= Solution
Use $\eta_{\mu\nu}=\operatorname{diag}(1,-1,-1,-1)$ and define the physical electromagnetic source by $\partial_\nu F^{\nu\mu}=J^\mu$. For a constant infinitesimal phase $\delta\varphi=ie\epsilon\varphi$, the matter <Noether current> is
$$
\boxed{J^\mu=ie\left[\varphi(D^\mu\varphi)^*-\varphi^*D^\mu\varphi\right]
=ie(\varphi\partial^\mu\varphi^*-\varphi^*\partial^\mu\varphi)-2e^2|\varphi|^2A^\mu.}
$$
A definition using phase parameter $\delta\varphi=i\epsilon\varphi$ instead divides this expression by $e$; the Maxwell source includes the electric coupling shown here. Current signs must be tied to the printed $D_\mu=\partial_\mu-ieA_\mu$ convention.
To verify the equation of motion, vary the matter kinetic term with respect to $A_\mu$:
$$
\delta\mathcal L_{\rm matter}=-J^\mu\delta A_\mu.
$$
Varying $-F^2/4$ and integrating by parts gives $(\partial_\nu F^{\nu\mu})\delta A_\mu$. The stationary action therefore requires
$$
\boxed{\partial_\nu F^{\nu\mu}=J^\mu.}
$$
The potential has no explicit dependence on $A_\mu$ and contributes no extra source term. Antisymmetry of $F^{\nu\mu}$ implies $\partial_\mu J^\mu=0$ on the equations of motion; the same conservation law follows from the scalar equations. The source current is <gauge-invariant> although its expanded derivative and vector-potential pieces are not separately invariant.<Quantum Field Theory by David Tong>[§§6.1.1 and 6.3.2, gauge symmetry and scalar coupling; the sign here follows the source's minus-sign covariant derivative]{ref}
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