From Elliptic Curves: Function Theory, Geometry and Arithmetic by McKean and Moll:
Exercise 3.1.2. Discuss the general solution of the two identities (a) $f(x+2)=f(x)$ and (b) $f(x+2\omega)=e^{ax+b}f(x)$. Hint: Expand $f$ in a Fourier series $\sum{\hat{f}(n)e^{\pi \sqrt{-1}nx}}$ and use (b) to pin down $\hat{f}(n)$. You will see thetas coming out.
The question is, how do I use (b) to pin down $\hat{f}(n)$?
I can write a Fourier series using the information from (a): $$f(x)=\sum{\hat{f}(n)e^{\pi \sqrt{-1}nx}}$$ where $$\hat{f}(n)=\frac{1}{2}\int_0^2{f(x)e^{-\pi \sqrt{-1}nx}}dx$$
To use (b), I thought that I could perhaps integrate the function $\frac{1}{2} f(x)e^{-\pi \sqrt{-1}nx}$ around the perimeter of the fundamental cell (this is the line segment from $0$ to $2$, followed by $2$ to $2 +2\omega$, then $2 +2\omega$ back to $2\omega$, and $2\omega$ to $0$). Then the contribution from the integrals from $2$ to $2 +2\omega$ and from $2\omega$ to $0$ cancel out due to the $2$-periodicity of $\frac{1}{2} f(x)e^{-\pi \sqrt{-1}nx}$ and since the function is entire and using Cauchy's integral theorem we end up with: $$\int_0^2{\frac{1}{2}f(x)e^{-\pi \sqrt{-1}nx}}dx=\int_0^2{\frac{1}{2} f(x+2\omega)e^{-\pi \sqrt{-1}n(x+2\omega)}}dx$$
Then: $$\hat{f}(n)= e^{-2\sqrt{-1}\pi\omega n}\int_0^2{\frac{1}{2} e^{ax+b}f(x)e^{-\pi \sqrt{-1}nx}}dx$$
But now I think I have hit a snag: I know that $\hat{f}(n)$ should contain an $e^{\sqrt{-1}\pi\omega n^{2}}$ term, but it doesn't look like that is going to happen. Have I made a mistake somewhere or should I be using a different technique? I can't think of any other viable way to use (b) that will give me the magical $n^{2}$.