The range of complex variable integration

So if we have a two dimensional Gaussian function $\frac{1}{2\pi}e^{-\frac{x^2+y^2}{2}}$, then the following integration $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{1}{2\pi}e^{-\frac{x^2+y^2}{2}}dxdy=1$$

Now if we switch to complex variables $z,z^*$, what should be the range of those complex variables for the same integration above transformed into complex variable?

$$\int\int\frac{i}{2}\frac{1}{2\pi}e^{-\frac{|z|^2}{2}}dzdz^*=1$$

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You would have just one complex variable, and you'd integrate over the entire complex plane. I think. – Potato Aug 27 '12 at 4:02
The measure on the complex plane is not a product measure, and the integral of the probability density is not an iterated integral. It becomes one when $\mathbb{C}$ is mapped into $\mathbb{R}^2$. – Sasha Aug 27 '12 at 4:15
So how to evaluate this Gaussian function in complex plane and get the same result? – Rob Aug 27 '12 at 4:21
I'm not sure how this goes, but I think it is exactly the same since the function $$e^{-|z|^2/2}$$is real... – DonAntonio Aug 27 '12 at 4:25
One evaluates the integral by mapping to $\mathbb{R}^2$. – Sasha Aug 27 '12 at 4:29

First note that the complex Gaussian $Z$ can be defined as having the density $\frac{1}{2 \pi} e^{\frac{-|z|^2}{2}}$. (When defining it this way the complex Gaussian $Z$ is the sum $X+iY$ of two real standard Gaussian $X,Y$. Then the variance of $Z$ will be 2. Because of this one often defines a standard complex Gaussian $\tilde{Z}$ via the density function $\frac{1}{\pi}e^{{-|z|^2}}$. Then $Var[Z]=\mathbb{E}[|\tilde{Z}|^2]=1$ and $\tilde{Z}$ is the sum of two real centered Gaussian with variance $\frac{1}{2}.)$
Now if you set $z=x+iy$ the density will be the same as you write at the beginning of your question. To see this, note that - as Sasha pointed out in the comments - we have that $\mathbb{C}$ is isomorphic to $\mathbb{R}^2$. The Lebesque measure $\lambda^{(2)}$ on $\mathbb{R}^2$ is the same as $\lambda\otimes\lambda$ (where $\lambda$ is the Lebesque measure on $\mathbb{R}$) so we see:
$$\int_{\mathbb{C}}{\frac{1}{2 \pi} e^{\frac{-|z|^2}{2}}}dz=\int_{\mathbb{R}^2}{\frac{1}{2 \pi} e^{\frac{-(x^2+y^2)}{2}}}d(x,y)=\int_{\mathbb{R}}\int_{\mathbb{R}}{\frac{1}{2 \pi} e^{\frac{-(x^2+y^2)}{2}}}dxdy.$$
The double integral you write at the end of your question makes not too much sense: The two $"\int"$'s indicate that the integral is over $\mathbb{C}^2$ but $z^*$ doesn't occur in the integrand. So you would integrate 1 over $\mathbb{C}$ and the integral would be infinite.