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.

$|z_{n+1}-1| = \dfrac{|z_n-1|^2}{2|z_n|}\leq \dfrac{|z_n-1|^2}{2}$. Reiterate this until you reach $|z_0-1|$, and its supposed to be less than $1$ then you use squeeze theorem to conclude. But first you need to somehow show that : $|z_n| \geq 1$. You can show this by using AM-GM inequality, but note that you are working on $\mathbb{C}$ so you probably need to write $z_n = a_n+ib_n$, and to proceed.
