Consider a $N \times N$ matrix $A=B_1+B_2$, where $B_1$ is a diagonal matrix with all the diagonal entries between $0$ and $1$ and $B_2$ is a skew symmetric matrix which can be written as
$$B_2= \begin{bmatrix} 0_{N-1} & f \\ -f^T & 0 \end{bmatrix}$$
where $0_{N-1}$ is the $N-1 \times N-1$ zero matrix, $f$ is a $N-1$ by $1$ vector. So $B_2=-B_2^T$.
My question is how many complex eigenvalues (non-zero imaginary part) $A$ has?
I have run some numerical simulations. It seems that $A$ can at most have one pair of complex eigenvalues. And it is plausible to me because the skew symmetric $B_2$ is only rank $2$.
How can we prove this mathematically or come up with a counter example? If needed, we can put an upper bound on $|f|_2$. But I don't know that if it is necessary, as indicated from my simulations.