TITLE: Convergence of a recursive sequence $z_{n+1} = \frac{1}{2} ( z_n - \frac{1}{z_n}) $
QUESTION [5 upvotes]: Let $p$ be a complex number. Let $ z_0 = p $ and, for $ n \geq 1 $, define $z_{n+1} = \frac{1}{2} ( z_n - \frac{1}{z_n}) $ if $z_n \neq 0 $. Prove the following:
i) If $ \{ z_n \} $ converges to a limit $a$, then $a^2 + 1 = 0 $
ii) If $ p $ is real, then $ \{ z_n \} $, if defined, does not converge
iii) If $ p = iq $, where $ q \in \mathbb{R} \backslash \{0\} $, then $ \{ z_n \} $ converges.
I have been able to do the first two parts of this (the second is because the sequence would be real, but would have to have a complex limit). I am stuck on the third part, though. Any help would be greatly appreciated.
Thanks

REPLY [2 votes]: From the recursion relation we get, $$\left(\frac{z_{n+1}+i}{z_{n+1}-i}\right)=\left(\frac{z_{n}+i}{z_{n}-i}\right)^2, \forall n\geq1 $$ So iteratively, we get, $$\left(\frac{z_{n+1}+i}{z_{n+1}-i}\right)=\left(\frac{p+i}{p-i}\right)^{2^{n+1}}$$ Rearranging, we get, $$z_{n+1}=i\frac{1+w^{2^{n+1}}}{1-w^{2^{n+1}}}$$ where $$w=\frac{p-i}{p+i}$$ When $p=iq$ $$ w=\frac{q-1}{q+1}$$ Clearly, $|w|<1$, since $q\ne 0$. Hence $$\lim_{n\rightarrow \infty}w^{2^{n+1}}=0.
$$ Thus $$\lim_{n\rightarrow \infty}z_n=1 \hspace{0.6cm}\Box$$