= Solution
For $\Delta\phi=-i\phi$ and $\Delta\phi^*=i\phi^*$, the interaction is invariant under the constant phase because every product has net zero phase. Applying the Noether formula to the two independent fields gives
$$
\boxed{J^\mu=i(\phi^*\partial^\mu\phi-\phi\partial^\mu\phi^*)
-2eA^\mu|\phi|^2.}
$$
The sign corresponds to the printed transformation $\phi\to e^{-i\epsilon}\phi$. Reversing the phase parameter would reverse the entire current, with no change in conservation.
The derivative products in the <divergence> of the first term cancel pairwise:
$$
\partial_\mu J^\mu=i(\phi^*\Box\phi-\phi\Box\phi^*)
-2e\left[(\partial_\mu A^\mu)|\phi|^2+A^\mu\partial_\mu|\phi|^2\right].
$$
The two field equations from part (c) imply
$$
\phi^*\Box\phi-\phi\Box\phi^*
=-2ie\left[(\partial_\mu A^\mu)|\phi|^2
+A^\mu\partial_\mu|\phi|^2\right].
$$
Multiplication by $i$ exactly cancels the remaining background terms, proving \b[$\partial_\mu J^\mu=0$] without requiring $A_\mu$ to be constant or <divergence> free. The global phase symmetry survives even though this truncated interaction is not exactly invariant under arbitrary local <gauge transformations>.
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