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Section 3.4 Another look at conjugacy

In this section, we'll look at some consequences of conjugacy within the quadratic family.

Subsection 3.4.1 Representing all quadratics

One reason that the quadratic family is so important is that the dynamics of every quadratic is captured by the behavior. This is made precise in Checkpoint 3.4.1

Let \(g(z) = \alpha z^2 + \beta z + \gamma\) and let \(\phi(z) = a z + b\text{.}\) Show that \(f_c\circ\varphi = \varphi\circ g\) when \(a=\alpha\text{,}\) \(b=\beta/2\text{,}\) and

\begin{equation*} c = \frac{1}{4} \left(4 \alpha \gamma -\beta ^2+2 \beta \right). \end{equation*}

That is, \(\varphi\) conjugates \(f_c\) to \(g\text{.}\)

Let \(g(z) = (1+i)z^2 - z + i/4\text{.}\) Find values of \(a\text{,}\) \(b\text{,}\) and \(c\) such that \(\varphi(z)=az+b\) conjugates \(f_c\) to \(g\text{.}\) Use the computer to generate images of the Julia sets of both \(f_c\) and \(g\text{.}\)

Subsection 3.4.2 The chaotic squaring map

Let \(H=[0,1)\) denote the half-open/half-closed unit interval and let \(d:H\to H\) denote the doubling map

\begin{equation*} d(x) = 2x \mod 1. \end{equation*}

Show that \(\varpi(x) = e^{2\pi i x}\) defines a conjugacy between \(d\) and the complex function \(f_0(z)=z^2\) restricted to the unit circle.

Subsection 3.4.3 A line segment

We now let \(c=-2\) so that \(f_c(z)=z^2-2\text{.}\) From our work in real iteration, we already know this function maps \([-2,2]\to[-2,2]\) and that it is chaotic on that interval. On the complement, \(\mathbb C\setminus [-2,2]\text{,}\) it turns out that \(f_{-2}\) is stable. In fact, the iterates of \(f_{-2}\) diverge to \(\infty\) for every initial point \(z_0 \in \mathbb C\setminus [-2,2]\text{.}\)

To see this, let \(F=\mathbb C\setminus [-2,2]\) and let \(E\) denote the exterior of the unit disk, i.e. \(E = \mathbb C\setminus \{z:|z|\leq1\}\text{.}\) We'll show that the action of \(f_{-2}\) on \(F\) is conjugate to the action of the squaring function \(g(z)=z^2\) on \(E\text{.}\)

The conjugacy function will be \(\varphi(z) = z+1/z\text{.}\) The geometric action of \(\varphi\) is shown in figure Figure 3.4.4 We first show that \(\varphi\) maps the boundary of \(E\) (the unit circle) to the boundary of \(F\) (the interval \([-2,2]\)). Of course, the points on the unit circle are exactly those points of the form \(e^{it}\) for some \(t\in [0,2\pi)\text{.}\) Thus, we compute

\begin{align*} \varphi(e^{it}) & = e^{it} + e^{-it}\\ & = (\cos(t) + i\sin(t)) + (\cos(t) - i\sin(t))\\ & = 2\cos(t). \end{align*}

The expression \(2\cos(t)\) traces out the interval \([-2,2]\) (twice, in fact) as \(t\) ranges through \([0,2\pi)\) as claimed.

We next show that \(\varphi\) is one-to-one on \(E\text{.}\) To this end, suppose that

\begin{equation*} z+\frac{1}{z} = w+\frac{1}{w}. \end{equation*}

Then,

\begin{equation*} z-w = \frac{1}{w} - \frac{1}{z} = \frac{z-w}{wz}. \end{equation*}

Assuming that \(z\neq w\text{,}\) we can divide off the numerators and then take reciprocals to obtain \(wz=1\text{.}\) Thus, if \(z\) is a point in the exterior of the unit disk, there is exactly one other point \(w\) such that \(\varphi(w)=\varphi(z)\text{,}\) namely the reciprocal of \(z\) which lies in the interior of the unit disk.

Figure 3.4.4. The conjugacy from \(F\) to \(E\)