Solution to Cauchy-Riemann Differential Equation of Compact Support I'm working through Forster's $\textit{Lectures on Riemann Surfaces}$ and am struggling with the following problem:
Suppose $g \in \mathcal{E}(\mathbb{C})$ is of compact support. Prove there is a solution $f \in \mathcal{E}(\mathbb{C})$ of the equation $\partial f/\partial \bar{z} = g$ having compact support iff $$\int \int_{\mathbb{C}} z^n g(z)dz   \wedge d\bar{z} = 0$$ for all $n \in \mathbb{N}$.
Here, $\mathcal{E}(\mathbb{C})$ denotes the $\mathbb{C}$-algebra of functions differentiable with respect to the coordinate $x$ and $y$ (where $z=x+iy$). We already have a solution, namely $f(\zeta) = \int \int_{\mathbb{C}} \frac{g(z)}{z-\zeta} dz \wedge d\bar{z}$, but it is not necessarily of compact support. I really don't have a good approach to the problem; maybe just a hint would be nice, not a full solution.
 A: By Stokes, our condition is necessary.  The converse direction follows from Serre duality.
Consider  $$\omega_k=g(z)\,d\bar z\in\Gamma(\mathbb{P}_\mathbb{C}^1,\bar K\otimes \mathcal O(-k))$$ as well-defined forms with values in $\mathcal O(-k)$, by using the standard trivialization of $\mathcal O(-k)$ and the assumption that $g$ has compact support. By Serre duality, there exists a solution of
$$\bar\partial f_k=\omega_k$$
if and only if our integral condition is satisfied for all $n$ up to order $k-2$, as the $z^kdz$, $k=0$, $\ldots$ , $k-2$ span the holomorphic section $H^0(\mathbb {P}_{\mathbb C}^1,K\otimes \mathcal O(k)).$
The section $f_k$ is unique, for $k\geq1$, and identifying $f_k$ with a function by using the aforementioned trivialization of $\mathcal O(-k)$, all $f_k$ give rise to the same function, denoted by $f$. Therefore, there exists $f\colon\mathbb {P}_\mathbb{C}^1\to\mathbb C$ which vanishes up to arbitrary order at $\infty$ and which satisfies
$$\bar\partial f=g(z)\,d\bar z.$$
Why has $f$ compact support in $\mathbb C$? Let $U\subset \mathbb{P}_\mathbb{C}^1$, $\infty\in U$ be an open connected set such that $g(z)\,d\bar z$ is identically zero on there. Solutions of $\bar\partial \tilde f=g(z)\,d\bar z$ on $U$ are unique up to holomorphic $1$-forms on $U$. Hence, as $f$ vanishes up to arbitrary order at $\infty$, $f$ must vanish identically on $U$, which means that $f$ has compact support.
