= Solution
A uniform <Lorentz boost> preserves the <plane wave> form, while changing both its <frequency> and its vector <wave amplitude>. Decompose vectors parallel and perpendicular to $\mathbf v$, so $\mathbf k=\mathbf k_\parallel+\mathbf k_\perp$. In the observer's primed coordinates,
$$
\boxed{\mathbf E'(\mathbf r',t')=\mathbf E'_0
\exp\!\left\{i\left[\mathbf k_\perp\cdot\mathbf r'_\perp+
\gamma\left(\mathbf k_\parallel-\frac\omega{c^2}\mathbf v\right)\cdot\mathbf r'_\parallel
-\gamma(\omega-\mathbf v\cdot\mathbf k)t'\right]\right\}.}
$$
The transformed <wave amplitude> is specified in part (d). The unprimed “<rest frame>” in this question is the original coordinate frame, not a <rest frame> of a vacuum light wave, which has no inertial <rest frame>.
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