= Solution
Take the plane of incidence to be the $xy$-plane, so all three electric polarizations point along $z$. Write their signed scalar amplitudes as $E_I,E_R,E_T$. Tangential-$E$ continuity gives
$$
E_I+E_R=E_T.
$$
For each plane wave,
$$
H=\frac1{\mu\omega}k\times E.
$$
Continuity of its tangential component gives
$$
n_-\cos\theta_I(E_I-E_R)
=n_+\cos\theta_T E_T,
$$
where the common proportionality between $|k|$ and $n$ cancels. Solving the two equations gives the <transverse-electric Fresnel reflection coefficient>
$$
\frac{E_R}{E_I}
=\frac{n_-\cos\theta_I-n_+\cos\theta_T}
{n_-\cos\theta_I+n_+\cos\theta_T}.
$$
Using Snell's law to eliminate $\theta_T$,
$$
n_+\cos\theta_T
=\sqrt{n_+^2-n_-^2\sin^2\theta_I}.
$$
Since $\epsilon_+>\epsilon_-$ and $\mu$ is common, $n_+>n_-$. Therefore
$$
(n_+\cos\theta_T)^2-(n_-\cos\theta_I)^2
=n_+^2-n_-^2>0.
$$
The numerator is consequently strictly negative throughout
$0\leq\theta_I\leq\pi/2$, and
$$
\boxed{
\frac{|E_R|}{|E_I|}
=
\frac{
\sqrt{n_+^2-n_-^2\sin^2\theta_I}
-n_-\cos\theta_I}
{
\sqrt{n_+^2-n_-^2\sin^2\theta_I}
+n_-\cos\theta_I}
>0}.
$$
\b[Thus transverse-electric polarization has no Brewster angle in this setting.]
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