Infinity as a superattractive fixed point

Question

I'm looking into filled Julia sets of functions from a family of complex rational functions of type $R(z) = z^n + \frac{P(z)}{Q(z)}$, where $n \geq 2$ and $\text{deg}(P) < \text{deg}(Q)$, and I want to apply Böttcher's theorem to show that they behave like functions $z^n$. To apply Böttcher's theorem, I need to show that infinity is a superattractive fixed point for the function $R(z)$.

I saw the idea of substituing $z$ with $\frac{1}{z}$ and inserting $0$ after rearranging , but I'm still looking for some more rigorous insight and proof.

Thank you in advance and best regards

Answer 1

The mapping $u:z\mapsto 1/z$ is an isometry on the Riemann sphere $\Bbb C\cup\{\infty\}$, see Möbius transformation. It maps 0 (the south pole) to ∞ (the north pole) and vice-versa, and it's perfectly fine to use it; nothing is non-rigorous about it.

First use $u$ to map a neighborhood of 0 to a neighborhood of ∞, then apply $R$ to get the behaviour close to ∞, and then use $u^{-1}$ to map neighborhood of ∞ back to a neighborhood of 0.

This means $r=u^{-1}\circ R\circ u$ behaves in a neighborhood of 0 like $R$ behaves in a neighborhood of ∞:

$$\begin{align} r(z) = (u^{-1}\circ R\circ u)(z) &= \frac1{z^{-n}+\dfrac{P(1/z)}{Q(1/z)}} \\ %&= \frac1{z^{-n}+\dfrac{z^qP(1/z)}{z^qQ(1/z)}} \\ %&= \frac{z^n}{1+\dfrac{z^{n+q}P(1/z)}{z^qQ(1/z)}} \\ \end{align}$$

If $p$ and $q$ denote the degree of $P$ resp. $Q$, then $z^pP(1/z)$ is a polynomial of degree $p$, and $z^qQ(1/z)$ is a polynomial of degree $q$ provided $z\nmid P(z)$ and $z\nmid Q(z)$.

If it helps you can assume $z\nmid P(z)$ and $z\nmid Q(z)$ and write the rational part as $\dfrac{P(z)}{Q(z)}x^k$ with $k\in\Bbb Z$ and $\partial P+k < \partial Q$.

Answer 2

I tried just like you wrote it, then inserted general polynomials of degree $m$ (deg($Q$) $= m$) and $r$ (deg($R$) $=r$). The only concern I have is that I may be "cancelling" $0/0$?

For $r(0)$, $$(\frac{z^n Q(\frac{1}{z})}{Q(\frac{1}{z}) + z^n P(\frac{1}{z})}),$$ I factored out $\frac{1}{z^m}$ in the numerator and the denominator and then it remains after inserting $0:\,$ $\frac{0}{q_m}=0$.

For $r'(z)$, $$(\frac{nz^{n-1} Q^2(\frac{1}{z}) - z^{2n-2}(Q'(\frac{1}{z})P(\frac{1}{z})-Q(\frac{1}{z})P'(\frac{1}{z}))}{Q^2(\frac{1}{z}) + 2z^n Q(\frac{1}{z})P(\frac{1}{z}) + z^{2n}P^2(\frac{1}{z})}),$$ I factored out $\frac{1}{z^{2m}}$ in the numerator and the denominator and then it remains after inserting $z=0$: $\frac{0}{q_m^2}$.

Were the cancellations "legal"?

More formal way is maybe like this: we write $R(z) = \frac{p(z)}{q(z)}=\sum_{k=0}^{n}a_kz^k, n\geq2; \quad u(z) = u^{-1}(z) = 1/z,$

$$r(z) = u \circ R \circ u^{-1}(z) = \frac{1}{R(1/z)} = \frac{1}{\sum_{k=0}^{n}a_k\frac{1}{z^k}}=\frac{z^n}{\sum_{k=0}^{n}a_kz^{n-k}}: r(0) = \frac{0}{a_n}=0$$

$$r'(z) = \frac{nz^{n-1}\sum_{k=0}^{n}a_kz^{n-k} - z^n \sum_{k=0}^{n-1}(n-k)a_kz^{n-k-1}}{(\sum_{k=0}^{n}a_kz^{n-k})^2} : r'(0) = \frac{0}{a_n^2} = 0.$$

Is this maybe better or I cancelled out $0/0$ again?

Sorry for the long messy posts and thanks again for your time :)