# How can I prove that a complex variable does not follow a normal distribution from $R$ and $\Phi$ distributions

I am trying to prove that a complex variable $$Z = R.\exp(i.\Phi) = R.\cos(\Phi) + i.R.\sin(\Phi) = X + i.Y$$ does not follow a normal distribution when $$R\sim \mathcal{N}\left(\mu_R, \sigma_R^{2}\right)$$ and $$\Phi \sim \mathcal{N}\left(\mu_\Phi, \sigma_\Phi^{2}\right)$$.

Any ideas on how to go about this? I tried computing the expectation value E[z] but that seems to be a dead end. Thank you in advance.

## 1 Answer

Assuming $$R,\,\Phi$$ are independent, their joint pdf $$f_{R,\,\Phi}(r,\,\phi)$$ satisfies $$f_{R,\,\Phi}(r,\,\phi)drd\phi=f_{X,\,Y}(x,\,y)dxdy$$, with $$f_{X,\,Y}$$ the joint pdf of $$X,\,Y$$. Since $$dxdy=rdrd\phi$$, $$f_{X,\,Y}(x,\,y)=r^{-1}f_{R,\,\Phi}(r,\,\phi)=\frac{1}{2\pi r\sigma_R\sigma_\Phi}\exp-\left[\frac{(r-\mu_R)^2}{2\sigma_R^2}+\frac{(\phi-\mu_\Phi)^2}{2\sigma_\Phi^2}\right].$$ I'm sure you can show, be rewriting this in terms of $$x,\,y$$, it's not even close to being bivariate normal.