# sequence of complex numbers

I'm studying Complex Analysis on my own and the book I use (Complex Analysis, Eberhard Freitag, Rolf Busam) is a bit terse and gives no examples. How can I solve the following exercise? Geometrically, I can see what happens, but how should I proceed algebraically?

Let $z_0 = x_0 + i y_0\neq0$ be a given complex number. Define the sequence $(z_n)_{n\geq0}$ recursively defined by $$z_{n+1}=\frac{1}{2}\left(z_n+\frac{1}{z_n}\right),\quad n\geq0.$$ Show:

1. If $x_0>0$, then $\lim_{n\to\infty}z_n=1$.
2. If $x_0<0$, then $\lim_{n\to\infty}z_n=-1$.
3. If $x_0=0, y_0\neq0$, then $(z_n)_{n\geq0}$ is undefined or divergent.

Hint. Consider $w_{n+1}=\displaystyle\frac{z_{n+1}-1}{z_{n+1}+1}$.

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Following the hint, write $w_{n}=(z_n-1)/(z_n+1)$ and compute $$w_{n+1}=\frac{z_{n+1}-1}{z_{n+1}+1}=\frac{z_n^2+1-2z_n}{z_n^2+1+2z_{n}}=w_n^2$$
The sequence $\{w_n\}$ is much easier to study: if $|w_0|>1$, then $w_n\to\infty$, if $|w_0|<1$, then $w_n\to0$, if $|w_n|=1$, then $w_n$ is undefined.
Now, you just have to rewrite everything in terms of $z_n$: if $w_n\to\infty$, then $z_n\to-1$, if $w_n\to0$, then $z_n\to1$. If $w_n$ stays on the circle $\{|w|=1\}$, then $|z_n-1|=|z_n+1|$, i.e. $z_n\in i\mathbb{R}$.
On the other hand, $|w_0|>1$ iff $x_0>0$, $|w_0|<1$ iff $x_0<0$ and $|w_0|=1$ iff $x_0=0$.