Consider the functions $f_c(z) := z^2 + c$ for $c \in \mathbb C$. For each such function, we may form the associated Julia set. My question: If $c, c' \in \mathbb C$ produce in this way the same Julia set, does this imply $c = c'$?

Trivially this is the case if we "consider one more dimension" by taking orbits into account. But if we consider the Julia set only, I can't find the solution.


You can deduce a positive answer to your question from Theorem 1 in the following paper: P. Atela, J. Hu, Commuting polynomials and polynomials with same Julia set, Internat. J. Bifur. Chaos Appl. Sci. Engrg. 6 (1996), no. 12A, 24 27–2432 (preprint available here: https://arxiv.org/pdf/math/9504210.pdf)

The theorem says in particular that, for two centered polynomials of degrees $n,m \geq 2$ respectively, if these polynomials have the same Julia set which is not a circle or an interval, then (up to a symmetry) they are both iterates of the same polynomial. `Centered' means that there is no term of degree $n-1$, resp. $m-1$ (the corresponding coefficients are zero).

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    $\begingroup$ The comment under the question appeared just while I was typing the details... $\endgroup$ – Margaret Friedland Aug 21 '18 at 21:00
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    $\begingroup$ I was too lazy to write up a full answer - all the credit to you! $\endgroup$ – Jack Burkart Aug 21 '18 at 21:03
  • $\begingroup$ I'll accept the answer as soon as I have absorbed the paper. $\endgroup$ – AlgebraicsAnonymous Aug 22 '18 at 6:56
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    $\begingroup$ The result is also proved in acadsci.fi/mathematica/Vol12/vol12pp229-236.pdf, 9 years before Atela and Hu. $\endgroup$ – Alexandre Eremenko Aug 22 '18 at 13:25
  • $\begingroup$ @AlexandreEremenko, you could upgrade your comment to an answer. Things do slip out of memory. $\endgroup$ – Margaret Friedland Aug 22 '18 at 18:13

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