= Solution
{c}
Use time dependence $e^{-i\omega t}$ and assume the dopants supply $N$ mobile <electrons> per unit volume. The <Drude model> equation is
$$
m_e^*(\ddot{\mathbf x}+\gamma\dot{\mathbf x})=-e\mathbf E_0e^{-i\omega t},\qquad
\mathbf x_0=\frac{e\mathbf E_0}{m_e^*(\omega^2+i\gamma\omega)}.
$$
The free-carrier <electric polarization> is $\mathbf P_c=-Ne\mathbf x$, so, after adding the bound-electron background,
$$
\mathbf D=\epsilon_0\epsilon_\infty\mathbf E+\mathbf P_c
=\epsilon_0\left[\epsilon_\infty-\frac{Ne^2}{\epsilon_0m_e^*(\omega^2+i\gamma\omega)}\right]\mathbf E.
$$
Therefore
$$
\boxed{\epsilon_r(\omega)=\epsilon_\infty\left(1-\frac{\omega_p^2}{\omega^2+i\gamma\omega}\right),\qquad\omega_p^2=\frac{Ne^2}{\epsilon_0\epsilon_\infty m_e^*}.}
$$
The symbol $\omega_p$ here is the background-screened <plasma frequency>; the unscreened carrier frequency is $\sqrt{\epsilon_\infty}\,\omega_p$. For positive $\gamma$, the <imaginary part> of $\epsilon_r$ is positive in this time convention, consistent with absorption. Incomplete dopant ionization would replace the nominal doping density by the actual mobile <carrier density>.
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