An archived instance of discourse for discussion in undergraduate Complex Dynamics.

# Finding a quadratic parameter with specified periodicity and multiplier

mark

Suppose we'd like to find a function of the form $f_c(z)=z^2+c$ with an orbit of period 3 and a mulitplier $\lambda = 5e^{\pi i/4}/6$.

• Write down a system of equations that $\lambda$ and the periodic point must satisfy. (Definite in class question!)
• Solve the system (Definite take home question!)
• Illustrate the corresponding Julia set (fun, fun, fun!)
RedCrayon

Let $z_0$ be a fixed point of $f_c$. By the definition of the multiplier, $$f_c^\prime(z_0)=2z_0=\lambda$$

This implies $z_0=5e^{\pi i/4}/12$.

Since $z_0$ is a fixed point of $f_c$, $$f_c(z_0)=z_0^2+c=z_0$$

Thus, $c=5e^{\pi i/4}/12-25e^{\pi i/2}/144$.

This $c$ value gives the Julia set

The white dots correspond to $-0.723234 - 0.561024 i$, $0.502947 + 0.932521 i$, and $-0.322011 + 1.05903 i$, which form an orbit of period $3$.

The red dots correspond to $-0.858598 + 0.631395 i$, $0.633157 - 0.963212 i$, and $-0.232262 - 1.09871 i$, which form an orbit of period $3$.

The green dot is located at $0.705372 - 0.294628 i$ and the pink dot is located at $0.294628 + 0.294628 i$. These are both fixed points of $f_c$, $c=5e^{\pi i/4}/12-25e^{\pi i/2}/144$.

mark

Well, this just looks so cool that I had to give it a like! But I do think there's a fundamental issue. The question asks for an orbit of period 3 - not a fixed point. So the equations we need are
\begin{align}
f_c^3(z) &= z \\
(f_c^3)'(z) &= 5e^{\pi i/4}/6.
\end{align}
The first says that $z$ is a fixed point; the second says that $z$ has the desired multiplier. These expand out to
\begin{align}
\left(\left(z^2+c\right)^2+c\right)^2+c &= z \\
8 z \left(z^2+c\right) \left(\left(z^2+c\right)^2+c\right) &= 5e^{\pi i/4}/6.
\end{align}
Of course, there's no way I'd ask you to solve these on a test. We can do so with Mathematica to get:

f[c_][z_] = z^2 + c;
F[c_][z_] = Nest[f[c], z, 3]
eqs = {F[c][z] == z, F[c]'[z] == 5 Exp[Pi*I/4]/6};
c /. NSolve[eqs, {c, z}]

(* Out:
{-1.74917 + 0.00245296 I, -1.74917 + 0.00245296 I,
-0.0690922 +  0.688001 I, -0.18174 - 0.690454 I,
-1.74917 + 0.00245296 I, -0.18174 - 0.690454 I,
-0.0690922 + 0.688001 I, -0.332706 + 0.554094 I,
0.0699479 - 0.565179 I, -0.0690922 + 0.688001 I,
-0.18174 - 0.690454 I, 0.262759 + 0.0110854 I}
*)