I'm trying to show that:
For $$\mathbb{P}^1 := \mathbb{C} \cup \{\infty\}$$ it holds:
For $f \in Aut(\mathbb{P}^1)$ with $f(\infty) = w \in \mathbb{C}$ there exists a $a \in \mathbb{C} \setminus \{0\}$ and a $b \in \mathbb{C}$, s.t.:
$$\forall z \in \mathbb{P}^1: az+b = \frac{1}{f(z)-w}$$
I tried using that for $f \in Aut(\mathbb{C})$ we have that $f(z) = az + b$ but that does not really work since $f_{\big| \mathbb{C}} \notin Aut(\mathbb{C})$ in general. But can I somehow use that $f(1/z)$ has no essential singularity in 0?
Does anyone have an idea how to approach this?
For any $z \in \mathbb{C}$, $\dfrac{1}{f(z)-w} \in \mathbb{C}$, because $f(z)-w$ cannot equal $0$ and, if $z_\infty$ is the value of $z$ for which $f(z)=\infty$, we have $\dfrac{1}{f(z_\infty)-w} =0$. This means that $\dfrac{1}{f(z)-w}$ is an automrphism of $\mathbb{C}$, so equal to $az+b$.