Re: Is Mandelbrot Set symmetrical around the imaginary axis?

Thanks IQ.
Except one typing error (marked *** below), it seems to work for Mandel.


I will use this notation: Zn is the "n"-th iteration of the map (Zn+1 =
Zn^2 + c). Of course, Zn is a function of "c", so I will use Zn(c). The
symbol ' means
complex conjugation (symmetry around the horizontal axis, ie, negating
the imaginary component of the number). Xn and Yn are the real and
imaginary parts
of Zn, and "a" and "b" the components of c. "^" menas squating, and
"||" means modulus. So,

Let's first assume that Zn'(c) = Zn(c'):

Xn(c) = Xn(c')
Yn(c) = -Yn(c')

Now, let's see what would happen to Zn+1(c') under that assumtion:

Zn+1(c') = [ Xn+1(c') , Yn+1(c') ]

Xn+1(c') = Xn(c')^2 - Yn(c')^2 + a
Yn+1(c') = 2*Xn(c')*Yn(c')^2 - b

Note how I also negated the imaginaaty part of c. Now, by using the
assumptions, we can substitute to see that

Xn+1(c') = Xn(c)^2 - Yn(c)^2 + a
Yn+1(c') = -2*Xn(c)*Yn(c)^2 - b

thus

Xn+1(c') = Xn(c)
Yn+1(c') = -Yn(c)

***> typing error, it should be;

Xn+1(c') = Xn+1(c)
Yn+1(c') = -Yn+1(c)

<***

So, if our assumptions holds for a given "n", it will hold also for
"n+1", and thus for all the remaining "n"s too. You can trivially check
that
the equations hold for n=1 (Z=c), so we can safely say that

Zn'(c) = Zn(c') for every n

Now, since the modulus of a number is the same if you conjugate this
number (the distance to the origin remains the same under mirroring on
the real axis),
we can write that

|Zn(c')| = |Zn(c)|

Since the Mandelbrot set is defined based of the modulus of Zn, it's
clear that the set will be invariant under conjugation (it's
symetrical).

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- -

For the julia sets you can do something similar I guess, by searching
for a symmetry of the form |Zn(zo)| = |Zn(-zo)| since the sets are
symmetrical around y=x...

Iñigo Quilez


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