Let $\Omega:=\{(x_1,y_1,z_2,\bar z_2)\;:\;\underbrace{y_1-f(z_2,\bar z_2)+g(z_2,\bar z_2)}_{=:r(x_1,y_1,z_2,\bar z_2)}<0\}\subset\Bbb C^2$ (we confuse consciously $\Bbb C$ with $\Bbb R^2$) where $f(z_2,\bar z_2):=-2x_2^2+y_2^2$ and $g$ is a real polynomial expression in $z_2,\bar z_2$ whose monomials are of degree $\ge3$.
We define $M:=\partial\Omega$, it's a real hypersurface. Let $z_0=0\in\Bbb C^2$. If my computations are right, the hessian matrix of $r$ in $z_0=0$ is $$ \operatorname{hess}_{r(0)}=\left(\begin{matrix} 0 & 0 & 0 & 0 \\ 0 & \frac32 & 0 & \frac12 \\ 0 & 0 & 0 & 0 \\ 0 & \frac12 & 0 & \frac32 \end{matrix}\right) $$ Thus the Levi matrix of $r$ is
$$ L_{r(0)}= \left(\begin{matrix} 0 & 0 \\ 0 & \frac12 \end{matrix}\right) $$ and being the Levi form of $M$ in $z_0=0$ $L_M(0):={L_{r(0)}}_{|T_0^{\Bbb C}M}$ by definition, we have that $L_M(0)=\frac12$ (it's a $1\times1$ matrix), in fact (up to holomorphic coordinate change) we are allowed to think at $T_0^{\Bbb C}M$ as $\{(0,z_2)\;,\;z_2\in\Bbb C\}\simeq\Bbb C$.
Thus the Levi form of $M$ works as follows: given $z:=(0,z_2)\in T_0^{\Bbb C}M$, we have that $L_M(0)(z,z)=\frac12z_2^2$.
In particular if we take $z=e_2=(0,1)$, we get $L_M(0)(e_2,e_2)=\frac12$.
My problem comes here: if we use the well known formula $$ L_{r(0)}(u,u)=\frac12[\operatorname{hess}_{r(0)}(u,u)+\operatorname{hess}_{r(0)}(iu,iu)] $$ since $$ \operatorname{hess}_{r(0)}(e_2,e_2):= (0,1,\bar 0,\bar1)\operatorname{hess}_{r(0)}\left(\begin{matrix} 0\\ 1\\ \bar0\\ \bar1 \end{matrix}\right)=4 $$ and similarly $ \operatorname{hess}_{r(0)}(ie_2,ie_2)=i^2+i^2=-2$ I would obtain $L_M(0)(e_2,e_2)=1\neq\frac12$.
I suppose $1/2$ is right, $1$ is wrong, because the book writes so.
Where did I got wrong?
Many thanks!
It was a misunderstanding: the quadratic form defined by the hessian matrix is to be taken divided by $2$ because we think at it as the quadratic part of the Taylor series. Hence the $4-2=2$ above becomes $1$ and all adds up!