Symmetries of Mandelbrot sets with integer exponents

Question

I have been experimenting with recursive formulas of the form:

\begin{equation} \forall c \in \mathbb{C} , z_{n+1} = z_{n}^\alpha + c \tag{1} \end{equation}

as well as:

\begin{equation} \forall c \in \mathbb{C}, z_{n+1} = \overline{z_{n}}^\alpha + c \tag{2} \end{equation}

where $z_0 = c$, $\alpha \in \mathbb{Z}$ and $|\alpha| > 1$.

I made the following observations:

1. In case $(1)$, the resulting structure has $\alpha-1$ symmetries when $\alpha \geq 2$ and $|\alpha|+1$ symmetries when $\alpha \leq -2$.
2. In case $(2)$, the resulting structure has $\alpha+1$ symmetries when $\alpha \geq 2$ and $|\alpha|-1$ symmetries when $\alpha \leq -2$.

If you'd like to experiment with the software I used to gain more insights, I made it publicly available: https://github.com/AidanRocke/TensorFlow-Fractals

Here are a few visualisations of case $(1)$ for $\alpha \in {2,4,-2,-4}$:

So far I don't have an explanation for all four observations but I have a stability argument for case $(1)$ where $\alpha > 1$:

\begin{equation} z_{n+1} \sim z_0^{\alpha(n+1)} \end{equation}

which may be deduced by multiplying and dividing all terms in the series expansion by $z_0^{\alpha(n+1)}$. As a result, near the boundary of the circumscribing disk with radius $R$ the most stable(and therefore most distant) points are near the roots of unity:

\begin{equation} \mathcal{U}_n = \{e^{\frac{2ik\pi}{\alpha}}: k \in [0,\alpha-1]\} \end{equation}

I think this argument is sufficient but feel free to correct me if I'm wrong. As for the three other cases, I don't have a good explanation yet.

Answer 1

Given the observed geometry of these figures, we are looking for reflection and rotation symmetries where the axes of reflection and rotation coincide. Now, a rotation in the complex plane $\mathbb{C}$ is equivalent to multiplication by some $R \in \mathbb{C}$ where $|R|=1$ whereas a reflection in $\mathbb{C}$ about the axis $R=e^{i\theta}$ is equivalent to the transformation $z \mapsto \overline{z}R^2$.

1. Given the first recurrence relation:

\begin{equation} \forall c \in \mathbb{C},z_{n+1} = z_n^\alpha + c \tag{1}\end{equation}

we are looking for $R$ that simultaneously satisfy:

\begin{equation} \begin{cases} (Rz_n)^\alpha + Rc=R(R^{\alpha-1}z_n^\alpha + c)=R(z_n^\alpha + c) \\ \lvert(\overline{z_n}R^2)^\alpha+\overline{c}R^2\rvert^2=\lvert \overline{z_n}^{\alpha}R^{2(\alpha-1)}+\overline{c}\rvert^2=\lvert \overline{z_n}^{\alpha}+\overline{c}\rvert^2= \lvert z_n^{\alpha}+c \rvert^2 \end{cases} \end{equation}

So we must solve for $R$ where $R^{\alpha-1}=1$ and we find that:

\begin{equation} R \in \mathcal{U}_{\alpha-1}=\{e^{\frac{2 k\pi i}{\alpha-1}}:k \in [0,..,\alpha-2\} \end{equation}

This corresponds to $\alpha−1$ symmetries when $\alpha \geq 2$ and $|\alpha|+1$ symmetries when $\alpha \leq -2$.

2. Given the second recurrence relation:

\begin{equation} \forall c \in \mathbb{C},z_{n+1} = \overline{z_n}^\alpha + c=\left(\frac{z_n^{-1}}{|z_n|^2}\right)^\alpha + c \tag{2}\end{equation}

we are looking for $R$ that simultaneously satisfy:

\begin{equation} \begin{cases} \left(\frac{R^{-1}z_n^{-1}}{|Rz_n|^2}\right)^\alpha + Rc=R\left(\frac{R^{-\alpha-1}z_n^{-\alpha}}{|z_n|^{2\alpha}}+c \right)=R\left(\left(\frac{z_n^{-1}}{|z_n|^2}\right)^\alpha + c\right) \\ \big\lvert \left(\frac{R^{-2}\overline{z_n}^{-1}}{|z_n|^2}\right)^\alpha + \overline{c}R^2 \big\rvert^2=\big\lvert z_n^\alpha R^{-2\alpha} + \overline{c}R^2 \big\rvert^2=\big\lvert z_n^\alpha R^{-2(\alpha+1)} + \overline{c} \big\rvert^2=\big\lvert \overline{z_n}^\alpha + c \big\rvert^2 \end{cases} \end{equation}

So we must solve for $R$ where $R^{-\alpha-1}=1$ and we find that:

\begin{equation} R \in \mathcal{U}_{\alpha+1}^{-1}=\{e^{\frac{-2 k \pi i}{\alpha+1}}:k \in [0,..,\alpha]\} \end{equation}

This corresponds to $\alpha+1$ symmetries when $\alpha \geq 2$ and $|\alpha|-1$ symmetries when $\alpha \leq -2$