Krdytkyu

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

$$
I = int_Omega frac{1}{z-zeta}Delta u(zeta) dzeta = -2i int_{partial Omega} frac{1}{z-zeta}partial u(zeta) dzeta.
$$

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?

Identifying the complex plane with $mathbb R^2$ with $z=x+iy$ the definition of $partial$ and $bar{partial}$ is
$$partial=frac{1}{2} left(partial_x+ipartial_y right)$$
$$bar{partial}=frac{1}{2} left(partial_x-ipartial_y right).$$
Then you indeed have
$$partial bar{partial}= partial_x^2+partial_y^2 = Delta$$

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

In these terms Green's theorem can be rewritten as
$$int_{partial Omega} g(zeta) d zeta = int_Omega 2i bar{partial} g(zeta) d zeta$$

As $zeta mapsto frac{1}{z-zeta}$ is holomorphic you have $bar{partial} frac{1}{z-zeta}=0$ so
$$bar{partial} left( frac{1}{z-zeta} partial f(zeta) right)=frac{1}{z-zeta} bar{partial} partial f(zeta)=frac{1}{z-zeta}4 Delta f(zeta)$$
from where you can obtain your inequality.