The fixed points of analytic self-maps of $\mathbb{D}$

Question

So far, I have assumed that $z_1$ is a fixed point of an analytic self map of $\mathbb{D}$. Then, I summoned the conformal self map of $\mathbb{D}$, $\phi$ to take $z_1\to 0$. It follows from Schwarz Lemma that the composition $\phi \circ f \circ \phi^{-1}$ is either a rotation or $|(\phi \circ f \circ \phi^{-1} )(z)|<|z|$. In either case, $\phi \circ f \circ \phi^{-1}$ has exactly one fixed point: zero. How do I show that if $\phi \circ f \circ \phi^{-1}$ has at most one fixed point in $\mathbb{D}$ then $f$ also has at most one fixed point?

Answer 1

Suppose that $f$ had two fixed points $z_1$ and $z_2$; find the preimages $w_1$ and $w_2$ under $\phi^{-1}$; that is, $\phi^{-1}(w_1) = z_1$ and $\phi(z_1) = w_1$. Then we compute

$$(\phi \circ f \circ \phi^{-1})(w_1) = \phi(f(\phi^{-1}(w_1))) = \phi(f(z_1) = \phi(z_1) = w_1$$

Identically, $w_2$ is a fixed point of $\phi \circ f \circ \phi^{-1}$, so $f$ can have at most $1$ fixed point.