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 $\lim_{n\to \infty} z_n =  1$.

 A: 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.
A: If the limit exists, then 
$$\lim_{n\to\infty}z_{n+1}=\lim_{n\to\infty}z_n$$
Substitute it in, you can get 
$$\lim_{n\to\infty}z_n=\pm1$$
Since there is one more constrain on arg$(z)$, you can conclude that the limit must be $1$.
It is not hard to prove the existence of limit.
Assume $|z_n|>1$, then $|\frac{1}{z_n}|<1$. By the triangle inequality, $|z_n+\frac{1}{z_n}|\leq|z_n|+|\frac{1}{z_n}|<2|z_n|$.
Therefore, $|z_{n+1}|<|z_n|$.
A: Let $f(z) = \frac12\left(z+\frac1z\right)$.  Clearly $f(1)=1$ and $f'(1)=0$.  So suppose $z=1+\Delta z$.  Then
$$
f(z) = 1 +f'(1)\,\Delta z + \text{higher-degree terms in $\Delta z$},
$$
so $f(z)$ is closer to $1$ than $z$ is.  You have an attractive fixed point.
Later edit:
Or put it this way: $f'(1)=0$ and $f'(z)$ is close to $0$ when $z$ is close to $1$.  In particular, certainly $f'(z)$ is between $\pm1/2$ when $z$ is close enough to $1$.  Therefore $f(z)$ is changing less than half as fast as $z$ is changing when $z$ is close to $1$.  That means $f(z)$ is less than half as far from $1$ as $z$ is.  If you keep cutting the distance in half, the distance approaches $0$ as a limit.
