proof of Jacobi’s identity for $\vartheta$ functions

We start with the Fourier transform of $f(x)=e^{i\pi \tau x^2+2ixz}$:

$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{i\pi \tau x^2+2ixz}{e}^{2\pi ixy}𝑑x={(-i\tau )}^{-1/2}{e}^{-i\frac{{(z+\pi y)}^2}{\pi \tau }}$$

Applying the Poisson summation formula, we obtain the following:

$$\sum _{n=-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{i\pi \tau n^2+2inz}={(-i\tau )}^{-1/2}\sum _{n=-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-i\frac{{(z+\pi n)}^2}{\pi \tau }}$$

The left hand equals ${\vartheta }_3(z\mid \tau )$. The right hand can be rewritten as follows:

$$\sum _{n=-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-i\frac{{(z+\pi n)}^2}{\pi \tau }}=e^{-i\frac{z^2}{\pi \tau }}\sum _{n=-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-i\frac{\pi n^2}{\tau }-\frac{2inz}{\tau }}=e^{-i\frac{z^2}{\pi \tau }}{\vartheta }_3(z/\tau \mid -1/\tau )$$

Combining the two expressions yields

$${\vartheta }_3(z\mid \tau )=e^{-i\frac{z^2}{\pi \tau }}{\vartheta }_3(z/\tau \mid -1/\tau )$$