Expected Value of product Complex Normal R.V. and its conjugate, different powers I came across something while looking into Complex Gaussian Random Variables and I've been trying to understand why it is true.
Let z ~ CN(0,1) and $m \neq n $ then $E[z^m\overline{z}^k]=0$
I thought maybe this has to do with the circular symmetry but I couldn't see how to proceed. 
I saw this question, but also wasn't sure if it really points me in a helpful direction: Few questions of circularly symmetric complex random variables
To be clear, this is not homework, this is self-study and there seems to be very little on Complex Normal variables around so I'm truly stumped how to proceed. I would really appreciate a full explanation, if possible.
Thanks in advance
 A: Okay, on closer sketching, this isn't actually that bad. For the sake of convenience, let's assume instead that $Z\sim \mathcal{N}(0,I)$ when viewed as an $\mathbb{R}^2$-valued random variable. Denote its density $f$.
Define $P:\mathbb{R}^2\setminus \{x\in (0,\infty),y=0\}\to (0,\infty)\times (0,2\pi)$ to be the standard polar coordinate transformation. Then, since $\{x\in (0,\infty),y=0\}$ is a $\mathcal{N}(0,I)$-null set, we can apply the Jacobi Coordinate Transformation theorem to get that $(R,\Theta):=P(Z)$ has density
$$rf(P^{-1}(r,\theta))=rf(r(\cos(\theta)+i\sin(\theta))=\frac{r}{2\pi}\exp(-\frac{r^2}{2})=\frac{1}{2\pi}\cdot r\exp(-\frac{r^2}{2}),$$ 
which is clearly a factorisation of the density, implying that $R$ and $\Theta$ are independent, and $\Theta$ is uniformly distributed on $(0,2\pi)$. Note that $R$ and $\Theta$ clearly have moments of all orders.
Accordingly, we get, by applying independence coordinate-wise, that
$$
E Z^m \overline{Z^k}=E(R^{m+k} e^{i (m-k)\Theta})=E(R^{m+k})E(e^{i(m-k)\Theta}),
$$
and
$$
E(e^{i(m-k)\Theta})=\frac{1}{2\pi}\left(\int_0^{2\pi} \cos((m-k)\theta)\textrm{d}\theta+i\int_0^{2\pi}\sin((m-k)\theta)\textrm{d}\theta\right)=0,
$$
since $m\neq k$. This yields the desired.
