Limit of complex sequence. I have the following limit:
If $z_{n+1}=\dfrac{1}{2}\left(z_n+\dfrac{1}{z_n}\right)$ for $n \in \mathbb{N}\cup \{0\}$ and $-\dfrac{\pi}{2}<arg(z_0)<\dfrac{\pi}{2},$ 
then $$\lim_{n \rightarrow \infty} z_n=1.$$
I have tried to look for a recurent formula, but every time it gets worst and worst. I have also tried to put $z_0=|z_0|e^{i \theta}$ for $-\dfrac{\pi}{2}<\theta<\dfrac{\pi}{2},$ but I'm really stuck.
Any hint would be appreciated.

Hint:
A key relation is $$\frac{z_{n+1}-1}{z_{n+1}+1}=\frac{z_n^2-2z_n+1}{z_n^2+2z_n+1}=\left(\frac{z_n-1}{z_n+1}\right)^2,$$
and by recurrence
$$\left(\frac{z_0-1}{z_0+1}\right)^{2^n}.$$
Consider the modulus
$$\left|\frac{z_0-1}{z_0+1}\right|.$$
