Some Julia sets 4

Instead of a preliminary remark

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

z/(z^4+6z+1,001)

f(z)=z/(z4+6z+1,001), shown on [-5;5]x[-5;5].

(z^3-z)/(dz^2+1) mit d=-0,7+i

f(z)=(z3-z)/(dz2+1) mit d=-0,7+i, dargestellt auf [-10;10]x[-10;10].

The function has four fixed points. 0 and Infinity are two of them. Infinity is a repelling fixed point, and the function behaves approximately as z->z/d around it. As a consequence there is a wide spiral around Infinity. The fixed point 0 has derivative -1. It is a rational indifferent fixed point, in whose neighborhood there are two "petals". So in the neighborhood of 0 there is no spiral.

(z^3-z)/(dz^2+1) mit d=-0,003+0,995i

f(z)=(z3-z)/(dz2+1) mit d=-0,003+0,995i, dargestellt auf [-7;7]x[-7;7].

Here the absolute value of d is smaller than 1, so that Infinity is in the Fatou set. Near 0 there are again to petals. Near Infinity the function behaves as z->-iz, so that the number of spirals, which run direction Infinity is exactly four. Probably the four turning points of the spirals form a repelling cycle of period 4.

(z^3-z)/(dz^2+1) mit d=1,001*e^(2pi/30)

f(z)=(z3-z)/(dz2+1) mit d=1,001· e2Pi/30, shown on [-9;9]x[-9;9].

The factor 1,001· e2Pi/30 causes exactly 30 lines running in direction Infinity, but ending somewhere in the nowhere. Because of these and the wormlike shape the Julia-set looks like a ??? (i.e. a worm of the order Errantia).

(z^3-z)/(dz^2+1) mit d=1,0001*e^(2pi*0,61)

f(z)=(z3-z)/(dz2+1) mit d=1,0001· e2Pi· 0,61, shown on [-4;4]x[-4;4].

0,61 can be written as reduced fraction 61/100. So there should be exactly 100 lines running towards Infinity. But I didn't count them.

(z^5-z)/(20z^2+1)

f(z)=(z5-z)/(20z2+1), shown on [-5;5]x[-5;5].

z^2/(z^9-z+0,025)

f(z)=z2/(z9-z+0,025), shown on [-2;2]x[-2;2].

z^2/(z^9+2z+0,05)

f(z)=z2/(z9+2z+0,05), shown on [-2;2]x[-2;2].

z^2/(z^9+2z+0,001)

f(z)=z2/(z9+2z+0,001), shown on [-3,5;3,5]x[-3,5;3,5].

(z^2+a)/(z^2+b) mit a=-0,2+0,7i und b=0,917

f(z)=(z2+a)/(z2+b) mit a=-0,2+0,7i und b=0,917, shown on [-2;2]x[-2;2].

(az^+b)/(z^2+b) mit a=1,005+0,002i und b=0,1

f(z)=(az3+b)/(z2+b) mit a=1,005+0,002i und b=0,1, shown on [-5;5]x[-5;5].

This is the proof, that Julia-sets may be clearly male.

(z^3-0,99)/(0,01z^3+0,999z^2-0,999i)

f(z)=(z3-0,99)/(0,01z3+0,999z2-0,999i), shown on [-8;8]x[-12;4].

Some Julia-sets are very wet.

by Michael Becker, 10/2004. Last modification: 10/2004.