Why $\iint_D f=\frac{1}{2i} \oint_\gamma \overline{z} f(z) dz$? It is proved here that $\mathrm{Area}(D)=\frac{1}{2i} \oint_\gamma \overline{z}  dz.$ Is it true more generally that
$$\iint_D f=\frac{1}{2i} \oint_\gamma \overline{z} f(z) dz$$
for analytic functions? Heuristically it's replacing $dz$ with $dF(z)$ where $F'=f$. Can that be made rigorous?
 A: Your guess is correct.
This is best understood in terms of differential forms. The area element in $\mathbb R^2$ is $dx \wedge dy$. Note that $dz = dx+idy, d \bar{z} = dx -idy$ so $dz \wedge d\bar z = -2idx \wedge d y$. By Stokes theorem,
$$
\frac{1}{2i}\int_\gamma \bar z f(z) dz = \frac{1}{2 i}\int_D  \frac{\partial}{\partial z}(\bar z f(z) )dz \wedge dz + \frac{\partial }{\partial \bar z }(\bar z f(z)) d \bar z \wedge d z
$$
Since $dz \wedge dz=0$ and $f$ analytic implies $\frac{\partial f}{\partial \bar z} = 0$. Using the product rule, one obtains that
$$
\frac{1}{2i}\int_\gamma \bar z f(z) dz = \frac{1}{2i} \int_D f(z) d\bar z \wedge dz = \int_D f dx \wedge dy,
$$which is Stokes theorem in 2 dimensions in a similar fashion, but I think thi is the natural way to think about this kind of formulas for holomorphic functions.
In cas you don't know what the symbols $\frac{\partial}{\partial \bar z}$ and $\frac{\partial }{\partial z}$ stand for, they are the Wirtinger derivatives. You can get some insight here. The formula
$$
dg = \frac{\partial g}{\partial \bar z}dz +\frac{\partial g}{\partial z}d \bar z
$$
Follows from standard computation using the definition of the $dz, d \bar z$ and the Wirtinger derivatives, but you should understand it as stating that the basis $\frac{\partial}{\partial \bar z},\frac{\partial }{\partial z}$ is dual to $dz,d\bar z$.
