Proof of Isoperimetric Inequality The authors of "Introduction to the Heisenberg Group and the Sub-Riemannian Isoperimetric Inequality" give several proofs of the classical isoperimetric inequality in the plane in the first chapter. There is one I cannot understand. 
Let us assume $\Omega \subset \mathbb{C}$ has a $C^1$ Jordan curve as its boundary. Denote the area enclosed by $A$ and the perimeter as $L$. We know the area form in complex notation is $i/2 dz \wedge d\bar{z}$ so
$$4\pi A = \int_\Omega 2 \pi i dz \wedge d\bar{z} = \int_{\Omega} \int_{\partial \Omega} \frac{d\zeta dz \wedge d\bar{z}}{\zeta-z} =^{?} \int_{\partial \Omega} \int_{\partial \Omega}\frac{\bar{\zeta}-\bar{z}}{\zeta-z}dzd\zeta \le L^2$$ 
I am very puzzled by the last equality. It seems like Green's theorem is being used but I cannot work out in what way. 

It is Stoke's theorem after swapping the order of integration.
As differentials on the $z$-plane (i.e., fixed $\zeta$),
\begin{align*}
\mathrm{d}\left(\frac{\bar{\zeta}-\bar{z}}{\zeta-z}\mathrm{d}z\right)
&=\frac{\partial}{\partial\bar{z}}\left(\frac{\bar{\zeta}-\bar{z}}{\zeta-z}\right)\,\mathrm{d}\bar{z}\wedge\mathrm{d}z\\
&=-\frac1{\zeta-z}\,\mathrm{d}\bar{z}\wedge\mathrm{d}z\\
&=\frac1{\zeta-z}\,\mathrm{d}z\wedge\mathrm{d}\bar{z}
\end{align*}
So
$$
\begin{align*}
\int_\Omega\int_{\partial\Omega}\frac1{\zeta-z}\,\mathrm{d}\zeta\,\mathrm{d}z\wedge\mathrm{d}\bar{z}
&=\int_{\partial\Omega}\color{red}{\int_\Omega\frac1{\zeta-z}\,\mathrm{d}z\wedge\mathrm{d}\bar{z}}\,\mathrm{d}\zeta\\
&=\int_{\partial\Omega}\color{red}{\int_{\partial\Omega}\frac{\bar{\zeta}-\bar{z}}{\zeta-z}\mathrm{d}z}\,\mathrm{d}\zeta.
\end{align*}
$$
