= Solution
The source-free <Maxwell equations> in vacuum are
$$
\nabla\cdot E=\nabla\cdot B=0,\qquad \nabla\times E=-\partial_t B,\qquad \nabla\times B=c^{-2}\partial_t E.
$$
Their real and imaginary parts are exactly
$$
\nabla\cdot W=0,\qquad \nabla\times W=(i/c)\partial_t W,\qquad W=E+icB.
$$
Put $\theta=\omega t-K\cdot x$. Differentiating gives $\nabla\cdot W=(\Omega\cdot K)\sin\theta$ and $\nabla\times W=-(\Omega\times K)\sin\theta$, while $(i/c)\partial_tW=-i\omega\Omega\sin\theta/c$. Thus the necessary and sufficient conditions are
$$
\Omega\cdot K=0,\qquad \Omega\times K=i\omega\Omega/c.
$$
Here the complex dot product is bilinear, without <complex conjugation>. Taking the dot product of the second equation with $\Omega$ gives $\Omega\cdot\Omega=0$. Crossing it with $K$ gives
$$
-|K|^2\Omega=(i\omega/c)^2\Omega,
$$
so, since $\Omega\ne0$, $|K|^2=\omega^2/c^2$. Writing $\Omega=e+icb$, the real conditions are
$$
\boxed{e\cdot K=b\cdot K=e\cdot b=0,\quad |e|=c|b|,\quad K\times e=\omega b,\quad K\times b=-\omega e/c^2}.
$$
Both physical fields are \b[linearly polarized]: $E=e\cos\theta$ and $B=b\cos\theta$ each oscillate along a fixed real direction. Combining two fields as $E+icB$ does not turn the electric field into a circularly polarized field.
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