# Zeros of the Jacobi Theta function

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?

• It is "quasi-periodic" with period $\tau$: $\theta(z+\tau)=G(z)\theta(z)$ with $G(z)$ some simple non-zero function. Apr 16, 2020 at 12:19
• Of course... How could I forget :-( Apr 16, 2020 at 16:01

• 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..).