= Solution
Write $\mu=\mu_0\mu_r$ and $\epsilon=\epsilon_0\epsilon_r$ for homogeneous scalar material constants, and use $\mathbf J=\sigma\mathbf E$. Taking the <curl> of <Faraday's law of induction> and substituting the <Ampere-Maxwell law> gives
$$
\nabla\times(\nabla\times\mathbf E)=-\mu\partial_t(\sigma\mathbf E+\epsilon\partial_t\mathbf E).
$$
The <vector calculus identity> for a double <curl> therefore yields
$$
\boxed{\nabla^2\mathbf E-\nabla(\nabla\cdot\mathbf E)=\mu\epsilon\,\partial_t^2\mathbf E+\mu\sigma\,\partial_t\mathbf E.}
$$
The second term represents <electric current> response, rather than <displacement current>. A dispersive $\sigma(\omega)$ is used in the <frequency domain>; its general time-domain counterpart is a temporal response convolution.
For the convention $e^{i(\mathbf k\cdot\mathbf r-\omega t)}$, the <plane wave> equation is
$$
\left[\mathbf k\mathbf k^T+\left(\mu\epsilon\omega^2+i\mu\sigma\omega-k^2\right)\mathsf I\right]\mathbf E_0=0.
$$
Decompose the <electric field> into components parallel and perpendicular to $\mathbf k$. With $S=\mu\epsilon\omega^2+i\mu\sigma\omega$, the longitudinal eigenvalue is $S$ and each transverse eigenvalue is $S-k^2$; thus the determinant is $S(S-k^2)^2$. For a transverse <electromagnetic wave>, $\mathbf k\cdot\mathbf E_0=0$ and
$$
\boxed{k^2=\frac{\omega^2}{c^2}+i\mu_0\sigma\omega}
$$
when $\mu_r=1$ and the background $\epsilon_r\simeq1$. A real dissipative <electrical conductivity> generally makes $k$ complex, with the decaying branch chosen for a passive medium.
For a cold, collisionless, unmagnetized <plasma>, treat ions as stationary and write $e>0$ for the magnitude of the <elementary charge>. The electron equation of motion is $-i\omega m_e\mathbf v=-e\mathbf E$, so
$$
\mathbf v=-\frac{ie}{m_e\omega}\mathbf E,\qquad
\mathbf J=-N_e e\mathbf v=\sigma(\omega)\mathbf E,\qquad
\boxed{\sigma(\omega)=\frac{iN_e e^2}{m_e\omega}.}
$$
This reactive response includes no collision-induced <Joule heating>. Substitution into the transverse <dispersion relation> gives
$$
\boxed{k^2c^2=\omega^2-\omega_p^2,\qquad \omega_p=\sqrt{\frac{N_ee^2}{m_e\epsilon_0}},\qquad n=\frac{ck}{\omega}=\sqrt{1-\frac{\omega_p^2}{\omega^2}}.}
$$
The quantity $\omega_p$ is the electron <plasma frequency>. The <refractive index> is real above this cutoff; below it, the <wave> is evanescent. The electron response is represented here by $\sigma(\omega)$, so it must not also be counted a second time in the background <permittivity>.
For a nonzero longitudinal component, taking the <dot product> of the matrix equation with $\mathbf k$ instead gives $S(\mathbf k\cdot\mathbf E_0)=0$. The same cold <plasma> response makes $S=(\omega^2-\omega_p^2)/c^2$, hence
$$
\boxed{\omega^2=\omega_p^2,\qquad \omega=\omega_p\quad\hbox{for positive frequency}.}
$$
At nonzero $k$, the remaining equation forces $\mathbf E_0$ to be parallel to $\mathbf k$, and <Faraday's law of induction> gives $\mathbf B_0=0$. These are electrostatic <Langmuir waves>, with no $k$ dependence of <angular frequency> in the zero-temperature model. Equilibrium neutrality does not forbid their perturbed <charge density>; <Gauss's law> requires such a density when $\mathbf k\cdot\mathbf E_0\ne0$.
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