Consider , .

Show that if , then the sequence z, f(z), f(f(z), ... diverges, i.e. is unbounded.

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- Apr 3rd 2011, 02:49 PMEinStoneDivergence of sequence
Consider , .

Show that if , then the sequence z, f(z), f(f(z), ... diverges, i.e. is unbounded. - Apr 3rd 2011, 05:21 PMDrexel28
- Apr 4th 2011, 02:38 AMEinStone
Yes good point! Now consider and and show that the sequence is unbounded, i.e. leaves the disk of radius 2.

- Apr 4th 2011, 03:43 AMchisigma
The sequence is alternatively defined by the recursive relation...

(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 the sequence converges to -1

b) if the sequence converges to 2

c) if the sequence diverges...

Kind regards

- Apr 4th 2011, 04:31 AMEinStone
Ok, but I think this is only for real , now I want to consider the case that is complex and the imaginary part is non-zero.