= Solution
Use <momentum conservation> and <conservation of energy> in the laboratory <inertial frame>. The <photon> has <momentum> magnitude $E/c$, while the outgoing <Electron> satisfies the <relativistic energy-momentum relation>. Its total <energy> and <momentum> obey
$$
E_e=mc^2+E_1-E_2,\qquad c^2|\mathbf p_e|^2=E_1^2+E_2^2-2E_1E_2\cos\theta.
$$
Substitute these into $E_e^2=m^2c^4+c^2|\mathbf p_e|^2$. Cancelling $m^2c^4$ and $(E_1-E_2)^2$ gives $mc^2(E_1-E_2)=E_1E_2(1-\cos\theta)$. Both photon <energies> are positive, so division yields
$$
\boxed{\frac1{E_2}-\frac1{E_1}=\frac{1-\cos\theta}{mc^2},\qquad E_2=\frac{E_1}{1+(E_1/mc^2)(1-\cos\theta)}.}
$$
This is <Compton scattering>. The energies include the <Electron>'s <rest energy>; using only its <kinetic energy> in the mass-shell equation would give the wrong result. The outgoing <photon> has unchanged <energy> at $\theta=0$ and smaller <energy> for a nonzero scattering angle.
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