= Solution
Treat $\phi$ and $\phi^*$ as independent variables in the <Euler-Lagrange equations>. Write $s=\phi^*\phi$, and assume the <scalar potential> $f(s)$ is real and differentiable. Variation with respect to $\phi^*$ gives
$$
\frac{\partial\mathcal L}{\partial\phi^*}=-f'(s)\phi,\qquad
\frac{\partial\mathcal L}{\partial(\partial_\mu\phi^*)}=\partial^\mu\phi.
$$
Consequently the equations for the <complex scalar field> and its <complex conjugate> are
$$
\boxed{\Box\phi+f'(\phi^*\phi)\phi=0,\qquad
\Box\phi^*+f'(\phi^*\phi)\phi^*=0.}
$$
No polynomial assumption is needed: only the first derivative of the <scalar potential> enters. For $f(s)=m^2s$, both reduce to the free <Klein-Gordon equation>.
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