Zeros of the Jacobi Theta function

Question

How do you obtain all the zeros in $z$ of the Jacobi Theta function $$\vartheta(z) = \sum_{n} e^{\pi i n^2 \tau + 2\pi i n z} \, ?$$

Probably the easiest way is to just read them of the Jacobi-Triple product, but I'm pretty sure they can also be derived from the series representation. The zeros are $$z=a\tau + b + \frac{\tau + 1}{2} \, ,$$ where $a,b \in {\mathbb Z}$, of which I lack to find the term $a\tau$. Since $\vartheta(z+1)=\vartheta(z)$, it is periodic with perdiod $1$ in $z$. So any zero $z_0$ will lead to a zero $b+z_0$ for any integer $b$. It can be seen that $z_0=\frac{\tau+1}{2}$ is a zero since $$\vartheta(z_0) = \sum_{n} e^{\pi i n^2 \tau + \pi i n (\tau+1) } \stackrel{n\rightarrow -n-1}{=} \sum_{n} e^{\pi i n^2 \tau + 2\pi i n \tau + \pi i\tau - \pi i (n+1)(\tau+1)} \\ = -\sum_{n} e^{\pi i n^2 \tau + \pi i n \tau - \pi i n} = -\sum_{n} e^{\pi i n^2 \tau + \pi i n (\tau + 1)} = - \vartheta(z_0) \, .$$

$\vartheta(z)$ has a period of $2$ in $\tau$, but that doesn't help to obtain the term $a\tau$. Any idea?

Answer 1

• You don't need the full Jacobi triple product, only that $$\prod_{m\ge 1}\left( 1 + x^{2m-1} y^2\right) \left( 1 +\frac{x^{2m-1}}{y^2}\right) = c_0(x)\sum_{n=-\infty}^\infty x^{n^2} y^{2n}, \qquad c_0(x)\ne 0, |x|< 1 $$ Which is elementary.

Finding $c_0(x)$ is harder.

• Otherwise just from the transformation law of $\theta$ you can locate its zeros and prove there are no more with the residue theorem (integrating over the boundary of a parallelogram, finding there is one more zero than pole, and since $\theta$ has no poles..).