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Currently I am struggling to understand a solution to 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 $\lim_{n \to \infty} \ z_n = 1$.

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We notice that the only possible values for a limit would be $\pm 1$. We further observe that since $z_0$ is in the right half plane all $z_n$ are in there too. Next we observe the sequence $w_n$ defined by

$$ w_n := \frac{z_n-1}{z_n+1}$$

It holds that $w_{n+1} = w_n^2$. And since $|w_n| < 1 $ we see that $w_n$ converges to $0$. From $|z_n+1| \ge 1$ we deduce that $z_n$ converges to $1$.

I do not understand the part

From $|z_n+1| \ge 1$ we deduce that $z_n$ converges to $1$.

Could you explain that to me ?

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    $\begingroup$ FWIW, I do not understand "From $|z_n+1| \ge 1$ we deduce that $z_n$ converges to $1$" either. From $w_n\to0$ and $$z_n=\frac{1+w_n}{1-w_n}$$ it is direct that $z_n\to1$ and the argument that $|z_n+1| \ge 1$ seems quite unrelated and not needed at all to get that $z_n\to1$. $\endgroup$ – Did Dec 17 '18 at 17:59
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    $\begingroup$ @Did It seems... circumvoluted. I understood it as "the denominator is bounded away from $0$, so we don't have to worry about anything at all anyway." $\endgroup$ – Clement C. Dec 17 '18 at 18:00
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    $\begingroup$ @ClementC. It seems... wrong, actually. If one knows that $w_n=u_n/v_n\to0$, what is needed to deduce that $u_n\to0$ is that $(v_n)$ stays bounded, not that $(v_n)$ stays bounded away from $0$ (since, well... obviously, $u_n=w_nv_n$). $\endgroup$ – Did Dec 17 '18 at 18:03
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    $\begingroup$ @Did You're right. (Again, I don't know where this proof is from, but my best-case assumption is that the person who wrote the argument wanted to say that that $w_n$ was well-defined since no division by zero occurred, and put it in the wrong place, in the most confusing manner) $\endgroup$ – Clement C. Dec 17 '18 at 18:06
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    $\begingroup$ Probably the intention was to say that $g:z\mapsto\frac{z-1}{z+1}$ is holomorphic in $B_1(-1)^c$ and the only root is $1$, so $g(z_n)\to 0$ iff $z_n\to1$. Very poorly phrased $\endgroup$ – Federico Dec 17 '18 at 18:26

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