= Solution
Use <Gaussian units>, <plane waves> proportional to $e^{i\mathbf k\cdot\mathbf r-i\omega t}$, and the \b[signed] <cyclotron frequency> $\Omega_s=q_sB_0/(m_sc)$. Write $\Omega_e=eB_0/(m_ec)>0$ for its <electron> magnitude, so the signed <electron> value is $-\Omega_e$. Define the species <plasma frequency> by $\omega_{ps}^2=4\pi n_{0s}q_s^2/m_s$. The <linearization> of the cold <Lorentz force> equations gives
$$
-i\omega u_{xs}-\Omega_su_{ys}=\frac{q_s}{m_s}E_x,\qquad
\Omega_su_{xs}-i\omega u_{ys}=\frac{q_s}{m_s}E_y,\qquad
-i\omega u_{zs}=\frac{q_s}{m_s}E_z.
$$
Solving this <linear system> and using the species <current density> $\mathbf j_s=n_{0s}q_s\mathbf u_s$ gives the <conductivity tensor>
$$
\boxed{\boldsymbol\sigma_s=\frac{n_{0s}q_s^2}{m_s}
\begin{pmatrix}
\dfrac{i\omega}{\omega^2-\Omega_s^2}&-\dfrac{\Omega_s}{\omega^2-\Omega_s^2}&0\\
\dfrac{\Omega_s}{\omega^2-\Omega_s^2}&\dfrac{i\omega}{\omega^2-\Omega_s^2}&0\\
0&0&\dfrac{i}{\omega}
\end{pmatrix}.}
$$
The <magnetized cold-plasma dielectric tensor> is $\boldsymbol\epsilon=\mathbf I+(4\pi i/\omega)\sum_s\boldsymbol\sigma_s$. Thus
$$
\boldsymbol\epsilon=\begin{pmatrix}S&ig&0\\-ig&S&0\\0&0&P\end{pmatrix},\quad
S=\epsilon_\perp=1-\sum_s\frac{\omega_{ps}^2}{\omega^2-\Omega_s^2},\quad
P=\epsilon_\parallel=1-\sum_s\frac{\omega_{ps}^2}{\omega^2},\quad
g=-\sum_s\frac{\omega_{ps}^2\Omega_s}{\omega(\omega^2-\Omega_s^2)}.
$$
At the stated <electron> <frequency> scale, the <ion> contribution is smaller by $O(Zm_e/m_i)$ for $q_i=Ze$ and <quasineutrality>. To this accuracy,
$$
\boxed{S=1-\frac{\omega_{pe}^2}{\omega^2-\Omega_e^2},\qquad
P=1-\frac{\omega_{pe}^2}{\omega^2},\qquad
g=\frac{\omega_{pe}^2\Omega_e}{\omega(\omega^2-\Omega_e^2)}.}
$$
These expressions require negligible <collisions> and the cold ordering $kv_{\mathrm{th},s}\ll|\omega|$; sufficiently close to a <cyclotron resonance>, thermal broadening or <collisions> must also be retained.
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