Fractals/Multibrot sets

Julia sets for polynomial functions

z^2+c
See complex quadratic polynomials

z^3 + c
$$z_{n+1} = z^3_n + c$$

It can be computed by :

$$  X_{n+1} = X^3_n - 3X_n Y^2_n + C_x$$

$$  Y_{n+1} = 3 X_n^2 Y_n - Y_n^3 + C_y$$

Examples:
 * c= -0.040000000000000036 + I * -0.78

zn + 1/c
It is an inverted parameter plain of z^n + c.

Number of vertices : V = (n - 1)

=z^n + m*z^(-d)=

McMullen maps :

$$z^n + \frac{m}{z^d}$$

where : n and d are >=1

"These maps are known as `McMullen maps', since McMullen 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."

=z^n + m*z =

Description

z^2+m*z
See complex quadratic polynomial

z^3 + z
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 :


 * $$z^3 + z = z$$


 * $$z^3 = 0 $$

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 a flower with 2 petals around fixed point z=0.

Compare figures :
 * by Alessandro Rosa
 * by Xavier Buff and Adam L. Epstein

z^4 + mz
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 ( see main menu / New / 4. Quartic polynomials / 4.1 )

Here c = -m/4 and Mandelbrot set is rotated by 180 degrees.

Period 1 components
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 :
 * one with radius = 1 and center =0 ( m=w )
 * second with radius 1/3 and center=4/3 ( m=-(w-4)/3 )

z^4+z


Finding roots and its multiplicity :


 * $$z^4 + z = z$$


 * $$z^4 = 0 $$

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

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

z^4-iz
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

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 :
 * k=13 and n=m*q = k-1 = 12
 * a = -76*%i-84

Attracting vectors satisfy :

$$nav^n = -1$$

so here :

$$-12(76i + 84)v^{12} = -1$$

$$v^{12} = \frac{1}{912i + 1008}$$

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

Find the 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]

Compute the multiplier of fixed points : (%i7) d:diff(f,z,1); (%o7) 4*z^3-%i

Check the stability of fixed points : (%i9) for z in s do disp(abs(ev(d))); 4.999999999999998 5.0 4.999999999999999 1 (%o9) done

Point z=0 is a parabolic point.

z^4 -z
It is a special case of polynomial from family :

$$f_{\lambda}(z) = z^4 + \lambda z$$

Here

$$\lambda = -1 = e^{\pi i} $$

so internal angle $$\theta $$ is :

$$\theta = \frac{p}{q} = \frac{1}{2} $$

(%i2) m:exp(2*%pi*%i/2); (%o2) -1

Because :

$$|\lambda| = 1 $$

it is a parabolic Julia set. Point $$\lambda = -1 = e^{\pi i} $$ is between two period one components ( root point ).

Periodic points

Point z=0 is a root of multiplicity seven

$$k = m*q + 1 = 3*2 +1 = 7$$

for equation :

$$f^2_{\lambda}(z) = z$$

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

$$m*q = 3*2 = 6$$

Atracting vectors  Denominator of internal angle $$\theta = \frac{p}{q} = \frac{1}{2} $$ is $$q = 2 $$ 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 :
 * k = 7 and n=m*q = k-1 = 6
 * a = -4

Attracting vectors satisfy :

$$nav^n = -1 $$

so here :

$$-24v^6 = -1 $$

$$v^6 = \frac{1}{24} $$

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

How to compute $$f_{\lambda}(z) $$: (%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

z^5 + z


Finding roots and its multiplicity :


 * $$z^5 + z = z$$


 * $$z^5 = 0 $$

so root z=0 has multiplicity 5. It means that there is a flower with 4 petals 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 :

It can be optimized

"...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

"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)"

z^6+A*z+c
c=(-6145144-20171676*i) * 2^-25 = (-6145144 - 20171676 i)/2^25 = -0.1831395626068115234375 - 0.60116279125213623046875 i

Critical points:
 * using WolframAlfa

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

z^6+m*z
dynamical plane

z6+z on plane [-1.2;1.2]x[-1.2;1.2]. It has 5 petals

z^14 - z


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.

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]

z^15-z
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))= 28.99999999999996 z= 0.82151462051137*%i+0.65513604843564 ; abs(d(z))= 28.99999999999998 z= 1.024411975933374*%i+0.23381534859391 ; abs(d(z))= 28.99999999999999 z= 1.024411975933374*%i-0.23381534859391 ; abs(d(z))= 28.99999999999997 z= 0.82151462051137*%i-0.65513604843564 ; abs(d(z))= 29.00000000000001 z= 0.45590621928146*%i-0.94669901916834 ; abs(d(z))= 28.99999999999995 z= -1.050756638653219 ;                  abs(d(z))= 29.00000000000003 z= -0.45590621928146*%i-0.94669901916834; abs(d(z))= 28.99999999999995 z= -0.82151462051137*%i-0.65513604843564; abs(d(z))= 29.00000000000001 z= -1.024411975933374*%i-0.23381534859391 ; abs(d(z))= 28.99999999999997 z= 0.23381534859391-1.024411975933374*%i abs(d(z))= 28.99999999999999 z= 0.65513604843564-0.82151462051137*%i ; abs(d(z))= 28.99999999999998 z= 0.94669901916834-0.45590621928146*%i ; abs(d(z))= 28.99999999999996 z= 1.050756638653219 ;                   abs(d(z))= 29.00000000000003 z= 0.0 ;                                  abs(d(z))= 1.0

So only z=0 is parabolic fixed points, the rest of them are repelling

=Other =
 * multibrot
 * z^2 +cz^5

m = 08+0.8i
Description
 * map : $$f(z)=z^5+(0.8+0.8i)z^4+z$$
 * coefficients in ascending order ( from ao to an): 0,1,0,0, 0.8+0.8i, 1
 * ListOfCriticalPoints [(- 0.7558074500261052 %i) - 0.7558074500261052, 0.2793534499540583 %i - 0.5310341598343944, 0.2793534499540583 - 0.5310341598343943 %i, 0.3674881599064412 %i + 0.3674881599064413]
 * fixed points : [(- 0.8 %i) - 0.8, 0.0] with stability : [0.6384000000000008, 1.0]

m=0.8+0.4i
=Inside dynamics under generalized power setting=

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

=See also=
 * commons:Category:Complex polynomial maps
 * Derivative of iterated fuction
 * Mark McClure : visualization of polynomial_julia_sets, coefficients a0, a1, a2, a3, .... for f(z) = an*z^n+ ... + a2*z^2 + a1*z + a0, example  : 1.42+0.37i,-1.5,0,1
 * Exploring the 4-Dimensional Mandelbrot Set
 * Elliptic curve julia sets by Clifford A. Reiter
 * FF: Julia sets: True shape and escape time

=References=