What is the distribution of it?

Question

If $\theta$ is uniformly distributed in $(0, 2\pi),$ then what is the distribution of $e^{i\theta},$ where $i = \sqrt{-1}?$ And what are the statistical properties of $\left[e^{i0\theta}\, e^{i1\theta}\, e^{i2\theta}\dots\, e^{i(N-1)\theta}\right],$ where $\theta$ is a single observation and $N$ some integer?

Answer 1

$e^{ik\theta}$, where $k$ is any non-zero integer, is uniformly distributed on the complex plane unit circle. The expected value of $e^{ik\theta}$ is $\displaystyle\frac{\int_0^{2\pi} e^{ik\theta} d\theta}{2\pi} = \frac{i - ie^{2\pi i k}}{k} = \frac{i - i}{k} = 0.$ The variance of $e^{ik\theta}$ is $1,$ because all values that may be observed are at a distance of $1$ from the expected value $0.$

$e^{i0\theta} = 1$ is constant.

If you by notation $(0, 2\pi)$ mean that values $0$ and $2\pi$ are never observed, then one may observe $e^{ik\theta} = 1$ only if $|k| > 1$, because the distribution wraps around passing the point $e^{ik\theta} = 1$ at $\theta = 2\pi/k.$

Answer 2

A more general solution can behad from the Von Mises Distribution $$ f(x\mid\mu,\kappa)=\frac{e^{\kappa\cos(x-\mu)}}{2\pi I_0(\kappa)} $$ When $\kappa=0$, it is equal to the uniform on $-\pi \le x \le \pi$

The circular moments $$ m_n = E\{ e^{\jmath n x} \}= \int_{-\pi}^{\pi} e^{\jmath n x}f(x\mid\mu,\kappa) dx =\frac{I_{|n|}(\kappa)}{I_0(\kappa)}e^{i n \mu} $$

https://en.wikipedia.org/wiki/Von_Mises_distribution