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The Symmetries of
the Fully General Mandelbrot Set

Fully general means that the Mandelbrot sets are generated from iteration schemes of the form

zn+1 = znw + c

where c, z and w are complex numbers. (The integers and the real numbers are included within the set of complex numbers.) A symmetry is such that when a value c is transformed, such as by rotation, the transformed value has the same characteristics as does c with respect to inclusion in a set,

A rotation in the complex plane is equivalent to multiplication by a complex number, say t. Thus c→tc, z→tz etc. Let {zn} be the iteration sequence for c. Consider the transformation of c and zn for some n. Because multiplication of complex numbers is associative and commutative

(tzn)w + tc = twznw +tc
= t(tw-1znw + c)

If tw-1=1 then

tznw + c = t(znw + c)
= t(zn+1)

Thus if tw-1=1 then the sequence associated with tc is just {tzn}. If |t|=1 then c and tc would have the same sequence of absolute values and thus the same classification as c with respect to inclusion or exclusion in a set or other aspects such as escape time which is used in coloring the Mandelbrot set images.

If w is just a positive integer m then t would have to be an (m-1) root of unity and the Mandelbrot set has the symmetries involving rotations by integral multiples of an angle equal to 2π/(m-1).

Symmetry Under Conjugation

Consider c and its associated iteration sequence {zn}. Now consider the complex conjugates of c and the some member of the sequence zn. For typographic convenience these will be denoted as c and zn. Let zn=re. Then the conjugate of zn is zn=re−iφ. Therefore

znw = znw = re−wφ
and hence
znw + c = znw + c
= znw + c = zn+1

This means that c generates the sequence made up of the conjugates of {zn}. Since |zn=||zn| the conjugate sequence will have the same characteristics as {zn} and c will be classified in the category as c. Thus any Mandelbrot set is symmetric under conjugation.

The Conditions for Symmetry
Under a General Power Transformation

The condition that the iteration sequence for c be mapped into the transformed sequence when c is transformed to tc then z is mapped into tz. This means that

tw-1 = 1

Without considering the matter of the absolute values of the transformed sequence let us find the values of t corresponding to the exponent in the iteration scheme. Let w=u+iv and express t in polar form as re. Then

(re)(u-1)+iv = ru-1eei(u-1)φriv
= e(u-1)ln(r)−vφei[(u-1)φ+vln(r)] = 1 = 1eik(2π)

where k is an integer. This requires that

(u-1)ln(r) − vφ = 0
vln(r) + (u-1)φ = k(2π)

This is simply two linear equations in two unknowns and the solution is

ln(r) = vk(2π)/[(u-1)² + v²]
φ = (u-1)k(2π)/[(u-1)² + v²]

For k=0 the only solution is φ=0 and ln(r)=1 and hence r=1. This is the identity transformation, a trivial form of symmetry.

For k=1 and u=m, an integer, and v=0 the solution is ln(r)=0 (r=1) and φ=2π/(m-1). The Mandelbrot set for an exponent equal to an integer m is symmetric under rotations of an angle equal to a multiple of 2π/(m-1).

Without the condition that |t|=1 the symmetries are not strictly rotational, but there is a symmetry which is in the nature of a spiral. When c and {zn} are transformed into tc and {tzn} the convergence nature of the sequence is preserved. That is to say, if {zn} converges to a fixed point z* then {tzn} converges to a fixed point tz*. If {zn} converges to a two-cycle then {tzn} converges to the corresponding two-cycle. Thus tc will have the same classification as c. The only difference is that tc is located at a different distance from the origin than is c.

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