Understanding the chain rule in the Wirtinger calculus The Wirtinger differential operators are defined by:
\begin{equation}
\frac{\partial}{\partial z} = \frac{1}{2}\left(\frac{\partial}{\partial x} - i\frac{\partial}{\partial y}\right) \\
\frac{\partial}{\partial \bar{z}} = \frac{1}{2}\left(\frac{\partial}{\partial x} + i\frac{\partial}{\partial y}\right)
\end{equation}
These satisfy the following chain rule:
\begin{equation}
\frac{\partial}{\partial z}(f \circ g) = \left(\frac{\partial f}{\partial z}\circ g\right)\frac{\partial g}{\partial z} + \left(\frac{\partial f}{\partial \bar{z}}\circ g\right)\frac{\partial \bar{g}}{\partial z}.
\end{equation}
Usually I think about partial derivatives as forming the components of the Jacobian (a.k.a differential/total derivative) and the chain rule for them as a matrix representation of the relation:
\begin{equation}
 \mathbf{D}(f \circ g) = (\mathbf{D}f \circ g)\cdot\mathbf{D}g.
\end{equation}
How can one interpret the chain rule for the Writinger differential operators in this light? I would particularly enjoy a formalism that allows me to understand why $\frac{\partial{f}}{\partial \bar{z}} = 0$ iff $f$ is analytic, or makes this seem like a really natural definition to begin investigation of the properties of analytic functions. Alternatively I would like formalism that is closely related to the idea of complexifying the tangent bundle of $\mathbb{R}^2$ with its standard complex structure, with an explanation of the connection.
 A: Think of $f\colon \mathbb{C} \longrightarrow \mathbb{C}$ as a map $f\colon \mathbb{R}^2 \longrightarrow \mathbb{R}^2$. The jacobian matrix $Jf(p)$ is just the matrix of $Df_p \colon T_p\mathbb{R}^2 \longrightarrow T_{f(p)}\mathbb{R}^2$ with respect to the frame $\frac{\partial}{\partial x}, \frac{\partial}{\partial y}$. Complexify the tangent bundle and change the frame to $\frac{\partial}{\partial z}, \frac{\partial}{\partial \bar z}$.  In this new frame we will have 
$$ { \large
[Df_p] = \begin{pmatrix}
\frac{\partial f}{\partial z}(p) & \frac{\partial f}{\partial \bar z}(p)\\
 \frac{\partial \bar f}{\partial z}(p)& \frac{\partial\bar f}{\partial \bar z}(p)
\end{pmatrix}}
$$
Then, by chain rule, $ [D(f\circ g)_p] =[Df_{g(p)}][Dg_p]$ which reads
$$  
\begin{pmatrix}
\frac{\partial f\circ g}{\partial z}(p) & \frac{\partial f\circ g}{\partial \bar z}(p)\\
 \frac{\partial \overline{f\circ g}}{\partial  z}(p) & \frac{\partial \overline{f\circ g} }{\partial \overline z}(p)
\end{pmatrix}
 = \begin{pmatrix}
\frac{\partial f}{\partial z}(g(p)) & \frac{\partial f}{\partial \bar z}(g(p))\\
 \frac{\partial \bar f}{\partial z}(g(p)) & \frac{\partial\bar f}{\partial \bar z}(g(p))
\end{pmatrix}
\begin{pmatrix}
\frac{\partial g}{\partial z}(p) & \frac{\partial g}{\partial \bar z}(p)\\
 \frac{\partial \bar g}{\partial z}(p) & \frac{\partial\bar g}{\partial \bar z}(p)
\end{pmatrix}
$$
And the result follows comparing the entries of the matrices. 
Added: In the frame $\frac{\partial}{\partial x}, \frac{\partial}{\partial y}$ we have for $f(x,y) = (u(x,y) , v(x,y))$
$$
Jf = \begin{pmatrix}
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\
\frac{\partial v}{\partial x} &\frac{\partial v}{\partial y}
\end{pmatrix}
$$
And the base change matrix to $\frac{\partial}{\partial z}, \frac{\partial}{\partial \bar z}$ is
$$
P = \frac{1}{2}\begin{pmatrix}
1 & -i \\
1 & i
\end{pmatrix}
$$
From the relations $\frac{\partial}{\partial z} = \frac{1}{2}\left(\frac{\partial}{\partial x} - i\frac{\partial}{\partial y}\right)$ and $\frac{\partial}{\partial \bar{z}} = \frac{1}{2}\left(\frac{\partial}{\partial x} + i\frac{\partial}{\partial y}\right)$. Hence the matrix for $Df$ in the $\frac{\partial}{\partial z}, \frac{\partial}{\partial \bar z}$ frame is $P \cdot Jf\cdot  P^{-1}$,
$$
\frac{1}{2}\begin{pmatrix}
1 & -i \\
1 & i
\end{pmatrix} 
\begin{pmatrix}
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\
\frac{\partial v}{\partial x} &\frac{\partial v}{\partial y}
\end{pmatrix}
\begin{pmatrix}
1 & 1 \\
i & -i
\end{pmatrix} =
\frac{1}{2}\begin{pmatrix}
1 & -i \\
1 & i
\end{pmatrix}
\begin{pmatrix}
\frac{\partial u}{\partial x} + i\frac{\partial u}{\partial y} &  \frac{\partial u}{\partial x} - i\frac{\partial u}{\partial y}\\
\frac{\partial v}{\partial x} + i\frac{\partial v}{\partial y} &  \frac{\partial v}{\partial x} - i\frac{\partial v}{\partial y}
\end{pmatrix} =
\frac{1}{2}\begin{pmatrix}
\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + i\left( \frac{\partial u}{\partial y} - \frac{\partial v}{\partial x}  \right) & \frac{\partial u}{\partial x} - \frac{\partial v}{\partial y} - i\left( \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x}  \right) \\
\frac{\partial u}{\partial x} - \frac{\partial v}{\partial y} + i\left( \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x}  \right) & \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} - i\left( \frac{\partial u}{\partial y} - \frac{\partial v}{\partial x}  \right)
\end{pmatrix}
$$
that agree with the matrix stated if we write $f = u+iv$ and $z=x+iy$.
