I found in Mumford (tata Lectures on Theta) that every elliptic function can be written as fraction of theta functions
$$\prod\frac{\vartheta_{a_{j}}\left(z-z_{j}\right)}{\vartheta_{b_{i}}\left(z-z_{i}\right)}$$
Does anybody have some reference on that? Is it simple to proof or involved?
It is immediate that for $q=e^{i \pi \tau}, |q| < 1$ with $$f(z) = \prod_{m \ge 1} (1+q^{2m-1} e^{2i \pi z})(1+q^{2m-1} e^{-2i \pi z})$$ then $f$ is entire satisfying $$f(z+1)=f(z), \qquad f(z + \tau)=\frac{1+q^{-1} e^{-2i \pi z}}{1+q e^{2 i \pi z}}f(z)= q^{-1} e^{-2i \pi z} f(z)$$ and with simple zeros at $\frac{\tau+1}{2}+\Bbb{Z}\tau + \Bbb{Z} $.
Thus if $a_1+a_2 = b_1+b_2$ then $$\frac{f(z-\frac{\tau+1}{2}-a_1)f(z-\frac{\tau+1}{2}-a_2)}{f(z-\frac{\tau+1}{2}-b_1)f(z-\frac{\tau+1}{2}-b_2)}$$ is meromorphic $\Bbb{Z+\tau Z}$ periodic with simple poles at $b_1,b_2$ and simple zeros at $a_1,a_2$.
Whence for any non-zero $\Bbb{Z+\tau Z}$ elliptic function $g$, with $a_j,b_j$ its poles/zeros (integrating $z \frac{g'(z)}{g(z)}$ on the boundary of the fundamental parallelogram to show $d=\sum_j a_j-\sum_j b_j \in \Bbb{Z+\tau Z}$, replacing $a_1$ by $a_1-d$) then $$g(z) \prod_{j=1}^J \frac{f(z-\frac{\tau+1}{2}-a_j)}{f(z-\frac{\tau+1}{2}-b_j)}$$ is entire and doubly periodic thus bounded and constant.
Finally, writing $f$'s Fourier series and using $f(z + \tau) =q^{-1} e^{-2i \pi z} f(z)$ we find $$f(z) = h(\tau) \sum_n q^{n^2} e^{2i\pi nz} = h(\tau) \vartheta(z,\tau)$$ The only difficult step is to find the product formula for $h(\tau)$ which we don't need here.