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For fixed $\mathrm{Im}\tau >0$, 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 givesSince ${\theta }_4(z)={\theta }_3(z+\pi /2)$, the corresponding second multiplier for ${\theta }_4$ has the opposite sign: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 $z_{\ast }=(\pi +\pi \tau )/2$,because the terms with indices $n$ and $-n-1$ cancel. The absolute convergence justifies that pairing. Quasi-periodicity propagates this zero to every $z_{\ast }+m\pi +n\pi \tau$.

For exhaustiveness and simplicity, let $v={\theta }_3^{\prime }/{\theta }_3$. 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 giveThe argument principle therefore counts one zero with multiplicity in every cell. Since a known zero class already exists, all zeros are simple and exactlyThe denominator's simple zeros are instead $\pi \tau /2+{\Lambda }_0$. These two cosets are disjoint: $\pi /2$ cannot lie in ${\Lambda }_0$ because $\mathrm{Im}\tau >0$. 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:The second lattice is for the squared function; no additional translation can preserve its zero set.