Why is $ I = \int_\Omega \frac{1}{z-\zeta}\Delta u(\zeta) d\zeta = -2i \int_{\partial \Omega} \frac{1}{z-\zeta}\partial u(\zeta) d\zeta. $? I am unsure of some notation and also how a particular identity was derived. I read in a paper that because $\Delta = 4 \partial \bar \partial$ we have
$$
I = \int_\Omega \frac{1}{z-\zeta}\Delta u(\zeta) d\zeta = -2i \int_{\partial \Omega} \frac{1}{z-\zeta}\partial u(\zeta) d\zeta.
$$
First of all, what is $\partial \bar \partial$ and how is the Laplacian $\Delta = 4 \partial \bar \partial$?
Secondly, how has the identity been derived..it seems to be due to some integration by parts type formula, but what exactly?
 A: Identifying the complex plane with $\mathbb R^2$ with $z=x+iy$ the definition of $\partial$ and $\bar{\partial}$ is
$$\partial=\frac{1}{2} \left(\partial_x+i\partial_y \right)$$
$$\bar{\partial}=\frac{1}{2} \left(\partial_x-i\partial_y \right).$$
Then you indeed have
$$4\partial \bar{\partial}= \partial_x^2+\partial_y^2 = \Delta$$

Note that you can avoid using complex number as in $\mathbb R^2$ a formulation can be:
$$\partial=\frac{1}{2} \begin{pmatrix} \partial_x \\ \partial_y\end{pmatrix}=\frac {1}{2}\nabla$$
$$\bar{\partial}=\frac{1}{2} \begin{pmatrix} \partial_x \\ -\partial_y\end{pmatrix}=\frac {1}{2} J \nabla^\perp=\frac{1}{2} J \operatorname{curl}$$
with $ J =\begin{pmatrix} 0&-1\\1&0\end{pmatrix}$ correspond to $i$.

In these terms Green's theorem can be rewritten as
$$\int_{\partial \Omega} g(\zeta) d \zeta =  \int_\Omega 2i \bar{\partial} g(\zeta) d \zeta$$
We use this formula with $g(\zeta)=\frac{1}{z-\zeta} \partial u(\zeta)$.
As $\zeta \mapsto \frac{1}{z-\zeta}$ is holomorphic you have $\bar{\partial} \frac{1}{z-\zeta}=0$ so
$$\bar{\partial} \left( \frac{1}{z-\zeta} \partial u(\zeta) \right)=\frac{1}{z-\zeta} \bar{\partial} \partial u(\zeta)=\frac{1}{z-\zeta}\frac{1}{4} \Delta u(\zeta)$$
from where you obtain:
$$\int_{\partial \Omega} \frac{1}{z-\zeta} \partial u(\zeta) d \zeta= \frac{2i}{4} \int_\Omega \frac{1}{z-\zeta} \Delta u(\zeta) d\zeta$$
whic is the claimed result as $(i/2)^{-1}=-2i$.
