Complex chain rule for complex valued functions Let $f=f(z)$ and $g=g(w)$ be two complex valued functions which are differentiable in the real sense, $h(z)=g(f(z))$. Prove the complex chain rule. 
All partial derivatives:
$$ 
\frac{\partial h}{\partial z} = \frac{\partial g}{\partial w}\frac{\partial f}{\partial z} + \frac{\partial g}{\partial \bar w}\frac{\partial \bar f}{\partial z}
$$
and
$$
\frac{\partial h}{\partial \bar z} = \frac{\partial g}{\partial w}\frac{\partial f}{\partial \bar z} + \frac{\partial g}{\partial \bar w}\frac{\partial\bar f}{\partial \bar z}
$$
Are we supposed to arrive at this through Cauchy-Riemann?
 A: This has nothing to do with Cauchy-Riemann's equations. Use the regular chain rule (for functions on $\mathbb{R}^2$) and the definition of the Wirtinger derivatives:
$$
\frac{\partial}{\partial z} = \frac12 \left( \frac{\partial}{\partial x}  - i\,\frac{\partial}{\partial y}  \right)
\qquad\text{and}\qquad
\frac{\partial}{\partial \bar z} = \frac12 \left( \frac{\partial}{\partial x}  + i\,\frac{\partial}{\partial y}  \right)
$$
It all boils down to a fairly long and tedious algebraic manipulation
(See also: Wikipedia)
A: \begin{gather*}
\text{Let }f(z)=w=u(z)+iv(z)\\
\frac{\partial g}{\partial w} \frac{\partial f}{\partial z} +\frac{\partial g}{\partial \bar{w}} \frac{\partial \bar{f}}{\partial z}=\frac{1}{2}\left(\frac{\partial g}{\partial u}+\frac{\partial g}{i\partial v}\right)\left(\frac{\partial u}{\partial z}+\frac{i\partial v}{\partial z}\right)
+\frac{1}{2}\left(\frac{\partial g}{\partial u}-\frac{\partial g}{i\partial v}\right)\left(\frac{\partial u}{\partial z}-\frac{i\partial v}{\partial z}\right)\\
=\frac{\partial g}{\partial u}\frac{\partial u}{\partial z}+\frac{\partial g}{\partial v}\frac{\partial v}{\partial z}=\frac{\partial g}{\partial z}
\end{gather*}
Proceed similarly for the second chain rule
