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The Mandelbrot set is the set of all complex numbers $c$ that cause the function $z_{n+1} = z_n^2 + c$ to remain bounded when iterated from $z_0 = 0$. Plotted in the complex plane it includes a main cardioid, formed by rolling a circle of radius $0.25$ around another circle of the same radius centered on the origin with its cusp at $0.25 + 0i$; and a main bulb, also a circle of radius $0.25$, centered on the value $-1 + 0i$ and touching the main cardioid at the value $-0.75 + 0i$.

Animated GIF showing the main cardioid and main bulb outlined in green, with the circles that generate the main cardioid shown in blue.

Taking values of $c$ on the perimeter of the main cardioid causes $z$ to converge on or orbit around fixed point values lying on a circle of radius $0.5$ centered on the origin. This circle is the path taken by the center of the rolling circle that generates the shape of the main cardioid. The fixed point value is specifically the location of the midpoint of the rolling circle where it generates the chosen value of $c$. It is easy to find the attracting value for any point on the cardioid given the angle of that point, or conversely, to find the point on the cardioid for any attracting value given the angle of that value.

Animated GIF showing values of c on the main cardioid and the corresponding orbits of z.

Taking values of $c$ on the perimeter of the main bulb causes $z$ to converge on or orbit around pairs of fixed point values lying on a lemniscate of Bernoulli with foci at $0 + 0i$ and $-1 + 0i$. The fixed points are symmetrical about the value $-0.5 + 0i$. I have been unable to find an equation that gives the exact attracting values for any point on the circle, or conversely, to find the point on the circle for any attracting values.

Animated GIF showing values of c on the main bulb and the corresponding orbits of z.

What is the relationship between the point on the circle and the points on the lemniscate of Bernoulli?

I've gotten close by representing the lemniscate with the parametric equations $$x = \frac{\sqrt{2}}{2} * \frac{-\sin(θ)} {\cos^2(θ) + 1} - \frac{1}{2}$$ $$y = \frac{\sqrt{2}}{2} * \frac{\sin(θ) \cos(θ)} {\cos^2(θ) + 1}$$ Where $θ$ is the angle around the main bulb, starting with 0° at the point where the bulb joins the main cardioid, but that only lines up exactly for angles 0° and 180° (and all further rotations mod 360°). For other angles it is either too far around, or not far enough.

Lawton
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    Just a comment: the sentence "Taking values of c on the perimeter of the main cardioid causes z to converge on values lying on a circle of radius 0.5 centered on the origin" is not quite true. It is in fact only true for rational angles. – Albert Jan 09 '21 at 19:22
  • @Glougloubarbaki are you sure? Can you elaborate? – Lawton Jan 09 '21 at 19:40
  • @Glougloubarbaki choosing irrational angles still seems to produce orbits converging on a point on that circle, but that may be a limitation of the software I'm using. – Lawton Jan 09 '21 at 19:44
  • @Glougloubarbaki is correct. At most points on the main cardioid, the Julia set is a Siegel disk. At rational points, the fixed point is technically not attractive but parabolic; the critical point converges to the fixed point but very slowly. – Mark McClure Jan 10 '21 at 01:45
  • https://en.wikibooks.org/wiki/Fractals/Iterations_in_the_complex_plane/Parameter_plane#How_to_move_on_parameter_plane_? – Adam Jul 25 '21 at 16:02
  • Please refer to an older post of mine here. – Ng Chung Tak Aug 09 '24 at 10:18

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You are looking for solutions of $z = (z^2 + c)^2 + c$, which expanded gives:

$$ z^4 + 2cz^2 - z + c^2 + c = 0 = c^2 + (2 z^2 + 1) c + z^4 - z $$

Solving for $c$ given $z$ is easier than solving for $z$ given $c$, but setting $c = -1 + e^{2\pi i t}/4$ and doing so we find that $$ z = \frac{1}{2} \left(-1\pm\sqrt{1-e^{2 i \pi t}}\right). $$ You can plot this to generate your lemniscate.

Mark McClure
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Claude
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