A Submanifold $M$ of $\Bbb C^N$ I have a Proposition in my book, and I write here:

For every $p \in M$, with $M$ be a hypersurface in $\Bbb C^N$ the following hold.
\begin{align*}
\mathcal V_p &= \left \{ X \in \Bbb C T_p M \colon J(X)=-iX \right \},\tag{1}\\
\mathcal V_p \oplus \overline{\mathcal V_p} &=\Bbb C T_{p}^{c}M, \tag{2} \\
\mathcal V_p &=\left \{ X+iJ(X) \colon   X \in  T_{p}^{c}M  \right \},\tag{3} \\
\operatorname{Re} \mathcal V_p &=T_{p}^{c}M, \tag{4}
\end{align*}
where $\operatorname{Re} \mathcal V_p=\left \{ X+\overline{X} \colon  X \in \mathcal V_p \right \}.$
With a real linear mapping $J$ ($J^2=-Id$) from $T_p\Bbb C^N$ into itself determined by 
  $$J \left (\dfrac{\partial }{\partial x_j} \mid_p   \right )= \dfrac{\partial }{\partial y_j} \mid_p , J \left (\dfrac{\partial }{\partial y_j} \mid_p   \right )= -\dfrac{\partial }{\partial x_j} \mid_p ,\ ~~~~ j=\overline{1,N}.$$

PROOF

A direct calculation shows that
  $$J\left ( \dfrac{\partial }{\partial Z_j} \right )=i\dfrac{\partial }{\partial Z_j},\ J\left ( \dfrac{\partial }{\partial \overline{Z_j}} \right )=-i\dfrac{\partial }{\partial \overline{Z_j}}. \tag{5}$$
  Thus, we obtain
  $$
J\left ( \operatorname{Re}\left (\sum_{j=1}^{N}c_j\dfrac{\partial }{\partial \overline{Z_j}}  \right ) \right )=\operatorname{Im}\left (\sum_{j=1}^{N}c_j\dfrac{\partial }{\partial \overline{Z_j}}  \right ).\tag{6}$$


And here's my trouble: I don't understand Why do we have $(5),\ (6)$?
On the other hand, Why the Proposition easily follows from these relations? 
Can anyone write the details? 
Thanks!
 A: The $\frac{\partial}{\partial Z_j},\,\frac{\partial}{\partial\overline{Z}_j}$ are the Wirtinger derivatives,
$$\frac{\partial}{\partial Z_j} = \frac{1}{2}\left(\frac{\partial}{\partial x_j} - i \frac{\partial}{\partial y_j}\right),\quad \frac{\partial}{\partial \overline{Z}_j} = \frac{1}{2}\left(\frac{\partial}{\partial x_j}+i\frac{\partial}{\partial y_j}\right).$$
Then by linearity of $J$ we compute
$$J\left(\frac{\partial}{\partial Z_j}\right) = \frac{1}{2}\left(J\left(\frac{\partial}{\partial x_j}\right) - i J\left(\frac{\partial}{\partial y_j}\right)\right) = \frac{1}{2}\left(\frac{\partial}{\partial y_j} + i\frac{\partial}{\partial x_j}\right) = i\frac{\partial}{\partial Z_j},$$
and
$$J\left(\frac{\partial}{\partial \overline{Z}_j}\right) = \frac{1}{2}\left(J\left(\frac{\partial}{\partial x_j}\right) + i J\left(\frac{\partial}{\partial y_j}\right)\right) = \frac{1}{2}\left(\frac{\partial}{\partial y_j} - i\frac{\partial}{\partial x_j}\right) = -i\frac{\partial}{\partial \overline{Z}_j}.$$
From that, we obtain
$$\begin{align}
J\left(\operatorname{Re}\sum c_j\frac{\partial}{\partial\overline{Z}_j}\right)
&= \frac{1}{2} J\left(\sum c_j\frac{\partial}{\partial\overline{Z}_j} + \sum \overline{c}_j \frac{\partial}{\partial Z_j}\right)\\
&= \frac{1}{2i} \left(\sum c_j\frac{\partial}{\partial \overline{Z}_j}-\overline{c}_j\frac{\partial}{\partial Z_j}\right)\\
&= \sum \operatorname{Im} c_j\frac{\partial}{\partial \overline{Z}_j}.
\end{align}$$
