Differential in Huybrechts the differential $df|x:T_x\mathbb{R}^2\to T_{f(x)}\mathbb{R}^2$ of a differentiable map $f:\mathbb{R}^2\to \mathbb{R}^2$ is given by the usual jacobian matrix.
However, if we complexify the tangent space, the differential is a complex linear map $df|x_\mathbb{C}:T_x\mathbb{R}^2\otimes \mathbb{C}\to T_{f(x)}\mathbb{R}^2\otimes \mathbb{C}$.
The tangent space can be given the $\mathbb{C}$-basis $\{\frac{\partial}{\partial z}:=\frac{1}{2}(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}), \frac{\partial}{\partial \bar z}:=\frac{1}{2}(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y})\}. $ This I understand. But the following is not clear to me:
With respect to this basis the differential $df|x_\mathbb{C}$ is given by the matrix $$\small \begin{pmatrix} \frac{\partial f}{\partial z} & \frac{\partial f}{\partial \bar z} \\     \frac{\partial \bar f}{\partial z} & \frac{\partial \bar f}{\partial \bar z} \end{pmatrix}$$
I would very much appreciate your help explicitely calculating this matrix! I have spent several hours on it, but didn't get a good result.
I found this statement in Huybrechts which I link here in case my explanation was not clear enough.
Thanks a lot
 A: $\frac{\partial f}{\partial z} = \frac{\partial f_1}{\partial z} + i\frac{\partial f_2}{\partial z} = \frac12\left(\frac{\partial f_1}{\partial x}-i\frac{\partial f_1}{\partial y} + i\frac{\partial f_2}{\partial x}+\frac{\partial f_2}{\partial y}\right)$,
$\frac{\partial f}{\partial \bar z} = \frac{\partial f_1}{\partial \bar z} + i\frac{\partial f_2}{\partial \bar z} = \frac12\left(\frac{\partial f_1}{\partial x}+i\frac{\partial f_1}{\partial y} + i\frac{\partial f_2}{\partial x}-\frac{\partial f_2}{\partial y}\right)$,
$\frac{\partial \bar f}{\partial z} = \frac{\partial f_1}{\partial z} - i\frac{\partial f_2}{\partial z} = \frac12\left(\frac{\partial f_1}{\partial x}-i\frac{\partial f_1}{\partial y} - i\frac{\partial f_2}{\partial x}-\frac{\partial f_2}{\partial y}\right)$,
$\frac{\partial \bar f}{\partial \bar z} = \frac{\partial f_1}{\partial \bar z} - i\frac{\partial f_2}{\partial \bar z} = \frac12\left(\frac{\partial f_1}{\partial x}+i\frac{\partial f_1}{\partial y} - i\frac{\partial f_2}{\partial x}+\frac{\partial f_2}{\partial y}\right)$.
Then $\begin{pmatrix}\frac{\partial  f}{\partial z} & \frac{\partial  f}{\partial \bar z} \\ \frac{\partial \bar f}{\partial z} & \frac{\partial \bar f}{\partial \bar z}\end{pmatrix}
= 
\frac12\begin{pmatrix} 
\frac{\partial f_1}{\partial x}-i\frac{\partial f_1}{\partial y} + i\frac{\partial f_2}{\partial x}+\frac{\partial f_2}{\partial y} & \frac{\partial f_1}{\partial x}+i\frac{\partial f_1}{\partial y} + i\frac{\partial f_2}{\partial x}-\frac{\partial f_2}{\partial y} \\ 
\frac{\partial f_1}{\partial x}-i\frac{\partial f_1}{\partial y} - i\frac{\partial f_2}{\partial x}-\frac{\partial f_2}{\partial y} &
\frac{\partial f_1}{\partial x}+i\frac{\partial f_1}{\partial y} - i\frac{\partial f_2}{\partial x}+\frac{\partial f_2}{\partial y}
\end{pmatrix} 
=
\frac12\begin{pmatrix} 1&i\\ 1 &-i\end{pmatrix}
\begin{pmatrix} 
\frac{\partial f_1}{\partial x} &
\frac{\partial f_1}{\partial y} \\
\frac{\partial f_2}{\partial x} &
\frac{\partial f_2}{\partial y}
\end{pmatrix}
\begin{pmatrix} 1&1 \\ -i & i\end{pmatrix}.
$
This may help. 
Note that $\frac12\begin{pmatrix} 1&i\\ 1 &-i\end{pmatrix}$ is the matrix of change of basis in the tangent space, and $\begin{pmatrix} 1&1 \\ -i & i\end{pmatrix}$ is its inverse.
