= Solution
Use the <Fourier transform> convention $\widehat h(\xi)=\int_{\mathbb R}h(x)e^{-2\pi ix\xi}\,dx$. For a <Schwartz function> $h$, the <Poisson summation formula> is
$$
\boxed{\sum_{n\in\mathbb Z}h(n)=\sum_{m\in\mathbb Z}\widehat h(m).}
$$
To prove it, periodize $h$ as $P(x)=\sum_{n\in\mathbb Z}h(x+n)$. Rapid decay of all derivatives makes this a smooth periodic function, with uniform termwise differentiability. Its $m$th <Fourier coefficient> is
$$
\int_0^1P(x)e^{-2\pi imx}\,dx=\int_{\mathbb R}h(u)e^{-2\pi imu}\,du=\widehat h(m),
$$
where <absolute convergence> justifies splitting the real-line integral into unit intervals. Integration by parts gives rapid decay of the coefficients of $P$, so its <Fourier series> converges absolutely and uniformly to $P$. Evaluate it at zero to obtain the formula.
For $\operatorname{Im}\tau>0$ and $z\in\mathbb C$, take the complex <Gaussian function> $h(x)=e^{\pi i\tau x^2+2\pi izx}$. It is a Schwartz function of the real variable $x$. Its Fourier transform is
$$
\widehat h(\xi)=(-i\tau)^{-1/2}\exp\!\left(-\frac{\pi i(z-\xi)^2}{\tau}\right).
$$
The real Gaussian integral proves this first for $\tau=iy$, $y>0$, and real $z$. Normal convergence of the defining integrals makes both sides entire in $z$ and holomorphic in $\tau\in\mathbb H$, so the identity theorem extends it to the stated domain. Choose the square-root branch holomorphic for $-i\tau$ in the right half-plane and positive when $\tau=iy$.
Applying <Poisson summation> gives the transformation of the <Jacobi theta function>:
$$
\vartheta_{00}(z,\tau)=(-i\tau)^{-1/2}e^{-\pi iz^2/\tau}\sum_{m\in\mathbb Z}e^{-\pi im^2/\tau+2\pi imz/\tau}.
$$
The series on the right is $\vartheta_{00}(z/\tau,-1/\tau)$. Rearranging therefore gives
$$
\boxed{\vartheta_{00}(z/\tau,-1/\tau)=\left(\frac\tau i\right)^{1/2}e^{\pi iz^2/\tau}\vartheta_{00}(z,\tau).}
$$
The branch choice is part of the formula; at $\tau=i$ its square-root factor equals one.
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