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I am currently struggling with the following exercise:

Let $z_0 = x_0+iy_0 \ne 0$ be a complex number and let the sequence $(z_n)_n$ be recursively defined as

$$z_{n+1} = \frac{1}{2} \left( z_n+\frac{1}{z_n} \right)$$

for $n \ge 0$. Show that if $x_{0} > 0$ then $\lim_{n \to \infty} \ z_n = 1$.

$\\$

I do not know how to work with complex recursive sequences, but I suppose that, as in the real-valued case, a solution will require proving that the limit exists and then to solve the recurrency $$l = \frac{1}{2} \left( l + \frac{1}{l} \right)$$ In the real valued case I would try to prove that $(z_n)_n$ is monotone and bounded, but since there is no order relation in $\mathbb{C}$ I can not use this apporoach here. Could you help me?

Edit: Thanks to Ahmad Bazzi's answer I now have an idea on how to prove that $(z_n)_n$ converges: If I can prove that both sequences $(x_n)_n$ and $(y_n)_n$ (these sequences are the real and imaginary parts of each $z_n$) converge I could deduce that $(z_n)_n$ converges. To prove that both $(x_n)_n$ and $(y_n)_n$ converge I would show that these sequences are both bounded and monotone (via induction). But unfortunately I do not get far with this approach because the definitions of these sequences is interwined, i.e.: \begin{align} x_{n+1} &= \frac{1}{2}(x_n + \frac{x_n}{x_n^2 + y_n^2})\\ y_{n+1} &= \frac{1}{2}(y_n - \frac{y_n}{x_n^2 + y_n^2}) \end{align} and I do not know how to go on from there. Any help would be appreciated.

3nondatur
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1 Answers1

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The way I see it is that you got two recursions going on \begin{equation} z_n = x_n + iy_n \end{equation} So \begin{equation} x_{n+1} + iy_{n+1} = \frac{1}{2} (x_n + iy_n + \frac{1}{x_n + iy_n}) \end{equation} The right hand side is \begin{equation} \frac{1}{2} (x_n + iy_n + \frac{1}{x_n + iy_n}) = \frac{1}{2} (x_n + iy_n + \frac{x_n - iy_n}{x_n^2 + y_n^2}) \end{equation} which is \begin{equation} \frac{1}{2} (x_n + iy_n + \frac{1}{x_n + iy_n}) = \frac{1}{2}\big( x_n + \frac{x_n}{x_n^2 + y_n^2} \big) + i\frac{1}{2}\big(y_n - \frac{y_n}{x_n^2 + y_n^2} \big) \end{equation} This means that \begin{align} x_{n+1} &= \frac{1}{2}(x_n + \frac{x_n}{x_n^2 + y_n^2})\\ y_{n+1} &= \frac{1}{2}(y_n - \frac{y_n}{x_n^2 + y_n^2}) \end{align} Solve the system

\begin{align} X &= \frac{1}{2}(X + \frac{X}{X^2 + Y^2})\\ Y &= \frac{1}{2}(Y - \frac{Y}{X^2 + Y^2}) \end{align} where the limit $L$ is \begin{equation} L = X + iY \end{equation} First equation tells you that $X - \frac{1}{2}X = \frac{1}{2} \frac{X}{X^2 +Y^2}$, which means that $$X^2 + Y^2 = 1$$Replace the above condition in the $Y$ equation, you will get $Y = \frac{1}{2}(Y-Y) = 0$, hence $X^2 = 1$ so $X = \pm 1$, i.e. $L = \pm 1$. So, the limit in your problem will depend on the initial value $z_0$.

Ahmad Bazzi
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