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I am doing a small project in school, which is dedicated to exploring what shapes might Julia sets of rational functions take. However, as we've started investigating into the results we've got before presenting them, it turned out that what we actually did was studying filled Julia sets for rational functions, i.e. $$\mathcal{K}(R) = \{z \in \mathbb{C} \,|\, R^n(z) \not\to \infty \text{ as } n \to \infty\};$$ and obtaining Julia sets as $$\mathcal{J}(R) = \partial \mathcal{K}(R).$$ However, it seems to be incorrect as it works only in case of $R$ being polynomial. But is it? I got totally confused at this point because all the Internet has to offer about filled Julia sets applies only to polynomials.

So my questions is whether the following statement is true: For a rational function $\mathbf{R : \hat{\mathbb{C}} \mapsto \hat{\mathbb{C}}}$ with $\mathbf{\infty}$ being an attracted fixed point of $\mathbf{R}$, the following two definitions are the same: $$\mathcal{J}(R) = \partial \mathcal{K}(R) \text{ with } \mathcal{K}(R) = \{z \in \mathbb{C} \,|\, R^n(z) \not\to \infty \text{ as } n \to \infty\}$$ and (the canonical one) : $$\mathcal{J}(R) = \hat{\mathbb{C}} \setminus \mathcal{F}(R) \text{ with } \mathcal{F}(R) \text{ being the Fatou set of } R.$$

The answer seems to be yes, but actually proving (or disproving) that requires more knowledge in area of complex dynamics than I have now. I would be really happy if somebody had already determined that (and somebody surely did, because it is crucial for approximations and computations of Julia sets) and if the answer is yes, I would just give a reference to the proof because my school does not require all proofs to be written out explicitly.

I will appreciate any help answering the question above. Thank you in advance!

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    $\begingroup$ There are rational functions whose Julia set is the whole plane. The first example was given by Lattès: $R(z)=\frac{(z^2+1)^2}{4z(z^2-1)}$. $\endgroup$
    – lhf
    Dec 18, 2019 at 10:52
  • $\begingroup$ Thank you for such an interesting example! But my assumption still seems to be correct as $\infty$ is not an attracting fixed point of $R$, since on the real number line $$\frac{(z^2+1)^2}{4z(z^2-1)} - z = \frac{-3z^4+4z^3+2z^2+1}{4z(z^2-1)} < 0 \text{ for big enough } z.$$ So in any neighborhood of infinity there is a point moving away from it and $\infty$ cannot be attracting. Is my reasoning correct? $\endgroup$
    – Aleksei Kubanov
    Dec 19, 2019 at 13:59

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Yes, this is correct. This follows from Theorem 5.2.1 of Alan Beardon's Iteration of Rational Functions which states that the boundary any component of the Fatou set which is both forward and backward invariant is exactly the Julia set. Of course, the basin of attraction of any attractive orbit is both forward and backward invariant so that your assertion generalizes. In fact, if any open set intersects the Julia set and one basin of attraction, then it must intersect all regions of attraction.

As you seem to have realized, there are rational functions for which $\infty$ isn't even fixed - let alone attractive. Generally (though not always), there is some attractive behavior. Thus a reasonable strategy is to iterate until an attractive orbit is discovered and to shade the starting point based on that. That's the strategy taken in this little rational function web app.

That strategy can't be full-proof, though. As @lhf points out, though, all periodic orbits can be repelling. It's also feasible to have parabolic behavior without attractive behavior.

Here's a fun example built into the web app illustrating the strategy: $$f(z) = \frac{0.01+z^{5}}{z^{3}}.$$ The basins of attraction looks like so:

enter image description here

There are four basins of attraction:

  • The green region converges to an attractive orbit of period 2
  • The red and yellow regions converge to different attractive orbits of period 3
  • The blue region converges to the super-attractive fixed point at $\infty$.

The filled Julia set would be the union of the red, green, and yellow regions.

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    $\begingroup$ Thank you very much for the answer! $\endgroup$
    – Aleksei Kubanov
    Dec 19, 2019 at 13:41

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