Why does the sequence $w_n = T^n(0)$ tend to infinity if $|z_0| > 2$?

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Here is this week's POTW:

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Let $z_0$ be complex constant. Consider the quadratic map $T : \Bbb C \to \Bbb C$ given by $T(w) = w^2 + z_0$. Show that the sequence $w_n = T^n(0)$ tends to infinity if $|z_0| > 2$.

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No one answered this week's problem. You can find my solution below.

Spoiler

Let $\lambda = |z_0| - 1$. Then $|w_n| \ge \lambda^{n-1}|z_0|$ for all $n \ge 2$ by induction on $n$. Indeed, $|w_1| = |c| = |c|r^{1-1}$, and if $|w_n| \ge \lambda^{n-1}|z_0|$ for some $n \ge 2$, then

$$|w_{n+1}| \ge |w_n|^2 - |c| \ge \lambda^{2n-2}|z_0|^2 - |z_0| = (\lambda^{2n-2}|z_0| - 1)|z_0| \ge (\lambda^{2n-2}|z_0| - \lambda^n)|z_0| = (\lambda^{n-2}|z_0| - 1)\lambda^n|z_0|.$$

Since $\lambda > 1 > \frac{2}{|z_0|}$ and $n \ge 2$, then $\lambda^{n-2} > \frac{2}{|z_0|}$. Thus $\lambda^{n-2}|z_0| - 1 \ge 1$. Therefore $|w_{n+1}| \ge \lambda^n|z_0|$.

Now as $\lambda > 1$, $\lambda^{n-1}|z_0| \to \infty$ as $n\to \infty$. Hence $w_n \to \infty$ as $n\to \infty$.