= Solution
Write $a=s+ib$ with real $s,b$, and first translate the real integration variable by $s$. It remains to integrate $e^{-z^2}$ along the horizontal line $\operatorname{Im}z=b$.
For $b>0$, apply the <Cauchy integral theorem> to the rectangle with vertices $-R,R,R+ib,-R+ib$. The <integrand> is an <entire function>, so
$$
\int_{-R}^{R}e^{-x^2}\,dx-\int_{-R}^{R}e^{-(x+ib)^2}\,dx
$$
is the negative of the two vertical-edge integrals. On either vertical edge, $z=\pm R+iy$ and
$$
|e^{-z^2}|=e^{-R^2+y^2}\leq e^{-R^2+b^2}.
$$
Their combined absolute value is at most $2b e^{-R^2+b^2}$ and tends to zero. The rectangle below the real axis gives the same estimate with $|b|$ when $b<0$; $b=0$ needs no deformation. Absolute convergence follows from $|e^{-(x+ib)^2}|=e^{b^2}e^{-x^2}$. Taking $R\to\infty$ proves the <Gaussian contour translation> identity
$$
\boxed{\int_{-\infty}^{\infty}e^{-(u+a)^2}\,du=\sqrt\pi,\qquad a\in\mathbb C}.
$$
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