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the symmetry of the Mandelbrot set

Intro


When looking to pictures of the generalize Mandelbrot one is tempted to make at least two observations. First, that the set seems to be always symmetric around the real axis. Second, the sets are symmetric under angle rotation. We will quickly show that both are indeed true.

Sets for k=2, 5 and 8, drawn with a distance estimation algorithm




Some pre math


We are going to concentrate on the Mandelbrot sets for monic one dimensional polynomials of the form . Note that the critical point is the same for all of them (zero), and that infinity is an super-attractive fixed point. The Mandelbrot set is defined as for the standard case k=2.





The set of point created by the iteration of with = 0 is called the orbit of . Each point in the orbit is thus



and when applied to = 0 it expands the next polynomials in c: . So, and so on. Then, a point c belongs to if stays bounded as



The Mandelbrot set for k=4




Vertical Symmetry


Vertical symmetry translates to saying that if a given point c belongs to the set, then it's conjugate does also: . The result comes from the fact that the iteration of develops a polynomial in c with only real coefficients. But, let's show it anyway by induction.

Let's assume that for one of the iterations we have


meaning that if c belongs to then will also do, since the modulus of a complex number and its conjugate is the same. So, this is symmetrical. Let's examine now what happens to the next iteration both for c

and for

From the property that for any complex number w and integer a we have , we conclude


and thus the next iterate is symmetrical too. In other words

is symmetrical is symmetrical


We just need to check that is symmetric to arrive to the conclusion that all the iterates are symmetrical.


Therefore the complete set is symmetrical around the real axis:



Rotational Symmetry


For the rotational symmetry we proceed in a similar way. Let's call a n-symmetry to a rotational symmetry of radians, for any integers m and n. Now, we first assume that one iterate is (k-1)-symmetric as the pictures suggest. That translates to


and now we try to demonstrate that next iterate will also be symmetrical. For that we analyze the iteration:


By the assumption we made,



So, again,

Because we can easily check that for n=1 the assumption holds, and because we can say that


Or put it in other way,