At the end of spontaneous symmetry breaking I get these mass terms:

$$W_{\mu}^{\pm}=\frac{1}{\sqrt{2}}\bigl(W_{\mu}^{1} \mp i W_{\mu}^{2} \bigr )$$

$$\mathcal{L}_{mass}=\frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{+}{W^{\mu}}^{-} + \frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{-}{W^{\mu}}^{+}$$

So I have $$M_{W^+}=g \frac{v}{2} \quad M_{W^-}=g \frac{v}{2} $$

Is it right? Or there are too many terms and it is enough:

$$\mathcal{L}_{mass}= \frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{-}{W^{\mu}}^{+} $$