The automorphisms of the open unit disk $\mathbb{D}=\{z \in \mathbb{C}:|z|<1\}$ are precisely the analytic bijections on $\mathbb{D}$ which have the form $$\varphi(z)=\lambda \frac{a-z}{1-\bar{a} z}.$$
where $\lambda$ is a unimodular constant and $a$ is a point in $\mathbb{D}$. These automorphisms form a group under composition denoted by Aut(D). Every element of Aut(D) has two fixed points (counting multiplicity), and thus can be classified by the location of the fixed points:
- elliptic: one fixed point in $\mathbb{D}$ and one in the complement of $\overline{\mathbb{D}}$;
- parabolic: one fixed point on the unit circle $\partial \mathbb{D}$ (of multiplicity 2);
- hyperbolic: two distinct fixed points on $\partial \mathbb{D}$.
Two disk automorphisms $\varphi$ and $\psi$ are conformally equivalent if there exists a disk automorphism $\tau$ for which $\psi=\tau \circ \varphi \circ \tau^{-1}$.
As Composition Operators on Spaces of Analytic Functions by C.Cowen says in Page59, exercise 2.3.5:
Let $\varphi$ be an elliptic disk automorphism with fixed point $a$ in $\mathbb{D}$. Then $\varphi$ is conformally equivalent to $\psi(z)=\lambda z$ where $\lambda=\varphi^{\prime}(a)$.
Let $\varphi$ be a hyperbolic disk automorphism. Then $\varphi$ is conformally equivalent to $\psi(z)=\frac{z+r}{1+r z}$ for some $r \in(0,1)$.
Let $\varphi$ be a parabolic disk automorphism. Then $\varphi$ is conformally equivalent to either $\psi_1(z)=\frac{(1+i) z-1}{z+i-1}$ or $\psi_2(z)=\frac{(1-i) z-1}{z-i-1}$.
As for the elliptic type, since the fixed point belongs to the open disk, it's not that hard to prove. But for the other two types, I have no idea how to prove them because the fixed points belong to $\partial \mathbb{D}$ and I have searched on google and StackExchange without finding any articles or books which are relevant to these two propositions.
Any hints or any articles or books (which is available online) will be highly appreciated! Thanks.