Phase and rescale transformation of k-g field

[ChrisVer]

We have the KG Lagrangian density:
$\mathcal{L}= \partial_{\mu} \phi \partial^{\mu} \phi^* - m^2 \phi \phi^*$

Suppose instead of taking $\phi \rightarrow e^{ia}\phi$ with $a \in \mathbb{R}$ we take $a \in \mathbb{C}$.
Then is it true to say that $\mathcal{L}'= e^{-2~ Im(a)} \mathcal{L}$ and so gives the same equations of motion?
What is to happen if I also allow for local transformations $a \equiv a(x)$? I guess this again depends on the KG Lagrangian:

$|\partial_\mu \phi|^2 \rightarrow \Big( e^{ia} \partial_\mu \phi + i e^{ia} \phi (\partial_\mu a) \Big)\Big( e^{-ia^*} \partial^\mu \phi^* - i e^{-ia^*} \phi^* (\partial^\mu a^*) \Big)$
$=e^{-2 ~Im(a)} \Big( \partial_\mu \phi \partial^\mu \phi^* + i \partial_\mu a (\phi \partial^{\mu} \phi^* -\phi^* \partial^{\mu} \phi) + \phi \phi^* \partial_\mu a \partial^\mu a \Big) = e^{-2 ~Im(a)} [D_\mu \phi) (D^\mu \phi)^*]$

The mass term is also going to give a $e^{-2~ Im(a)}$ as a common factor, so I can drop it out the Lagrangian? I think I cannot, because in this case the kinetic term of the E-L equations will act on this exponential as well...

Where in this case the $D_\mu$ shall differ from the standard one, since it will have to give a complex vector field as a connection and additionally the vector's transformation is going to bring also some rescaling it it: $A'_\mu \propto A_\mu - \partial(Im(a)) + i \partial(Re(a))$
Is that correct?

 

[LightHero]

Yes, that is correct. The kinetic term of the Euler-Lagrange equation will act on the exponential factor as well. The D_μ will differ from the standard one since it will have to give a complex vector field as a connection and additionally its transformation will bring a rescaling: A'_μ ∝ A_μ - ∂(Im(a)) + i∂(Re(a)).