# Divergence of sequence

• Apr 3rd 2011, 03:49 PM
EinStone
Divergence of sequence
Consider $f: \mathbb{C} \rightarrow \mathbb{C}$, $f(z) = z^2 - 2$.
Show that if $|z| > 2$, then the sequence z, f(z), f(f(z), ... diverges, i.e. is unbounded.
• Apr 3rd 2011, 06:21 PM
Drexel28
Quote:

Originally Posted by EinStone
Consider $f: \mathbb{C} \rightarrow \mathbb{C}$, $f(z) = z^2 - 2$.
Show that if $|z| > 2$, then the sequence z, f(z), f(f(z), ... diverges, i.e. is unbounded.

Isn't this solved almost immediately by the reverse triangle inequality?
• Apr 4th 2011, 03:38 AM
EinStone
Yes good point! Now consider $|z| \leq 2$ and $Im(z) \neq 0$ and show that the sequence is unbounded, i.e. leaves the disk of radius 2.
• Apr 4th 2011, 04:43 AM
chisigma
Quote:

Originally Posted by EinStone
Consider $f: \mathbb{C} \rightarrow \mathbb{C}$, $f(z) = z^2 - 2$.
Show that if $|z| > 2$, then the sequence z, f(z), f(f(z), ... diverges, i.e. is unbounded.

The sequence is alternatively defined by the recursive relation...

$a_{n+1} = a^{2}_{n}-2\ , \ a_{0}=z$ (1)

The procedure for solving (1) has been described a lot of times, the last in...

http://www.mathhelpforum.com/math-he...ce-176658.html

Following the procedure you find that...

a) if $|z|=|a_{0}|< 2$ the sequence converges to -1

b) if $|z|=|a_{0}|= 2$ the sequence converges to 2

c) if $|z|=|a_{0}|> 2$ the sequence diverges...

Kind regards

$\chi$ $\sigma$
• Apr 4th 2011, 05:31 AM
EinStone
Ok, but I think this is only for real $a_0$, now I want to consider the case that $a_0$ is complex and the imaginary part is non-zero.