= Solution
For a cold, unmagnetized, collisionless <plasma>, take an equilibrium <Electron> density $n_e$, <Electrons> of <mass> $m_e$ and charge $-e$, and approximately stationary ions. Use fields proportional to $e^{-i\omega t}$. The <Electron> <displacement> satisfies
$$
m_e\ddot{\mathbf x}=-e\mathbf E,
\qquad \mathbf x=\frac{e\mathbf E}{m_e\omega^2}.
$$
Its <electric polarization> is $\mathbf P=-n_ee\mathbf x$, hence the <cold collisionless plasma dielectric response>
$$
\boxed{\epsilon_r(\omega)=1-\frac{\omega_p^2}{\omega^2},
\qquad \omega_p^2=\frac{n_ee^2}{\epsilon_0m_e}.}
$$
The negative susceptibility arises from the inertial <Electron> response, rather than a restoring <force> binding an <Electron> to a particular atom.
A transverse <electromagnetic wave> then obeys
$$
k^2c^2=\omega^2\epsilon_r=\omega^2-\omega_p^2.
$$
For $\omega>\omega_p$, the <plasma> supports propagation, with <refractive index> $n_r=\sqrt{1-\omega_p^2/\omega^2}<1$. The <phase velocity> is greater than $c$, while the <group velocity> is $c\sqrt{1-\omega_p^2/\omega^2}<c$; their product is $c^2$, and no superluminal <signal> is implied. Below the <plasma frequency>, the bulk <wavenumber> is imaginary and the field is evanescent. A sufficiently thick <plasma> largely reflects such radiation in the lossless model.
Including an <Electron> momentum-relaxation rate $\nu_c$ replaces the denominator by $\omega(\omega+i\nu_c)$:
$$
\epsilon_r=1-\frac{\omega_p^2}{\omega(\omega+i\nu_c)}.
$$
Its <imaginary part> describes absorption with the stated <time> convention. Finite <Electron> <temperature> also produces <plasma screening> of slowly varying electrostatic fields, with $\lambda_D=\sqrt{\epsilon_0k_BT_e/(n_ee^2)}$. That spatial, longitudinal screening problem is distinct from simply taking $\omega=0$ in the cold transverse-wave formula. Magnetic bias, ion motion and kinetic resonances require more general <dielectric> responses.
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