What, if anything, is remarkable about this transfer of structure from $\Bbb Z[\frac12]$ to $\Bbb C$?
$Y=\Bbb Z[\frac12]\setminus0$
$X=\Bbb Z[\frac12]\cap(\frac12,1]$
$N=\{\ldots2,1,\frac12,\frac14,\ldots\}$
For any given $y\in Y$, let $x\in X$ and $n\in N$ be the unique pair satisfying $y=x\cdot n$
Uniqueness follows from the fact that every odd numerator appears precisely once in $X$.
$f:Y\to\Bbb C$
$f(x\cdot n)=\dfrac{\exp{((4x-2)\cdot\pi i)}}{x\cdot n}$
I don't want to be too concerned with $0$ at this point as I'm certain we have something like $f:0\mapsto0$ - it's a detail I'm happy to reserve to later.
What's remarkable about this ring extension structure transferred to $\Bbb C$?
I've engineered this function for certain purposes but often when I do such a thing, it's been done before and there's a wealth of insight to be learnt by retracing the footsteps of others.
I can begin to answer by describing what I've set out to do, and what I've engineered to be remarkable. The object was to impose on $\Bbb Z[\frac12]$ (and also $\Bbb Z[\frac16]$ for that matter), the relation $x\sim y\iff\exists i\in\Bbb Z:2^ix=y$.
This function sends the odd numbers to a dense set of roots of unity, so viewed as a set $(r,\theta)\in\Bbb C$, the value of $\theta$ is a dense set in $S^1$ representing the odd numbers and representing equivalence classes of the odd numerators throughout $\Bbb Z[\frac12]$. The interval $0<\theta\leq2\pi$ corresponds to the odd numerators in $X$. These might be the Prufer 2-group $\Bbb Z(2^\infty)$ or some subset thereof, but I'm not too clear on that
Meanwhile if we allow $r$ to vary, there's a unique Fibonacci spiral representing every power of $2$. Each point on a spiral whose $\theta$ co-ordinate equals an appropriate root of unity, corresponds to an element of $\Bbb Z[\frac12]$
I'm interested in the analytic continuation of these spirals and in the spaces in-between. Is it easy to define a continuous (holomorphic?) function in $\Bbb C$ that defines these spirals?
Am I right in thinking that if we give appropriate thought to the distances between the spirals, measured either orthogonal to them, or radial to the centre, there's an analytic continuation of this function which defines a much richer valuation than $\lvert x\rvert_2$ which gives an alternative completion of $\Bbb Z$ than $\Bbb Z_2$? Moreover, what structure do we come back to in $\Bbb R\supset\Bbb Z[\frac12]$ when we map this enriched structure back?
Looking at the inverse function $f^{-1}$ from $\Bbb C$ back to $\Bbb Z[\frac12]$, or even to $\Bbb R$ it looks necessary to think about branch cuts and the Lambert $W$ function - something I know nothing about, but I think the function is defined in the forward direction such that the possible cuts in reverse are all superimposed, so the inverse function is bijective when we might not expect it to be.
With regards to your comment - the uniqueness in $Y=\mathbb{Z}[\frac{1}{2}]$ can be done in two ways (Note that the following factorizations only consider $Y_{>0}$, but it's not hard to work with the sign). The first way is how you explained it: $$ x\in\mathbb{Z}[\tfrac{1}{2}]\cap (\tfrac{1}{2},1]\\ n\in\{2^N\:|\:N\in \mathbb{Z}\}. $$ If you fix $n$ than you get $y\in\mathbb{Z}[\frac{1}{2}]\cap (\frac{n}{2},n]$.
On the other hand you can write $y$ as a product of two elements $n',x'$ in the following way: $$ y=2^N p_1^{a_1} \cdots p_s^{a_s} $$ where $N\in\mathbb{Z}$, $s\geq 0$, $a_i\geq 1$, $p_i$ odd primes. Uniqueness comes from the unique factorization of $\mathbb{Z}$ or $\mathbb{Q}$. Now you have $y=x'\cdot n'$ where $$ n'=2^{N'}\\ x'=p_1^{a_1} \cdots p_s^{a_s} $$ so $n'$ is a power of two, and $x'$ is an odd number.
These two unique factorizations are different! The value $x$ doesn't have to be an odd number, in fact you can have $x=\frac{3}{4}$, it can have some powers of two in its factorization, it doesn't have to be an integer, but $x'$ on the other hand is always an odd integer.
The map $$ f:y\mapsto \tfrac{e^{(4x-2)i \pi}}{x\cdot n} $$ can be continued analytically to $\mathbb{R}$ for each $n$. For example if you fix $n$ you get the spiral you were talking about defined as $$ \text{Spiral}_n=\{\tfrac{e^{(4x-2)i \pi}}{x\cdot n}\: | \:x\in\mathbb{Z}[\tfrac{1}{2}]\cap (\tfrac{1}{2},1] \} $$ from which you create a full connected spiral by taking $x\in (\tfrac{1}{2},1]$. It seems easy to study these spirals as full connected paths, but I'll leave this to you. The main thing is that you're not using any special properties of $\mathbb{Z}[\tfrac{1}{2}]$, so I don't believe there's much to expect from this.
On the other hand if you consider the map $$ f':y\mapsto \tfrac{e^{(4x'-2)i \pi}}{x'\cdot n'} $$ you get a completely different thing. Now if you fix $n'$, it's true that $x'$ are odd numbers. So fixing $n'=\tfrac{1}{8}$ as you said, you get $y=\tfrac{1}{8}, \tfrac{3}{8},\tfrac{5}{8},\ldots$. But now the image $$ \text{Image}_{n'}(f')=\{\tfrac{e^{(4x'-2)i \pi}}{x'\cdot n'}\: | \:x'\text{ odd} \} $$ becomes a crazy scattered set of points without much order. Maybe you get something better in the whole image of $f'$, but I doubt so. I'll give it more thought.
I hope this is helpful.