Both calculations are correct. 

In general, there is an isomorphism $$H^2(G, \, \mathbb{C}^{\times})=H_2(G, \, \mathbf{Z})^*$$
Now, it can happen that an infinite group is *not* isomorphic to its dual. For instance, in your case we have $$\mathbf{C}^{\times}=H^2(\mathbf{Z} \times \mathbf{Z}, \, \mathbf{C}^{\times})=H_2(\mathbf{Z} \times \mathbf{Z}, \, \mathbf{Z})^*=\mathrm{Hom}_{\mathbf{Z}}(\mathbf{Z}, \, \mathbf{C}^{\times}).$$