Integrating $z^n$ as a complex valued function

Question

Let $B_1(0) \subset \mathbb{C}$ be the unit ball in the complex plane and $f:B_1(0) \to \mathbb{C}, z \mapsto z^n$. I want to know how one would calculate the integral $$\int_{B_1(0)}f(z) \text{d}z$$ I expect the integral to be 0 because of the orthogonality in $L^2$ of the normed monomials. But I want to calculate the integral "by hand". Would one use the lebesgue measure $\lambda^2$? It seems that something like polar coordinates seem to work better but it is not that obious how to proceed. I also thought about splitting $f$ into its real and imaginary part. But using the binomial theorem for $z^n=(a+ib)^n$ does not get me any further.

Answer 1

One could/would indeed use the lebesgue measure. With the substitution for multiple variables, fubini and the usual function for polar coordinates $\Phi(r,\phi) = (r \cos(\phi), r \sin(\phi))$, as such $\det(D \Phi) = r$, one would get: \begin{align*} \int_{B_1(0)} z^n d z &=^1 \int_{[0,1) \times [0, 2\pi)} (r e^{i \phi})^n r d \lambda^2 \\ &= \int_{[0,1) \times [0, 2\pi)} r^{n+1} e^{n i \phi} r d \lambda^2\\ &=^2 \int_{[0,1)} r^{n+1} \int _{[0,2\pi)} e^{n i \phi} d \phi d r\\ &= \int_{[0,1)} r^{n+1} \int _{[0,2\pi)} e^{n i \phi} d \phi d r\\ &=^3 \int_{[0,1)} r^{n+1} \left[\frac{1}{n i} e^{n i \phi} \right]_0 ^{2 \pi} d r\\ &= \int_{[0,1)} r^{n+m+1} \frac{1}{n i} \left[ e^{n i \phi} \right]_0 ^{2 \pi} d r\\ &= \int_{[0,1)} r^{n+m+1} \frac{1}{n i} 0 \; d r\\ &= 0 \end{align*} $^1$ Multivariate substitution
$^2$ Fubini
$^3$ Note that $(e^x)' = e^x$ even for $x \in \mathbb{C}$.

Which is $0$ as you guessed.