 Solution

Put $r=v+\sigma /\sqrt{2}$. Direct differentiation of $\phi =r{e}^{i\pi /f}$ cancels the mixed radial-angular terms, leavingUse $\lambda =2m^2/v^2$ from part b to expand the potential. The complete local expression isThe radial field is canonically normalized, giving $m_{\sigma }^2=2m^2$. For arbitrary $f$, the canonically normalized angular field is ${\pi }_c=\sqrt{2}v\pi /f$ and has zero mass:The interactions comprise radial cubic and quartic self-couplings, a derivative $\sigma \pi \pi$ coupling and a derivative $\sigma \sigma \pi \pi$ coupling. In canonical variables the latter coefficients are $1/(\sqrt{2}v)$ and $1/(4v^2)$ respectively. The additive constant shifts the vacuum energy and does not affect these masses or interactions.

The original phase symmetry becomes the exact shift $\pi \mapsto \pi +f\alpha$, with $\sigma$ unchanged. Any undifferentiated polynomial ${\pi }^3$ or ${\pi }^4$ would violate this shift, so such terms are absent. This argument allows derivative interactions; it forbids an angular potential. The massless angular excitation is the Goldstone boson for the broken global $U(1)$.