Let $z_0 = x_0 + i y_0$ be a given complex number.
I previously calculated the limit of
$$ z_{n+1} = {1\over 2}( z_n + z_n^{-1})$$
for $x_0 <0$ and $x_0>0$ respectively. Now I am trying to do the case $x_0 = 0$ but the previous method does not work in this case and I am stuck again.
Please could someone explain to me how to calculate this limit when $z_0$ is purely imaginary?
Because of the recurrence relation all terms in the sequence will be purely imaginary, and hence the limit, if it exists, must be purely imaginary. Let us call it $z = iy$ with $y \in \mathbb{R}$. Since it is a limit from
$$ \lim_{n \to \infty} z_n z_{n+1} = \lim_{n \to \infty} \frac{1}{2} (z_n^2 + 1) $$ we find
$$ z^2 = \frac{1}{2}(z^2 +1) \quad \mbox{or}\quad z^2 =1 $$ but this implies $y^2 =-1$ which is impossible because $y$ was real. Hence the limit does not exist.
Notice that, since $\frac{1}i=-i$, we can write: $$z_1=\frac{1}2\left(iy_0+\frac{-i}{y_0}\right)=\frac{i}2\cdot\left(y_0-\frac{1}{y_0}\right).$$ which makes it clear that the value stays purely imaginary as we iterate this process. In particular, we, at each step, average $y_n$ with $-\frac{1}{y_n}$. However, it's fairly easy to see that such an iteration is doomed to cycle about forever - if $y_n$ is positive, then $-\frac{1}{y_n}$ is negative and $y_{n+1}$ will be less than $\frac{y_n}2$. Similarly, if $y_n$ is negative, then $y_{n+1}$ will be at least $\frac{y_n}2$. However, when $y_n$ gets to be near zero, then $\frac{1}{y_n}$ will have large absolute value, so $y_{n+1}$ will be catapulted far away from the origin - and we end up forever cycling high values of $y_n$ decreasing towards $0$, being flung to a negative value, increasing towards $0$, being flung to positive value, and so on - with no limit possible.
Algebraically, this boils down to the fact that $f(y)=y-\frac{1}y$ has no fixed point, as presented in the other answer.
