Let $\Omega\subseteq\Bbb C^n$ open, $z_0\in\Omega$, $r:\Omega\to\Bbb R$ twice real differentiable.
We know that $\Bbb C^n\simeq\Bbb R^{2n}$ is an isomorphism of vect.sp. So we think $\Bbb C^{n}$ as an $\Bbb R$-space, so it has dimension $2n$. A base for it is given by $z_1,\dots,z_n,\bar z_1,\dots,\bar z_n$. So $\Bbb C^n\simeq\Bbb C^n\times\overline{\Bbb C}^n$.
The Hessian matrix of $r$ in $z_0$ is the $2n\times2n$ matrix $$ \operatorname{hess}r_{z_0}= \left[ \begin{array}{cc} \partial_{z_i}\partial_{z_j}r(z_0) & \partial_{z_i}\partial_{\bar z_j}r(z_0) \\ \partial_{\bar z_i}\partial_{z_j}r(z_0) & \partial_{\bar z_i}\partial_{\bar z_j}r(z_0) \end{array} \right]_{i,j=1,\dots,n}\;\;\;. $$ Now we define the Levi form of $r$ as the block $(1,2)$ (or $(2,1)$, they are equal), i.e. the $n\times n$ matrix $$ L_r(z_0)= \left[ \begin{array}{c} \partial_{z_i}\partial_{\bar z_j}r(z_0) \end{array} \right]_{i,j=1,\dots,n}\;\;\;. $$ My book, writes as follows: $$ \operatorname{hess}r_{z_0}=\sum_{i,j=1}^n\partial_{z_i}\partial_{z_j}r(z_0)\,dz_i\otimes dz_j+\\ +\sum_{i,j=1}^n\partial_{\bar z_i}\partial_{\bar z_j}r(z_0)\,d\bar z_i\otimes d\bar z_j+ 2\sum_{i,j=1}^n\partial_{z_i}\partial_{\bar z_j}r(z_0)\,dz_i\otimes d\bar z_j $$ (he thinks the hessian as the quadratic term in Taylor series of $r$ so he divides by $2$, but is not relevant) in this manner, both forms work on the same space (the matrix form should force the Levi form to work on $n$-tuple, but the hessian works on $2n$-tuples). I wrote the matrix form to be more clear.
My problem is the proposition 1.6.3. below. if I compute by hand, $\operatorname{hess}r(iu,iu)=-\operatorname{hess}r(u,u)$, so the proposition would be senseless. Here I think $u\in\Bbb C^n$, but the hessian works on the $2n$-tuples, so I think the couple of vector he "eats" is $[(u,\bar u),(u,\bar u)]$ with $(u,\bar u)\in\Bbb C^{2n}$. Where did I wrong? How can I see it by hand, without passing thru $\operatorname{hess}r=LP_r+L_r$ (otherwise it wouldn't be really explicative)?
Many thanks
Compute by hand again. You must have just miscalculated. When you plug $(iu,iu)$ into the real hessian the $(z,z)$ and $(\bar{z},\bar{z})$ part of the hessian gets a minus sign, the complex hessian acquires a $-i$. So there is no direct relationship.