For all $\xi \in \mathbb{C}$ we have $e^{-\pi\xi^2}=\int_{-\infty}^\infty \! e^{-\pi x^2}e^{2\pi ix\xi}\ \mathrm{d}x.$ This is Exercise 2.4 in Stein & Shakarchi's Complex Analysis. Prove that for all $\xi \in \mathbb{C}$
$$e^{-\pi\xi^2}=\int_{-\infty}^\infty \! e^{-\pi x^2}e^{2\pi ix\xi}\ \mathrm{d}x.$$
They prove it for the real case, so I assume that I'm supposed to use that. All I can think to do is write $\xi=a+ib$, which gives a term $e^{2\pi i xa}$ in the integral which becomes 1, since the power is a real multiple of $2\pi$ and $i$. That leaves
$$e^{-\pi\xi^2}=\int_{-\infty}^\infty \! e^{-\pi x^2}e^{-2\pi xb}\ \mathrm{d}x$$
but I don't have any idea where to go from there, or if this is even a fruitful direction. Any hints would be appreciated.

Note that the integral
$$
\int_{-\infty}^{\infty} e^{-\pi x^2}e^{-2\pi i x\xi}\,\mathrm{d}x
$$
is the Fourier transform of $e^{-\pi x^2}$. Now simplify
$$
e^{-\pi x^2}e^{-2\pi ix\xi} = e^{-\pi(x^2+2\pi i x)}
$$
and complete the square of $x^2 + 2\pi i x$ to deduce that
$$
x^2 + 2\pi i x = x^2 + 2\pi i x -\xi^2 + \xi^2 = (x+i\xi) + \xi^2.
$$
Substituting, this gives
$$
e^{-\pi(x^2 + 2\pi i x)} = e^{-\pi(x+i\xi)^2-\pi\xi^2} = e^{-\pi(x+i\xi)^2} e^{- \pi\xi^2}
$$
and hence
$$
\int_{-\infty}^{\infty} e^{-\pi(x^2+2\pi ix\xi)}\,\mathrm{d}x= e^{-\pi\xi^2}\int_{-\infty}^{\infty} e^{-\pi(x+i\xi)^2}\,\mathrm{d}x.
$$
Making the substitution $u = x+i\xi$ so that $\mathrm{d}x = \mathrm{d}u$ then yields
$$
e^{-\pi\xi^2}\int_{-\infty}^{\infty} e^{-\pi(x+i\xi)^2}\,\mathrm{d}x = e^{-\pi\xi^2}\int_{-\infty}^{\infty}e^{-\pi u^2}\,\mathrm{d}u= e^{-\pi\xi^2}\frac{\sqrt{\pi}}{\sqrt{\pi}} = e^{-\pi\xi^2},
$$
as was to be shown. Note that this shows that
$$
\mathscr{F}(e^{-\pi x^2}) = e^{-\pi \xi^2},
$$
where $\mathscr{F}$ is the Fourier transform on $L^2(\mathbb{R})$.
