Show that orbits of quadratic polynomial $Q_{-2}$ tend to infinity (problem 7.4 -8 in Elaydi's "Discrete Chaos")

Question

The following sets and mappings are considered in chapter Fractals in Elaydi's Discrete Chaos: $B= \mathbb{C}\setminus [-2,2]$, $A= \{ z \in \mathbb{C} |\quad |z|>1\}$. Mapping $Q_{-2}:B \rightarrow B$ is defined as $Q_{-2}(z)=z^2-2$ and $Q_0: A \rightarrow A$ is $Q_0(z)=z^2$. I have to show that $Q_{-2}(z)\rightarrow \infty$, $\forall z \in B$.

My attempt was: if $z \in B$ and $|z|>2$, then by Escape criterion (Elaydi, Discrete Chaos, page 359) we have $|Q_{-2}^n (z)| \rightarrow \infty$, $n \rightarrow \infty$. But, what if $|z|<2$? All I know is that there exists the homeomorphism $h: A \rightarrow B$, $h(z)= z + \frac{1}{z}$ and that $Q_0(z) \rightarrow \infty$, $\forall z \in A$. Homeomorphism sends orbits of one system to orbits of another system. How can I prove that orbits of $Q_{-2}$ tend to infinity? Thank a lot in advance.

Answer 1

You know that $$ Q_{-2}(h(z))=h(z^2) $$ so that in consequence $$ Q_{-2}^{\circ n}(h(z))=h(z^{2^n})=z^{2^n}+z^{-2^n}. $$ This means that for $0<|z|<1$ and for $1<|z|$ the iteration with $Q_{-2}$ starting at $h(z)$ diverges to infinity. Note that the image of the unit circle under $h$ is the interval $[-2,2]$. What remains to show can be concluded from a formula like $$ w\in B\implies z=\frac{w}{2}\left(1+\sqrt{1-4w^{-2}}\right)\in A. $$

Answer 2

The map $h:A\to B$ maps circles to ellipses with focal point $-2,2$. This observation can be used: Take $z\in B$. This is equivalent to $|z-2|+|z+2|=:c>4$. Then show $|z^2-2 -2|+|z^2-2+2|>c$. Therefore points on a ellipse will be mapped to bigger ellipses. This leads to $\lim Q_{-2}^n(z) =\infty$

$h$ maps circles to ellipses. Take $z\in\mathbb{C}$ with $|z|=r$. Then $$|h(z)-2|+|h(z)+2|=\ldots=\frac{1}{|z|}(|z^2|-2(z+\overline{z})+1+|z^2|+2(z+\overline{z})+1)=\frac{2r^2+2}{r}\text{.}$$ Therefore the image of all points with $|z|=r$ are on the same ellipse with $|w-2|+|w+2|=\frac{2r^2+2}{r}$.