$f$ is holomorphic iff $df$ is $\Bbb C$-linear Let $\Omega\subseteq\Bbb C^n$ open connected, $f:\Omega\to\Bbb C$ differentiable in the real sense. We know that $f$ is holomorphic iff $\partial_{\bar z_j}f=0\;\;\forall j=1,\dots,n$ .
We know also that $df=\partial_zfdz+\partial_{\bar z}f d\bar z=:\partial f+\bar{\partial}f$.
How can I prove from this that $f$ is holomorphic iff $df$, which is in general $\Bbb R$-linear, is now $\Bbb C$-linear?
We know by definition that such an $f$ is differentiable in the complex sense in $w\in\Omega$ iff $\;\exists\; T:\Bbb C^n\to\Bbb C$ $\Bbb C$-linear such that
$$
\lim_{z\to w,z\in\Omega}\frac{f(z)-f(w)-T(z-w)}{||z-w||_{\Bbb C^n}}=0
$$
and being $T=df$ we can conclude. But I would like to prove it directly, using that $df=\partial f+\bar{\partial}f$, showing that $df$ is $\Bbb C$-linear iff $\bar{\partial}f=0$.
Even because trying in different ways it seems that $df$ be always $\Bbb C$-linear, and this is wrong, so showing what I've asked I'd shed some lights on this.
Can somebody give me some hints? Many thanks!

The form $dz$ is identified with $\begin{bmatrix}1&0\\0&1\end{bmatrix}$ and the form $d\bar{z}$ is identified with $\begin{bmatrix}1&0\\0&-1\end{bmatrix}$. Also, $i$ is identified with $\begin{bmatrix}0&0\\0&-1\end{bmatrix}$. You can check via multiplication of the identified matrices that $dz\cdot i = i\cdot dz$ and $d\bar{z} = -i\cdot d\bar{z}$, which establish the $\Bbb C$-linearity of $dz$ and anti-linearity of $d\bar{z}$, respectively. Then the $\Bbb C$-linearity of $\partial f$ and anti-linearity of $\bar{\partial} f$ follow. Thus, $df$ is $\Bbb C$-linear $\iff$ $\bar{\partial} f$ is $\Bbb C$-linear $\iff $ $\bar{\partial}f = 0$  $\iff$ $\partial_{\bar{z}}f = 0$ $\iff$ $f$ is holomorphic.
