Let $f(z)=\frac{az+b}{cz+d}$ be a Möbius Transformation.

1. Then $f$ is not defined at the point $-\frac{d}{c}$. In addition, we see that $f(z)\ne \frac{a}{c}$ for any choosable $z$ (because of our nontrivial condition). So $f$ is a holomorphic function $f:\mathbb{C}-\{-\frac{d}{c}\}\to \mathbb{C}-\left\{\frac{a}{c}\right\}$. If $c=0$, then the möbius transformation is just an affine map.
2. $f$ depends only on 1;;three complex parameters, since given $(a,b,c,d)$, we can multiply each coordinate by a nonzero scalar $\lambda$, and we get the same möbius transformation.
3. We note that the set of (non-trivial) Möbius transformations forms a monoid under composition, and by observing the coefficients we construct a homomorphism: $R:G{L}_2(\mathbb{C})\to Mö(\mathbb{C})$ which sends 1;;$R(\left[\begin{array}{cc}a& b\\ c& d\end{array}\right])=\frac{az+b}{cz+d}$. This map is surjective (look at the construction of the möbius transformation). And so this actually turns $Mö(\mathbb{C})$ into a group. One can check that the kernel of this homomorphism is $\{\left[\begin{array}{cc}\lambda & 0\\ 0& \lambda \end{array}\right]|\lambda \in {\mathbb{C}}^{\ast }\}$, and the quotient $G{L}_2(\mathbb{C})/ker(R)$ is commonly denoted 1;;$PS{L}_2(\mathbb{C})$. We can also view this as $\frac{S{L}_2(\mathbb{C})}{\{\pm 1\}}$.