I'm reading Milnor's Dynamical Systems in One Complex Variable and I'm stuck with a detail in one of his proofs and would appreciate some help!
Let $F$ be an automorphism of $\overline{\mathbb{D}}$ such that $F$ is an involution with two fixed points. We have that $F'(z) = -1$ for the two fixed points. Thus, if $F$ maps $\mathbb{D}$ onto itself, neither of these fixed points can be on the boundary circle.
This is what the book says. I don't quite follow why neither of the fixed points can be on the boundary circle. I know that any automorphism of the closed disk is of the form $e^{i\theta} \frac{z-a}{1-\bar{a}z}$. So, $F'(z) = \frac{ad-bc}{(cz+d)^2} = \frac{e^{i\theta}(1+a\bar{a})}{(1-\bar{a}z)^2}=-1$. Since $z$ is a fixed point, $e^{i\theta} \frac{z-a}{1-\bar{a}z}=z$. I divide the two equations to get that $-z = \frac{(z-a)(1-\bar{a}z)}{1+a\bar{a}}$. Suppose that $|z|=1$ so that $z$ is in the boundary circle. We have $1 = |\frac{(z-a)(1-\bar{a}z)}{1+a\bar{a}}|= \frac{|\bar{z}||z-a||1-\bar{a}z|}{|1+a\bar{a}|} = \frac{|1-\bar{z}a||1-\bar{a}z|}{|1+a\bar{a}|} = \frac{|1-\bar{z}a|^2}{|1+a\bar{a}|}$ so that $|1-\bar{z}a|^2 =1+a\bar{a}$. I tried using reverse triangle inequality, but it failed.
If $z_0\in\partial \mathbb{D}$ is a fixed point of an automorphism $f$ of $\mathbb{D}$, then $f'(z_0) > 0$.
For an automorphism of the unit disk is a Möbius transformation, hence an automorphism of the entire Riemann sphere, and it maps the unit circle to itself.
So $f'(z_0)$ must map the tangent to the unit circle in $z_0$ to itself, hence $f'(z_0) \in\mathbb{R}$. If we had $f'(z_0) < 0$, $f$ would map points $(1-\varepsilon)z_0$ outside the unit circle, contradicting the assumption that it be an automorphism of the unit disk.