= Solution
For the positive-<frequency> branch write $s=1+\delta$, $\delta>0$, and $Q=k^2V_e^2/(3\omega_{\mathrm pe}^2)=\kappa^2/3$. The <degenerate-plasma dielectric response> is equivalent to
$$
1+\frac{1+\delta}{2}\log\frac{\delta}{2+\delta}=-Q.
$$
At large $Q$, a logarithmic divergence is needed to balance the right-hand side; hence $\delta\ll1$. Keeping the constant as well as the large term gives
$$
\log\frac\delta2=-2(1+Q)+O(\delta|\log\delta|),\qquad
\boxed{\delta\sim2\exp\left[-2-\frac{2k^2V_e^2}{3\omega_{\mathrm pe}^2}\right].}
$$
More precisely the relative correction is $O((1+Q)e^{-2(1+Q)})$. The additive constant $-2$ in the exponent fixes the leading prefactor $2e^{-2}$ and should not be discarded.
The negative-<frequency> branch follows from the even real <dielectric function> outside the occupied interval. Thus
$$
\boxed{\omega_\pm=\pm kV_e(1+\delta),\qquad
\delta\sim2\exp\left[-2-\frac29(k\lambda_{\mathrm De}^{\rm exam})^2\right].}
$$
Using the physical <Thomas-Fermi screening length> instead gives $\delta\sim2e^{-2-2(k\lambda_{\rm TF})^2}$. This is <degenerate-plasma zero-point sound>: the <phase velocity> lies exponentially close to, but strictly beyond, the edge of the <Fermi sphere> for every finite $k$ in this approximation. The limit is a property of the supplied classical-trajectory <kinetic theory>; applying it at microscopic <wavelengths> would require checking quantum recoil, which that equation omits.
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