I am trying to prove that the Mandelbrot set defined as the set $\mathcal M$ of complex numbers $c$: the recursive sequence defined as $$z_0=c, \space \space \space z_{n+1}={z_n}^2+c$$ is bounded.
Prove that $\mathcal M \subset \{|z|\leq 2\}$.
I've tried to show this by the absurd: take $c \in \mathcal M$ : $|c|>2$. Now I would like to show that the sequence $\{z_n\}_{n \in \infty}$ is unbounded. I need to show that for an arbitrary $M \in \mathbb R_{\geq 0}$ there is $N \in \mathbb n$ such that if $n\geq N \implies |z_n|>M$. I am stuck, what could I do?
To simplify notation, put $\gamma = |c|$. Suppose $\gamma > 2$. If $|z_n| \geq 2^n \gamma$, then $$|z_{n+1}| \geq |z_n|^2 - \gamma \geq 2^{2n} \gamma^2 - \gamma \geq \left(2^{2n} - \frac{1}{2}\right)\gamma^2 \geq 2^{2n-1}\gamma^2 \geq 2^{n+1}\gamma.$$ Thus all $|z_n| \geq 2^{n}\gamma$ by induction. It follows that $c\not\in {\cal M}$.