# Perfect circles in the Mandelbrot set?

It is known that the boundary of the period 2 hyperbolic component of the Mandelbrot set is a perfect circle of radius $\frac{1}{4}$ centered at $-1$. Moreover it is known that the boundaries of the circle-like period 3 hyperbolic components are not perfect circles ("A Parameterization of the Period 3 Hyperbolic Components of the Mandelbrot Set", Dante Giarrusso, Yuval Fisher).

Question: is the period 2 component the only hyperbolic component in the Mandelbrot set whose boundary is a perfect circle?

Similarly, is the period 1 component the only hyperbolic component in the Mandelbrot set whose boundary is a perfect cardioid?

I suspect the answer to both questions is "yes", but haven't found any conclusive references.

• For the first question, it appears that there is a sequence of circular components converging to the so-called Myreberg point. I don't recall seeing a proof that they are circular, though, anywhere. Jul 12 '16 at 17:26
• @YiannisGalidakis An easy numerical experiment shows that the disk-like component of period 4 centered at $c\approx1.32$ just to the left of the period two bulb is not a perfect circle. I'm confident that Claude is correct in his conjecture but I don't have a definitive argument either. :) Jul 12 '16 at 18:19
• @MarkMclure: I hear you. Can you please elaborate on the numerical experiment which shows that the period 4 bulb is not a circle? Jul 12 '16 at 18:46
• @YiannisGalidakis lpaste.net/7577354575468822528 is my program performing the numerical experiment, output at i.imgur.com/J7uFX5y.png Jul 12 '16 at 19:01
• Maybe is it possible to show that degree of equation defining boundary is greater then 1 so it cannot be a circle.
Jul 18 '16 at 16:09

Here's something I tried, thanks to Adam's comment for the basic idea for $p \ge 3$. This answer is missing some details, comments suggesting improvements are welcome, as would be other answers that fill in the gaps. It also relies on the fact (?) that:

$$\forall 0 \neq a \in \mathbb{C}, 0 \neq b \in \mathbb{C}, 1 < m \in \mathbb{N} . \exists x . |a e^{i x} + b e^{i m x}| \neq 1$$

Let $F(z, c) = z^2 + c$ with $F^{p+1}(z, c) = F^p(F(z, c), c)$. Now the boundary of a hyperbolic component can be parameterized by $\theta \in \mathbb{R}$ by the solution of the equation system: $$F^p(z,c) = z \\ \frac{\partial}{\partial z}F^p(z,c) = e^{i \theta}$$ Now the question reduces to showing $c$ is of the form $c = c_0 + r_0 e^{i \phi}$ where $c_0 \in \mathbb{C}$ and $r_0 \in \mathbb{R}$ are constants and $\phi \in \mathbb{R}$.

$F^p(z,c) = z$ defines a polynomial of even degree $P(z) = 0$, whose constant coefficient is the product of its roots and is a polynomial in $c$ of degree $2^{p-1}$. The roots include those of $F^q(z, c) = z$ where $q | p$. Also, $\frac{\partial}{\partial z}F^p(z, c) = 2^p \Pi z_k$ where the $z_k$ are the $p$ roots in the periodic orbit of the desired solution $z$ (all $z_k$ are roots of $F^p(z, c) = z$, the remaining roots have lower period).

Case $p = 1$: $$z^2 + c_0 + r_0 e^{i \phi} = z \\ \therefore z = \frac{1 \pm \sqrt{1 - 4(c_0 + r_0 e^{i \phi})}}{2} \\ \frac{\partial}{\partial z} = 2 z = e^{i \theta}$$

Now $|e^{i \theta}| = 1$ but $\exists x . |2 z| = |1 \pm \sqrt{x}| \neq 1$, so conclude that period $1$ component is not a perfect circle.

Case $p = 2$:

The equations reduce to $$4(1 + c_0 + r_0 e^{i \phi}) = e^{i \theta}$$ with obvious solution $c_0 = -1, r_0 = \frac{1}{4}, \phi = \theta$, so conclude that the period 2 component is a perfect circle.

Case $p = 3$:

The equations reduce to $$8 (c^3 + 2c^2 + c + 1) = e^{i \theta} \text{ where } c = c_0 + r_0 e^{i \phi}$$ For this to hold, the coefficients of $e^{i k \phi}$ must be zero for all $k > 1$. But setting $k = 3$ implies $r_0^3 = 0$ but we know that $r_0 > 0$ as hyperbolic components have non-empty interior. Contradiction, conclude that no period 3 component is a perfect circle.

Case $p > 3$:

Similarly to the $p = 3$ case, get a polynomial of degree $m > 1$ in $e^{i \phi}$ whose highest term has coefficient $r_0^m$. It remains to show that the polynomial really does have degree greater than $1$. The constant coefficient (product of roots) is a polynomial of degree $2^{p-1}$ in $c$, divided by the corresponding constant coefficient of all smaller divisors of the period gives:

$$m = 2^{p-1} - \sum_{q | p, q < p} 2^{q-1}$$

which solved numerically gives:

\begin{aligned} p & & & m \\ 1 & & & 1 \\ 2 & & & 1 \\ 3 & & & 3 \\ 4 & & & 5 \\ 5 & & & 15 \\ 6 & & & 25 \\ \vdots \end{aligned}

Finally, $m > 1$ for all $p \ge 3$ because $\exists q > 1 . q \nmid p$.