How to find limit of imaginary recursive sequence? 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?

 A: 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.
A: 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.
