$\int_{-\infty}^{\infty}e^{-zt^2 + 2wt} \ dt = e^{w^2 / z} \sqrt{\frac{\pi}{z}}$ for $z,w \in \mathbb{C}, Re(z) > 0$ I'm doing complex analysis out of Gamelin, and I'm stuck on a problem in section V.7 on zeros of analytic functions.
Problem:  Show that $\int_{-\infty}^{\infty}e^{-zt^2 + 2wt} \ dt = e^{w^2 / z} \sqrt{\frac{\pi}{z}}$ for $z,w \in \mathbb{C}, Re(z) > 0$.
Gamelin gives a hint to start with $z$ as a real positive number, and I was able to show this is true when $z$ is a positive real number.  But I do not see how to extend this result to general $z$.
 A: First we complete the square and write $-zt^2+2wt=-z(t-w/z)^2+w^2/z$.  Thus, we see that
$$\int_{-\infty}^\infty e^{-zt^2+2wt}\,dt=e^{w^2/z}\int_{-\infty}^\infty e^{-z(t-w/z)^2}\,dt \tag 1$$
Enforcing the substitution $t\to t/\sqrt{z}+w/z$ into $(1)$ yields
$$\int_{-\infty}^\infty e^{-zt^2+2wt}\,dt=\frac{e^{w^2/z}}{\sqrt z}\lim_{L\to \infty}\int_{-\sqrt{z}(L+w/z)}^{\sqrt{z}(L-w/z)} e^{-t^2}\,dt\tag2$$
Denote $\text{Re}(\sqrt{z})=a>0$.  From Cauchy's Integral Theorem, we have
$$\begin{align}
0&=\oint_C e^{-t^2}\,dt\\\\
&=\int_{-\sqrt{z}(L+w/z)}^{\sqrt{z}(L-w/z)}e^{-t^2}\,dt+\int_{\sqrt{z}(L-w/z)}^{aL} e^{-t^2}\,dt+\int_{aL}^{-aL}e^{-t^2}\,dt+\int_{-aL}^{-\sqrt{z}(L+w/z)}e^{-t^2}\tag3
\end{align}$$
As $L\to \infty$, the second and fourth integrals on the right-hand side of $(3)$ approach $0$ where we have tacitly used the fact that since $\text{Re}(z)>0$, then $\text{Im}(\sqrt{z})<\text{Re}(\sqrt{z})$.  Therefore, 
$$\lim_{L\to \infty}\int_{-\sqrt{z}(L+w/z)}^{\sqrt{z}(L-w/z)} e^{-t^2}\,dt=\int_{-\infty}^\infty e^{-t^2}\,dt=\sqrt \pi\tag4$$
Substituting $(4)$ into $(2)$ yields 

$$\int_{-\infty}^\infty e^{-zt^2+2wt}\,dt=\sqrt{\frac{\pi}{z}}e^{w^2/z}$$

as was to be shown!
