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
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
Then,
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.
