I came over a calculation which used the identity
$$
\frac{\partial(u \circ f)}{\partial \overline{z}} = \left( \frac{\partial u}{\partial \overline{z}} \circ f \right) \overline{f'}.
$$
I tried to confirm this for myself, but I got lost in the notation. How can one justify the above identity?
The assumptions in the context are that $u : \mathbb{C} \to \mathbb{R}$ has continuous second partial derivatives and $f : \mathbb{C} \to \mathbb{C}$ has a complex derivative and continuous partial derivatives.
Best Answer
My preferred way of avoiding losing track of notation is by using linear approximation to $u$ and $f$. That is: express the given derivatives as $$ f(a+h)=f(a)+hf'(a) + o(|h|) \tag{1} $$ and $$ u(x+h_1,y+h_2)=u(x,y)+h_1 \frac{\partial u}{\partial x}(x,y)+h_2 \frac{\partial u}{\partial y}(x,y)+o\left(\sqrt{h_1^2+h_2^2}\right) \tag{2} $$ Before even doing anything with $f$, notice that $(2)$ can be equivalently written in complex form using $h = h_1+ih_2$ (hence $\bar h=h_1-ih_2$). Namely, $$ u(b+h )=u(b)+h \frac{\partial u}{\partial z}(b)+\bar h \frac{\partial u}{\partial \bar z}(b)+o\left(|h|\right) \tag{3} $$ (In fact, it is the identity $(3)$ that motivates the definition of Wirtinger derivatives.)
It remains to plug (1) into (3) (so that $b=f(a)$) and clean up: $$ u(f(a+h))=u(f(a))+ h f'(a) \frac{\partial u}{\partial z}(f(a))+\bar h \overline{f'(a)}\frac{\partial u}{\partial \bar z}(f(a))+o\left(|h|\right) \tag{4} $$ An advantage of this approach is that we get both Wirtinger's derivatives of the composition $v=u\circ f$ at once: (4) gives both $$ \frac{\partial v}{\partial z}(a) = f'(a) \frac{\partial u}{\partial z}(f(a)) $$ and $$ \frac{\partial v}{\partial \bar z}(a) = \overline{f'(a)} \frac{\partial u}{\partial \bar z}(f(a)) $$