The fixed points of a rational function of multiplicity 2 or more are in the Julia Set

Question

Definition: The multiplicity of a fixed point $z_0$ of $f$ is the order of the zero the the function $f(z)-z$ at $z_0$.

I have already shown the iff statement:

Claim: The multiplicity of fixed $z_0$ is $m\geq 2$ $\iff$ $f'(z_0)=1$

and somehow I am to show that if $f$ is rational and fixed $z_0$ has multiplicity $m\geq 2$, then $z_0$ is in $\mathcal{J}$, the Julia set of $f$.

Going with the what I thought was obvious, I have that

$$f(z)-z = (z-z_0)^mg(z),\text{ $g(z)$ is analytic and $g(z_0)\neq 0$.}$$

but also, since $f$ is rational RHS must be rational. So $g(z)$ is a rational function $P/Q$ and the factorization of $P$ contains no copies of $(z-z_0)$.

I have tried computing spherical derivatives of the iterates of $f$ to show that the are not bounded on a neighborhood of $z_0$, but that gets ugly. I have tried conjugating the RHS by $\phi(z) = z+z_0$ to make $$\phi^{-1}(f(\phi(z))-\phi(z)) = z^mg(z+z_0)+z$$ But, I don't see how this is useful immediately. One thing I notice is that my approaches have failed to really make use of the fact that $f$ is rational. I feel as though there is a key fact that I am missing and that this statement should be obvious, so I am reaching frustration. A hint would be much appreciated.

Answer 1

This is a consequence of the Leau-Fatou Flower theorem, which is covered very well in section 7 of this version of Milnor's famous notes. I'm not certain if there is a more elementary way to prove it and application of the flower theorem is overkill, but it definitely does the job.

Answer 2

I have a result.

Without loss of generality let $0$ be the fixed point of $f$ with multiplicity $\geq 2$. Also, without loss of generality assume that $f'(z)\neq 0$ for $|z|<1$. For we know that $f'(0)=1$ and that $f$ is meromorphic. And so there is a neighborhood around $0$ such that $f'(z)\neq 0$ on this neighborhood. We could conjugate $f$ to make this neighborhood large enough to contain $\mathbb{D}$.

Suppose by way of contradiction that $f$ is a self-map of $\mathbb{D}$. Then $f$ is a self-map of $\mathbb{D}$, $f$ is conformal on $\mathbb{D}$, and $f$ fixes $0$. By a consequence of Schwarz Lemma it follows that $f$ is a rotation. We know that $f$ is not a rotation. So $f$ is not a self map of $\mathbb{D}$.

How does this look? Does this force $0$ to be in Julia set?