The famous Mandelbrot set is the bifurcation locus for the family of functions \(f_c(z)=z^{2}+c\), where \(c\in\mathbb C\) is a parameter. Let us consider the family \(g_c(z)=z^{-2}+c\). The bifurcation locus is then a partition of the complex parameter plane into disjoint regions, as shown in the figure on the left below.

The different colors in the bifurcation locus correspond to various dynamical behavior of \(g_c\). In fact, there are several relatively prominent components of the bifurcation locus:

- The period 1 domain: the large, yellow-ish, region exterior to the main hypocycloid
- The period 2 domain: the blue-ish, central, disk-like domain
- The period 3 domain: the three, dark cardioid shapes placed symmetrically around the central disk

Typically, the attractive orbits contain very large values outside of the viewing rectangle, as well as values in the viewing rectangle. This can make it difficult to see what's going on. To make it a little easier, the opacity of the orbit path is set quite low for the first few iterates so that we aren't distracted by the transient behavior. You can always click several times to darken the path.

The boundaries of the period 1 and 2 domains can be parametrized. The period 1 domain is fairly easy; it consists of all values of \(c\) such that \(g_c\) has an attractive fixed point and, on the boundary, there is a neutral fixed point. This yields two equations that \(c\) and \(z\) must satisfy so that \(g_c\) has a neutral fixed point at \(z\): $$ \begin{array}{ll} g_c(z) = z^{-2}+c=z & z \text{ should be fixed} \\ g_c'(z) = -2z^3 = e^{it} & z \text{ should be neutral} \end{array} $$ Solving this system for \(c\), we obtain $$c=(2+e^{it})e^{-it/3}.$$ The inner disk seems harder but I believe that it's just the disk of radius \(2^{-2/3}\) centered at the origin. Plotting these together, we get: