# How to Derive Rayleigh distribution using transformation formula

Consider a complex random variable $$Z=X+\jmath Y$$, where the probability density function of $$X$$ and $$Y$$ are given by $$p(x) = \frac{1}{\sqrt{2\pi\sigma^2}} {\rm e}^{-\frac{x^2}{2\sigma^2}}\quad\mbox{and}\quad p(y) = \frac{1}{\sqrt{2\pi\sigma^2}} {\rm e}^{-\frac{y^2}{2\sigma^2}},$$ respectively. Show that the magnitude of $$Z$$ is a Rayleigh distributed random variable with probability density function $$p(z) = \frac{z}{\sigma^2} {\rm e}^{-\frac{z^2}{2\sigma^2}}, \quad z \geq 0$$ and that the phase is uniformly distributed in $$[0,2\pi]$$.

I am able to prove this using CDF method but unable to get to same result using the transformation formula: $$f_{Y_1,Y_2}(y_1,y_2) = f_{X_1,X_2}(w_1(y_1,y_2),w_2(y_1,y_2))\cdot \left|\mathbf J\right|,$$ where $$\mathbf J$$ is the Jacobian of the transformation, defined via $$\mathbf J = \begin{bmatrix} \frac{\partial x_1}{\partial y_1} & \frac{\partial x_1}{\partial y_2} \\ \frac{\partial x_2}{\partial y_1} & \frac{\partial x_2}{\partial y_2} \end{bmatrix}.$$