We are asked to show that the following Lagrangian is invariant under the three $SU(2)$ transformations $\Phi \rightarrow \exp{({\frac{i}{2}{\alpha_j\sigma^j}}) \Phi}$, where $\Phi$ is a doublet complex scalar field$$\Phi =( \phi_1 \phi_2)^{T}$$ The given Lagrangian is
$$ \mathcal{L} = \partial_\mu\Phi^{\dagger}\partial_\mu\Phi-m^2\Phi^{\dagger}\Phi $$
I have re-written the Lagrangian as
$$ \mathcal{L} = g^{\mu\nu}\partial_\mu{\phi_i}^\dagger\partial_\nu{\phi_i}-m^2\phi^{\dagger}_i\phi_i $$
Where $i = 1,2$.
I derived $\delta\mathcal{L}=0$ using the fact that $\vec{\sigma}_j = \vec{\sigma}_j^\dagger$. My problem is finding $\delta\phi_i$ and $\delta\phi_i^\dagger$. How can I expand the exponential? And also, will I get 3 conserved currents for each complex field?
How would I calculate $$\alpha_j(j^\mu)_i = \frac{\partial\mathcal{L}}{\partial({\partial_\mu\phi_i})}\delta\phi_i+\frac{\partial\mathcal{L}}{\partial({\partial_\mu\phi^\dagger_i})}\delta\phi^\dagger_i$$
Edit 1:
Attempting to calculate $\delta\phi_i$ I expand the exponential to linear order since the parameter $\alpha_j \ll 1$. Hence $\delta\phi_i= I + i\alpha_j\sigma^j$. Assuming this is true, how does that summation look like? Would each j have a different charge?