= Solution
For a constant real parameter, $\partial_\mu(e^{i\alpha}\phi)=e^{i\alpha}\partial_\mu\phi$. The phases therefore cancel separately in the <kinetic term> and in $\phi^*\phi$, leaving any $f(\phi^*\phi)$ unchanged. This is the <global phase symmetry of a complex scalar field>, with <circle group> $U(1)$.
Use infinitesimal variations per unit parameter, $\delta\phi=i\phi$, $\delta\phi^*=-i\phi^*$. The <Lagrangian density> is exactly invariant, so no total-derivative correction is needed in <Noether's theorem>. The <Noether current> is
$$
\boxed{J^\mu=i(\partial^\mu\phi^*)\phi-i\phi^*\partial^\mu\phi.}
$$
It depends only on derivatives of the <Lagrangian density> with respect to <field> derivatives, so the derivative-free <scalar potential> cannot alter its form. Direct differentiation cancels the mixed first-derivative terms:
$$
\partial_\mu J^\mu=i(\Box\phi^*)\phi-i\phi^*\Box\phi
=-if'(s)\phi^*\phi+if'(s)\phi^*\phi=0.
$$
Thus it is conserved on solutions of the <Euler-Lagrange equations>. Its spatial integral is a conserved <Noether charge> when the <flux> through the boundary vanishes. Its overall sign is fixed here by $\delta\phi=+i\phi$; reversing the orientation of the phase transformation reverses the current.
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