 Solution

In vacuum the source-free equations areAdding the divergence equations with weights $1,ic$ gives the complex divergence equation. Adding the curl equations in the same way gives $\nabla \times \mathbf{W}=(i/c){\mathbf{W}}_t$. Conversely real and imaginary parts of these two complex equations recover all four real equations.

For the cosine ansatz, differentiation yields divergence $(\mathbf{\Omega }\cdot \mathbf{K})\sin (\omega t-\mathbf{K}\cdot \mathbf{x})$ and curl $-(\mathbf{\Omega }\times \mathbf{K})\sin (\omega t-\mathbf{K}\cdot \mathbf{x})$. The time derivative is $-\omega \mathbf{\Omega }\sin (\omega t-\mathbf{K}\cdot \mathbf{x})$. Since that sine is not identically zero, the required conditions are preciselyDotting the second equation with $\mathbf{\Omega }$ gives $\mathbf{\Omega }\cdot \mathbf{\Omega }=0$ because $\omega \ne 0$. This dot product is complex bilinear, not Hermitian. Taking the cross product again with $\mathbf{K}$ and using transversality gives $-|\mathbf{K}{|}^2\mathbf{\Omega }=-{\omega }^2\mathbf{\Omega }/c^2$. Nonzero amplitude therefore requires $\boxed{|\mathbf{K}{|}^2={\omega }^2/c^2}$.

Taking real and imaginary parts yields the full conditionsThey imply $\mathbf{e}\cdot \mathbf{b}=0$ and $|\mathbf{e}|=c|\mathbf{b}|$, together with the dispersion relation. Conversely a nonzero transverse $\mathbf{e}$, this dispersion relation and $\mathbf{b}=\mathbf{K}\times \mathbf{e}/\omega$ satisfy all the conditions.

The physical fields are $\mathbf{E}=\mathbf{e}\cos (\omega t-\mathbf{K}\cdot \mathbf{x})$ and $\mathbf{B}=\mathbf{b}\cos (\omega t-\mathbf{K}\cdot \mathbf{x})$. Each oscillates along one fixed real direction, so both fields are linearly polarized. Complex encoding of the two fields here does not mean either real field is circularly polarized.