Circularly symmetric complex Gaussian

Special case: scalar variable \(Z\sim\mathcal{CN}(0,1)\)
\begin{aligned}
f_Z(z) &= \frac{1}{\pi}e^{-|z|^2} \\
f_{|Z|,\Theta}(|z|,\theta) &= \frac{1}{\pi} |z| e^{-|z|^2}
\end{aligned}
Note that the distribution is a function of the magnitude of \(Z\), and therefore constant over all angles \(\theta\).  To get the distribution of the magnitude of Z, we integrate over all angles \( -\pi \le \theta \le \pi\).
\begin{aligned}
f_{|Z|}(|z|) &= \int f_{|Z|,\Theta}(|z|,\theta) d\theta \\
                  &= \int_{-\pi}^\pi d\theta \; \frac{1}{\pi} |z| e^{-|z|^2} \\
                  &= 2 |z| e^{-|z|^2} ,\; |z| > 0
\end{aligned}
Define \(Y=|Z|\) then,
\[ f_Y(y) = 2 y e^{-y^2}, \quad y > 0\]
For magnitude squared distribution, we can perform a change of variable, with \(X=Y^2=g(Y)\), since \( P( \{-\sqrt{x}, x > 0\} ) = 0 \), we only have one region (the set \(\{ \sqrt{x}, x>0 \}\) ) to worry about.
\begin{aligned}
f_X(x) &= f_Y(g^{-1}(x)) \left|\frac{ \partial g^{-1}(x) }{ \partial x} \right| \\
            &= f_Y(\sqrt{x})  \left|\frac{ \partial \sqrt{x} }{ \partial x} \right|  \\
            &= (2x^{\frac{1}{2}}e^{-x}) \frac{1}{2} x^{-\frac{1}{2}} \\
            &= e^{-x}, \; x>0
\end{aligned}
Note:
Here we assume the probability spaces \((X,\mathcal{X},P_X)\), \((Y,\mathcal{Y},P_Y)\)and that a mapping T exists that maps \(A\in\mathcal{X}\) to \(B\in\mathcal{Y}\).
Start from the joint cdf in the X-space, define \(A\equiv\{\sqrt{X^2+Y^2}\leq |z|, \tan^{-1}(Y/X) \leq \theta_0\}\)
\[F_{|Z|,\Theta}(|z|,\theta_0) = P(A)  = \underset{(x,y)\in A}{\iint} f(x,y) dx \; dy \]
With a change of variable, define \(B\equiv\{|Z|\leq |z|, \Theta \leq \theta_0\}\)
\begin{aligned}
F_{|Z|,\Theta}(|z|,\theta_0) = P(B) &=  \underset{(r,\theta)\in B}{\iint} f(x(r,\theta), y(r,\theta)) r \; dr \; d\theta \\
&= \int_0^{|z|} \int_0^{\theta_0} f(x(r,\theta), y(r,\theta)) r \; dr \; d\theta
\end{aligned}
Differentiate wrt \(|z|\) and \(\theta\) to get the pdf.

Theorem.  Given any function \(g:\mathbb{R}^2 \rightarrow \mathbb{R}\) and T that maps the \(A\in\mathcal{X}\) to \(B\in\mathcal{Y}\) then
\[\iint_A g(x_1,x_2) dx_1 \; dx_2 = \iint_B g(x_1(y_1,y_2), x_2(y_1,y_2))  |J(y_1, y_2)| dy_1\; dy_2\] where \(J(y_1,y_2)\) is the Jacobian.