The arithmetic mean of two complex numbers $a,b$ is $tfrac12(a+b)$, and the geometric mean is $pmsqrt{ab}$, which is two-valued. The arithmetic-geometric mean is the limit of the pair of sequences formed by iterating

$$mathbb C^2tomathbb C^2:;(a,b)mapsto(a’,b’)=left(frac{a+b}{2},;sqrt{ab}right)$$

(and choosing one of the two possible square roots at each step). See Cox’s paper The Arithmetic-Geometric Mean of Gauss.

Instead of taking one path and indexing by $mathbb N$, let’s consider the set $Ssubseteqmathbb C^2$ of all pairs of points which can be reached in this way, taking both square roots, and also iterating backwards:

$$(a,b)mapsto(a’,b’)=left(frac{a+b}{2},quadpmsqrt{ab}right)$$

$$(a’,b’)mapsto(a,b)=left(a’pmsqrt{a’^2-b’^2},quad a’mpsqrt{a’^2-b’^2}right).$$

$S$ could be indexed by some directed graph where each vertex has two successors and two predecessors. The two successors of $(a,b)$ (given by the two choices of geometric mean) share their two predecessors: $(a,b)$ itself and $(b,a)$. The two predecessors of $(a’,b’)$ share successors: $(a’,b’)$ and $(a’,-b’)$.

With that in mind, we can define equivalence classes

$$(a,b)={(a,b),(a,-b),(-b,a),(-b,-a),(-a,-b),(-a,b),(b,-a),(b,a)};$$

note that $(a,b)in S$ if and only if $(a,b)subseteq S$. Our multi-valued functions act on these by

$$(a,b)mapstoleft(frac{a+b}{2},;sqrt{ab}right),;left(frac{a-b}{2},;isqrt{ab}right)$$

$$(a,b)mapstoleft(a+sqrt{a^2-b^2},;a-sqrt{a^2-b^2}right),;left(b+isqrt{a^2-b^2},;b-isqrt{a^2-b^2}right).$$

Let’s also define $A={amidexists b:(a,b)in S}={bmidexists a:(a,b)in S}subseteqmathbb C$, and $X^*$ as the derived set (the set of accumulation points, or limit points) of a set $X$. Now for the questions.

Given a single pair $(a_0,b_0)$ to generate $S$, in general, what does $S$ or $A$ look like? Where does it accumulate?

If $a^*in A^*$, is there some $b^*$ such that $(a^*,b^*)in S^*$?

If $(a^*,b^*)in S^*$, must $a^*=pm b^*$?

If $(a^*,b^*)in S^*$, is it actually the limit of some sequence $(a_n,b_n)to(a^*,b^*)$ following the two-valued function and its inverse? that is, a sequence where, for all $ninmathbb N$, $(a_{n+1},b_{n+1})$ is one of the four values obtainable from $(a_n,b_n)mapsto(a_{n+1},b_{n+1})$?

The case $a_0=b_0=0$ is trivial: $S={(0,0)}$.

The case $a_0neq b_0=0$ is tractable:

$$S={(a_0/2^n,0),,(ia_0/2^n,ia_0/2^n)mid ninmathbb Z}$$

(I know, this is abusing set notation to have $(a,b)in S$ rather than $(a,b)subseteq S$)

$$A={0,,pm a_0/2^n,,pm ia_0/2^nmid ninmathbb Z}$$

$$S^*={(0,0)},quad A^*={0}.$$

The case $a_0=pm b_0neq0$ has the same form.

It remains to consider the general case $a_0b_0(a_0!^2-b_0!^2)neq0$.