I am looking at the following equation:
$$\partial_z \left\lbrace (1-z)^2 f (z,\bar{z}) \right\rbrace = 0, \tag{1}$$
with $z = x+ i y$ and $\bar{z} = x - i y$. Can I directly conclude without any caveat that
$$f(z,\bar{z}) = \frac{c(\bar{z})}{(1-z)^2}, \tag{2}$$
as in the real case?
I think that I could solve it myself using Wirtinger derivatives. The complex derivative is defined as:
$$\partial_z := \frac{1}{2} \left(\partial_x - i \partial_z \right) \tag{3}$$
Hence my partial differential equation becomes:
$$\partial_x \left\lbrace (1-x-iy)^2 f(x,y) \right\rbrace = i\partial_y \left\lbrace (1-x-iy)^2 f(x,y) \right\rbrace \tag{4}$$
where by abuse of notation I wrote $f(x,y)$ instead of $f(z,\bar{z})$. The meaning should be clear though. This is now a real PDE, and the solution is:
$$f(x,y) = \frac{c(x-iy)}{(1-x-iy)^2} \tag{5}$$
which is precisely $(2)$. Actually, it seems that the Wirtinger derivative was defined such that it has the behavior of ordinary derivatives.