= Solution
Use the time dependence actually specified, $e^{i\omega t}$, and $\mathbf B=B\hat{\mathbf z}$. The component equations are
$$
(\tau^{-1}+i\omega)v_x=-\frac e{m^*}E_x-\omega_cv_y,\qquad
(\tau^{-1}+i\omega)v_y=-\frac e{m^*}E_y+\omega_cv_x,
\quad\omega_c=\frac{eB}{m^*}.
$$
Adding the second equation multiplied by $\pm i$ diagonalizes the response. With the literal definitions $v_\pm=v_x\pm iv_y$, $E_\pm=E_x\pm iE_y$, and $J_\pm=-ne v_\pm$,
$$
\boxed{\sigma_\pm=\frac{ne^2/m^*}{\tau^{-1}+i(\omega\mp\omega_c)}
=\frac{\sigma_0}{1+i(\omega\mp\omega_c)\tau},\qquad\sigma_0=\frac{ne^2\tau}{m^*}.}
$$
\b[The source's printed $\omega\pm\omega_c$ reverses the helicity labels for these definitions.] It is recovered by naming the two circular components oppositely, reversing $B$, or changing the matching time/<polarization> convention. As a concrete check, with $E_+\ne0$, $E_-=0$ and positive $B$, the collisionless component equation resonates at $\omega=\omega_c$. The printed plus-labeled denominator would instead remain proportional to $2i\omega_c$. The physical result is unambiguous: only the <circular polarization> rotating with the electron orbit has positive-frequency <cyclotron resonance>.
For the convention used here, the dielectric eigenvalues are $\epsilon_\pm=\epsilon_\infty+\sigma_\pm/(i\epsilon_0\omega)$. In the lossless limit,
$$
\boxed{\epsilon_\pm=\epsilon_\infty-\frac{\omega_p^2}{\omega(\omega\mp\omega_c)},\qquad
\omega_p^2=\frac{ne^2}{\epsilon_0m^*}.}
$$
These eigenvalues determine propagation and reflection in the Faraday geometry; changing the helicity names swaps their subscripts without changing any optical prediction.
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