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Derivative at infinity on the Riemann sphere

  1. Apr 12, 2012 #1
    I am reading a recent (2003) paper, "Fatou and Julia Sets of Quadratic Polynomials" by Jerelyn T. Watanabe. A superattracting fixed point is a fixed point where the derivative is zero. The polynomial P(z) = z2 has fixed points P(0) = 0 and P(∞) = ∞ (note we are working in [itex]\hat{\mathbb{C}} = \mathbb{C}\cup\infty[/itex]). The derivative at 0 is P'(0) = 0. The author then writes "the derivative of P at ∞ is given by 1/P'(1/ζ) = ζ/2. Evaluating at ζ = 0 gives P'(∞) = 0."

    What's going on here? We seem to have conjugated P' by the Mobius map [itex]\zeta\mapsto1/\zeta[/itex]. Why? I would think, if anything, we'd have to conjugate P itself, in order to find properties of P...
    Last edited: Apr 12, 2012
  2. jcsd
  3. Apr 12, 2012 #2
    Oh, I think I've gotten somewhere.

    Suppose P(a) = a. Let φ be a Mobius map. With b = φ(a), the derivative of φPφ-1 at b is φ'(P(φ-1(b))) P'(φ-1(b)) / φ'(φ-1(b)) = φ'(P(a)) P'(a) / φ'(a) = φ'(a) P'(a) / φ'(a) = P'(a).

    So we can get P' at a fixed point a by evaluating the derivative of φPφ-1 at b = φ(a). Let P(z) = z2 + c with fixed point ∞, and φ(z) = 1/z = φ-1(z) so that φ(∞) = 0. Then φPφ(z) = φP(1/z) = φ(1/z2 + c) = z2/(cz2+1), whose derivative is 2z/(cz2+1)2, which evaluates to 0 at z =0. So P'(∞) = 0, apparently.

    I don't quite understand this. And this still doesn't explain the conjugation of P' by 1/z in the author's example. Can somebody shed some light?
  4. Apr 12, 2012 #3


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    Aside: you have the fixed points of P wrong, unless you were assuming [itex]c=0[/itex].
  5. Apr 12, 2012 #4
    Hurkyl: oops, thanks! I'll fix that.
  6. Apr 12, 2012 #5
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