Show that if $deg(R)=1$, then every indifferent cycle is in $F(R)$

Question

Let $R$ be a rational function with $deg(R)=1$. Then every indifferent cylce is in the Fatou set $F(R)$.

If $R(z)=az+b$, then $z=az+b \Leftrightarrow (a-1)z+b=0$. So in the first case $a=1, b=0$, so $R$ is the identity and obvious normal. However I am not sure how to deal with the case $R(z)=az+b+\frac{c}{z}$. Thanks for your help.

Answer 1

The degree of a rational map is defined to be the maximum of the degrees of the numerator and the denominator when it is put in the form $R(z)=p(z)/q(z)$. Thus, your map $$R(z) = az+b+c/z = \frac{a z^2+b z+c}{z}$$ is not of degree 1. Indeed, if $a=2$, $b=2$, and $c=1$ then $-1$ is a neutral fixed point in the Julia set.

The rational maps of degree 1 are exactly the Mobius transformations $$\varphi(z) = \frac{az+b}{cz+d}.$$ To analyze the dynamics of a general Mobius transformation $\varphi$, consider two cases:

1. $\varphi$ has exactly one fixed point on the Riemann sphere.
   In this case, conjugate $\varphi$ to send the fixed point to $\infty$ to see that $\varphi$ is conjugate to a shift.
2. $\varphi$ has exactly two fixed points on the Riemann sphere.
   Now, conjugate $\varphi$ to send one fixed point to zero and the other to $\infty$. Conclude that $\varphi$ is conjugate to $z\to az$.

In both cases, it's not hard to show that the family of iterates of the canonical version is normal.

I suppose we could extend this result on fixed points to cycles using the fact that $\varphi^n$ is again a Mobius transformation.

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One other comment: The dynamics of degree one maps are fairly easy to characterize completely and really display no chaotic behavior. For this reason, the theory of Fatou and Julia typically considers maps of only degree two and above. Thus, it sounds a little odd to say that indifferent cycles lie in the Fatou set.