# Why does the attenuation constant not affect the wavelength?

Let's say I have a propagation constant ($$\gamma$$), where $$jk_z = \gamma = \alpha +j \beta$$. Why does the attenuation constant have no role in determining the wavelength, especially in a waveguide where TE/TM modes exists peacefully?

Background

1. According to the reference we can write wavelength as $$\lambda = \frac{2 \pi}{Im\{ \gamma \}}$$ (see reference)
2. Constant equation in a waveguide is $$\gamma ^2 = h_{cut}^2 - k^2$$

No Problem in TEM Mode because $$h_{cut} = 0$$, hence, $$\gamma = jk$$ and $$im \{ \gamma \} = im\{jk\} = k$$. Conclusion: $$\lambda = \frac{2 \pi}{k}$$

Problem in TE/TM Mode because $$h_{cut} \ne 0$$, hence, $$\gamma = \sqrt{ h_{cut}^2 - k^2}$$ and $$im\{\gamma \} = Im\{ \sqrt{ h_{cut}^2 - k^2} \}$$. Conclusion: $$\lambda = \frac{2 \pi}{ Im\{ \sqrt{ h_{cut}^2 - k^2} \} }$$ and sources I am finding define the phase constant ($$\beta$$) as $$\beta = Im\{ \sqrt{ h_{cut}^2 - k^2} \}$$

So why is the attenuation constant being ignored, even in wave guides?

I think your question should be asked in a different way, and the answer to that simpler question will also answer your question implicitly, namely: what is the frequency of a damped oscillation, such as $$f(t)=e^{-\alpha t } sin(\omega_0 t)$$?
First of all, this $$f(t)$$ is an exponentially unbounded signal and has no Fourier transform, so one cannot really analyze it over the full time axis but it can be analyzed for $$\alpha >0$$ as a transient from $$t=0$$ to $$t\to \infty$$,
$$F(\omega) = \mathcal F [f] = \int_0^{\infty}e^{-\mathfrak j \omega t}e^{-\alpha t } sin(\omega_0 t)dt\\ =\frac{\omega_0}{(\alpha +\mathfrak j \omega)^2+\omega_0^2}$$ This $$|F(\omega)|$$ has frequency content everywhere but where is its peak? Obviously the peak is where the denominator's modulus $$|(\alpha +\mathfrak j \omega)^2+\omega_0^2|$$ is the smallest, that is we have to find the minimum of $$(\alpha^2 + \omega_0^2 - \omega^2)^2+2\alpha^2\omega^2$$. Simple differentiation shows that happens at the frequency $$\omega = \hat \omega$$ where $$\hat \omega = \sqrt{\omega_0^2 - \alpha^2}.$$ If $$\omega_0 <\alpha$$ then the oscillation is so heavily damped that there is no clear frequency at which we can meaningfully say that it oscillates. If $$\omega_0 >> \alpha$$ then $$\hat \omega \approx \omega _0\left(1-\frac{\alpha}{2\omega_0}\right)$$, and you see that indeed the peak is dependent on the attenuation, but its relative importance decreases with their ratio and in a waveguide with metallic walls this is the normal propagating situation.