Eigenvalues of complex Hessian and real Hessian Let $f:\mathbb{C}^n\rightarrow \mathbb{R}$ be a smooth function. The complex Hessian is given by
$$\left(\frac{\partial^2f}{\partial z_i\partial \bar{z}_j}\right)_{ij}$$
and the real Hessian by
$$\begin{pmatrix}
\dfrac{\partial^2f}{\partial x_i\partial x_j} & \dfrac{\partial^2f}{\partial x_i \partial y_j}\\
\dfrac{\partial^2f}{\partial x_j\partial y_i} & \dfrac{\partial^2f}{\partial y_i \partial y_j}
\end{pmatrix}.$$
Is it true that any (real) eigenvalue of the complex Hessian is an eigenvalue of the real one? If yes why?
Edit: In light of Giuseppe's comments and checking with $f=|z|^2$ this seems to be straight up wrong. What I am really interested in is to show that if the complex Hessian has $n$ negative (or positive) eigenvalues, so does the real Hessian.
EDIT 2.0: I may have something, I think this is correct:
Denote by $L=\left(\dfrac{\partial f}{\partial z_i\partial \bar{z_j}}\right)$ the Levi matrix of a smooth function $f$, by $H(f)$ its Hessian and by $X\in M_{2n,n}(\mathbb{C})$ the matrix
\begin{equation*}
X=\begin{pmatrix}
Id\\
iId
\end{pmatrix}.
\end{equation*}
We then have
\begin{equation*}
L=\overline{X}^tH(f)X
\end{equation*}
It follows that if $v\in\mathbb{C}^n$ verifies $\bar{v}^tLv = \lambda |v|^2$ for $\lambda\in\mathbb{R}\setminus \{0\}$ then
\begin{eqnarray*}
\overline{(Xv)}^tH(f)(Xv) &=& \bar{v}^tLv \\
                          &=& \lambda |v|^2\\
                          &=& \dfrac{\lambda}{2}|Xv|^2
\end{eqnarray*}
We deduce that if $L$ is negative (positive) then $H(f)$ has $n$ negative (positive) eigenvalues.
 A: Think about one dimension.  You can write the real Hessian in terms of the $z$ and the $\bar{z}$ as
$$
\begin{bmatrix}
\frac{\partial^2 f}{\partial z^2} & \frac{\partial^2 f}{\partial z\partial \bar{z}}\\
\frac{\partial^2 f}{\partial z \partial \bar{z}} & \frac{\partial^2 f}{\partial \bar{z}^2}
\end{bmatrix}
$$
The real hessian in terms of $z$ and $\bar{z}$ is just a change of variables of the form $T^t X T$, there $X$ is the real hessian in terms of $x$ and $y$ as you give it and $T = \begin{bmatrix}1/2&1/2\\
-i/2&i/2\end{bmatrix}$.  Note that $T^t$ is not quite the inverse of $T$.
Anyway, suppose the real Hessian is $X = \begin{bmatrix}a& c\\ c& b\end{bmatrix}$.  Then the complex Hessian, the upper right (or lower left) corner of the above matrix is $\frac{b+a}{4}$.  If the real Hessian is is positive definite, then $a+b > 0$ (trace of $X$).  If it is negative definite, then $a+b < 0$.  So the complex Hessian "sees" that.  But if the real Hessian has mixed eigenvalues, then the complex Hessian might be positive, negative, or even zero.
More generally (in any dimension) you can prove that if the real Hessian is positive definite then the complex Hessian is also.  It is not necessarily just the existence of positive eigenvalues, it is also how their eigenspaces interact with the complex structure of $\mathbb{C}^n$.  Suppose the real hessian has $k > n$ positive eigenvalues, then the there is a $k$ dimensional real subspace on which the real Hessian is positive definite.  That subspace must contain a $k-n$ dimensional complex subspace, and on that subspace the complex Hessian is positive definite.  That means that the complex Hessian has at least $k-n$ positive eigenvalues.  But it could have more as the above simple example illustrates.
A: TOO LONG FOR A COMMENT TO THE MAIN QUESTION.
In the comments to the main question I casually suggested that the Hessian of $f\colon \mathbb C\to \mathbb C$, that is,
$$\tag{1}
\begin{bmatrix} \partial^2_{x} f & \partial_x\partial_{y} f \\ 
\partial_{y} \partial_x f & \partial^2_{y} f\end{bmatrix},
$$
is similar to the Hessian in the $\partial_z, \partial_{\overline z}$ basis, which is
$$\tag{2}
\begin{bmatrix} \partial^2_{z} f & \partial_z\partial_{\overline z} f \\ 
\partial_{\overline{z}} \partial_z f & \partial^2_{\overline z} f\end{bmatrix}.$$
This is wrong and the function $f(z)=z\overline z = x^2+y^2$ provides an immediate counterexample. The first matrix is
$$
\begin{bmatrix} 2 & 0 \\ 0 & 2\end{bmatrix}, $$
while the second is
$$
\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, $$
and they cannot be similar, since they have different eigenvalues; the first has the double eigenvalue $2$ while the second has the simple eigenvalues $1, -1$. This is in accordance with Jiri’s answer; (1) and (2) are congruent, that is, they are related by a change of variable of the form $T^t X T$, but $T$ is not a unitary matrix, so this is not the same as matrix similarity.
A consequence of this fact is that you cannot use (2) to determine the convexity of $f$. The function $f(z)=|z|^2$ is manifestly convex, but the matrix (2) has sign-changing eigenvalues, as we just saw. Interesting.
