It' all, simply, about the signature of a matrix.
Let $\Omega\subseteq\Bbb C^n$ open, $r:\Omega\to\Bbb R$ twice differentiable (real differentiable, not necessarely complex differentiable, i.e. not necessarely holomorphic), $z_0\in\Omega$.
The Hessian matrix of $r$ in $z_0$ is the $2n\times2n$ matrix $$ H[r]_{z_0}= \left[ \begin{array}{cc} \partial_{z_i}\partial_{z_j}r(z_0) & \partial_{z_i}\partial_{\bar z_j}r(z_0) \\ \partial_{\bar z_i}\partial_{z_j}r(z_0) & \partial_{\bar z_i}\partial_{\bar z_j}r(z_0) \end{array} \right]_{i,j=1,\dots,n}\;\;\;. $$ Now we define the complex Hessian of $r$ as the block $(1,2)$ (or $(2,1)$, they are equal), i.e. the $n\times n$ matrix $$ \Bbb CH[r]_{z_0}= \left[ \begin{array}{c} \partial_{z_i}\partial_{\bar z_j}r(z_0) \end{array} \right]_{i,j=1,\dots,n}\;\;\;. $$
I have reason to believe the signature of $\Bbb CH[r]_{z_0}$ behaves in a strange way. Call it $(s^+,s^-,s^0)$; we know that $s^++s^-+s^0=n$. If $n\ge2$, I think the following conditions are true: $$ \left\{ \begin{array}{lll} s^+\ge1\\ s^++s^-\le n-1 \end{array} \right. \;\;. $$
The reason I believe this is that these are necessary to make sum's index work in 1.6.6 of https://math.stackexchange.com/questions/1215207/normal-coordinates-theorem-for-a-hypersurface.
Is this true? How can I prove it?