Wirtinger derivative of composition of functions So I have a very basic question : let $h : \mathbb{R} \rightarrow \mathbb{R}$ be a $C^1$ function, and let $g : \mathbb{C} \rightarrow \mathbb{R}$ be defined by $g(z)=h(z \overline{z})$. I want to compute $\frac{\partial g}{\partial \overline{z}}$ (which is defined because $g$ is defined on $\mathbb{C}$). I know the chain rule for $\overline{\partial}$. However, it involves terms like $\frac{\partial h}{\partial z}$ and $\frac{\partial h}{\partial \overline{z}}$, which confuses me because $h$ is only defined on $\mathbb{R}$. 
Any help appreciated ! 
 A: You're absolutely right to be skeptical of the expressions $\partial h/\partial z$ and $\partial h/\partial\overline{z}$. Just like you say, since $h$ is only defined on $\mathbb{R}$, it isn't possible to take these derivatives. There are a couple of things you might do here in order to solve your problem.
(1) You can extend $h$ to a function $\tilde{h}$ on $\mathbb{C}$ in some way, and then define $g(z) = \tilde{h}(z\overline{z})$. This of course doesn't change the function $g$ in any way, but now you'll be able to take derivatives $\partial\tilde{h}/\partial z$ and $\partial\tilde{h}/\partial \overline{z}$. The most natural extension $\tilde{h}$ is the one given by $\tilde{h}(z) := h(\Re z)$.
(2) You can use the ordinary real chain rule as follows. Let $f(z) = z\overline{z}$, or, in terms of $x$ and $y$, this is $f(x,y) = x^2 + y^2$. Then $g = h\circ f$, so $$\frac{\partial g}{\partial x} = (h'\circ f)\frac{\partial f}{\partial x}\,\,\,\,\,\mbox{ and }\,\,\,\,\,\frac{\partial g}{\partial y} = (h'\circ f)\frac{\partial f}{\partial y}.$$ You can then use these expressions to compute $\partial g/\partial z$ and $\partial g/\partial\overline{z}$.
