 Period lattice of theta3 over theta4

Use the radian spatial convention ${\theta }_3(z,\tau )={\sum }_n{e}^{\pi i{n}^2\tau +2inz}$ and ${\theta }_4(z,\tau )={\theta }_3(z+\pi /2,\tau )$. Their ratio is unchanged by $\pi$ and changes sign under $\pi \tau$. The numerator has simple zeros at $\pi (1+\tau )/2+\pi \mathbb{Z}+\pi \tau \mathbb{Z}$, while the denominator has simple zeros at $\pi \tau /2+\pi \mathbb{Z}+\pi \tau \mathbb{Z}$. The two sets are disjoint, so any period must lie in $\pi \mathbb{Z}+\pi \tau \mathbb{Z}$, by translating the zero set. The sign rule then gives the displayed exact lattice. Squaring the ratio removes that sign and has precisely the smaller lattice $\pi \mathbb{Z}+\pi \tau \mathbb{Z}$.