 Solution

Use $\hslash =c=1$, signature $(+---)$, and define $D_{\mu }={\partial }_{\mu }+ie{A}_{\mu }$. Under a local phase transformation,the derivative of the phase cancels the change of the potential, giving $D_{\mu }^{\prime }{\varphi }^{\prime }=e^{-ie\alpha }{D}_{\mu }\varphi$. Therefore a gauge-invariant scalar-electromagnetic Lagrangian density isThe mass term is invariant, the covariant-derivative product has canceling phases, and $F_{\mu \nu }^{\prime }=F_{\mu \nu }$ because mixed partial derivatives commute. A potential depending only on ${\varphi }^{\ast }\varphi$ may be added without changing the symmetry. The Maxwell term makes the electromagnetic field dynamical; it can be omitted when $A_{\mu }$ is a prescribed external background.

Expanding the covariant-derivative term gives the interaction linear in $e$ plus $e^2{A}_{\mu }{A}^{\mu }|\varphi {|}^2$. This quadratic term is required for exact local gauge invariance. The later part (c) deliberately gives only the linear interaction, and must be solved as that different stated Lagrangian rather than silently adding the missing term.