function,it is bounded.By Liouville's theorem it must be a constant,say $c$.Then,we have,$$\theta'(z)=c\theta(z)$$ and solving it we get $$\theta(z)=be^{cz}$$.