quantum field theory - How to calculate the charges of W and Z bosons?

In the Standard Model, electroweak unification is based on the symmetry breaking of $\text{SU(2)}\times \text{U(1)}_Y \rightarrow \text{U(1)}_{EM}$ by the VEV of a complex doublet $H$ with hypercharge $\frac{1}{2}$.

The Lagrangian is

$$\mathcal{L}=-\frac{1}{4}(W^a_{\mu\nu})^2-\frac{1}{4}B^2_{\mu\nu}+(D_\mu H)^\dagger(D_\mu H)+m^2H^\dagger H-\lambda(H^\dagger H)^2,$$

where $W^a_\mu$ are the $\text{SU(2)}$ gauge bosons and $W^a_{\mu\nu}$ are given by $W^a_{\mu\nu}=\partial_\mu W^a_\nu-\partial_\nu W^a_\mu+gf^{abc}W^b_\mu W^c_\nu$, and $B_\mu$ is the $\text{U(1)}$ gauge bosons and $B_{\mu \nu}=\partial_\mu B_\nu-\partial_\nu B_\mu$.

The covariant derivative is $$D_\mu H=\partial_{\mu}H-igW^a_\mu \tau^aH-\frac{1}{2}ig'B_\mu H.$$

After $H$ gets its VEV $$H_0=\begin{pmatrix} 0 \\ \frac{v}{\sqrt{2}} \end{pmatrix},$$ the mass term of the gauge bosons comes from $$(D_\mu H_0)^\dagger(D_\mu H_0)=\frac{v^2}{8}\left[ g^2(W^1_\mu)^2+g^2(W^2_\mu)^2+g'^2(g'B_\mu-gW^3_\mu)^2 \right].$$

My questions are why we define W bosons as $W^\pm_\mu=\frac{1}{\sqrt{2}}(W^1_\mu\mp iW^2_\mu)$ rather than just $W^1_\mu$ and $W^2_\mu$ respectively, and how to see the charges of $W^\pm$ and $Z$ bosons?