= Solution
{c}
For a <photon>, $m=0$ and $u=c$, so the result from (i) reduces to
$$
\boxed{V=c\cos\phi,\qquad \boldsymbol v_O=-c\cos\phi\,\mathbf e_x.}
$$
This is <relativistic aberration> for a photon initially transverse to the boost. It also follows directly from $p'_x=\gamma VE/c^2$ and $E'=\gamma E$: the observed direction cosine is $cp'_x/E'=V/c$. The transverse observed speed is $c/\gamma=c\sin\phi$, and the total photon speed is still $c$. If desired the observed photon energy is $E'=E/\sin\phi$ for $0<\phi\le\pi/2$.
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