Limit of complex sequence.

Question

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.

Answer 1

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 $$\left|\frac{z_0-1}{z_0+1}\right|.$$

Answer 2

$|z_{n+1}-1| = \dfrac{|z_n-1|^2}{2|z_n|}\leq \dfrac{|z_n-1|^2}{2}$. 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.