I want to solve this kind of eigenvalue problem
$$\Bbb{A}(\omega) \vec{T} = k\Bbb{B}(k)\vec{T}$$
where $\Bbb{A}$ and $\Bbb{B}$ are both nonsymmetric, complex matrices. $\omega$ is real, and the eigenvalue $k$ is complex. $\vec{T}$ is the eigenvector. Note that $\Bbb{B}$ is dependant on $k$.
The condition is equivalent to
$$ \begin{align} &\Bbb{A}(\omega) \vec{T} - k\Bbb{B}(k)\vec{T} = 0 \\ \iff &[\Bbb{A}(\omega)- k\Bbb{B}(k)]\vec{T} = 0. \end{align} $$ This happens if the kernel $\textrm{ker}(\Bbb{A}(\omega)- k\Bbb{B}(k))$ is non-empty, hence we must have $$ \det(\Bbb{A}(\omega)- k\Bbb{B}(k)) = 0 $$ which you can solve numerically or analytically depending on your application.
Simple example
Let's take $\Bbb A(\omega)$ to be the rotation of angle $\omega$ and $$ \Bbb B(z) = \begin{pmatrix}0 & 0 \\ -z & 0 \end{pmatrix}. $$ The equation to solve is $$ \det\begin{pmatrix} \cos \omega & - \sin \omega \\ \sin \omega + z^2& \cos \omega \end{pmatrix} = 0, $$ which becomes $$ \begin{align} \cos^2 \omega - (z^2 + \sin \omega)\sin \omega = 0 \\ \iff z^2 + \sin \omega = \cot(\omega)\cos(\omega) \\ \iff z^2 = \cot(\omega)\cos(\omega) - \sin \omega \\ \iff z^2 = \frac{\cos 2\omega}{\sin \omega} \end{align} $$ For $\omega = \pi/6$ we get $z^2 = 1$ and you can then see by inspection that $$ \begin{pmatrix}\sqrt 3 /2 & 1/2 \\ 3/2 & \sqrt 3 / 2 \end{pmatrix} $$ has $$ \textrm{ker}(\Bbb{A}(\omega)- k\Bbb{B}(k)) = \{(-w, \sqrt{3}w ) : w\in \mathbb C\} $$ which in turn implies that $T=(-1,\sqrt{3})$ satisfies $$ \Bbb A(\pi/6) T = \Bbb B(1) T $$ but also $$ \Bbb A(\pi/6) T = -\Bbb B(-1) T. $$