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 $-0.75 + 0i$.
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.
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.
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.
Best Answer
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.