# Understanding a proof that a given recursive complex sequence converges

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 $$\lim_{n \to \infty} \ z_n = 1$$.

$$\\$$

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 ?

• 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 $z_n\to1$. – Did Dec 17 '18 at 17:59
• @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." – Clement C. Dec 17 '18 at 18:00
• @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, $u_n=w_nv_n$). – Did Dec 17 '18 at 18:03
• @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) – Clement C. Dec 17 '18 at 18:06
• 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 – Federico Dec 17 '18 at 18:26