TITLE: Does iterating the complex function $z\mapsto\frac{2\sqrt z}{1+z}$ always converge?
QUESTION [11 upvotes]: Given $z_0\in\mathbb C\setminus\{-1\}$, define the sequence
$$z_{n+1}=\frac{2\sqrt{z_n}}{1+z_n}$$
where the square root is the one with positive real part (or, if that's not possible, non-negative imaginary part).
This is always defined; if ever $z_{n+1}=-1$, then
$$-(1+\sqrt{z_n}^2)=2\sqrt{z_n}$$
$$0=(1+\sqrt{z_n})^2$$
which implies $\sqrt{z_n}=-1$, a contradiction. So $z_n\neq-1$ for any $n$.
If it converges at all, then it converges to $0$ or $1$:
$$z=\frac{2\sqrt z}{1+z}$$
$$\sqrt z^2(1+\sqrt z^2)=2\sqrt z$$
$$\sqrt z\big(\sqrt z+\sqrt z^3-2\big)=0.$$
The cubic factor has roots $\sqrt z=1$ and $\sqrt z=-\tfrac12\pm\tfrac i2\sqrt7$, but the latter have negative real part so must be discarded.
If $z_n$ is near $0$, then $1+z_n\approx1$, and $|z_{n+1}|\approx2\sqrt{|z_n|}>2|z_n|$; the sequence gets pushed away from $0$. But of course if $z_n=0$ exactly, then $z_{n+1}=0$ and it converges trivially.
If $z_n=1+\varepsilon$ is near $1$, then $1+z_n\approx2$, and $z_{n+1}\approx\sqrt{z_n}\approx1+\tfrac12\varepsilon$ is even closer to $1$.
So, given $z_0\in\mathbb C\setminus\{-1,0\}$, if the sequence converges it must converge to $1$. Consider the distance from $1$:
$$1-z_{n+1}=\frac{1+z_n-2\sqrt{z_n}}{1+z_n}=\frac{(1-\sqrt{z_n})^2}{1+z_n}=\frac{(1-z_n)^2}{(1+z_n)(1+\sqrt{z_n})^2}.$$
Since a square root has non-negative real part, $|1+\sqrt{z_n}|>1$, and thus
$$|1-z_{n+1}|<\frac{|1-z_n|^2}{|1+z_n|}.$$
Now consider the distance from $-1$:
$$1+z_{n+1}=\frac{1+z_n+2\sqrt{z_n}}{1+z_n}=\frac{(1+\sqrt{z_n})^2}{1+z_n}$$
$$|1+z_{n+1}|>\frac{1}{|1+z_n|}.$$
We also have
$$1-z_{n+1}\!^2=(1-z_{n+1})(1+z_{n+1})=\frac{(1-z_n)^2}{(1+z_n)^2}$$
and
$$\frac{1-z_{n+1}}{1+z_{n+1}}=\frac{(1-\sqrt{z_n})^2}{(1+\sqrt{z_n})^2}.$$
I don't know where this is going. Can we show that $\lim_{n\to\infty}|1-z_n|=0$?

Here's another approach:
$$z_{n+1}=\frac{2\sqrt{z_n}}{1+z_n}=\frac{2\sqrt{z_n}(1+z_n^*)}{|1+z_n|^2}$$
$$=\frac{2}{|1+z_n|^2}\big(\sqrt{z_n}+|z_n|\sqrt{z_n}^*\big).$$
Both $\sqrt{z_n}$ and its conjugate $\sqrt{z_n}^*$ have non-negative real part. This expression shows that $z_{n+1}$ is a conical combination of them, so its angle is between their angles, which are half of the original angle of $z_n$. Thus, with $z_n=r_ne^{i\theta_n}$,
$$|\theta_{n+1}|\leq\frac{|\theta_n|}{2}$$
$$|\theta_n|\leq\frac{|\theta_0|}{2^n}\leq\frac{\pi}{2^n}$$
$$\lim_{n\to\infty}\theta_n=0.$$

I found something interesting, but maybe not useful. If $|z_{n+1}|=1$ then
$$|1+z_n|^2=4|z_n|;$$
this equation represents a limacon, with shape parameters $a=4,\,b=\sqrt8$.
If $|z_{n+1}|<1$ then $z_n$ is outside of the curve, or in the tiny loop containing $0$. If $|z_{n+1}|>1$ then $z_n$ is in the larger inside part of the curve (which includes the unit circle $|z_n|=1$).

Going in the other direction, if $|z_n|=1,\,z_n=e^{i\theta_n}$, then
$$z_{n+1}=\frac{2}{|1+e^{i\theta_n}|^2}\big(e^{i\theta_n/2}+1e^{-i\theta_n/2}\big)$$
$$=\frac{2}{1+2\cos\theta_n+1}\big(2\cos(\theta_n/2)\big)$$
$$=\frac{\cos(\theta_n/2)}{\cos^2(\theta_n/2)}=\sec(\theta_n/2)>1.$$
And if $z_n=r_n>0$, then
$$z_{n+1}=\frac{2\sqrt{r_n}}{1+r_n}<1$$
because
$$2\sqrt{r_n}<1+\sqrt{r_n}^2$$
$$0<\big(1-\sqrt{r_n}\big)^2.$$

My angle argument shows that $z_n$ is in the right half of the plane for $n\geq1$, so
$$|1+z_n|>1$$
and thus
$$|1-z_{n+1}|<\frac{|1-z_n|^2}{|1+z_n|}<|1-z_n|^2$$
$$<|1-z_{n-1}|^4<\cdots<|1-z_1|^{2^n}$$
which clearly converges to $0$, provided that $|1-z_1|<1$. So we only need to show that the sequence eventually comes within the unit circle around $1$.

REPLY [3 votes]: This function at $z$ is the same as at $1/z$, excluding the case $\mathbb R\ni z<0$, which is the only case where $\sqrt{1/z}\neq1/\sqrt z$.
$$\frac{2\sqrt{1/z}}{1+1/z}=\frac{2\cdot1/\sqrt z}{1/z+1}\cdot\frac zz=\frac{2\sqrt z}{1+z}.$$
Since $|\theta_2|\leq\pi/4$, if also $|z_2|\leq1$, then the circular sector represented by these inequalities is contained in the unit disk centred at $1$, so the sequence converges. (See the last part of the Question.) On the other hand, if $|z_2|>1$, then we can replace $z_2$ with $1/z_2$ without affecting the later terms, so again the sequence converges.

That answers the Question as asked.
Now I'll also answer a related question: Is there a well-defined arithmetic-geometric mean of two complex numbers?
Given $a_0,b_0\in\mathbb C$, define the sequences
$$a_{n+1}=\text{AM}(a_n,b_n)=\frac{a_n+b_n}{2},\quad b_{n+1}=\text{GM}(a_n,b_n)=\begin{cases}0,&a_n=0\\a_n\sqrt{\frac{b_n}{a_n}},&a_n\neq0\end{cases}$$
(again using the principal square root). This particular form of GM, rather than $\sqrt{ab}$, ensures that the result is "between" $a$ and $b$; if $a$ and $b$ are linearly independent over $\mathbb R$, then GM$(a,b)$ is a conical combination of them. (That's a linear combination with non-negative coefficients.) It behaves nicely with rotations: $\text{GM}(ka,kb)=k\,\text{GM}(a,b)$ for any $k\in\mathbb C$.
In any of the cases where $a_0b_0(a_0\!^2-b_0\!^2)=0$, it's easy to show that both sequences converge, to $0$ or to $a_0=b_0$. So let's assume that $a_0b_0(a_0\!^2-b_0\!^2)\neq0$.
Define $z_n=b_n/a_n$; then the formula for $z_{n+1}$ is exactly that in the OP, and we've already shown that $\lim_{n\to\infty}z_n=1$.
From the triangle inequality,
$$|a_{n+1}|\leq\frac{|a_n|+|b_n|}{2}\leq\frac{\max(|a_n|,|b_n|)+\max(|a_n|,|b_n|)}{2}=\max(|a_n|,|b_n|),$$
and similarly $|b_{n+1}|=\sqrt{|a_n||b_n|}\leq\max(|a_n|,|b_n|)$, so
$$\max(|a_{n+1}|,|b_{n+1}|)\leq\max(|a_n|,|b_n|)\leq\max(|a_{n-1}|,|b_{n-1}|)\leq\cdots\leq\max(|a_0|,|b_0|).$$
Thus, we see that both sequences are bounded, so
$$|a_n-b_n|=|a_n||1-z_n|\leq\max(|a_0|,|b_0|)\,|1-z_n|\to0;$$
if either one converges, they must both converge to the same value.
Indeed one of them does converge: $|a_{n+1}-a_n|=\tfrac12|a_n-b_n|\to0$, and for $k>1$,
$$|a_{n+k}-a_n|=\left|\sum_{j=0}^{k-1}(a_{n+j+1}-a_{n+j})\right|$$
$$\leq\sum_{j=0}^{k-1}|a_{n+j+1}-a_{n+j}|$$
$$=\frac12\sum_{j=0}^{k-1}|a_{n+j}-b_{n+j}|$$
$$\leq\frac12\max(|a_0|,|b_0|)\sum_{j=0}^{k-1}|1-z_{n+j}|$$
$$\leq\frac12\max(|a_0|,|b_0|)\sum_{j=0}^\infty|1-z_{n+j}|.$$
From the last part of the Question and the first part of this Answer, for $n\geq3$ we have $|1-z_n|<1$ and $|1-z_{n+j}|\leq|1-z_n|^{2^j}$, so
$$\sum_{j=0}^\infty|1-z_{n+j}|\leq\sum_{j=0}^\infty|1-z_n|^{2^j}$$
$$\leq\sum_{l=1}^\infty|1-z_n|^l$$
$$=\frac{|1-z_n|}{1-|1-z_n|}\to0.$$
Therefore, by Cauchy's criterion, $a_n$ converges. We may call the common limit
$$\lim_{n\to\infty}a_n=\lim_{n\to\infty}b_n=\text{AGM}(a_0,b_0).$$