Distribution of $e^{j\theta}$ Is there a canonical or analytic expression for the probability distribution for the circularly-symmetric complex random variable $Z$:
$$
Z = e^{j\theta},
$$
where $\theta \sim \mathcal U(0, 2\pi)$?

Side notes:

It is known that the real and imaginary parts, i.e.:
$$
\Re(Z) = \cos \theta \\
\Im(Z) = \sin \theta
$$
have marginal densities given by:
$$
f_{\Re(Z)}(z) = f_{\Im(Z)}(z) = \frac{1}{\pi\sqrt{1-z^2}}, \quad -1 &lt; z &lt; 1,
$$
but because they are not independent, computing their joint PDF is nontrivial. 

EDIT: $Z$ is different from a complex normal in that here, the amplitude $|Z|$ is deterministic and identically 1, whereas if $Z$ were complex normal, $|Z|$ would be Rayleigh distributed.
Avoid complicated calculations, let $X$ and $Y$ be i.i.d. standard normal random variables, your random variable $Z$ has the same distribution of $V$ 
$$V \triangleq \left(\frac{X}{\sqrt{X^2+Y^2}},\frac{Y}{\sqrt{X^2+Y^2}} \right)$$
(easy to see $\lVert V\rVert=1$ and the angle of $V$ is equivalent to the angle of a circularly symmetric Normal hence uniform).

This kind of $V$ is one of the constructions of a point uniformly distributed on circle (which can be generalized to $(n-1)$-sphere, see Sphere Point Picking and for example this answer).

Thus the PDF of $Z$ is simply the reciprocal of the unit circle's circumference. 
For $Z_\rho = \rho e^{j\Theta}$ with fixed $\rho$ and uniform $\Theta$,

in polar coordinates (where infinitesimal area is $r dr d\theta$),
$$f_{R,\Theta}(r,\theta)=\frac{1}{2\pi} \delta(r - \rho)$$