### Intro

When looking to pictures of the generalize Mandelbrot f

_{c,k}= z

^{k}+ c 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 2π⋅ m/(k-1) 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 f

The set of point created by the iteration of z

and when applied to z

So,

T

and so on. Then, a point c belongs to M

_{c,k}= z^{k}+ c. Note that the critical point f '_{c,k}(z)=0 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 z

_{n+1}= f_{c,k}(z_{n}) with z_{0}= 0 is called the orbit of z_{0}. Each point in the orbit is thusand when applied to z

_{0}= 0 it expands the next polynomials in c:So,

T

_{k}^{0}(c) = 0, T_{k}^{1}(c) = c, T_{k}^{2}(c) = c^{k}+ cand so on. Then, a point c belongs to M

_{k}if T_{k}^{n}(c) stays bounded as n → ∞*The Mandelbrot set for k=4*

### Vertical Symmetry

Vertical symmetry translates to saying that if a given point c belongs to the M

_{k}set, then it's conjugate does also: . The result comes from the fact that the iteration of f

_{c,k}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 M

_{k}then c

^{*}will also do, since the modulus of a complex number and its conjugate is the same. So, this T

_{k}

^{n}(c) is symmetrical. Let's examine now what happens to the next iteration both for c

and for c

^{*}

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

T

_{k}

^{n}(c) is symmetrical is symmetrical

We just need to check that T

_{k}

^{1}(c) = c is symmetric to arrive to the conclusion that all the iterates T

_{k}are symmetrical.

Therefore the complete M

_{k}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 2π⋅ m/n radians, for any integers

*m*and

*n*. Now, we first assume that one iterate T

_{k}

^{n}(c) 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 |w⋅ e

^{iα}| = |w| we can say that

Or put it in other way,