 Solution

The source-free Maxwell equations in vacuum areTheir real and imaginary parts are exactlyPut $\theta =\omega t-K\cdot x$. Differentiating gives $\nabla \cdot W=(\Omega \cdot K)\sin \theta$ and $\nabla \times W=-(\Omega \times K)\sin \theta$, while $(i/c){\partial }_{t}W=-i\omega \Omega \sin \theta /c$. Thus the necessary and sufficient conditions areHere 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 $K$ givesso, since $\Omega \ne 0$, $|K{|}^2={\omega }^2/c^2$. Writing $\Omega =e+icb$, the real conditions areBoth physical fields are 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.