Green's function for $\frac{\partial}{\partial\bar{z}}$ I've read some complex analysis texts and often there is some appeals to Green's theorems when proving facts about contour integrals of holomorphic functions yet there seems to be a lack of appeals to Green's functions (at least explicitly). We know that for a holomorphic function $f$, we have
$$\frac{\partial f}{\partial\bar{z}} = 0$$
and also that
$$f(z) = \frac{1}{2\pi i}\int_{\gamma} \frac{f(\zeta)}{\zeta-z}\,d\zeta,$$
where $\gamma$ is a circular contour around $z$ (with a winding number of one). Given the duality between partial differential operators and integral operators, this has made me a bit curious as to what the Green's function for $\dfrac{\partial}{\partial \bar{z}}$ is (when one restricts to holomorphic functions). Intuitively, I would hazard to guess that the Green's function is related to $\dfrac{1}{z}$ based on the above observations but I haven't figured out a way to show this. Solving
$$\frac{\partial}{\partial\bar{z}}G = \delta(z)$$
for $G$ seems to be a bit of a tall task. Distributions seem to be a little bit complicated on $\mathbb{C}$. Ultimately, I want to use this understanding to tackle some aspects of Green's functions in several complex variables. Any help would be greatly appreciated!
 A: Actually, it's not so hard to find a solution of $\frac{\partial}{\partial\overline{z}}G =\delta$. You have the right thing there already, modulo normalisation. We have, in the sense of distributions, for any $\varphi\in \mathscr{D}(\mathbb{C})$,
$$\begin{align}
\left(\frac{\partial}{\partial\overline{z}}\frac{1}{z-w}\right)[\varphi]
&= - \int_{\mathbb{C}} \frac{\partial\varphi}{\partial\overline{z}}(z)\frac{1}{z-w}\,d\lambda\\
&= -\lim_{\varepsilon\to 0} \int_{\varepsilon < \lvert z-w\rvert < R} \frac{\partial\varphi/\partial\overline{z}(z)}{z-w}\,d\lambda
\end{align}$$
where $\lambda$ is the Lebesgue measure on $\mathbb{C}$ and $R$ is sufficiently large, so that the support of $\varphi$ is contained in $D_R(w)$. Now write the integral with differential forms, that is, replace $d\lambda$ with $dx\wedge dy$, and write the latter as $\frac{1}{2i}d\overline{z}\wedge dz$. We get
$$\begin{align}
\left(\frac{\partial}{\partial\overline{z}}\frac{1}{z-w}\right)[\varphi] &= -\lim_{\varepsilon\to 0} \frac{1}{2i}\int_{\varepsilon < \lvert z-w\rvert < R} \frac{\partial\varphi/\partial\overline{z}(z)}{z-w}\,d\overline{z}\wedge dz\\
&= -\frac{1}{2i}\lim_{\varepsilon\to 0} \int_{\varepsilon < \lvert z-w\rvert < R} d\left(\frac{\varphi(z)}{z-w}\, dz\right) \tag{Stokes}\\
&= \frac{1}{2i}\lim_{\varepsilon\to 0} \int_{\lvert z-w\rvert = \varepsilon} \frac{\varphi(z)}{z-w}\,dz\\
&= \frac{1}{2i}\lim_{\varepsilon\to 0} \left(\int_{\lvert z-w\rvert = \varepsilon} \frac{\varphi(w)}{z-w}\,dz + \int_{\lvert z-w\rvert =\varepsilon} \frac{\varphi(z)-\varphi(w)}{z-w}\,dz \right)\\
&= \pi \varphi(w),
\end{align}$$
since by the differentiability of $\varphi$ in $w$, the integrand of the second integral in the penultimate line remains bounded as $\varepsilon\to 0$, and so the second integral is $\leqslant C\cdot 2\pi\varepsilon$ in absolute value by the ML-inequality. That means
$$\frac{1}{\pi}\frac{\partial}{\partial\overline{z}} \frac{1}{z-w} = \delta_w.$$
