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As Lasse suggests, you can draw pictures of the parameter space to help you explore the possible dynamic behavior of cubic polynomials. Any cubic polynomial is conjugate to some polynomial of the for …

Implementing Alexandre's suggestion, I found that for $\lambda\approx 0.483096 + 1.00504\ i$, we have $$ \begin{align} P_{\lambda}(z) &= (z - 1)\ (z + 1/2 - \lambda)\ (z + 1/2 + \lambda) \\ &\approx …

I don't think your initial assertion is accurate. Consider, for example, $f(z)=z^5-z-1$. If you iterate the Newton's method function $N(z) = z-f(z)/f'(z)$ from $z_0=0$, you'll quickly find an attracti …

Joe's answer and the cited paper are, indeed, quite nice. I think the basic idea is fairly self-contained and probably easier than the reference. Two complex polynomials $f$ and $g$ of the same degre …

I think it is very unusual for the $h_n$s to have a simpler formulation. Even in the case of a quadratic polynomial, the conjugating function $h$ need not be a polynomial. I'm afraid I don't have a ge …

As Alexandre points out, if every critical point of a rational function is pre-periodic, then the Julia set of the function is the entire Riemann sphere. This is stated as Theorem 4.3.1 in Beardon's …

Julia sets are all very closely related to self-similar sets - each one can be thought of as the invariant set of something like an iterated function system. Specifically, the Julia set of $f(z)=z^2 + …

Computer generated images of Julia sets tend to agree strongly with theory so we expect there must be some reason. I believe there are two main forces at work that generally mitigate the problem of r …

In spite of the potential issues arising from the fact that this function is not entire, there is a standard way to describe the components that you see in these types of pictures. Suppose that we are …