 Solution

For bremsstrahlung, the electron scatters from the nucleus by virtual-photon exchange and emits a real photon: $e^{-}Z\to e^{-}\gamma Z$. The two lowest-order graphs have emission before or after the Coulomb scattering. One is drawn in the figure; interchanging the emission and exchange vertices gives the other. Their sum is needed for the full gauge-consistent amplitude. Reversing the charged-lepton arrows gives Positron bremsstrahlung, with the same underlying electromagnetic mechanism.