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Question regarding Julia sets

  1. Nov 23, 2006 #1
    I have a question regarding Julia sets. As far as I know, they are made by graphing functions on the imaginary plane, so when you get the final graph, you have an image that seems to have a plane of symmetry; i.e, as if one part of the graph was reflected about that plane to get the whole graph.

    Is this because complex numbers occur in pairs, so if one part of the graph is the root of the equation, its mirror is the conjugate root? If anyone out there has the answer, I would appreciate it if they could enlighten me. :wink:
  2. jcsd
  3. Nov 24, 2006 #2


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    It's not clear to me what you mean by a "plane" of symmetry. Julia sets are, as you say, in the complex plane and are 2 dimensional, not 3 dimensional as they would have to be to have a "plane" of symmetry. Perhaps you meant "line of symmetry" but that is also not true. Julia sets are symmetric about the origin. That is, if (x,y) is in the Julia set, corresponding to x+ iy, then so is (-x,-y), corresponding to -(x+ iy).
    That's obvious from the definition of Julia sets: Jc is the set of complex numbers z0 such that the sequence defined by zn+1= zn2+ c, starting with z0, converges.
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