$X \sim \mathcal{U}\left( { - \pi ,\pi } \right)$, $\mathcal{U}$ is uniform distribution, $Y = {e^{jX}}$, $j$ is unit imaginary number. How do I get the distribution of $Y$ , the expected value of $Y$ ($EY = ?$) and the variance of $Y$ ($DY = ?$)
I calculate it as $EY = \int_{ - \pi }^\pi {{e^{jx}}dx} = 0$, $DY = E{Y^2} - {\left( {EY} \right)^2} = \int_{ - \pi }^\pi {{e^{j2x}}dx = 0}$. But $EY = DY = 0$ seems strange. Can someone help me ?
$E[Y]=0$ makes sense from a geometric perspective: your points are uniformly distributed on the unit circle in $\mathbb{C}$, so they average out to nothing.
Your variance calculation is wrong. A better way to do the calculation rather than splitting is to fall back on $E[|Y-E[Y]|^2]$, where in this case the absolute value bars denote the modulus. So in this case you need to compute $\frac{1}{2\pi} \int_{-\pi}^\pi |e^{ix}|^2 dx$, which is not zero.
By the way, your title would be clearer if it referenced the fact that you're dealing with a complex exponential rather than a real exponential.