I want to know the distribution of $z = \exp(j\varphi)$, with $\varphi \sim \mathcal{U}[-\pi;+\pi]$.
From the book "Probability, Random Variables and Stochastic Processes" by Papoulis and Pillai I figured the following: If $y=g(x)$ is a function of random variable $x$ with the density function $f_x(x)$, one can retrieve the density function $f_y(y)$ of $y$ by $f_y(y)=f_x(x)/|g'(x)|$. Therefore, if $g(x)=y=exp(x)$, this yields $f_y(y)=f_x(ln(y))/|exp(x)|$. In my case with $f_x(x)$ being a uniform distribution.
Unfortunately, I haven't figured out yet how to include this uniform distribution into the equation and even more important how to transfer the solution to complex variables.
You are confusing real and complex exponentials, and the Jacobian formula you cite applies to real exponentials. Here $z$ is a complex-valued random variable and whether $z$ has a density or not depends on the target space one considers and the reference measure one puts on it. As a random variable with values in $\mathbb C\sim\mathbb R^2$ whose Borel sigma-algebra is endowed with the two-dimensional Lebesgue measure $\lambda_2$, $z$ has no density since $z\in S^1$ almost surely where the unit circle $S^1$ is such that $\lambda_2(S^1)=0$. But one can also consider $z$ as a random variable with values in $S^1$, whose Borel sigma-algebra is endowed with its own one-dimensional uniform measure $\sigma_1$. Then $z$ has density $1$ with respect to $\sigma_1$.