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How would I find the filled Julia set for $f(z)=z^3$? I know it should be the filled unit circle, but I don't quite understand the math. This is what I have so far: Fixed points $z^3=z$ so $z=1,-1,0,\infty$ $f'(1)=3>1$--repelling $f'(-1)=3>1$--repelling $f'(0)=0<1$--super attracting $f'(\infty)=0<1$--super attracting

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One definition says that the filled Julia set is the set of points that have bounded orbit.

It is clear that $z$ has bounded orbit under $f$ iff $|z|\le1$.

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  • $\begingroup$ For this example it is clear, but what about for something like f(z)=z^3-3z? $\endgroup$ – user121955 Dec 12 '15 at 3:02
  • $\begingroup$ @user121955, cubic Julia sets have been studied by Branner and Hubbard. For a quick picture, ask WA. See also mathoverflow.net/questions/177270/examples-of-cubic-julia-sets. $\endgroup$ – lhf Dec 12 '15 at 9:28
  • $\begingroup$ @lhf Would you believe that the Julia set of $f(z)=z^3-3z$ is the interval $[-2,2]$! In fact, the dynamics of $f$ on $\mathbb C\setminus [-2,2]$ are semi-conjugate to $g(z)=z^3$ on the exterior of the unit disk via $\varphi(z)=z+1/z$. Of course, finding such a simple Julia set is highly unusual and you are completely correct in pointing out that we should often use a computer to sketch it. $\endgroup$ – Mark McClure Dec 17 '15 at 21:49

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