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 {z_n} be the iteration sequence for c. Consider the transformation of c and z_n for some n. Because multiplication of complex numbers is associative and commutative

(tz_n)^w + tc = t^wz_n^w +tc
= t(t^w-1z_n^w + c)

If t^w-1=1 then

tz_n^w + c = t(z_n^w + c)
= t(z_n+1)

Thus if t^w-1=1 then the sequence associated with tc is just {tz_n}. 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 {z_n}. Now consider the complex conjugates of c and the some member of the sequence z_n. For typographic convenience these will be denoted as c and z_n. Let z_n=re^iφ. Then the conjugate of z_n is z_n=re^−iφ. Therefore

z_n^w = z_n^w = re^−wφ
and hence
z_n^w + c = z_n^w + c
= z_n^w + c = z_n+1

This means that c generates the sequence made up of the conjugates of {z_n}. Since |z_n=||z_n| the conjugate sequence will have the same characteristics as {z_n} 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

t^w-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^iφ. Then

(re^iφ)^(u-1)+iv = r^u-1e^-φe^i(u-1)φr^iv
= e^{(u-1)ln(r)−vφ}e^{i[(u-1)φ+vln(r)]} = 1 = 1e^ik(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 {z_n} are transformed into tc and {tz_n} the convergence nature of the sequence is preserved. That is to say, if {z_n} converges to a fixed point z* then {tz_n} converges to a fixed point tz*. If {z_n} converges to a two-cycle then {tz_n} 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.