In David Pine's Theory Of Quantum Liquids: Normal Fermi Liquids, it's said that we can find charged Fermi liquid has plasmon modes easily from Eq. (3.40), replicated as follows: $$ (\boldsymbol{q} \cdot \boldsymbol{v}_p - \omega) \delta n_{\boldsymbol{p}} + \boldsymbol{q} \cdot \boldsymbol{v}_p \delta(\varepsilon_p - \mu) \sum_{\boldsymbol{p'}} \left( f_{\boldsymbol{p} \boldsymbol{p'}} + \frac{4 \pi e^2}{q^2} \right) \delta{n}_{\boldsymbol{p}'} = 0. $$ Here $\delta {n}_{\boldsymbol{p}}$ is the quasiparticle distribution function, $f_{\boldsymbol{p} \boldsymbol{p}'}$ the forward-scattering strength between quasiparticles, $\boldsymbol{q}$ the wave vector of the collective mode, and $\boldsymbol{v}_{\boldsymbol{p}}$ is the group velocity of the quasiparticle.
But I don't really know how to proceed: the expected answer is when $\boldsymbol{q} \to 0$, we have $$ \omega = \sqrt{\frac{4 \pi N e^2}{m}}, $$ but the left-hand side of the above quantum Boltzmann equation only contains $\omega$, and not $\omega^2$, so it confuses me how we are able to get the square root. Also when $\boldsymbol{q} \to 0$, the interaction correction term is $\sim \boldsymbol{q} \cdot \boldsymbol{v}_p / q^2 \sim 1/q$ so it's divergent?
So my question is, how can we get the plasmon frequency from Eq. (3.40) in the book? This is actually problem 1 in Chapter 3, but no answers are attached to the book.