We now let \(f(z)=z^2-2\text{.}\) From our work in real iteration, we already know that \(f:[-2,2]\to[-2,2]\) and that \(f\) is chaotic on that interval. On the complement, \(\mathbb C\setminus [-2,2]\text{,}\) it turns out that \(f\) is stable. In fact, the iterates of \(f\) 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 \(f\) 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 1 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

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.