Preliminary remarks and some short explanations can be found on the page
**Some Julia Sets 1**.

f(z)=z^{4}/(z^{8}+c) with c=0.01, shown on
[-1.4;1.4]x[-1.4;1.4].

The function has attracting fixed points in 0 and infinity. For very small
c it can be expected, that the Julia sets looks approximately like the Julia
set of 1/z^{4}, i.e. the unit circle. Indeed the concentric Jordan
curves get thinner and more circular, when c is decreased.

On the other hand the Julia sets gets "thicker" and its components less circular, if the parameters are changed. This can be seen on the next picture.

f(z)=z^{4}/(0.4z^{12}+z^{8}+0.635z^{4}+0.001),
shown on [-1.8;1.8]x[-1.8;1.8].

If the parameters are increased, the bulges in this Julia set finally grow together and the annular components of the Fatou set disappear. (Cmp. next image!)

f(z)=z^{4}/(0.4z^{12}+z^{8}+0.2i),
shown on [-1.7;1.7]x[-1.7;1.7].

f(z)=(z^{3}+0.37z-0.04)/z, shown on
[-1.7;1.7]x[-1.7;1.7].

I only bring this set here, because it looks like an accumulation of gnome feet.

f(z)=z^{6}+1.01*z, shown on [-1.2;1.2]x[-1.2;1.2].

The Julia set of the Funktion z^{6}+z is nearly identical. This is
an example for the "petal theorem": If a function has the Taylor series
z+z^{p+1}+... around the origin, the Fatou set has p components
around the origin, which approximately look like flower petals. Details can
be read in the section about "rational indifferent periodic points" of the
mathematical part of this site. (But only in german.)

In this example I made the function numerical better conditioned, increasing the derivation slightly over 1. If the original derivation hadn't been 1, but another complex number with modulus 1, this small factor would have been enough to result in a totally other picture around the origin.

f(z)=z^{5}+a*e^{it}*z with a=1.0001 and t=Pi/1000
shown on [-1.2;1.2]x[-1.2;1.2].

Here one can see, what happens, if the derivation at the fixed point is only slightly changed: The picture around the origin is totally different.

f(z)=z^{5}+a*e^{2Pi*i/g}*z with a=1.0001 und
g=(sqrt(5)-1)/2,
shown on [-1.2;1.2]x[-1.2;1.2].

f(z)=(z^{3}-z+a)/z = (z^{2}-1)+a/z with a=0.0003i,
shown on [-1.7;1.7]x[-1.7;1.7].

This Julia sets looks like the one of z^{2}-1. But because of the
small disruptive term a/z the interiors of those components of the Fatou set,
which are eventually mapped onto the component containing the 0, get filled
with a fine tissue, similar as on
some pictures on page 1. Those components are in this case all components
except the unbounded component.

f(z)=(z^{8}+z^{4}+0.01)/z=
z^{7}+z^{3}+0.01/z, shown on
[-1.2;1.2]x[-1.2;1.2].

This too is an example with such a disruptive term.

by Michael Becker,
8/2003. Last modification: 8/2003.