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This question is motivated by the coloring schemes of the Mandelbrot fractal, namely that the color is determined by the points outside the set, and is proportional to the number of iterations $n_c$ it takes for the mapping, $z_{n+1} = z_{n}^2 + c$ to escape the closed disk $|z_{n_c}|>2$ for a fixed $c\in\mathbb{C}, z_0=0$.

It's observed (and perhaps easy to show?), that the closer one gets to the boundary of the set the larger $n_c$ is. Is there an explicit construction for a complex number $c$ that has an arbitrarily large $n_c$? If one exists, is there one that doesn't involve a perturbation off the real axis?

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If one exists, is there one that doesn't involve a perturbation off the real axis?

Why perturb at all? It's easier to analyse if you stay on the real axis.

By consideration of the roots of $z^2 - z + c$, the easy construction is $c = \tfrac14 + \varepsilon$. Without the $\varepsilon$ it would converge to $z \to \tfrac12$; $$(\tfrac12 - \delta + \alpha\varepsilon + O(\varepsilon^2))^2 + \tfrac14 + \varepsilon = \tfrac12 - (\delta - \delta^2) + (\alpha + 1 - 2\delta\alpha)\varepsilon + O(\varepsilon^2)$$ so $n_c \approx \tfrac{3}{2\varepsilon}$

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  • $\begingroup$ Thanks for the answer! Why did you consider the roots to the polynomial $z^2 -z+c$ instead of $z^2+c$? $\endgroup$
    – Hooked
    Commented Aug 20, 2019 at 13:35
  • $\begingroup$ @Hooked, one way of seeing it is that $z = z^2 + c$ rearranges to $z^2 - z + c = 0$. Another way is to start by looking at the iteration as bouncing between the lines $y = z^2 + c$ and $y = z$. We want a tiny gap between the lines so that it takes a lot of small steps to get past it. $\endgroup$
    – Peter Taylor
    Commented Aug 20, 2019 at 13:38
  • $\begingroup$ Thanks for the comment, the answer makes a lot more sense now! $\endgroup$
    – Hooked
    Commented Aug 21, 2019 at 16:17

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