Gaussian integral over complex variables I am trying to learn how to compute a Gaussian integral over complex variables. I am struggling to understand under what conditions the integral exist. I have not found such identities on wikipedia, hence my question on this website.
How does one compute the following integral?
$$\int \frac{\mathrm{d} z \mathrm{d} \bar{z}}{2 \mathrm{i} \pi} \exp \{a\bar{z} z-\bar{b} z-b \bar{z}\}$$
with $a\in\mathbb{C}$ and $b\in\mathbb{C}$. $\bar{z}$ represents the complex conjugate of $z$.
What constraints do $a$ and $b$ need to hold in order for this integral to exist?
(There is a paper written by Sommer et al. where they perform such an integral over complex variables: Sommers, H. J., Crisanti, A., Sompolinsky, H., & Stein, Y. (1988). Spectrum of large random asymmetric matrices. Physical review letters, 60(19), 1895.)
 A: Writing $z=x+iy,\,\overline{z}=x-i y$, the double integral over all $z,\overline{z}$ becomes 
\begin{align}
\int \frac{dz\,d\overline{z}}{2\pi i}\exp(&a z\overline{z}-\overline{b}z-b\overline{z})\\
&=\int_{-\infty}^\infty \int_{-\infty}^\infty \frac{(2i) \,dy\,dx}{2\pi i}\exp\left[a(x^2+y^2)-\overline{b}(x+i y)-{b}(x-i y)\right]\\
&=\frac{1}{\pi}\int_{-\infty}^\infty \int_{-\infty}^\infty\,dy\,dx \exp\left[a(x^2+y^2-2(\text{Re }b)x)-2 (\text{Im b})y\right]\\
&=\frac{1}{\pi}\int_{-\infty}^\infty dy\, \exp[a y^2-2(\text{Im b})y]\int_{-\infty}^\infty\,dx \exp\left[ax^2-2(\text{Re }b)x)\right].
\end{align}
These Gaussian integrals will converge so long as $\text{Re }a<0$, in which case we can use the usual complete-the-square trick to compute them. Hence under this assumption the product evaluates to 
$$\frac{1}{\pi} \exp\left(-\frac{(\text{Im }b)^2}{a}\right)\sqrt{-\frac{\pi}{a}}\exp\left(-\frac{(\text{Re }b)^2}{a}\right)\sqrt{-\frac{\pi}{a}}=-\frac{1}{a}\exp\left(-\frac{|b|^2}{a}\right).$$
A: Let's do it in the contour integration method. since it's a double integral integrating out the first variable
$$\int \frac{dzd\bar{z}}{2\pi i}e^{az\bar{z}-\bar{b}z-b\bar{z}}\to\oint \frac{dz}{2\pi i} \frac{e^{az\bar{z}-\bar{b}z-b\bar{z}}}{az-b}$$
Now we can see the pole emerging $$z\to\frac{b}{a}$$ hence the contour integral
$$\oint \frac{dz}{2\pi i} \frac{e^{az\bar{z}-\bar{b}z-b\bar{z}}}{az-b}=\frac{2i\pi}{2i\pi a}Res(f_{z=\frac{b}{a}})=\frac{1}{a}e^{b\bar{z}-\frac{\bar{b}b}{a}-b\bar{z}}=\frac{1}{a}e^{-|b|^2/a}$$
