The Question Let $c$ be a complex number. The complex numbers $z_n(c)$ are defined recursively by
$z_1(c)=c$, $z_{n++1}(c)=(z_n(c))^2+c$ for $n\geq1$
The Mandelbrot set is defined by
$M=\{c\in\mathbb{C} |$ the sequence $(z_n(c))$ is bounded $\}$
Show that each of $-2, -1, 0, i$ lies in $M$ but that $1 \notin M$
For -2, -1, and 0 I just did some math and showed by example that they have an upper bound. I am stuck on how to prove that $i\in M$. I'd imagine that it's an induction, but not sure how to go about it.
Additionally, it's pretty obvious (if you go through it a bit) that for $c=1$ the set of recursive equations are unbounded. Could someone help me figure out how to to show this inductively?
The sequence for $i$ goes like $$i\mapsto -1+i\mapsto -i\mapsto -1+i\mapsto\ldots $$ so it is eventually periodic, hence bounded. The sequence for $1$ starts $$1\mapsto 2\mapsto 5\mapsto\ldots $$ so seems to diverge. Indeed, this follows from proposition 2 below.
Proposition 1. If $|c|>2$, then $c\notin M$.
Proof. Now if $|c|>2$, then we show by induction that $|z_n(c)|$ is $\ge |c|$ and strictly increasing by a factor $>1$: $$|z_{n+1}(c)|\ge |z_n(c)|^2-|c|\ge (|z_n(c)|-1)|c|>|c|$$ as $|z_n(c)|\ge |c|>2$. And then $$|z_{n+1}(c)|\ge |z_n(c)|^2-|c|\ge |z_n(c)|\cdot|c|-|z_n(c)|=|z_n(c)|(|c|-1).$$ We conclude $|z_{n}(c)|\ge (|c|-1)^n$ for $n\ge 1$, so the sequence is unbounded. $_\square$
Proposition 2. If $|z_m(c)|>2$ for some $m$, then $c\notin M$.
Proof. By proposition 1, we may assume $|c|\le 2$. $$ |z_{n+1}(c)|\ge |z_n(c)|^2-2$$ Now if $|z_n(c)|=2+h$, then $|z_{n+1}(c)|\ge 2+2h$. Thus from $n=m$ on, $|z_n(c)|-2$ grows exponentially. $_\square$