Divergence of sequence

**EinStone** · April 3rd 2011, 02:49 PM

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.

**Drexel28** · April 3rd 2011, 05:21 PM

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?

**EinStone** · April 4th 2011, 02:38 AM

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.

**chisigma** · April 4th 2011, 03:43 AM

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

$\chi$ $\sigma$

**EinStone** · April 4th 2011, 04:31 AM

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.

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