To get cutoff frequencies and eigenmode field distributions for a waveguide, one uses following equation: $$1/\epsilon  \times 1/\mu  \times E = \omega^2 E$$With $ \omega^2 $ as eigenvalues. This works fine. But what if I would like to calculate the wave number $k_z$ ?If the waveguide is not changing in $z$ direction, then the E-Field can be described as: $E(x,y,z) = E(x,y)e^{jk_zz}$. Following that, the upper equation becomes:$$1/\epsilon  \times 1/\mu  \times E(x,y)e^{jk_zz} = \omega^2 E(x,y)e^{jk_zz}$$But I need an eigenvalue equation whre $k_z$ is my eigenvalue. And in this case it is in the exponential, which is not good for me. 
FEM or FD eigenvalue equation to get wave number instead of cutoff frequency
eigenvalues;electromagnetism
Assuming that you have scalar permitivity and permeability, you can relate your wavenumber with your angular frequency by$$\omega = kc$$where $c = 1/(\epsilon_0 \mu_0)$ is the speed of the wave in vacuum. Then your equation would read$$1/\epsilon_r\nabla \times 1/\mu_r\nabla \times E_n(x,y) = \alpha_n^2 E_n(x,y)$$where $\epsilon_r$ and $\mu_r$ are the relative permitivity and permeability, and the wavenumber for your propagating wave is given by$$k_n^2 = k^2 - \alpha_n^2 \, .$$Then, when $k^2 > \alpha^2$ you have propagating waves. And you need to find the eigenvalues for the transverse section.