 Solution

There is no Landau damping of this branch within the completely degenerate collisionless model. Its phase velocity obeys $\omega /k>V_e>V_i$, while each parallel marginal distribution vanishes outside $|u|\le V_{\alpha }$. There are no particles at the resonant velocity, and $F_{\alpha }^{\prime }(\omega /k)=0$. The Landau contour consequently contributes no imaginary part to the dielectric function; a real root is allowed with $\boxed{\gamma =0}$.

If a small nonzero temperature supplied a tail beyond the Fermi surface, the conclusion would change: the resonant derivative would generally be nonzero. For the convention $p=\gamma -i\omega$ and a simple weakly damped root, expanding the continued dielectric function would give $\gamma =-{ϵ}_I(\omega ,k)/{\partial }_{\omega }{ϵ}_R(\omega ,k)$. This conditional formula is not needed to manufacture damping for the exact cutoff. Neither collisional attenuation nor nonlinear evolution is included in the given dispersion relation.