Integration by parts of Wirtinger derivative As a physics student, I am currently trying to use Wirtinger derivatives in everyday work.
While the basics are clear and so far seem to make life considerably simpler, I am having trouble to see if integration by parts is possible as in the real case.
Transforming to real two-dimensional coordinates, I have come to the following conclusions:
$$
\int_D \frac{\partial f}{\partial z} \, g \, \mathrm{d}^2 z
= \int_{\partial D} f \, g \, \mathrm{d} \bar z
- \int_D f \, \frac{\partial g}{\partial z} \, \mathrm{d}^2 z
$$
$$
\int_D \frac{\partial f}{\partial \bar z} \, g \, \mathrm{d}^2 z
= \int_{\partial D} f \, g \, \mathrm{d} z
- \int_D f \, \frac{\partial g}{\partial \bar z} \, \mathrm{d}^2 z
$$
Are these correct? I would be glad for any help, and especially for references that cover this topic, as I could not find any.

First of all, there is no way that your suggested formulas can be true. Your integrals over $D$ contain $1$-forms, which is nonsense since you're integrating over the 2-dimensional $D$.
On the other hand, Stokes' theorem is still valid in the complex setting:
$$
\int_{\partial D} \omega = \int_{D} d\omega.
$$
Furthermore, $d = \partial + \bar\partial$ so if $\omega = f(z)g(z)\,dz$ you get
\begin{align*}
\int_{\partial D} f(z)g(z)\,dz &= \int_{D} d\big( f(z)g(z)\,dz \big) \\
&= \int_{D} \bar\partial\big( f(z)g(z)\,dz \big) \\
&= \int_D \Big( \frac{\partial f}{\partial \bar z}(z) g(z) + f(z)\frac{\partial g}{\partial \bar z}(z) \Big) d\bar z \wedge dz,
\end{align*}
where the second equality it true since $dz \wedge dz = 0$ (the $\partial$-term picks up a $dz$).
If you want derivatives with respect to $\bar z$ in the integral over $D$, apply Stokes' to $\int_{\partial D} f(z)g(z)\,d\bar z$ instead.
