Julia sets for polynomial functions
" The dynamics of polynomials is much better understood than the dynamics of general rational maps" due tothe Bottcher’s theorem[1]
Map has d petals. "Although the parabolic point at z = 0 has only one petal, the map f also has a preparabolic critical point b = −1 of local degree d. Thus f has d petals at b. "[2]
For example "the filled Julia set of f (z) = z(1 + z) 3 has three petals at z = −1." ( Curtis T McMullen )
"The condition
describes a set of lines through the origin. Note that the map g
leaves these lines invariant. Then
is the composition of g with the periodic rotation "[3]
Example
Z = Zn + cp
L = (m - 1) * p
"the mulitbrot family, ... they all have a single critical point, namely the origin" Mark McClure[7]
L = m - 1
Compare with it's inverse parameter plane : Z^n + 1/c[12]
How to compute powers of z :
Fortan source code :
! Fortran program by P.M.J. Trevelyan! http://philiptrevelyan.co.uk/PROGRAMFRACTALIMPLICIT NONEINTEGERI,J,ITERATION,N,MPARAMETER(N=2000,M=50)REAL*8U,V,X,Y,P,QOPEN(99,FILE='Fractal_quad.dat')25FORMAT(2F9.5,I3)DO10,I=1,NDO20,J=1,NCDefinefirstpointz(n)=U+iVandk=X+iYU=0.D0V=0.D0X=I*3.2D0/(N-1.D0)-2.1D0Y=J*2.8D0/(N-1.D0)-1.4D0DO30,ITERATION=1,MCCalculatez(n+1)=z(n)**2+kwherez(n+1)=P+iQP=U**2-V**2+XQ=2.D0*U*V+YU=PV=QCIf|z|>2stopiteratingIf(U**2+V**2.GT.4.D0)GOTO10030CONTINUE100WRITE(99,25)X,Y,ITERATION20CONTINUE10CONTINUE STOP END
Parameter planes for positive integer powers :
Negative powers
Whend is negative the set surrounds but does not include the origin. There is interesting complex behaviour in the contours between the set and the origin, in a star-shaped area with(1 −d)-fold rotational symmetry. The sets appear to have a circular perimeter, however this is just an artifact of the fixed maximum radius allowed by the Escape Time algorithm, and is not a limit of the sets that actually extend in all directions to infinity.
Seecomplex quadratic polynomials
It can be computed by :[13]
Examples:
It is aninverted parameter plain of z^n + c.
Number of vertices : V = (n - 1)
McMullen maps :
where : n and d are >=1
"These maps are known as `McMullen maps', since McMullen[15] first studied these maps and pointed out that when (n: d) = (2: 3) and m is small, the Julia set is a Cantor set of circles."[16]
Seecomplex quadratic polynomial
It can be found using Maxima CAS :
(%i2) z:zx+zy*%i;(%o2) %i*zy+zx(%i6) realpart(z+z^3);(%o6) -3*zx*zy^2+zx^3+zx(%i7) imagpart(z+z^3);(%o7) -zy^3+3*zx^2*zy+zy
Finding roots and its multiplicity :
so root z=0 has multiplicity 3.
(%i1) z1:z^3+z;(%o1) z^3+z(%i2) solve(z1=z);(%o2) [z=0](%i3) multiplicities;(%o3) [3]
It means that there is aflower with 2 petals around fixed point z=0.
Compare figures :
How to compute iteration :
(%i1) z:zx+zy*%i;(%o1) %i*zy+zx(%i2) m:mx+my*%i;(%o2) %i*my+mx(%i3) z1:z^4+m*z;(%o3) (%i*zy+zx)^4+(%i*my+mx)*(%i*zy+zx)(%i4) realpart(z1);(%o4) zy^4-6*zx^2*zy^2-my*zy+zx^4+mx*zx(%i5) imagpart(z1);(%o5) -4*zx*zy^3+4*zx^3*zy+mx*zy+my*zx
See also f(z) = c(z^4-4z). It is a family 4.1 in program Mandel by Wolf Jung[21] ( see main menu / New / 4. Quartic polynomials / 4.1 )
Here c = -m/4 and Mandelbrot set is rotated by 180 degrees.

Maxima CAS code :
(%i1) f:z^4+m*z;(%o1) z^4+m*z(%i2) e1:f=z;(%o2) z^4+m*z=z(%i3) d:diff(f,z,1);(%o3) 4*z^3+m(%i4) e2:d=w;(%o4) 4*z^3+m=w(%i5) s:eliminate ([e1,e2], [z]);(%o5) [-(m-w)*(w+3*m-4)^3](%i6) s:solve([s[1]], [m]);(%o6) [m=-(w-4)/3,m=w]
It means that there are 2 period 1 components :

Finding roots and its multiplicity :
so root z=0 has multiplicity 4.
(%i1) z1:z^4+z;(%o1) z^4+z(%i2) solve(z1=z);(%o2) [z=0](%i3) multiplicities;(%o3) [4]
It means that there are 3 petals around fixed point z=0[22]
How to compute iteration :
(%i17) z:x+y*%i;(%o17) %i*y+x(%i18) realpart(z+z^4);(%o18) y^4-6*x^2*y^2+x^4+x(%i19) imagpart(z+z^4);(%o19) -4*x*y^3+4*x^3*y+y

First compute multiplier for internal angle=3/4 :
(%i1) m:exp(2*%pi*%i*3/4);(%o1) -%i
Then find how to compute iteration :
(%i1) z:x+y*%i;(%o1) %i*y+x(%i2) z1:z^4-%i*z;(%o2) (%i*y+x)^4-%i*(%i*y+x)(%i3) realpart(z1);(%o3) y^4-6*x^2*y^2+y+x^4(%i4) imagpart(z1);(%o4) -4*x*y^3+4*x^3*y-x
It is a parabolic Julia set with 12 petal flower[23]
Critical points :
(%i12) s:GiveListOfCriticalPoints(f(z))(%o12) [0.31498026247372*%i-0.54556181798586,-0.62996052494744*%i,0.31498026247372*%i+0.54556181798586](%i13) multiplicities(%o13) [1,1,1](%i14) length(s)(%o14) 3
with arguments in turns :
[0.41666666666667,0.75,0.083333333333334] = [5/12 , 9/12, 1/12]
Attracting vectors
Multiplier of fixed point −i is a fourth root of unity ( q=4), thus we examine 4-th iteration :
(%i1) z1:z^4-%i*z;(%o1) z^4-%i*z(%i2) z2:z1^4-%i*z1;(%o2) (z^4-%i*z)^4-%i*(z^4-%i*z)(%i3) z3:z2^4-%i*z2;(%o3) ((z^4-%i*z)^4-%i*(z^4-%i*z))^4-%i*((z^4-%i*z)^4-%i*(z^4-%i*z))(%i4) z4:z3^4-%i*z3;(%o4) (((z^4-%i*z)^4-%i*(z^4-%i*z))^4-%i*((z^4-%i*z)^4-%i*(z^4-%i*z)))^4-%i*(((z^4-%i*z)^4-%i*(z^4-%i*z))^4-%i*((z^4-%i*z)^4-%i*(z^4-%i*z)))(%i6) taylor(z4,z,0,20);(%o6)/T/ z+(-76*%i-84)*z^13+(-36*%i+720)*z^16+(1812*%i-2556)*z^19+...
Next term after z is a :
(-76*%i-84)*z^13
so here :
Attracting vectors satisfy :
so here :
One can solve it in Maxima CAS :
(%i14) s:map('float,s);(%o14) [1.007236559448514*%i+1.521106958434882,1.632845927320289*%i+0.81369898815363,1.820935547602145*%i-0.11173896888541,1.521106958434882*%i-1.007236559448514,0.81369898815363*%i-1.632845927320289,-0.11173896888541*%i-1.820935547602145,-1.007236559448514*%i-1.521106958434882,-1.632845927320289*%i-0.81369898815363,0.11173896888541-1.820935547602145*%i,1.007236559448514-1.521106958434882*%i,1.632845927320289-0.81369898815363*%i,0.11173896888541*%i+1.820935547602145]With arguments in turns :
[0.093087406197659,0.17642073953099,0.25975407286433,0.34308740619766,0.42642073953099,0.50975407286433,0.59308740619766,0.67642073953099,0.75975407286433,0.84308740619766,0.92642073953099,0.009754072864326]
different then arguments of critical points. Thus critical orbits form distorted 12-arms star
Findthe fixed points :
(%i1) f:z^4-%i*z;(%o1) z^4-%i*z(%i2) s:solve(f=z);(%o2) [z=((%i+1)^(1/3)*(sqrt(3)*%i-1))/2,z=-((%i+1)^(1/3)*(sqrt(3)*%i+1))/2,z=(%i+1)^(1/3),z=0](%i4) multiplicities;(%o4) [1,1,1,1](%i3) s:map(rhs,s);(%o3) [((%i+1)^(1/3)*(sqrt(3)*%i-1))/2,-((%i+1)^(1/3)*(sqrt(3)*%i+1))/2,(%i+1)^(1/3),0](%i5) s:map('float,s);(%o5) [0.5*(%i+1.0)^(1/3)*(1.732050807568877*%i-1.0),-0.5*(%i+1.0)^(1/3)*(1.732050807568877*%i+1.0),(%i+1.0)^(1/3),0.0](%i6) s:map(rectform,s);(%o6) [0.7937005259841*%i-0.7937005259841,-1.084215081491351*%i-0.29051455550725,0.29051455550725*%i+1.084215081491351,0.0]Computethe multiplier of fixed points :
(%i7) d:diff(f,z,1);(%o7) 4*z^3-%i
Checkthe stability of fixed points :
(%i9) for z in s do disp(abs(ev(d)));4.9999999999999985.04.9999999999999991(%o9) done
Point z=0 is a parabolic point.



It is a special case of polynomial from family :
Here
so internal angle is :
(%i2) m:exp(2*%pi*%i/2);(%o2) -1
Because :
it isa parabolic Julia set. Point is between two period one components ( root point ).
Periodic points
Point z=0 is a root of multiplicity seven
for equation :
One can check it in Maxima CAS using numerical :
(%i1) z1:z^4-z;(%o1) z^4-z(%i2) z2:z1^4-z1;(%o2) (z^4-z)^4-z^4+z(%i3) eq2:z2-z=0;(%o3) (z^4-z)^4-z^4=0(%i4) allroots(eq2);(%o4) [z=0.0,z=0.0,z=0.0,z=0.0,z=0.0,z=0.0,z=0.0,z=1.259921049894873,z=0.7937005259841*%i-0.7937005259841,z=-0.7937005259841*%i-0.7937005259841,z=1.084215081491351*%i-0.29051455550725,z=-1.084215081491351*%i-0.29051455550725,z=0.29051455550725*%i+1.084215081491351,z=1.084215081491351-0.29051455550725*%i,z=1.091123635971722*%i-0.62996052494744,z=-1.091123635971722*%i-0.62996052494744](%i5) expand(eq2);(%o5) z^16-4*z^13+6*z^10-4*z^7=0(%i6) factor(eq2);(%o6) z^7*(z^3-2)*(z^6-2*z^3+2)=0
and symbolic methods :
(%i1) z1:z^4-z;(%o1) z^4-z(%i2) solve(z1=z);(%o2) [z=(2^(1/3)*sqrt(3)*%i-2^(1/3))/2,z=-(2^(1/3)*sqrt(3)*%i+2^(1/3))/2,z=2^(1/3),z=0](%i3) multiplicities;(%o3) [1,1,1,1](%i4) z2:z1^4-z1;(%o4) (z^4-z)^4-z^4+z(%i5) solve(z2=z);(%o5) [z=(2^(1/3)*sqrt(3)*%i-2^(1/3))/2,z=-(2^(1/3)*sqrt(3)*%i+2^(1/3))/2,z=2^(1/3),z=((%i+1)^(1/3)*(sqrt(3)*%i-1))/2,z=-((%i+1)^(1/3)*(sqrt(3)*%i+1))/2,z=(%i+1)^(1/3),z=(sqrt(3)*(1-%i)^(1/3)*%i-(1-%i)^(1/3))/2,z=-(sqrt(3)*(1-%i)^(1/3)*%i+(1-%i)^(1/3))/2,z=(1-%i)^(1/3),z=0](%i6) multiplicities;(%o6) [1,1,1,1,1,1,1,1,1,7]
Number of petals = 6[24]
Atracting vectorsDenominator of internal angle is so one have to check second iteration of function :
(%i5) z1:z^4-z;(%o5) z^4-z(%i6) z2:z1^4-z1;(%o6) (z^4-z)^4-z^4+z(%i8) expand(z2);(%o8) z^16-4*z^13+6*z^10-4*z^7+z
Next term after z is a -4z^7. Then :
Attracting vectors satisfy :
so here :
One can solve it using Maxima CAS :
(%i10) s:solve(z^6=1/24);(%o10) [z=(sqrt(3)*%i+1)/(2^(3/2)*3^(1/6)),z=(sqrt(3)*%i-1)/(2^(3/2)*3^(1/6)),z=-1/(sqrt(2)*3^(1/6)),z=-(sqrt(3)*%i+1)/(2^(3/2)*3^(1/6)),z=-(sqrt(3)*%i-1)/(2^(3/2)*3^(1/6)),z=1/(sqrt(2)*3^(1/6))](%i11) s:map(rhs,s);(%o11) [(sqrt(3)*%i+1)/(2^(3/2)*3^(1/6)),(sqrt(3)*%i-1)/(2^(3/2)*3^(1/6)),-1/(sqrt(2)*3^(1/6)),-(sqrt(3)*%i+1)/(2^(3/2)*3^(1/6)),-(sqrt(3)*%i-1)/(2^(3/2)*3^(1/6)),1/(sqrt(2)*3^(1/6))](%i12) s:map('float,s);(%o12) [0.29439796075012*(1.732050807568877*%i+1.0),0.29439796075012*(1.732050807568877*%i-1.0),-0.58879592150024,-0.29439796075012*(1.732050807568877*%i+1.0),-0.29439796075012*(1.732050807568877*%i-1.0),0.58879592150024](%i13) s:map(rectform,s);(%o13) [0.50991222566388*%i+0.29439796075012,0.50991222566388*%i-0.29439796075012,-0.58879592150024,-0.50991222566388*%i-0.29439796075012,0.29439796075012-0.50991222566388*%i,0.58879592150024](%i14) s:map(carg_t,s);(%o14) [0.5235987755983/%pi,1.047197551196598/%pi,1/2,1-1.047197551196598/%pi,1-0.5235987755983/%pi,0](%i15) s:map('float,s);(%o15) [0.16666666666667,0.33333333333333,0.5,0.66666666666667,0.83333333333333,0.0]So critical points lie on attracting vectors. Thus critical orbits tend straight to the origin under the iteration[25]
How to compute:
(%i2) z:x+y*%i;(%o2) %i*y+x(%i3) realpart(z^4-z);(%o3) y^4-6*x^2*y^2+x^4-x(%i4) imagpart(z^4-z);(%o4) -4*x*y^3+4*x^3*y-y
Critical points :
s:GiveListOfCriticalPoints(f(z))(%o8) [0.54556181798586*%i-0.31498026247372,-0.54556181798586*%i-0.31498026247372,0.62996052494744]
These points has arguments in turns : 1/3, 2/3, 0

Finding roots and its multiplicity :
so root z=0 has multiplicity 5. It means that there is aflower with 4 petals[26]
around fixed point z=0.
It How to compute :
(%i23) z:x+y*%i;(%o23) %i*y+x(%i24) realpart(z+z^5);(%o24) 5*x*y^4-10*x^3*y^2+x^5+x(%i25) imagpart(z+z^5);(%o25) y^5-10*x^2*y^3+5*x^4*y+y
In c programs one must use temporary variable so it can be :
tempx=5*x*y*y*y*y-10*x*x*x*y*y+x*x*x*x*x+x;// temporary variabley=y*y*y*y*y-10*x*x*y*y*y+5*x*x*x*x*y+y;x=tempx;
It can beoptimized
"...an escape time algorithm would take forever to generate that type of image, since the dynamics are so slow there. If you want resolution of 1/100, it would take roughly 2*10^8 iterates to move the point z0=0.01 to z=2 by iterating f(z)=z+z^5." ( Mark McClure[27]
"This picture shows the Julia Set of f(z) = z + z^5 which has an indifferent fixed point at z = 0. ( f(0) = 0 and f ' (0) = 1 .)
The 4 lines :
Re z = 0 and Im z = 0 and Re z = Im z and Re z = -Im z
are invariant under iteration of f.
On Im z = 0: f(x) = x + x^5
On Re z = 0: f(ix) = ix + (ix)^5 = ix + i^5 x^5 = i(x+x^5)
On Re z = Im z , f(z) = r e^(i pi/4) + r^5 e^(i 5 pi/4) = e^ (i pi/4)(r - r^5)
On Re z = -Im z, f(z) = = r e^(i 3pi/4) + r^5 e^(i 15 pi/4) = e^ (i 3pi/4)(r - r^5)
Using one dimensional analysis it is easily shown that f(x) = x + x^5 has a repelling fixed point at x = 0 and f(x) = x - x^5 has an attracting fixed point at x = 0. Thus along the four invariant lines 0 is attracting on the first two and repelling on the second two. The points repelled from 0 are shown in shades of blue, while those attracted to 0 are shown in shades of brown. 0 has four attracting petals, which are in shades of brown. (A simply connected region C is a petal for an indifferent fixed point p if p is contained in the boundary of C and for each z in C,
F^n(z) -> p (see Devaney - 1987)"[28]
c=(-6145144-20171676*i) * 2^-25 = (-6145144 - 20171676 i)/2^25 = -0.1831395626068115234375 - 0.60116279125213623046875 i
Critical points:
List:
z = -0.454407 + 0.0918858 I z = -0.227808 - 0.403772 I z = -0.0530308 + 0.460561 I z = 0.313614 - 0.341431 I z = 0.421632 + 0.192756 I
Higher precision
+0.4216319827875524+0.1927564710317439*%i-0.4544068504035357+0.09188580053693407*%i-0.2278080284907348 -0.4037723222177803*%i+0.3136137458861571-0.3414308193839966*%i-0.05303084977943909+0.4605608700330989*%i
dynamical plane
z6+z on plane [-1.2;1.2]x[-1.2;1.2]. It has 5 petals[29]


How to compute iteration :
/* Maxima CAS session */(%i1) z:x+y*%i;(%o1) %i*y+x(%i2) z1:z^14-z;(%o2) (%i*y+x)^14-%i*y-x(%i3) realpart(z1);(%o3) -y^14+91*x^2*y^12-1001*x^4*y^10+3003*x^6*y^8-3003*x^8*y^6+1001*x^10*y^4-91*x^12*y^2+x^14-x(%i4) imagpart(z1);(%o4) 14*x*y^13-364*x^3*y^11+2002*x^5*y^9-3432*x^7*y^7+2002*x^9*y^5-364*x^11*y^3+14*x^13*y-y
f(z)=z^14-z, on [-1,2;1,2]x[-1,2;1,2] has 26 petals. Compare with image by Michael Becker.[30]
How to find fixed points :
(%i1) z1:z^14-z;(%o1) z^14-z(%i2) solve(z1=z);(%o2) [z=2^(1/13)*%e^((2*%i*%pi)/13),z=2^(1/13)*%e^((4*%i*%pi)/13), z=2^(1/13)*%e^((6*%i*%pi)/13),z=2^(1/13)*%e^((8*%i*%pi)/13), z=2^(1/13)*%e^((10*%i*%pi)/13),z=2^(1/13)*%e^((12*%i*%pi)/13), z=2^(1/13)*%e^(-(12*%i*%pi)/13),z=2^(1/13)*%e^(-(10*%i*%pi)/13), z=2^(1/13)*%e^(-(8*%i*%pi)/13),z=2^(1/13)*%e^(-(6*%i*%pi)/13), z=2^(1/13)*%e^(-(4*%i*%pi)/13),z=2^(1/13)*%e^(-(2*%i*%pi)/13), z=2^(1/13),z=0](%i3) multiplicities;(%o3) [1,1,1,1,1,1,1,1,1,1,1,1,1,1](%i4) z2:z1^14-z1;(%o4) (z^14-z)^14-z^14+z(%i5) solve(z2=z);(%o5) [z=2^(1/13)*%e^((2*%i*%pi)/13),z=2^(1/13)*%e^((4*%i*%pi)/13), z=2^(1/13)*%e^((6*%i*%pi)/13),z=2^(1/13)*%e^((8*%i*%pi)/13), z=2^(1/13)*%e^((10*%i*%pi)/13),z=2^(1/13)*%e^((12*%i*%pi)/13), z=2^(1/13)*%e^(-(12*%i*%pi)/13),z=2^(1/13)*%e^(-(10*%i*%pi)/13), z=2^(1/13)*%e^(-(8*%i*%pi)/13),z=2^(1/13)*%e^(-(6*%i*%pi)/13), z=2^(1/13)*%e^(-(4*%i*%pi)/13),z=2^(1/13)*%e^(-(2*%i*%pi)/13), z=2^(1/13),z=0,0=z^78-7*z^65+21*z^52-35*z^39+35*z^26-21*z^13+7, 0=z^78-5*z^65+11*z^52-13*z^39+9*z^26-3*z^13+1](%i6) multiplicities;(%o6) [1,1,1,1,1,1,1,1,1,1,1,1,1,27,1,1]


How to compute iterations :
/* Maxima CAS session */(%i1) z:x+y*%i;(%o1) %i*y+x(%i2) z1:z^15-z;(%o2) (%i*y+x)^15-%i*y-x(%i3) realpart(z1);(%o3) -15*x*y^14+455*x^3*y^12-3003*x^5*y^10+6435*x^7*y^8-5005*x^9*y^6+1365*x^11*y^4-105*x^13*y^2+x^15-x(%i4) imagpart(z1);(%o4) -y^15+105*x^2*y^13-1365*x^4*y^11+5005*x^6*y^9-6435*x^8*y^7+3003*x^10*y^5-455*x^12*y^3+15*x^14*y-y
Critical points :
(%i1) m:-1;f:z^15+ m*z;d:diff(f,z,1);s:solve(d=0,z)$s:map(rhs,s)$s:map(rectform,s)$s:map('float,s);multiplicities;(%o1) -1(%o2) z^15-z(%o3) 15*z^14-1(%o7) [0.35757475986465*%i+0.74251163973317,0.64432745317147*%i+0.51383399763062,0.80346319222004*%i+0.1833852305369,0.80346319222004*%i-0.1833852305369,0.64432745317147*%i-0.51383399763062,0.35757475986465*%i-0.74251163973317,-0.8241257452789,-0.35757475986465*%i-0.74251163973317,-0.64432745317147*%i-0.51383399763062,-0.80346319222004*%i-0.1833852305369,0.1833852305369-0.80346319222004*%i,0.51383399763062-0.64432745317147*%i,0.74251163973317-0.35757475986465*%i,0.8241257452789](%o8) [1,1,1,1,1,1,1,1,1,1,1,1,1,1]It means that here are 14 critical points and 14 critical orbits.
Fixed points :
kill(all);remvalue(all);/*------------- functions definitions ---------*//* function */f(z):=z^15 -z;/* find fixed points returns a list */GiveFixedPoints():= block( [s], s:solve(f(z)=z), /* remove "z=" from list s */ s:map('rhs,s), s:map('rectform,s), s:map('float,s), return(s))$compile(all);ff:GiveFixedPoints();multiplicities;length(s);for i:1 thru length(ff) step 1 do (z:ff[i], disp("z= ",z, " abs(d(z))= ",abs(15*z^14-1)));Result is :
(%i12) ff:GiveFixedPoints()(%o12) [0.45590621928146*%i+0.94669901916834,0.82151462051137*%i+0.65513604843564,1.024411975933374*%i+0.23381534859391,1.024411975933374*%i-0.23381534859391,0.82151462051137*%i-0.65513604843564,0.45590621928146*%i-0.94669901916834,-1.050756638653219,-0.45590621928146*%i-0.94669901916834,-0.82151462051137*%i-0.65513604843564,-1.024411975933374*%i-0.23381534859391,0.23381534859391-1.024411975933374*%i,0.65513604843564-0.82151462051137*%i,0.94669901916834-0.45590621928146*%i,1.050756638653219,0.0](%i13) multiplicities(%o13) [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1](%i14) length(s)(%o14) 14(%i15) for i thru length(ff) do (z:ff[i],disp("z= ",z," abs(d(z))= ",abs(15*z^14-1)))z= 0.45590621928146*%i+0.94669901916834 ; abs(d(z))= 28.99999999999996z= 0.82151462051137*%i+0.65513604843564 ; abs(d(z))= 28.99999999999998z= 1.024411975933374*%i+0.23381534859391 ; abs(d(z))= 28.99999999999999z= 1.024411975933374*%i-0.23381534859391 ; abs(d(z))= 28.99999999999997z= 0.82151462051137*%i-0.65513604843564 ; abs(d(z))= 29.00000000000001z= 0.45590621928146*%i-0.94669901916834 ; abs(d(z))= 28.99999999999995z= -1.050756638653219 ; abs(d(z))= 29.00000000000003z= -0.45590621928146*%i-0.94669901916834; abs(d(z))= 28.99999999999995z= -0.82151462051137*%i-0.65513604843564; abs(d(z))= 29.00000000000001z= -1.024411975933374*%i-0.23381534859391 ; abs(d(z))= 28.99999999999997z= 0.23381534859391-1.024411975933374*%i abs(d(z))= 28.99999999999999z= 0.65513604843564-0.82151462051137*%i ; abs(d(z))= 28.99999999999998z= 0.94669901916834-0.45590621928146*%i ; abs(d(z))= 28.99999999999996z= 1.050756638653219 ; abs(d(z))= 29.00000000000003z= 0.0 ; abs(d(z))= 1.0So only z=0 is parabolic fixed points, the rest of them are repelling

Description[33]
The orbit dynamics of the set can become more complex when the power is something other than 2.0. The iterated function can becomemultivalued and the structure of the set is then affected by the'arbitrary' choice of which value is chosen.