Sinusoidal plane-wave solutions

Example

Let $c$ be the speed of light, let $k \in \R^3$ be a nonzero wave vector, and put $\omega = c \abs{k}$. Let $E_0 \in \C^3$ be a complex amplitude with $k \cdot E_0 = 0$. Define the electric field and the magnetic field $$E(x, t) = \mathrm{Re}\big( E_0 \, e^{i(k \cdot x - \omega t)} \big), \qquad B(x, t) = \frac{1}{\omega} \, k \times E(x, t).$$ These are sinusoidal plane waves that solve the source-free Maxwell's equations.

Check, writing $e = e^{i(k \cdot x - \omega t)}$:

1. $\nabla \cdot E = \mathrm{Re}(i \, k \cdot E_0 \, e) = 0$ and $\nabla \cdot B = 0$, because $k \cdot (k \times E) = 0$.
2. $\nabla \times E = \mathrm{Re}(i \, k \times E_0 \, e)$, and $-\partial_t B = -\frac{1}{\omega} k \times \mathrm{Re}(-i \omega E_0 e) = \mathrm{Re}(i \, k \times E_0 \, e)$.
3. $\nabla \times B = \frac{1}{\omega} \mathrm{Re}\big(i \, k \times (k \times E_0) \, e\big) = -\frac{\abs{k}^2}{\omega} \mathrm{Re}(i E_0 e)$, using $k \times (k \times E_0) = k (k \cdot E_0) - \abs{k}^2 E_0$; and $\frac{1}{c^2} \partial_t E = -\frac{\omega}{c^2} \mathrm{Re}(i E_0 e)$, which is the same since $\omega = c \abs{k}$.

So each component solves the electromagnetic wave equation, and $E$, $B$ and $k$ are mutually orthogonal with $\abs{B} = \abs{E} / c$.