Consider the following two subsets of the complex numbers $\mathbb{C} \newcommand{\R}{\operatorname{Re}}$:
$$\mathcal{D}(0,1) = \{z \in \mathbb{C} : \lvert z\rvert<1 \} \\ \mathbb{C}_+ = \{s \in \mathbb{C} : \R (s) > 0\}$$
Then the following relations hold: $$f: \mathcal{D}(0,1)\to \mathbb{C}_+ : z\mapsto \frac{1-z}{1+z}$$ $$g: \mathcal{D}(0,1) \to \mathbb{C}_+ : z\mapsto \frac{1+z}{1-z}$$ Both of them are valid mappings since $$\R\!\bigg(\frac{1-z}{1+z}\bigg) = \frac{1}{2}\bigg(\frac{1-z}{1+z}+\frac{1-\bar{z}}{1+\bar{z}}\bigg) = \frac{1-\lvert z\rvert^2}{\lvert 1+z\rvert^2} > 0$$ $$\R\!\bigg(\frac{1+z}{1-z}\bigg) = \frac{1}{2}\bigg(\frac{1+z}{1-z}+\frac{1+\bar{z}}{1-\bar{z}}\bigg) = \frac{1-\lvert z\rvert^2}{\lvert 1-z\rvert^2} > 0$$ as a consequence of having $z \in \mathcal{D}(0,1)$ and therefore $\lvert z\rvert < 1$.
However, the denominator is different, so I was wondering which is the difference between the mappings $f$ and $g$, are they both valid?
If $z\neq0$, let $\iota(z)=\frac1z$. Then, if $z=x+yi$, with $x,y\in\mathbb R$,$$\iota(x+yi)=\frac{x-yi}{x^2+y^2}=\frac x{x^2+y^2}-\frac y{x^2+y^2}i.$$But then $z\in\mathbb C_+\implies\iota(z)\in\mathbb C+$ and you have both $g=\iota\circ f$ and $f=\iota\circ g$.
In particular, each of the maps can be obtained as the composition of a bijection of $\mathbb C_+$ onto itself with the other map.