= Solution
For a transverse electromagnetic <plane wave> in a nonmagnetic medium, <Maxwell's equations> give $c^2k^2=\omega^2\epsilon_r(\omega)$. Set <damping> to zero and define $u=c^2k^2/\epsilon_\infty$. Substitution yields
$$
u(\omega_{\rm TO}^2-\omega^2)=\omega^2(\omega_{\rm LO}^2-\omega^2),
$$
so $\omega^4-(\omega_{\rm LO}^2+u)\omega^2+u\omega_{\rm TO}^2=0$. Therefore the two <phonon polariton> branches are
$$
\boxed{\omega_\pm^2(k)=\frac12\left[\omega_{\rm LO}^2+u
\pm\sqrt{(\omega_{\rm LO}^2+u)^2-4u\omega_{\rm TO}^2}\right].}
$$
Take the nonnegative square root for the positive physical frequencies. At $k=0$, $\omega_-=0$ and $\omega_+=\omega_{\rm LO}$. At small <wavevector> the lower branch is <photon>-like with velocity $c/\sqrt{\epsilon_r(0)}$; at large <wavevector> it approaches $\omega_{\rm TO}$ from below. The upper branch ultimately becomes <photon>-like with velocity $c/\sqrt{\epsilon_\infty}$. There is no real propagating transverse branch with frequency strictly between $\omega_{\rm TO}$ and $\omega_{\rm LO}$ in this lossless bulk model. Coupling changes the eigenmodes from a bare <photon> crossing a flat <oscillator> into mixed <normal modes>.
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