 Solution

An isolated free electron in vacuum cannot spontaneously emit a nonzero real photon. If $P=P_f+Q$, where $Q$ is the emitted photon four-momentum, $P^2=P_f^2=m^2{c}^2$ and $Q^2=0$, conservation would require $P_f\cdot Q=0$. A future-pointing timelike four-vector has strictly positive inner product with every nonzero future-pointing null four-vector, so this is impossible.

The same conclusion follows from the classical Cherenkov radiation condition $v>c/n$: in vacuum $n=1$, while the electron with $\gamma =100$ has $\beta =\sqrt{1-10^{-4}}<1$. Hence there is no vacuum emission angle for the specified free particle. Acceleration by an external field or scattering from another body would change the physical problem and can permit radiation.