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.

Write $\xi=a+bi$. The exponent inside the integrand looks like
$$\begin{array}{ll} -\pi x^2+2\pi ix\xi & =-\pi x^2+2\pi ix(a+bi) \\ & =-\pi x^2+2\pi iax-2\pi bx \\ & = -\pi (x^2+2bx)+2\pi iax \end{array}$$
The real part has both an $x$ and an $x^2$ term, but we may complete the square:
$$ \begin{array}{l} = -\pi(x^2+2bx+b^2)+\pi b^2+2\pi iax \\  =-\pi(x+b)^2+\pi b^2+2\pi iax. \end{array} $$
With the substitution $u=x+b$ ($\Leftrightarrow x=u-b$) this becomes
$$ \begin{array}{l} =-\pi u^2+\pi b^2+2\pi ia(u-b) \\ =-\pi u^2+2\pi iau+[\pi b^2-2\pi iab]. \end{array}$$
Therefore
$$ \begin{array}{ll} \displaystyle \int_{-\infty}^{\infty} \exp(-\pi x^2+2\pi ix\xi)\,\mathrm{d}x & \displaystyle =\exp(\pi b^2-2\pi iab)\int_{-\infty}^{\infty} \exp(-\pi u^2+2\pi iau)\,\mathrm{d}u \\ & \displaystyle =\exp(\pi b^2-2\pi iab)\exp(-\pi a^2) \\ & =\exp(-\pi(a+bi)^2) \\ & =\exp(-\pi \xi^2). \end{array} $$
