= Solution
For fixed $\operatorname{Im}\tau>0$, Gaussian decay of $|q|^{n^2}$ gives locally uniform convergence of the <theta function> series and all its $z$-derivatives. It is therefore entire. Shifting the series argument and reindexing gives
$$
\theta_3(z+\pi)=\theta_3(z),\qquad
\theta_3(z+\pi\tau)=q^{-1}e^{-2iz}\theta_3(z).
$$
Since $\theta_4(z)=\theta_3(z+\pi/2)$, the corresponding second multiplier for $\theta_4$ has the opposite sign:
$$
\theta_4(z+\pi)=\theta_4(z),\qquad
\theta_4(z+\pi\tau)=-q^{-1}e^{-2iz}\theta_4(z).
$$
Thus $h=\theta_3/\theta_4$ obeys $h(z+\pi)=h(z)$ and $h(z+\pi\tau)=-h(z)$. These give candidate <period lattices>; to show they are exact, first locate every zero.
At $z_*=(\pi+\pi\tau)/2$,
$$
\theta_3(z_*)=\sum_{n\in\mathbb Z}(-1)^nq^{n(n+1)}=0,
$$
because the terms with indices $n$ and $-n-1$ cancel. The absolute convergence justifies that pairing. Quasi-periodicity propagates this zero to every $z_*+m\pi+n\pi\tau$.
For exhaustiveness and simplicity, let $v=\theta_3'/\theta_3$. It satisfies $v(z+\pi)=v(z)$ and $v(z+\pi\tau)=v(z)-2i$. Take a <fundamental parallelogram> for $\Lambda_0=\pi\mathbb Z+\pi\tau\mathbb Z$, translated so its boundary has no zero. The two sloping edges cancel by the first identity; the bottom and reversed top give
$$
\oint v(z)\,dz=\int_{z_0}^{z_0+\pi}\bigl(v(z)-v(z+\pi\tau)\bigr)dz=2\pi i.
$$
The <argument principle> therefore counts one zero with multiplicity in every cell. Since a known zero class already exists, all zeros are simple and exactly
$$
\boxed{z=\frac\pi2+\frac{\pi\tau}{2}+m\pi+n\pi\tau,\qquad m,n\in\mathbb Z}.
$$
The denominator's simple zeros are instead $\pi\tau/2+\Lambda_0$. These two cosets are disjoint: $\pi/2$ cannot lie in $\Lambda_0$ because $\operatorname{Im}\tau>0$. Thus there are no cancelled zeros or <poles> in the quotient.
Any period of $h$ or $h^2$ must translate its <zero set> onto itself, so must belong to $\Lambda_0$. A translation $m\pi+n\pi\tau$ multiplies $h$ by $(-1)^n$; the quotient is nonconstant, so an odd $n$ cannot be a period of $h$. Squaring removes precisely that sign. This proves the <period lattice of theta3 over theta4> and its squared counterpart:
$$
\boxed{\operatorname{Per}\left(\frac{\theta_3}{\theta_4}\right)=\pi\mathbb Z+2\pi\tau\mathbb Z},\qquad
\boxed{\operatorname{Per}\!\left[\left(\frac{\theta_3}{\theta_4}\right)^2\right]=\pi\mathbb Z+\pi\tau\mathbb Z}.
$$
The second lattice is for the squared function; no additional translation can preserve its <zero set>.
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