 Solution

Choose the positive axis from the approaching source towards the observer. The photon that reaches the observer is emitted forward, with source-frame four-momentum $(E/c,E/c,0,0)$. Boosting to the observer frame gives $E_{\mathrm{obs}}=\gamma E(1+\beta )$, where $\beta =v/c$. Since photon energy is $h_{\mathrm{P}}\nu$,This is the approaching-source relativistic Doppler effect. A photon emitted in a general source-frame direction would instead have the factor $\gamma (1+\beta \cos {\theta }_{\ast })$; forward emission is the geometry of the photon directed to the observer here.