In particle physics, what is the motivation to have invariance of a defined lagrangian under a transformation of phase $e^{iα}$
of the fields?
Based on your comments on the other answers, I think you might just be looking for the straightforward answer: Charge conservation.
A derivation of Noether's theorem starts by considering a classical field transformation
$$
\phi(x) \to \phi(x) + \eta\Delta \phi(x)\;,
$$
where $\eta$ is "small," and where the transformation only changes the the Lagrange density by a total divergence:
$$
\mathcal{L}\to\mathcal{L}+\eta\partial_\mu K^\mu\;.
$$
Such a transformation is called a "symmetry," since the equations of motion are not changed.
Given the above type of symmetry transformation we have a conserved current:
$$
j^\mu = \frac{\partial \mathcal{L}}{\partial \partial_\mu\phi}\Delta \phi(x) - K^\mu\tag{1}\;.
$$
If you consider the specific transformation
$$
\phi(x)\to e^{i\alpha}\phi(x)\;,
$$
$$
\phi^*(x)\to e^{-i\alpha}\phi^*(x)\;,
$$
for a complex scalar field $\phi(x)$, and the specific Lagrange density
$$
\mathcal{L} = |\partial_\mu \phi|^2 - V(|\phi|^2)\;,
$$
then $K^\mu=0$ and the conserved current of Eq. (1) above is
$$
j^\mu = i\left(\phi\partial^\mu \phi^* - \phi^*\partial^\mu\phi\right)
$$
and the conserved charge is
$$
Q=\int d^3x j^0(x)\;,\tag{2}
$$
which is often interpreted as the electric change associated with the complex $\phi$ field. (E.g., in the context of "poor man's QED.")
The charge of Eq. (2) is conserved because
$$
\frac{dQ}{dt} = \int d^3 x \frac{dj^0}{dt} = \int d^3 x\vec \nabla \cdot \vec j = \oint d^2\vec S\cdot \vec j = 0\;,
$$
since we assume the current is localized somewhere (and so is zero far away at "infinity").