How to prove by induction that $|z_n| \geq |c|r^{n-1}$ in complex numbers

Question

Having $c \in \mathbb{C}$, $z_0 = 0$, $r=|c|-1$ and $z_n = z_{n-1}^2 + c$ for $n \geq 1$. Prove that if $|c| > 2$, then $|z_n| \geq |c|r^{n-1}$.

I have proved that it is true for $z_1$ and $z_2$.

Then I want to prove it for $n+1$: $|z_{n+1}| \geq |z_{n}|^2 + |c| \underset{by \; I.H.}{\geq} |c|^2r^{2n-2}+|c|$.

I tried to show that $|c|^2r^{2n-2}+|c| \geq |c|r^{n-1}$ but I do not know how.

Answer 1

It is equivalent to show that $$|c|u^2-u+1\ge 0$$with $u=r^{n-1}$. This inequality is quadratic with $\Delta=b^2-4ac=1-4|c|\le -7<0$, hence the inequality holds for every $u$ and the induction is complete.