Find $ \lim_{n \to \infty } z_n $ if $z_{n+1} = \frac{1}{2} \left( z_n + \frac{1}{z_n} \right) $ How do I approach the problem?

Q: Let $ \displaystyle z_{n+1} = \frac{1}{2} \left( z_n + \frac{1}{z_n} \right)$ where $ n =  0, 1, 2, \ldots $ and $\frac{-\pi}{2} < \arg (z_0) < \frac{\pi}{2} $. Prove that $\lim_{n\to \infty} z_n =  1$.


One common trick to deal with recurrence equation involving reciprocal is to rewrite
$z_n$ as ratio of two sequence $p_n, q_n$ to be determined. Notice,
$$\frac{p_{n+1}}{q_{n+1}} = z_{n+1} = \frac12 (z_n + \frac{1}{z_n}) = \frac12(\frac{p_n}{q_n} + \frac{q_n}{p_n}) = \frac{p_n^2 + q_n^2}{2p_nq_n}$$
We can take $p_{n+1} = p_n^2 + q_n^2$ and $q_{n+1} = 2 p_{n} q_{n}$. From this, we get
$$p_{n+1} \pm q_{n+1} = ( p_{n} \pm q_{n} )^2 \implies (p_{n} \pm q_{n}) = (p_{0} \pm q_{0})^{2^n} $$
Choose $p_0 = z_0$ and $q_0 = 1$, we get $p_n \pm q_n = (z_0 \pm 1)^{2^n}$ and hence:
$$z_n = \frac{(z_0 + 1)^{2^n} + (z_0 - 1)^{2^n}}{(z_0 + 1)^{2^n} - (z_0 - 1)^{2^n}}$$
For $z_0 \in \mathbb{C}\setminus\{0\}$ such that $-\frac{\pi}{2} < \arg( z_0 ) < \frac{\pi}{2}$, we have:
$$|z_0 + 1| > |z_0 - 1|   \implies |\frac{z_0-1}{z_0+1}| < 1$$
Let $\alpha = \frac{z_0-1}{z_0+1}$, we get:
$$\lim_{n\to\infty} z_n = \lim_{n\to\infty}\frac{1 + \alpha^{2^n}}{1 - \alpha^{2^n}} = 1$$
A similar argument shows that if you start with a $z_0$ such that $-\frac{\pi}{2} < \arg( -z_0 ) < \frac{\pi}{2}$, $|\alpha|$ will be greater than $1$ and $z_n$ converges to $-1$ instead.
