# The Julia sets symmetry

The Julia set J(c) is made of all points zj , which do not go to an attractor (it may be at infinity too) under iterations. It is evident, that iterations of the points fc(zj ) do not go to an atractor too. Therefore the Julia sets are invariant under fc .
The J-set is centrally symmetric as since fc(z) = z 2 + c is an even function.
For   z = r e the squared value is   z 2 = r 2 e 2iφ. Therefore the map fc wraps twice the complex plane z onto itself (with quadratic deformation of r and displacement by c).

This is the simplest Julia set for c = 0 + 0i . As since
for | zo | < 1,   zn converges to the fixed point z = 0,
for | zo | > 1,   zn go to infinity and
for | zo | = 1,   zn rotates and stays on the same circle | z | = 1.
The circle is the Julia set J(0). It is evident, that the circle is invariant under fo = z 2.

## The Julia sets self-similarity

Let f maps a point z1 into z2 = f(z1). For small enough ε it follows from the Taylor's theorem that
f(z1+ε) = z2 + f '(z1)ε + O(ε 2).
So small neighbourhood of z1 is mapped linearly (by scaling and rotation) into the z2 one. Therefore the Julia set is self-similar in these regions. As iterated preimages f o(-n)(z1) are everywhere dense in J therefore J is self-similalar in every point.
You can trace quadratic map dynamics here. The white square is mapped in the region with faded colors. You see thet Julia set is similar in both regions ("faded" square is deformed due to 2 and higher terms in the Taylor's formula).

Controls: Drag the white square by mouse to move it (its coordinates are shown).

You can see below self-similarity of "midget" Julia sets.

## More examples

 It is not difficult to imagine how f-1 maps points of the J(-1) set from the Re(z) > 0 (or Im(z) > 0) half-plane onto the whole J(-1). Squaring "moves" J(-1) to the right (the lower picture) and after addition of c = -1 the Julia set returns into its original position. Note, that the two points a are mapped into one point a'.
 It is easy to see, that the "cauliflower" J(0.35) set has the same "squaring" symmetry.

Contents     Previous: The Mandelbrot, Julia and Fatou sets   Next: Critical points and Fatou theorem
updated 11 Sep 2013