If both the Poincaré disc model and the upper half plane model are considered as subsets of $\mathbb{C}$ rather than as subsets of $\mathbb{R}^2$ (that is, the Poincaré disc model is $\{ z \in \mathbb{C} : |z|<1\}$ and the upper half plane model is $\{ z \in \mathbb{C} : \operatorname{Im}(z)>0\}$ ), then one can use Möbius transformations to convert between the two models. The entry unit disk upper half plane conformal equivalence theorem yields that $f \colon \mathbb{C} \cup \{ \infty \} \to \mathbb{C} \cup \{ \infty \}$ defined by $\displaystyle f(z)=\frac{z-i}{z+i}$ maps the upper half plane model to the Poincaré disc model, and thus its inverse, $f^{-1} \colon \mathbb{C} \cup \{ \infty \} \to \mathbb{C} \cup \{ \infty \}$ defined by $\displaystyle f^{-1}(z)=\frac{-iz-i}{z-1}$ , maps the Poincaré disc model to the upper half plane model.

Note that the Möbius transformation $f^{-1}$ gives another justification of including $\infty$ in the boundary of the upper half plane model (see the entry on parallel lines in hyperbolic geometry for more details): $1$ (or the ordered pair $(1,0)$ ) is on the boundary of the Poincaré disc model and $f^{-1}(1)=\infty$ .

Note also that lines in the Poincaré disc model passing through $1$ (or the ordered pair $(1,0)$ ) are in one-to-one correspondence with the lines that are vertical rays in the upper half plane model.