A Proof about $Aut(\Bbb{C})$ and Schwarz Lemma

Question

I have problems showing that the analytic automorphisms of $\Bbb{C}$ is the set of all linear functions by Schwarz lemma only. I assume $f \in Aut(\Bbb{C})$ satisfies $f(0)=0, f'(0)=1$ and I want to prove that $f(z)=z$. I get $\operatorname{lim}_{z \to \infty}f(z)=\infty$ and $\operatorname{inf}_{|z| \ge 1}|f(z)|=c>0$, but I got stuck here. How can I continue?

Answer 1

I don't know whether this is the correct answer or not; in my opinion, it is. I will write it below.

Let $f \in \text{Aut}(\mathbb{C})$, now let $K = \max \{ |f(z)-f(0)| : z \in \bar{\mathbb{D}}\}$. Then $\frac{f-f(0)}{K}$ is an automorphism of $\mathbb{C}$ too, and aslo an automorphism of $\mathbb{D}$. Hear I think is where you should use the Schawrz lemma, and deduce that on $\mathbb{D}$, $f$ is equal to $e^{i\theta}\frac{a-z}{1-\bar{a}z}$, for some real number $\theta$ and $ a \in \mathbb{D}$. So $f$ is the analytic continuation of that automorphism to $\mathbb{C}$. But know we know that $e^{i\theta}\frac{a-z}{1-\bar{a}z}$ also extends to holomorphic function of $\mathbb{C} \setminus \{\frac{1}{\bar{a}} \}$. So As this function and $f$ should be the same on $\mathbb{C} \setminus \{\frac{1}{\bar{a}} \}$ . thus $e^{i\theta}\frac{a-z}{1-\bar{a}z}$ should be a holomorphic function on all of $\mathbb{C}$, which implies that $a=0$. And you will get the desired result.