a Consider the recurrence relation defined by zn1 zn2 c fo

(a) Consider the recurrence relation defined by zn+1 = zn^2 + c for z0, c ? C. Show that if,for some n,we have|zn|>|c|>2,then|zn+1|>k|zn|forsomek>1. Why does this show the sequence is unbounded?

[Hint: The inequalities |x ? y| ? |x| ? |y| and |xy| = |x| |y| may be useful.]

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Solution

We have |z_{n+1}| >= |z_{n}^2 +c| >= |z_{n}^2| -|c|=|z_n||z_n| -|c| >= |c| |z_{n}| -|z_{n}|=(|c|-1)|z_{n}|.

Hence |z_{n+1}|>k|z_n|.

Now |z_{n+m}| >k |z_{n+m-1}| >... >k^m |z_{n}| , k>1 As m -> infinity, k^m -> infinity

Hence sequence is unbounded.

Hence M >= |z_[n+1} | > k |z_n| >k^2|z_{n-1}| >.....>k^{n+1} |z_0|

for all n