Iteration of complex rational functions^[1]^[2]^[3]

Examples

Newton fractals
commons:Category:Complex rational maps
The Boundedness Locus and baby Mandelbrot sets for some generalized McMullen maps by Suzanne Boyd, Alexander J. Mitchell 2023
Mark McClure : a-julia-set-on-the-riemann-sphere
f(z)=z2/(z9-z+0,025) ^[4]
f(z)=(z3-z)/(dz2+1) where d=-0,003+0,995i ^[5]
f(z)=(z3-z)/(dz2+1) where d=1,001· e2Pi/30 ^[6]
Multibrot sets by Xender^[7]
$f_{\lambda }(z)=z^{n}+{\frac {\lambda }{z^{d}}}$ ^[8]
Rational Julia Sets by Marc McClure
f(z) = (z^n+c)/(c^n+z), for n = -2 ^[9]
Jasper Weinrich Burd: A Thompson-Like Group for the Bubble Bath Julia Set
Abalone Fractals by Anders Sandberg, 2004
(z2-1)/(z2+1). It has simple zeros at ±1 and simple poles at ±i.
Shigehiro Ushiki = ComplexExplorer Page
Rational maps with a superattracting 2-cycle by Wolf Jung
Julia Sets of Cubic Rational Maps with Escaping Critical Points by Jun Hu, Arkady Etkin
ON HYPERBOLIC RATIONAL MAPS WITH FINITELY CONNECTED FATOU SETS by YUSHENG LUO, 2021
On geometrically finite degenerations I: boundaries of main hyperbolic components by Yusheng Luo, 2021
On geometrically finite degenerations II: convergence and divergence by Yusheng Luo, 2021
math.stackexchange question: how-to-compute-a-negative-multibrot-set
Exotische Juliamengen - Jürgen Meier
There are rational functions whose Julia set is the whole plane. The first example was given by Lattès: $R(z)={\frac {(z^{2}+1)^{2}}{4z(z^{2}-1)}}$ ^[10]
Rational maps whose Julia sets are Cantor circles by Weiyuan Qiu, Fei Yang, Yongcheng Yin
Petersen, C. L., and S. Zakeri. “On the Julia Set of a Typical Quadratic Polynomial with a Siegel Disk.” Annals of Mathematics, vol. 159, no. 1, 2004, pp. 1–52.
A Note on a Quadratic Rational Map with Two Siegel Disks by Liang SHEN, Sheng Jian WU. Acta Mathematica Sinica, English Series. Jul., 2010, Vol. 26, No. 7, pp. 1393–1402 Published online: June 15, 2010 DOI: 10.1007/s10114-010-6611-3 Http://www.ActaMath.com
MandelbrotMaps by Chris King
fractalforums.org : z-plus-c2-z3-plus-c $f(z)={\frac {\left(z+c\right)^{2}}{z^{3}+c}}$

Gallery

Rational functions
${\frac {z}{z-1}}$
${\frac {z^{2}}{z^{2}-1}}$
${\frac {z^{3}}{z^{3}-1}}$
${\frac {z^{4}}{z^{4}-1}}$
${\frac {z^{5}}{z^{5}-1}}$
${\frac {z^{7}}{z^{7}-1}}$

Rational functions
${\frac {z^{4}+0.05}{z^{4}-1}}$

${\frac {z^{3}}{z^{5}-1}}$
${\frac {z^{4}}{z^{5}-1}}$
${\frac {z^{5}}{z^{5}-1}}$
${\frac {z^{6}}{z^{5}-1}}$
${\frac {z^{7}}{z^{5}-1}}$

${\frac {(z-4i+3)(z-2-3i)z}{(z+2i+4)(z+3i-2)}}$

Blaschke fraction

Fatou sets for Blaschke fraction f(z) = rho * z^2 * (z-3) over (1-3z)

Analysis of critical points:

kill(all);
remvalue(all);
display2d:false;
ratprint : false; /* remove "rat :replaced " */



rho : -0.6170144002709304 +0.7869518599370003*%i;

define(f(z), rho * z^2 * (z-3)/(1-3*z));

/* first derivativa wrt z */
define( d(z), diff(f(z),z,1));


/* hipow does not expand expr, so hipow (expr, x) and hipow (expand (expr, x)) may yield different results */
n : hipow(num(expand(f(z))),z);
m : hipow(denom(expand(f(z))),z);

/* check if infinity is a fixed point */
limit(f(z),z,infinity);



/* finite critical points */

s:solve(d(z)=0)$
s : map(rhs,s)$
s : map('float,s)$
s : map('rectform,s)$

So there are 3 critical points :

2 finite critical points : z=1.0 i z= 0.0
infinity

Dynamical plane consist of 3 basins

basin of attraction of fixed point z = infinity ( superattracting) with inf many componnets
basin of attraction of fixed point z = 0 ( superattracting) with inf many componnets
basin of parabolic period 3 cycle ( with z= 1 critical point)

Finite Blaschke product

$B_{n}(z)$ is a finite Blaschke product of degree n.^[11] It is: ^[12]

a rational function
an analytic function on the open unit disc such that f can be extended to a continuous function on the closed unit disc that maps the unit circle to itself
have no poles in the open unit disc
In particular, if ƒ satisfies the condition above and has no zeros inside the unit circle, then ƒ is constant (this fact is also a consequence of the maximum principle for harmonic functions, applied to the harmonic function log(|ƒ(z)|)).
the Blaschke products B are rational perturbations of the doubling map of the circle R(z) = z^2 (equivalently given by θ → 2θ (mod 1)).
a finite Blaschke product may be uniquely described by the set of its critical points
rational map which fix a disc = which takes the closed unit disc D to itself
their iteration theory can be analyzed from the point of view of Fuchsian groups.
polynomials’ in the hyperbolic plane = hyperbolic polynomial
A finite Blaschke product, restricted to the unit circle, is a smooth covering map
the unit disk D, the unit circle ∂D and the complement of the closed unit disk C\D are all completely invariant sets for B

B_{n}(z)=\zeta \prod _{i=1}^{n}\left({{z-a_{i}} \over {1-{\overline {a_{i}}}z}}\right)^{m_{i}}

where

$\zeta$ is a unimodular constant. It is a point which lies on the unit circle: $|\zeta |=1$
$m_{i}$ is the multiplicity of the zero $a_{i}$
$(a_{i})_{i=1}^{n}$ is a finite sequence of n points in the open unit disc $|a_{i}|<1$

(a_{i})_{i=1}^{n}=(a_{1},a_{2},\ldots ,a_{n}).

The building blocks^[13] of Blaschke products are Mobius transformations of the form

$b_{k}(z)=e^{i\theta k}{\frac {a_{k}-z}{1-{\overline {a_{k}}}z}}$

where

ak ∈ D := {z ∈ C|, |z| < 1}
θk ∈ R.

A finite (infinite) Blaschke product has the form

$w=B(z)=\prod _{k=1}^{n}b_{k}(z)$

Examples:

Quadratic rational maps by Curtis T McMullen

classification

There is a classification of finite Blaschke products in analogy with Möbius transformations.^[14]

B is elliptic if the Denjoy-Wolff point z0 of B lies in D. |B' (z0)| < 1.
B is hyperbolic if the Denjoy-Wolff point z0 of B lies on ∂D and B'(z0) < 1,
B is parabolic if the Denjoy-Wolff point z0 of B lies on ∂D and B'(z0) = 1,

The Denjoy-Wolff point of B is a unique z0 ∈ D such that $B^{n}(z)\to z_{0}$ for every z ∈ D

Julia set

Let B be a finite Blaschke product of degree d > 1. Julia set $J_{B}$ , the set on which the iterates $B^{n}$ fail to be normal on any neighbourhood is either the unit circle $S^{1}$ or a Cantor subset^[15] ^[16]

if B is elliptic, J(B) = ∂D,
if B is hyperbolic, J(B) is a Cantor subset of D,
if B is parabolic and z0 ∈ ∂D is the Denjoy-Wolff point of B,
- J(B) = ∂D if B(z0) = 0
- J(B) is a Cantor subset of ∂D if $B''(z0)\neq 0.$

boudaries of BDM

critial orbit

McMullen maps

singularly perturbed maps, also called McMullen maps^[17]

  $R_{\lambda }(z)=z^{m}+\lambda /z^{d}$

degree 2

reversed Basilica Julia set

Function: $f(z)={\frac {z^{2}}{z^{2}-1}}$


maxima

Maxima 5.41.0 http://maxima.sourceforge.net
using Lisp GNU Common Lisp (GCL) GCL 2.6.12
Distributed under the GNU Public License. See the file COPYING.
Dedicated to the memory of William Schelter.
The function bug_report() provides bug reporting information.
(%i1) display2d:false;

(%o1) false
(%i2) f:z^2/(z^2-1);

(%o2) z^2/(z^2-1)
(%i3) dz:diff(f,z,1);

(%o3) (2*z)/(z^2-1)-(2*z^3)/(z^2-1)^2
(%i4) s:solve(f=z);

(%o4) [z = -(sqrt(5)-1)/2,z = (sqrt(5)+1)/2,z = 0]
(%i5) s:map('float,s);

(%o5) [z = -0.6180339887498949,z = 1.618033988749895,z = 0.0]
(%i6)

So fixed points $z:f(z)=z$ :

z = -0.6180339887498949
z = 1.618033988749895
z = 0.0

The Bubble bath Julia set

The quadratic rational function f:

f(z)={\frac {1-z^{2}}{z^{2}}}

The derivative wrt z is

f'(z)={\frac {-2}{z^{3}}}

The Julia set for f is called the Bubble Bath Julia set.^[18] It is called the Bubble Bath for its visual similarity to a tub of bubbles.

Function f is defined for all z in the Riemann sphere ${\hat {\mathbf {C} }}$ = it is defined on the whole Riemann sphere

f:{\hat {\mathbf {C} }}\to {\hat {\mathbf {C} }}

The Fatou set of f is the basin of attraction of the 3-cycle consisting of the points 0, −1, and infinity. It is the only one attracting cycle and it is superattracting
The Julia set J(f) is the set of points whose orbits are not attracted to the above 3-cycle
the only critical points of f are:
- z = 0 because it is the pole of order 3 of d(z), the zero of 1/d(z)
- z = infinity because it is the zero of function d(z)

Maxima CAS code :

kill(all);
remvalue(all);
display2d:false;

define(f(z), (1 -z^2)/(z^2));

(%o3) f(z):=(1-z^2)/z^2
define( d(z), ratsimp(diff(f(z),z,1)));

(%o13) d(z):=-2/z^3
(%i14) limit(d(z),z,infinity);

(%o14) 0
(%i15) limit(d(z),z,0);

(%o15) infinity

(%i2) f(-1);
(%o2)                                  0
(%i3) limit(d(z),z,0);
(%o3)                             limit  d(z)
                                  z -> 0
(%i4) limit(f(z),z,0);
(%o4)                                 inf
(%i5) limit(f(z),z,inf);
(%o5)                                 - 1

Satbility of periodic cycle:

kill(all);
display2d:false;
ratprint : false; /* remove "rat :replaced " */


define(f(z), (1 -z^2)/(z^2));

F(z0):= block(
	[z],
	if is(z0 = 0) then 	z: limit(f(z),z,0)
	elseif is(z0 = infinity) then z: limit(f(z),z,infinity)
	elseif is(z0 = inf) then z: limit(f(z),z,inf)
	else z:f(z0),
	
	return(z)
)$

define( dz(z), ratsimp(diff(f(z),z,1)));

Dz(z0) := block(

	[m,z],
	if is(z0 = 0) then m: limit(dz(z),z,0)
	elseif is(z0 = infinity) then m: limit(dz(z),z,infinity)
	elseif is(z0 = inf) then m: limit(dz(z),z,inf)
	else m:dz(z0),
	
	return(m)

)$

GiveStability(z0, p):=block(
	[z,d],
	
	/* initial values */
	d : 1,
	z : z0,
	
	for i:1 thru p step 1 do (
	
		d : Dz(z)*d,	
		z: F(z)
		/*print("i = ", 0, "  d =",d, "  z = ", z)*/
	),
	
	return (cabs(d))
)$

GiveStability(-1,3);

degree 2 by Michael Becker

1 period 4 basin

Julia set for f(z)=(z2+a) over (z2+b) a=-0.2+0.7i , b=0.917

degree 3

The components of the map $f(z)=z-(z^{3}-1)/3z^{2}$ contain the attracting points that are the solutions to $z^{3}=1$ . This is because the map is the one to use for finding solutions to the equation $z^{3}=1$ by Newton–Raphson formula. The solutions must naturally be attracting fixed points.

degree 3 by Michael Becker

1 period 2 basin

Julia set f(z)=1 over z3+z*(-3-3*I)

$f(z)={\frac {1}{z^{3}+z*(-3-3*I)}}$

2 critical points : { -0.4550898605622273*I -1.098684113467809, 0.4550898605622273*I+1.098684113467809}; Both critical points tend to the periodic cycle.

There is only one attractive period cycle : period 2 cycle = {0, infinity}.

Whole plane ( sphere) is a basin of attraction of period 2 cycle ( which is divided into 2 components ). Julia set is a boundary.

2 period 2 basins

only Julia set
Julia set, basins of attractions, critical orbits, attracting cycles

Function

  $f(z)={\frac {1}{z^{3}+a*z+b}}$

where

a = 2.099609375
b = 0.349609375

Derivative:

 d(z):=-(3*z^2+2.099609375)/(z^3+2.099609375*z+0.349609375)^2

Critical points:

  [-0.8365822085525526*%i,0.8365822085525526*%i]

One can check it also using Wolfram Alpha

 solve (3*z^2+2.099609375)/(z^3+2.099609375*z+0.349609375)^2=0

the result:

 z = ± (5/16)* i* sqrt(43/6))

These are 2 finite critical points.

Infinity is a critical point too, as the 1st derivative's denominator degree is strictly greater than the numerator's. In numerical computations one can use the critical value (an image of critical point)

  $f(\infty )=0$

There are two period 2 cycle:

{ +0.4101296722285255 +0.5079485669960778*I , +0.4101296722285255 -0.5079485669960778*I };
{ +1.6890328811664648 +0.0000000000000000*I , +0.1147519899962205 +0.0000000000000000*I };

Both finite critical points fall into first cycle. Infinity ( or it's image zero) falls into the second cycle ( on the horizontal axis)

Infinity is not a fixed point

remvalue(all);
display2d:false;
define(f(z), 1/(z^3+ 2.099609375*z +  0.349609375));
(%i5)limit(f(z),z,infinity);
(%o5) 0
(%i6) limit(f(z),z,0);
(%o6) 2.860335195530726

degree 5

by L. Javier Hernandez Paricio

the rational map $h(z)={\frac {1+4z^{5}}{5z^{4}}}$ has six fixed points:

∞ ( repelling)
−0,809017 − 0,587785i
−0,809017 +0,587785i
0,309017 − 0,951057i
0,309017 + 0,951057i
1

the basin of an end point associated to a fixed point (6= ∞) of f is the same that the attraction basin of the Newton-Raphson numerical method when it is applied to find the roots of the equation $z^{5}-1=0$ ^[19]

degree 6

Julia set of rational function f(z)=z^2(3 − z^4 ) over 2.png

The Julia set of the degree 6 function f :^[20]

$f(z)=z^{2}{\frac {3-z^{4}}{2}}$

There are 3 superattracting fixed points at :

z = 0
z = 1
z = ∞

All other critical points are in the backward orbit of 1.

How to compute iteration :

z:x+y*%i;
z1:z^2*(3-z^4)/2;
realpart(z1);
((x^2−y^2)*(−y^4+6*x^2*y^2−x^4+3)−2*x*y*(4*x*y^3−4*x^3*y))/2
imagpart(z1);
(2*x*y*(−y^4+6*x^2*y^2−x^4+3)+(x^2−y^2)*(4*x*y^3−4*x^3*y))/2

Find fixed points using Maxima CAS :

z1:z^2*(3-z^4)/2;
s:solve(z1=z);
s:float(s);

result :

[z=−1.446857247913871,z=.7412709105660023,z=−1.357611535209976*%i−.1472068313260655,z=1.357611535209976*%i−.1472068313260655,z=1.0,z=0.0]

check multiplicities of the roots :

multiplicities;
[1,1,1,1,1,1]

 z1:z^2*(3-z^4)/2;
 s:solve(z1=z)$
 s:map(rhs,s)$
 f:z1;
 k:diff(f,z,1);
 define(d(z),k);
 m:map(d,s)$
 m:map(abs,m)$
 s:float(s);
 m:float(m);

Result : there are 6 fixed point 2 of them are supperattracting ( m=0 ), rest are repelling ( m>1 ):

 [−1.446857247913871,.7412709105660023,−1.357611535209976*%i−.1472068313260655,1.357611535209976*%i−.1472068313260655,1.0,0.0]
 [14.68114348748323,1.552374536603988,10.66447061028112,10.66447061028112,0.0,0.0]

Critical points :

[%i,−1.0,−1.0*%i,1.0,0.0]

degree 9 by Michael Becker

Julia set for f(z)=z2 over (z9-z+0.025)

References

Describing Blaschke products by their critical points by Oleg Ivrii, Tel Aviv University: video, presentation
mathoverflow question: finding-the-critical-points-of-a-degree-5-blaschke-product

Fractals/Rational

Contents

Examples

Gallery

Blaschke fraction

Finite Blaschke product

classification

Julia set

critial orbit

McMullen maps

degree 2

The Bubble bath Julia set

degree 2 by Michael Becker

1 period 4 basin

degree 3

degree 3 by Michael Becker

1 period 2 basin

2 period 2 basins

degree 5

by L. Javier Hernandez Paricio

degree 6

degree 9 by Michael Becker

References

See also