Distribution of $e^{j\theta}$
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 $r dr d\theta$),
$$f_{R,\Theta}(r,\theta)=\frac{1}{2\pi} \delta(r - \rho)$$