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) = z(adsbygoogle = window.adsbygoogle || []).push({}); ^{2}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...

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# Derivative at infinity on the Riemann sphere

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