Finding super-attractive orbits in the Mandelbrot set

Suppose I want to find all the $c$ values such that $f_c(z)=z^2+c$ has a super-attractive orbit of period 6.

• What equation must I solve?
• Solve the equation (possible take home)
• Illustrate

Suppose I want to find all the cc values such that $f_c(z)=z^2+c$ has a super-attractive orbit of period 6.

What equation must I solve?

$(((((0^2+c)^2+c)^2+c)^2+c)^2+c)^2+c=0$.

I thought this is what we went over in class, but when I try to solve for this, I just get $c=0$ which seems too easy to be right. Someone tell me what I'm missing here.

@Yousername You've got the correct equation and, to be clear, I would not ask for more than that on the in class portion of the test. You're also correct that zero is a solution. It's not the only solution, however. In fact, the equation is of order 32 in $c$, so we expect 32 solutions. We can find these using Mathematica on a take home portion like so:

f[c_][z_] = z^2 + c;
cs = c /. NSolve[Nest[f[c], 0, 6] == 0, c]

(* Out:
  {-1.99638, -1.96677, -1.90728, -1.7729, -1.75488, -1.47601, 
  -1.28408 - 0.427269 I, -1.28408 + 0.427269 I, 
  -1.138 - 0.240332 I, -1.138 + 0.240332 I, -1., 
  -0.596892 - 0.662981 I, -0.596892 + 0.662981 I, 
  -0.217527 - 1.11445 I, -0.217527 + 1.11445 I, 
  -0.163598 - 1.09778 I, -0.163598 + 1.09778 I, 
  -0.122561 - 0.744862 I, -0.122561 + 0.744862 I, 
  -0.113419 - 0.860569 I, -0.113419 + 0.860569 I, 
  -0.0155704 - 1.0205 I, -0.0155704 + 1.0205 I, 0., 
  0.359893 - 0.684762 I, 0.359893 + 0.684762 I, 
  0.389007 - 0.215851 I, 0.389007 + 0.215851 I, 
  0.396535 - 0.604182 I, 0.396535 + 0.604182 I, 
  0.443326 - 0.372962 I, 0.443326 + 0.372962 I}
*)

And we can even show where these points lie in the Mandelbrot set:

MandelbrotSetPlot[Epilog -> {Yellow, Point[{Re[#], Im[#]} & /@ cs]}]

For the take home portion. Is it true that just using NSolve will return all of the roots which will include some values of c which will have lower (with less elements) orbits?

@Captain_Flapjack Yes, that's correct. If, for example, $z_0$ has period 3 so that
$$z_0 \to z_1 \to z_2 \to z_0,$$
then
$$z_0 \to z_1 \to z_2 \to z_0 \to z_1 \to z_2 \to z_0.$$
Thus, $z_0$ maps back to itself after six iterates - or $3n$ iterates for any $n$.

Of course, the easy way to deal with this is to delete the lower order orbits that you find.