# Smooth Julia set for quadratic polynomials

This question is related to a classification of rational maps in terms of properties of their Julia set.

Let $$f= z^2 + c$$, with $$c\in \mathbb{C}$$ be a quadratic polynomial such that its Julia set $$J(f)$$ is connected.

• Q1: If there exists a relatively open set of $$J(f)$$ that is (the support of) a smooth curve. Is $$f$$ conjugate (resp equal) to a Tchebychev polynomial or power map $$z^2$$?
• Q2: Is the answer to Q1 yes under the additional assumption that $$J(f)$$ is also locally connected?
• Q3: If the answer to Q1 is no, can one describe the set of such counterexamples in terms of the parameter $$c$$?

Thanks a lot.

Fatou showed that if the Julia set $$J$$ is a smooth curve, then either $$J$$ is the unit circle, or $$J$$ is a real interval. If $$J$$ is the circle, then $$f$$ is equivalent to $$z → z^d$$ , where $$d$$ is an integer with $$|d| ≥ 2$$; if $$J$$ is the interval, then $$f$$ is equivalent to a Chebyshev polynomial.
• They take $f$ arbitrary holomorphic. What's meant by "is the unit circle"? If one conjugates $z\mapsto z^2$ by an affine map, one can get any circle. – YCor Jun 21 '19 at 17:41