= Solution
Let $v=\sqrt{-\ln\lambda/\lambda}>0$ and write $\phi=v+\chi$. Then
$$
|\phi|^2=v^2+v(\chi+\chi^*)+\chi^*\chi.
$$
Since $V'(v^2)=0$ and $V''(v^2)=\lambda$, the quadratic potential is
$$
V=V_{\min}+\frac\lambda2v^2(\chi+\chi^*)^2+O(\chi^3).
$$
Therefore
$$
\boxed{\mathcal L^{(2)}=\partial_\mu\chi^*\partial^\mu\chi-\frac{\lambda v^2}{2}(\chi+\chi^*)^2-V_{\min}}.
$$
The constant can be dropped from the fluctuation equations. For canonical real fields, use $\chi=(h+i\pi)/\sqrt2$:
$$
\mathcal L^{(2)}=\frac12(\partial h)^2+\frac12(\partial\pi)^2-\frac12m_h^2h^2,
\qquad
\boxed{m_h^2=2\lambda v^2=-2\ln\lambda>0,\quad m_\pi^2=0}.
$$
These are the <radial and angular masses of a phase-invariant scalar potential>. The radial <Higgs mode> changes the field magnitude and costs <energy> to second order. The angular <Goldstone boson> changes the <phase> along the vacuum circle without a quadratic restoring <force>. Its linear dispersion is $\omega^2=|\mathbf k|^2$, whereas the radial mode has $\omega^2=|\mathbf k|^2+m_h^2$. The zero angular <mass> follows the continuous broken global <symmetry>, not from an accidental cancellation in a particular coordinate choice.
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