How to show $\int_D z^n \overline {z^m}dxdy= 0 $ when $n\ne m$? Assuming
$$
D=\{z\in  \mathbb C: |z|<1\},
$$
let $z=x+i y$, then how to show
$$
\int_D z^n \overline {z^m}  dx dy= 0  ,~~~n\ne m 
$$
PS: I just know how to show it is not zero when $n=m$, but fail to deal $n\ne m$.
 A: @ArcticChar's hint writes the integral as $\int_0^1r^{n+m+1}dr\int_0^{2\pi}e^{i(n-m)\theta}d\theta=\frac{\pi}{n+1}\delta_{mn}$ for $m-n\in\Bbb Z$.
A: We have that $z^n \overline{z}^m = r^n e^{in\theta} r^m e^{-im\theta}=r^{m+n}e^{i(n-m)\theta}$ and so doing the integral in polar coordinates yields
$$\int_0^1\int_0^{2\pi} r^{n+m+1}(\cos((n-m)\theta)+i\sin((n-m)\theta))d\theta dr$$
and the inside integral is $0$ when $n\neq m$.
However, another interesting approach is to recast this in terms of differential forms an use Stoke's theorem.
Note that $\frac{\partial z^n}{\partial \overline{z}}=0$ and similarly the other way, and since $x=(z+\overline z)/2$ and $y=(z-\overline z)/2i$, $dx \wedge dy = (1/2i)d\overline{z} \wedge dz$.  We can thus (up to a sign) write
$$\begin{align}\int_D z^n \overline{z}^m dxdy &=\int_D z^n \overline{z}^m dx\wedge dy \\
&=\frac{1}{2i}\int_D z^n \overline{z}^m d\overline{z}\wedge dz \\
&=\frac{1}{2i(m+1)}\int_D d(z^n \overline{z}^{m+1}  dz) \\
&=\frac{1}{2i(m+1)}\int_{\partial D} z^n \overline{z}^{m+1}  dz \\
&=\frac{1}{2i(m+1)}\int_{\partial D} z^{n-m-1} dz
\end{align},$$
where we have used the fact that on the boundary of $D$, $z\overline{z}=1$.  This is now a line integral, which is $0$ unless $n=m$.
