The Mandelbrot set may be generalized by quaternion numbers
Qn+1 = Qn2 + Qc, Q0 = 0, Qc = (cr , ci , cj , ck ).
As since (a + b i + c j + d k)2 = a2 - (b2 + c2 + d2) + 2ab i + 2ac j + 2ad k
due to terms like bc i j and cb j i cancel each other (bc i j = -cb j i). We get
a = a2 - (b2 + c2 + d2) + cr
b = 2ab + ci
c = 2ac + cj
d = 2ad + ck
a2 + b2 + c2 + d2 < BailOut2
The Quaternions Mandelbrot set is a 4 dimensional object. For Qc = (a, b, 0, 0 ) all points belongs to 2D (a, b ) plane and the set coincides with the classical M-set. For Qc =(a, b, c, 0 ) we get the 3D M-set.
We can rewrite the last equations as
a = a2 - |r|2 + cr
r = 2ar + c
a2 + |r|2 < BailOut2
where r = (b, c, d) and c = (ci , cj , ck) are 3 dimensional vectors. By rotations of coordinate system any 3D vector c may be put into (|c|, 0, 0) or Qc = (a, |c|, 0, 0) - the classical Mandelbrot plane. Rotations conserve |r|. As since dynamic of the a - component and exiting condition depend on |r| only, therefore the QM-set is spherical symmetric in (b, c, d) subspace.
Due to the spherical symmetry we can get (a, b, c) 3D QM-set rotating the classical M-set relatively the real axis. The 3D QMandelbrot cactus consists of spheres and toruses so its cross-sections are circles and ellipses. Here you see the QM-set cross sections by (Re, Im, c, 0 ) planes for different c.
Sorry that quaternions are a bit "alien" here. Further we will return to the complex quadratic mappings.