Some Julia sets

Preliminary remarks

What is meant here by Julia and Fatou sets?

If a function f maps a region G to itself, f can be iterated. The Julia set of f then is the set of all points of G, at which this sequence of iterated functions is not equicontinous. The Fatou set is its complement. Laxly said the action of the iterated functions on near points is examined. Places, where points, which are near enough, remain near during iterations, belong to the Fatou set. Places, where points, as near they may be, are teared apart, belong to the Julia set.

In the following I only consider functions, which map the Riemann sphere, i.e. the complex plane with an ideal point "infinity" added, to itself. The Julia sets are white, the Fatou sets black.

What has this to do with the "usual" Julia sets?

In the internet many pictures of Julia sets of polynomes can be found, especially of polynomes of the form z2+c with a complex constant c. In those pictures the Julia set ist defined as the set of all points, of which the iterated points do not converge to infinity, but remain bounded. The problem with this definition is, that it can be applied seriously only to functions, for which infinity is an attracting fixed point, i.e. for which there exist points, whose iterates go to infinity. Otherwise this definition won't generally yield a fractal set.

It is possible to show, that for polynomes the boundary of the sets such produced is identical to the Julia set as defined here.

Fixed points

An important help for the orientation of a viewer of Julia sets are the fixed points of the function f. One distinguishes between attracting fixed points (|f'|<1), repelling fixed points (|f'|>1) and indifferent fixed points (|f'|=1). Attracting fixed points always are in the Fatou set, and repelling fixed points in the Julia set. Indifferent fixed points may be in the one or in the other.

Other files with Julia sets

On this page, there are also 3 other files with some Julia sets:

(2cz^3+2z^2)/(z^3+3cz^2-z+c), c=(1+i)/2

f(z)=(2cz3+2z2)/(z3+3cz2-z+c) with c=(1+i)/2, shown on [-3.5;1]×[-2;2]

This function has 4 fixed points: -1, 0, 1 and -c. The first three are attracting (the derivative of f is null), the fourth is repelling. In this picture one can easily see, how any of the attracting fixed points is in its own component of the Fatou set, whereas the repelling fixed point is in the Julia set. The point infinity belongs to the catchment area of the fixed point 1.

z^2+c, c=-0.2-0.7i

f(z)=z2+c with c=-0.2-0.7i, shown on [-1.8;1.8]×[-1.8;1.8]

There is not much to say about this very nice Julia set, which reminds of a chinese dragon. The function has two repelling fixed points, at the right end (approximately 1.3+0.437i), and in the big spiral on the left side (approximately -0.3-0.437i). Infinity is, as for all polynomes, an attracting fixed point. All components of the Fatou set are simply connected.

z^3+1.55z^2

f(z)=z3+1.55z2, shown on [-2.5;1]×[-2;2].

This is an example for a Julia set, which connection components are Jordan curves. The function has 4 fixed points, two attracting at infinity and 0, and two repelling, which are on the real axis too. Those two are the most right and the most left point of the Julia set on the real axis.

(z^5-(0.1+0.07i)z+0.09)/z^3

f(z)=(z5-(0.1+0.07i)z+0.09)/z3, shown on [-1.5;1.5]×[-1.5;1.5].

The Julia sets of this and similar functions react very sensitively on changes of parameter. I am only bringing this set here, because it reminds me of some very muddled handwriting. As far as I can say the Julia set consists of one single Jordan curve.

(-4.0004z^5+0.005(1-i))/(-4iz^4+0.001i)

f(z)=(-4.0004z5+0.005(1-i))/(-4iz4+0.001i) shown on [-1;1]×[-1;1].

This function is via g(z)=(1+i)z conjugated to f2(z)=(1.0001z5+0.01)/(iz4+0.001i). Infinity is an attracting fixed point. That's why the four rays don't go to infinity, but really end somewhere.

f(z)=(z3+c)/z with c=0.001, shown on [-1.5;1.5]×[-1.5;1.5].

What looks like ramifying branches in this picture, in reality consists of a mosaic of smaller black regions. E.g. the origin is not in the Julia set, as it seems, but in the Fatou set, because there is a pole at it. Its component of the Fatou set starts on the negative real axis at approx. -c. But the resolution of the picture is not high enough to see this: The distance between -c and 0 is 1/4 pixel.

The point -c1/3 is not in the Julia set too, because there the function has a zero. It is the center of the slighly bigger knot a bit left of the origin. Right of it at approx. -sqrt(c) there is a repelling fixed point. So in this small intervall there are points of the Julia as of the Fatou set. Indeed this intervall is one of the inverse images of the whole negative real axis: Even if it seems white on the picture, in detail is looks like the negative real axis.

If the parameter c is increased, the white branches get broader and more mosaic like, and the black regions rounder, a bit like it can be seen on the picture after the next.

(z^3+c)/(dz), c=0.001, d=0.95-0.31225i

f(z)=(z3+c)/(dz) with c=0.001 and d=0.95-0.31225i, shown on [-1.5;1.5]×[-1.5;1.5].

d is a complex number with absolute value approx. 1. It makes the triangle knots on the last picture turn a bit. So their three branches cannot meet each other straightly, but run into spirals.

(z^2+c)/(z^2-c), c=0.7

f(z)=(z2+c)/(z2-c) with c=0.7, shown on [-2;2]×[-2;2].

This Julia set is remarkably only for seemingly consisting of nothing but smooth, nearly circular Jordan curves. I said "seemingly", because I don't know it.

(z^3+c)/(z^3-c), c=0.45

f(z)=(z3+c)/(z3-c) with c=0.45, shown on [-2;2]×[-2;2]

(z^5+c)/z^3, c=0.001(?)

f(z)=(z5+c)/z3 with c=0.001(?), shown on [-1.5;1.5]×[-1.5;1.5].

Up to this point we only had Fatou sets, which components were simply or infinitely connected. But there are also Fatou sets with doubly or multiple connected components. On this picture none of them can be seen of course. But if the parameter c is decreased, the smaller, completely visible components are getting bigger and bigger, until they join to ringlike components around the origin. (This is already the case e.g. for c=0.00001.) Finally the Julia set passes over to a Cantor set of concentric circles around the origin. The corresponding pictures are not very attractive in my opinion. So I am content here with a "normal" Julia set.

(z^3+(0.5+0.78i)z)/(z+1)

f(z)=(z3+(0.5+0.78i)z)/(z+1), shown on [-1.5;2]×[-2;1.5]

2/3*(z^3-2)/z

f(z)=2/3*(z3-2)/z, shown on [-2;2]×[-2;2]

Even a Sierpinsky triangle occurs as Julia set. The critical points are exactly the negative third roots of unity. These are the "side midpoints" of the triangle. Their function values are the corner points of the triangle. The right corner (2) is a repelling fixed point. The two other corners form a repelling cycle of period 2.

If the parameters occuring in f(z) are changed, the Julia set decays to a Cantor set, or the connecting points of the single triangles grow out.

2/3*(z^3+2z-2)/z

f(z)=2/3*(z3+2z-2)/z, shown on [-2.5;2.5]×[-2.5;2.5]

by Michael Becker, 6/2003. Last modification: 2/2004.