We start with the Fourier transform of $f(x) = e^{i \pi \tau x^2 + 2 i x z}$ $$\int_{-\infty}^{+\infty} e^{i \pi \tau x^2 + 2 i x z} e^{2 \pi ixy} \, dx = (- i \tau)^{-1/2} e^{-i {(z + \pi y)^2 \over \pi \tau}}$$

Applying the Poisson summation formula, we obtain the following: $$\sum_{n=-\infty}^{+\infty} e^{i \pi \tau n^2 + 2 i n z} = (- i \tau)^{-1/2} \sum_{n=-\infty}^{+\infty} e^{-i {(z + \pi n)^2 \over \pi \tau}}$$

The left hand side equals $\vartheta_3 (z \mid \tau)$ The right hand side can be rewritten as follows: $$\sum_{n=-\infty}^{+\infty} e^{-i {(z + \pi n)^2 \over \pi \tau}} = e^{-i {z^2 \over \pi \tau}} \sum_{n=-\infty}^{+\infty} e^{-i {\pi n^2 \over \tau} - {2 i n z \over \tau}} = e^{-i {z^2 \over \pi \tau}} \vartheta_3 (z / \tau \mid -1 / \tau)$$

Combining the two expressions yields $$\vartheta_3 (z \mid \tau) = e^{-i {z^2 \over \pi \tau}} \vartheta_3 (z / \tau \mid -1 / \tau)$$