Let $\Omega\subseteq\Bbb C^n$ open, $z_0\in\Omega$, $r:\Omega\to\Bbb R$ twice real differentiable.
We know $\Bbb C^n\simeq\Bbb R^{2n}$ 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 $$ \operatorname{hess}r_{z_0}= \left[ \begin{array}{cc} \partial_{z_i}\partial_{z_j}r(z_0) & \partial_{z_i}\partial_{\bar z_j}r(z_0) \\ \partial_{\bar z_i}\partial_{z_j}r(z_0) & \partial_{\bar z_i}\partial_{\bar z_j}r(z_0) \end{array} \right]_{i,j=1,\dots,n}=: \left[ \begin{array}{cc} A & B \\ C & D \end{array} \right] \;\;\;. $$ Clearly $A,B,C,D\in M_n(\Bbb R)$ (they are real since $r$ is such), they are all symmetric and $A=D, B=C$.
Now we define the Levi form of $r$ as the bilinear hermitian form whose matrix is given by block $B$ (or $C$, they are equal), i.e. the $n\times n$ matrix $$ L_r(z_0)=B= \left[ \begin{array}{c} \partial_{z_i}\partial_{\bar z_j}r(z_0) \end{array} \right]_{i,j=1,\dots,n}\;\;\;. $$ i.e. the form is given by $(v,u)\mapsto\;^TvB\bar u$, where $(v,u)\in\Bbb C^n\times\Bbb C^n$.
My book is very very cryptic, it doesn't say where this form comes out; it clearly comes out from the bilinear form induced by the hessian, which is defined on $\Bbb C^n\times\Bbb C^n$: \begin{align*} (v,u)\mapsto[^Tv,^T\bar v] \left[ \begin{array}{cc} A & B \\ C & D \end{array} \right] \left[ \begin{array}{cc} u \\ \bar u \end{array} \right] &=\;^TvAu+\;^TvB\bar u+\;^T\bar vCu+\;^T\bar vD\bar u \end{align*} so the Levi form is "a piece" of the hessian.
ATTENTION: I've deduced this last formula by myself!! Is that correct?
Now it should be immediate to see that $$ \;^TvAu+\;^T\bar vD\bar u=\;^TvAu+\;\overline{^TvAu}=2\Re(^TvAu) $$ and similarly $$ \;^TvB\bar u+\;^T\bar vCu=\;^TvB\bar u+\;^T\bar vBu=\;^TvB\bar u+\overline{\;^TvB\bar u}=2\Re(\;^TvB\bar u)\;. $$ In this way, the hessian form should be $$ [^Tv,^T\bar v] \left[ \begin{array}{cc} A & B \\ C & D \end{array} \right] \left[ \begin{array}{cc} u \\ \bar u \end{array} \right]= 2\Re(^TvAu+\;^TvB\bar u) $$
My book, writes all this as follows (he thinks the hessian as the quadratic term in Taylor series of $r$ so he divides by $2$, so what follows should be what I wrote but divided by $2$): $$ \operatorname{hess}r_{z_0}=\frac12\sum_{i,j=1}^n\partial_{z_i}\partial_{z_j}r(z_0)\,dz_i\otimes dz_j+\\ +\frac12\sum_{i,j=1}^n\partial_{\bar z_i}\partial_{\bar z_j}r(z_0)\,d\bar z_i\otimes d\bar z_j+ \sum_{i,j=1}^n\partial_{z_i}\partial_{\bar z_j}r(z_0)\,dz_i\otimes d\bar z_j $$ but to me is senseless because of the third term: if the book was right it should be $$ \;^TvB\bar u=\;^T\bar vCu $$ but this is not true! Where is the error?!
Moreover the hessian form is a piece of the Taylor expansion of $r$, which is a real function; writing as the book does, $r$ could be even complex!
Am I right? Many thanks!