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\title{\textbf{Origin-Centered M\"obius Circles and Dyadic Pairings of the Riemann Zeta Function on the Critical Line}}
\author{Ryan Hackbarth \\ \texttt{\href{mailto:hackbarthr16@hanover.edu}{hackbarthr16@hanover.edu}}}
\date{October 22, 2026}

\begin{document}

\maketitle

\begin{abstract}
We investigate the complex geometry of the normalized Dirichlet eta terms $t_n(s) = \frac{1}{1-2^{1-s}}\frac{(-1)^{n-1}}{n^s}$ for $s = a+bi$. Decomposing the series into base-$2$ resonant components, we demonstrate that the initial partial sum $S_2(s) = t_1(s) + t_2(s)$ forms an exact M\"obius circle. Remarkably, this base circle is centered \textit{strictly} at the origin $(0,0)$ if and only if $\operatorname{Re}(s) = \frac{1}{2}$. We establish a dyadic pairing theorem showing that every odd-even pair $P_m(s) = t_m(s) + t_{2m}(s)$ inherits this origin-centered circular symmetry on the critical line with strictly descending radii $\mathcal{R}_m = \frac{1}{\sqrt{2m}}$. In contrast to individual unpaired terms whose radii pulsate inside precessing annuli, the dyadic pairs eliminate prefactor amplitude oscillations entirely. Analyzing the phase distribution on these concentric circles reveals key geometric mechanisms that distinguish non-trivial Riemann zeta zeros $b = b_n$ from generic values of $b$, including right-half-plane slow-zone clustering of $P_1$ and opposing-hemisphere alignment of the leading inner circles $P_3, P_5, P_7$. Finally, we prove that removing all odd-even dyadic pairs leaves the multiples-of-$4$ subseries $\sum_{k=1}^\infty t_{4k}(s) = t_4(s)\zeta(s)$, whose partial sums spiral around a target center $C_{\mathrm{target}}(b)$ that coincides with the origin $(0,0)$ if and only if $b$ is a Riemann zeta zero.
\end{abstract}

\section{Introduction and Definitions}

The Dirichlet eta function $\eta(s)$ provides the standard analytic continuation of the Riemann zeta function $\zeta(s)$ across the critical strip $0 < \operatorname{Re}(s) < 1$ via the Euler factor normalization \cite{Titchmarsh1986,Edwards1974,Apostol1976}:
\begin{equation}
\zeta(s) = \frac{\eta(s)}{1 - 2^{1-s}}, \quad \text{where } \eta(s) = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}.
\end{equation}

To examine the local geometry of this summation in the complex plane $\mathbb{C} \cong \mathbb{R}^2$, we distribute the prefactor into each term, defining the normalized sequence:
\begin{equation}
\label{eq:tn}
t_n(s) = \frac{1}{1 - 2^{1-s}} \cdot \frac{(-1)^{n-1}}{n^s}, \quad s = a + bi.
\end{equation}

For any fixed real part $a$, treating $b \in \mathbb{R}$ as the curve parameter gives each term an instantaneous modulus:
\begin{equation}
\label{eq:modulus_general}
|t_n(a+bi)| = \frac{n^{-a}}{\sqrt{1 + 4^{1-a} - 2^{2-a}\cos(b \ln 2)}}.
\end{equation}
On the critical line $a = \frac{1}{2}$, this modulus breathes between a periapsis minimum $R_{n,\min} = \frac{\sqrt{2}-1}{\sqrt{n}}$ (when $\cos(b\ln 2) = -1$) and an apoapsis maximum $R_{n,\max} = \frac{\sqrt{2}+1}{\sqrt{n}}$ (when $\cos(b\ln 2) = +1$), spanning a nearly sixfold amplitude variation.

In this paper, we demonstrate that grouping terms into dyadic pairs eliminates this prefactor breathing entirely on the critical line $a = \frac{1}{2}$, collapsing the series into an infinite sequence of exact origin-centered concentric circles. We examine how these concentric circles geometrically govern zero-crossing alignments and analyze the complementary multiples-of-$4$ subseries.

\section{\texorpdfstring{The Base-2 Anchor: $S_2(s) = t_1(s) + t_2(s)$ and the Critical Line $a = \frac{1}{2}$}{The Base-2 Anchor: S2(s) = t1(s) + t2(s) and the Critical Line a = 1/2}}

For $n = 1$ and $n = 2$, the term frequencies $\ln 1 = 0$ and $\ln 2$ resonate with the prefactor frequency $\ln 2$. Letting $z = e^{-i b \ln 2}$ parametrize the unit circle $|z| = 1$, each individual term is a M\"obius transformation \cite{Needham1997} of $|z| = 1$:
\begin{equation}
t_1(a+bi) = \frac{1}{1 - 2^{1-a}z}, \qquad t_2(a+bi) = \frac{-2^{-a}z}{1 - 2^{1-a}z}.
\end{equation}
For any $a \neq 1$, $t_1(a+bi)$ and $t_2(a+bi)$ trace exact circles symmetric across the real axis with centers $C_1(a) = (h_1(a), 0)$, $C_2(a) = (h_2(a), 0)$ and radii $\mathcal{R}_1(a)$, $\mathcal{R}_2(a)$ given by:
\begin{align}
h_1(a) &= \frac{1}{1 - 4^{1-a}}, & \mathcal{R}_1(a) &= \frac{2^{1-a}}{|1 - 4^{1-a}|}, \\
h_2(a) &= \frac{-2^{1-2a}}{1 - 4^{1-a}}, & \mathcal{R}_2(a) &= \frac{2^{-a}}{|1 - 4^{1-a}|}.
\end{align}

On the critical line $a = \frac{1}{2}$, let $B(z) = \frac{\sqrt{2} - z}{1 - \sqrt{2}z}$ denote the unit-modulus Blaschke factor ($|B(z)| = 1$ for $|z| = 1$) \cite{Garnett2007}. Then $t_1$ and $t_2$ admit the explicit representations:
\begin{align}
t_1\left(\frac{1}{2} + bi\right) &= -1 + \sqrt{2}\,B(z), \quad \text{centered at } C_1 = (-1, 0), \ \mathcal{R}_1 = \sqrt{2} \approx 1.4142 \text{ (Figure \ref{fig:t1})}, \\
t_2\left(\frac{1}{2} + bi\right) &= +1 - \frac{1}{\sqrt{2}}\,B(z), \quad \text{centered at } C_2 = (+1, 0), \ \mathcal{R}_2 = \frac{1}{\sqrt{2}} \approx 0.7071 \text{ (Figure \ref{fig:t2})}.
\end{align}

\begin{figure}[H]
\centering
\begin{tikzpicture}[scale=1.55]
  \draw[step=0.5, gray!30, thin] (-2.6,-1.6) grid (1.1,1.6);
  \draw[->, thick] (-2.7,0) -- (1.2,0) node[right] {$\operatorname{Re}(z)$};
  \draw[->, thick] (0,-1.7) -- (0,1.7) node[above] {$\operatorname{Im}(z)$};
  \foreach \x in {-2.5,-2.0,-1.5,-1.0,-0.5,0.5,1.0}
    \draw (\x,2pt) -- (\x,-2pt) node[below, font=\scriptsize] {$\x$};
  \foreach \y in {-1.5,-1.0,-0.5,0.5,1.0,1.5}
    \draw (2pt,\y) -- (-2pt,\y) node[left, font=\scriptsize] {$\y$};
  \draw[blue!80!black, ultra thick] (-1,0) circle ({sqrt(2)});
  \filldraw[blue!80!black] (-1,0) circle (1.6pt) node[below=4pt, font=\small\bfseries] {$C_1 = (-1,0)$};
  \filldraw[black] ({-1-sqrt(2)},0) circle (1.6pt) node[above left, font=\small] {$b=0$};
  \filldraw[black] ({-1+sqrt(2)},0) circle (1.6pt) node[above right, font=\small] {$b=\frac{\pi}{\ln 2}$};
  \draw[dashed, thick, blue!80!black] (-1,0) -- ({-1+sqrt(2)*cos(45)}, {sqrt(2)*sin(45)}) node[midway, above, sloped, font=\small\bfseries] {$\mathcal{R}_1 = \sqrt{2} \approx 1.414$};
\end{tikzpicture}
\caption{Locus of $t_1(\frac{1}{2}+bi) = \frac{1}{1 - 2^{1/2-bi}}$ in the complex plane: an exact circle centered at $C_1 = (-1,0)$ with radius $\mathcal{R}_1 = \sqrt{2} \approx 1.4142$.}
\label{fig:t1}
\end{figure}

\begin{figure}[H]
\centering
\begin{tikzpicture}[scale=2.0]
  \draw[step=0.5, gray!30, thin] (-0.5,-1.0) grid (2.0,1.0);
  \draw[->, thick] (-0.6,0) -- (2.1,0) node[right] {$\operatorname{Re}(z)$};
  \draw[->, thick] (0,-1.1) -- (0,1.1) node[above] {$\operatorname{Im}(z)$};
  \foreach \x in {-0.5,0.5,1.0,1.5,2.0}
    \draw (\x,1.5pt) -- (\x,-1.5pt) node[below, font=\scriptsize] {$\x$};
  \foreach \y in {-1.0,-0.5,0.5,1.0}
    \draw (1.5pt,\y) -- (-1.5pt,\y) node[left, font=\scriptsize] {$\y$};
  \draw[red!80!black, ultra thick] (1,0) circle ({1/sqrt(2)});
  \filldraw[red!80!black] (1,0) circle (1.4pt) node[below=4pt, font=\small\bfseries] {$C_2 = (+1,0)$};
  \filldraw[black] ({1+1/sqrt(2)},0) circle (1.4pt) node[above right, font=\small] {$b=0$};
  \filldraw[black] ({1-1/sqrt(2)},0) circle (1.4pt) node[above left, font=\small] {$b=\frac{\pi}{\ln 2}$};
  \draw[dashed, thick, red!80!black] (1,0) -- ({1+cos(45)/sqrt(2)}, {sin(45)/sqrt(2)}) node[midway, above, sloped, font=\small\bfseries] {$\mathcal{R}_2 = \frac{1}{\sqrt{2}} \approx 0.707$};
\end{tikzpicture}
\caption{Locus of $t_2(\frac{1}{2}+bi) = \frac{-2^{-(1/2+bi)}}{1 - 2^{1/2-bi}}$ on the critical line $a = \frac{1}{2}$: an exact circle centered at $C_2 = (+1,0)$ with radius $\mathcal{R}_2 = \frac{1}{\sqrt{2}} \approx 0.7071$.}
\label{fig:t2}
\end{figure}

Grouping $t_1(s)$ and $t_2(s)$ yields the two-term partial sum:
\begin{equation}
S_2(s) = t_1(s) + t_2(s) = \frac{1 - 2^{-s}}{1 - 2^{1-s}}.
\end{equation}
Letting $z = e^{-i b \ln 2}$ parametrize the unit circle $|z|=1$, $S_2(s)$ acts as the M\"obius transformation $M_a(z) = \frac{1 - 2^{-a}z}{1 - 2^{1-a}z}$.

\begin{proposition}[Center and Radius of $S_2$]
\label{prop:S2}
For any fixed $a \in \mathbb{R} \setminus \{1\}$, the locus of $S_2(a+bi)$ is a circle centered on the real axis at $C(a) = (h(a), 0)$ with radius $\mathcal{R}(a)$, where:
\begin{equation}
h(a) = h_1(a) + h_2(a) = \frac{1 - 2^{1-2a}}{1 - 2^{2-2a}}, \qquad \mathcal{R}(a) = \frac{2^{-a}}{|1 + 2^{1-a}|}.
\end{equation}
\end{proposition}

\begin{theorem}[Unique Origin-Centering on the Critical Line]
\label{thm:origin}
The center of the circle $S_2(a+bi)$ lies exactly at the origin $(0,0)$ if and only if $a = \frac{1}{2}$.
\end{theorem}

\begin{proof}
Setting the center $h(a) = 0$ (for $a \neq 1$) requires $1 - 2^{1-2a} = 0$, which yields $1 - 2a = 0 \implies a = \frac{1}{2}$. At $a = \frac{1}{2}$, the centers $C_1 = (-1,0)$ and $C_2 = (+1,0)$ cancel, and the map reduces to a scaled Blaschke factor \cite{Garnett2007}:
\begin{equation}
S_2\left(\frac{1}{2} + bi\right) = \frac{1}{\sqrt{2}} \left(\frac{\sqrt{2} - z}{1 - \sqrt{2}z}\right),
\end{equation}
whose modulus is strictly constant at $\mathcal{R}\left(\frac{1}{2}\right) = \frac{1}{\sqrt{2}}$ for all $b \in \mathbb{R}$ (Figure \ref{fig:sum}).
\end{proof}

\begin{figure}[H]
\centering
\begin{tikzpicture}[scale=1.55]
  \draw[step=0.5, gray!30, thin] (-2.6,-1.6) grid (1.9,1.6);
  \draw[->, thick] (-2.7,0) -- (2.0,0) node[right] {$\operatorname{Re}(z)$};
  \draw[->, thick] (0,-1.7) -- (0,1.7) node[above] {$\operatorname{Im}(z)$};
  \foreach \x in {-2.5,-2.0,-1.5,-1.0,-0.5,0.5,1.0,1.5}
    \draw (\x,2pt) -- (\x,-2pt) node[below, font=\scriptsize] {$\x$};
  \foreach \y in {-1.5,-1.0,-0.5,0.5,1.0,1.5}
    \draw (2pt,\y) -- (-2pt,\y) node[left, font=\scriptsize] {$\y$};
  \draw[blue!75!black, dashed, very thick] (-1,0) circle ({sqrt(2)});
  \filldraw[blue!75!black] (-1,0) circle (1.5pt) node[above=3pt, font=\scriptsize\bfseries] {$C_1=(-1,0)$};
  \draw[red!80!black, dotted, very thick] (1,0) circle ({1/sqrt(2)});
  \filldraw[red!80!black] (1,0) circle (1.5pt) node[above=3pt, font=\scriptsize\bfseries] {$C_2=(+1,0)$};
  \draw[violet, ultra thick] (0,0) circle ({1/sqrt(2)});
  \filldraw[violet] (0,0) circle (1.8pt) node[below right=2pt, font=\footnotesize\bfseries] {$C=(0,0)$};
  \draw[violet, dashed, thick] (0,0) -- ({cos(45)/sqrt(2)}, {sin(45)/sqrt(2)}) node[midway, above, sloped, font=\scriptsize\bfseries] {$\mathcal{R}=\frac{1}{\sqrt{2}}$};
  \node[blue!80!black, font=\small\bfseries] at (-1.85, 1.32) {$t_1(\frac{1}{2}+bi)$};
  \node[red!80!black, font=\small\bfseries] at (1.45, 0.78) {$t_2(\frac{1}{2}+bi)$};
  \node[violet, font=\small\bfseries] at (-0.48, 0.82) {$S_2 = t_1 + t_2$};
\end{tikzpicture}
\caption{Superposition of $t_1(\frac{1}{2}+bi)$ (dashed blue, $C_1=(-1,0)$, $\mathcal{R}_1=\sqrt{2}$), $t_2(\frac{1}{2}+bi)$ (dotted red, $C_2=(+1,0)$, $\mathcal{R}_2=\frac{1}{\sqrt{2}}$), and their sum $S_2(\frac{1}{2}+bi) = t_1 + t_2$ (solid purple), which forms an exact circle centered at the origin $C=(0,0)$ with radius $\mathcal{R} = \frac{1}{\sqrt{2}} \approx 0.7071$.}
\label{fig:sum}
\end{figure}

When $a \neq \frac{1}{2}$, the circle $S_2(a+bi)$ immediately shifts its center $C(a) = (h(a), 0)$ off the origin along the real axis, as summarized in Table \ref{tab:shifts} and illustrated in Figure \ref{fig:offcenter}.

\begin{table}[H]
\centering
\caption{Center $C(a) = (h(a), 0)$ and radius $\mathcal{R}(a)$ of $S_2(a+bi) = t_1 + t_2$ across values of $a = \operatorname{Re}(s)$.}
\label{tab:shifts}
\small
\begin{tabular}{@{}llll@{}}
\toprule
\textbf{Real Part $a$} & \textbf{Center $C(a) = (h(a), 0)$} & \textbf{Radius $\mathcal{R}(a)$} & \textbf{Shift Relative to $(0,0)$} \\ \midrule
$a = 0$ ($a < \frac{1}{2}$) & $(+\frac{1}{3}, \, 0) \approx (+0.333, 0)$ & $\frac{1}{3} \approx 0.333$ & Shifted right ($h(a) > 0$) \\
\textbf{$a = \frac{1}{2}$ (Critical Line)} & \textbf{$(0, \, 0)$} & \textbf{$\frac{1}{\sqrt{2}} \approx 0.707$} & \textbf{Centered at origin $(0,0)$} \\
$a = 0.75$ ($\frac{1}{2} < a < 1$) & $(-\frac{1}{\sqrt{2}}, \, 0) \approx (-0.707, 0)$ & $2^{-1/4} + \frac{1}{2} \approx 1.341$ & Shifted left ($h(a) < 0$) \\
$a = 1.5$ ($a > 1$) & $(+\frac{3}{7}, \, 0) \approx (+0.429, 0)$ & $\frac{2\sqrt{2}}{7} \approx 0.404$ & Shifted right ($h(a) > 0$) \\ \bottomrule
\end{tabular}
\end{table}

\begin{figure}[H]
\centering
\begin{tikzpicture}[scale=1.55]
  \draw[step=0.5, gray!30, thin] (-2.2,-1.5) grid (1.1,1.5);
  \draw[->, thick] (-2.3,0) -- (1.2,0) node[right] {$\operatorname{Re}(z)$};
  \draw[->, thick] (0,-1.6) -- (0,1.6) node[above] {$\operatorname{Im}(z)$};
  \foreach \x in {-2.0,-1.5,-1.0,-0.5,0.5,1.0}
    \draw (\x,2pt) -- (\x,-2pt) node[below, font=\scriptsize] {$\x$};
  \foreach \y in {-1.5,-1.0,-0.5,0.5,1.0,1.5}
    \draw (2pt,\y) -- (-2pt,\y) node[left, font=\scriptsize] {$\y$};
  \draw[violet, ultra thick] (0,0) circle (0.7071);
  \filldraw[violet] (0,0) circle (1.8pt);
  \draw[green!60!black, dashed, very thick] (0.3333,0) circle (0.3333);
  \filldraw[green!60!black] (0.3333,0) circle (1.8pt);
  \draw[orange!90!black, dashdotted, very thick] (-0.7071,0) circle (1.3409);
  \filldraw[orange!90!black] (-0.7071,0) circle (1.8pt);
  
  \node[violet, font=\footnotesize\bfseries] at (0.48, 0.80) {$a=0.5: C=(0,0)$};
  \node[green!50!black, font=\footnotesize\bfseries] at (0.62, -0.45) {$a=0: C=(\frac{1}{3},0)$};
  \node[orange!85!black, font=\footnotesize\bfseries] at (-1.15, 1.42) {$a=0.75: C=(-\frac{1}{\sqrt{2}},0)$};
\end{tikzpicture}
\caption{Off-center shift of the circle $S_2(a+bi) = t_1 + t_2$ when $a \neq \frac{1}{2}$: $a = 0.5$ is centered at $(0,0)$ (solid purple, $\mathcal{R} \approx 0.707$), $a = 0$ shifts right to $C = (\frac{1}{3}, 0)$ (dashed green, $\mathcal{R} = \frac{1}{3}$), and $a = 0.75$ shifts left to $C = (-\frac{1}{\sqrt{2}}, 0)$ (dash-dot orange, $\mathcal{R} \approx 1.341$).}
\label{fig:offcenter}
\end{figure}

\section{The Dyadic Pairing Theorem}

For individual unpaired terms $t_n(s) = (-1)^{n-1}n^{-s}t_1(s)$ with $n \ge 3$ ($n \neq 2^k$), the incommensurable rotation $e^{-i b \ln n}$ spins an off-center circle around the origin, sweeping out a thick annular rosette. In contrast, grouping every odd integer $m \in \{1, 3, 5, \dots\}$ with its even double $2m$ locks the prefactor phase to the base-$2$ term, defining the $m$-th dyadic pair:
\begin{equation}
P_m(s) = t_m(s) + t_{2m}(s) = \frac{1}{m^s} \left(\frac{1 - 2^{-s}}{1 - 2^{1-s}}\right) = \frac{1}{m^s} S_2(s).
\end{equation}

\begin{corollary}
\label{cor:dyadic}
On the critical line $a = \frac{1}{2}$, every dyadic pair $P_m\left(\frac{1}{2}+bi\right)$ traces an \textbf{exact circle centered at the origin $(0,0)$} with descending radius $\mathcal{R}_m = \frac{1}{\sqrt{2m}}$:
\begin{equation}
\left|P_m\left(\frac{1}{2} + bi\right)\right| = m^{-1/2} \left|S_2\left(\frac{1}{2} + bi\right)\right| = \frac{1}{\sqrt{m}} \cdot \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2m}}.
\end{equation}
For $a \neq \frac{1}{2}$, these circles shift off-center into annular rosettes of thickness $|h(a)|m^{-a}$.
\end{corollary}

\begin{proof}
Direct evaluation from Theorem \ref{thm:origin} shows that $|S_2(\frac{1}{2}+bi)| = \frac{1}{\sqrt{2}}$ is strictly constant for all $b \in \mathbb{R}$. In particular, beyond $m = 1$ ($P_1 = t_1 + t_2$, $\mathcal{R}_1 = \frac{1}{\sqrt{2}} \approx 0.7071$), the higher odd-even pairs $m = 3$ ($P_3 = t_3 + t_6$, $\mathcal{R}_3 = \frac{1}{\sqrt{6}} \approx 0.4082$) and $m = 5$ ($P_5 = t_5 + t_{10}$, $\mathcal{R}_5 = \frac{1}{\sqrt{10}} \approx 0.3162$) trace exact concentric circles centered at $(0,0)$, as shown in Figure \ref{fig:cor41}.
\end{proof}

\begin{figure}[H]
\centering
\begin{tikzpicture}[scale=2.8]
  \draw[step=0.2, gray!30, thin] (-0.85,-0.85) grid (0.85,0.85);
  \draw[->, thick] (-0.9,0) -- (0.9,0) node[right] {$\operatorname{Re}(z)$};
  \draw[->, thick] (0,-0.9) -- (0,0.9) node[above] {$\operatorname{Im}(z)$};
  \foreach \x in {-0.8,-0.6,-0.4,-0.2,0.2,0.4,0.6,0.8}
    \draw (\x,1pt) -- (\x,-1pt) node[below, font=\scriptsize] {$\x$};
  \foreach \y in {-0.8,-0.6,-0.4,-0.2,0.2,0.4,0.6,0.8}
    \draw (1pt,\y) -- (-1pt,\y) node[left, font=\scriptsize] {$\y$};
  \draw[violet, ultra thick] (0,0) circle (0.7071);
  \draw[violet, dashed, thick] (0,0) -- ({0.7071*cos(45)}, {0.7071*sin(45)});
  \draw[teal!85!black, ultra thick] (0,0) circle (0.4082);
  \draw[teal!85!black, dashed, thick] (0,0) -- ({0.4082*cos(135)}, {0.4082*sin(135)});
  \draw[orange!90!black, ultra thick] (0,0) circle (0.3162);
  \draw[orange!90!black, dashed, thick] (0,0) -- ({0.3162*cos(-45)}, {0.3162*sin(-45)});
  \filldraw[black] (0,0) circle (0.9pt) node[below left=1pt, font=\scriptsize\bfseries] {$C=(0,0)$};
  
  \node[violet, font=\footnotesize\bfseries, fill=white, inner sep=1pt] at (0.52, 0.62) {$P_1 = t_1+t_2 \ (\mathcal{R}_1=\frac{1}{\sqrt{2}}\approx 0.707)$};
  \node[teal!85!black, font=\footnotesize\bfseries, fill=white, inner sep=1pt] at (-0.45, 0.45) {$P_3 = t_3+t_6 \ (\mathcal{R}_3=\frac{1}{\sqrt{6}}\approx 0.408)$};
  \node[orange!90!black, font=\footnotesize\bfseries, fill=white, inner sep=1pt] at (0.42, -0.35) {$P_5 = t_5+t_{10} \ (\mathcal{R}_5=\frac{1}{\sqrt{10}}\approx 0.316)$};
\end{tikzpicture}
\caption{Illustration of Corollary \ref{cor:dyadic} on the critical line $a = \frac{1}{2}$: the odd-even pairs $P_1 = t_1+t_2$ ($\mathcal{R}_1 = \frac{1}{\sqrt{2}} \approx 0.7071$), $P_3 = t_3+t_6$ ($\mathcal{R}_3 = \frac{1}{\sqrt{6}} \approx 0.4082$), and $P_5 = t_5+t_{10}$ ($\mathcal{R}_5 = \frac{1}{\sqrt{10}} \approx 0.3162$) trace exact concentric circles centered at the origin $(0,0)$.}
\label{fig:cor41}
\end{figure}

\section{\texorpdfstring{Distinguishing Riemann Zeta Zeros via Odd-Even Dyadic Pairs}{Distinguishing Riemann Zeta Zeros via Odd-Even Dyadic Pairs}}

To understand how the concentric origin-centered circles $P_m(\frac{1}{2}+bi)$ distinguish non-trivial Riemann zeta zeros $b = b_n$ from other choices of $b \in \mathbb{R}$, we write $S_2(\frac{1}{2}+bi) = \frac{1}{\sqrt{2}}e^{i\Phi_2(b)}$, where the base-$2$ Blaschke phase $\Phi_2(b)$ is given by:
\begin{equation}
\cos\Phi_2(b) = \frac{2\sqrt{2} - 3\cos(b\ln 2)}{3 - 2\sqrt{2}\cos(b\ln 2)}, \qquad \sin\Phi_2(b) = \frac{-\sin(b\ln 2)}{3 - 2\sqrt{2}\cos(b\ln 2)}.
\end{equation}
For any odd integer $m \in \{1, 3, 5, \dots\}$, the point $P_m(\frac{1}{2}+bi)$ lies on the concentric circle of radius $\mathcal{R}_m = \frac{1}{\sqrt{2m}}$ at polar angle $\Psi_m(b) = \Phi_2(b) - b\ln m$. Summing over all odd $m$ yields:
\begin{equation}
\sum_{m \text{ odd}} P_m(s) = S_2(s) \sum_{m \text{ odd}} \frac{1}{m^s} = S_2(s)\left(1 - 2^{-s}\right)\zeta(s).
\end{equation}
Since neither $S_2(\frac{1}{2}+bi)$ nor $(1 - 2^{-1/2-bi})$ ever vanishes for $b \in \mathbb{R}$, the sum of the concentric dyadic circles vanishes (in the analytic continuation sense) if and only if $\zeta(\frac{1}{2}+bi) = 0$. Dividing by the non-zero anchor $P_1(\frac{1}{2}+bi) = S_2(\frac{1}{2}+bi)$ shows that $b$ is a Riemann zeta zero if and only if the inner circles ($m = 3, 5, 7, \dots$) satisfy two simultaneous relative-phase balance equations:
\begin{equation}
\label{eq:zero_conditions}
\underbrace{\sum_{m=3,5,7,\dots} \frac{\cos(b\ln m)}{\sqrt{m}} = -1}_{\text{Radial Opposition to } P_1}, \qquad \underbrace{\sum_{m=3,5,7,\dots} \frac{\sin(b\ln m)}{\sqrt{m}} = 0}_{\text{Transverse Balance}}.
\end{equation}

\subsection{Four Geometric Patterns Distinguishing Zeta Zeros from Generic $b$}

Comparing true zeta zeros $b = b_n$ against generic values of $b \in \mathbb{R}$ reveals four structural mechanisms on the concentric circles:

\begin{enumerate}
    \item \textbf{Right-Half-Plane (``Slow-Zone'') Clustering of the Outer Circle $P_1$:}
    The angular velocity of $P_1(\frac{1}{2}+bi)$ along its circle of radius $\frac{1}{\sqrt{2}}$ is highly non-uniform:
    \begin{equation}
    \left|\frac{d\Phi_2}{db}\right| = \frac{\ln 2}{3 - 2\sqrt{2}\cos(b\ln 2)}.
    \end{equation}
    Near $b\ln 2 \equiv \pi \pmod{2\pi}$ (where $\Phi_2 = 0^\circ$ and $P_1 = (+\frac{1}{\sqrt{2}}, 0)$), $P_1$ moves at its minimum speed $\frac{\ln 2}{3+2\sqrt{2}} \approx 0.1189\text{ rad/unit }b$. Near $b\ln 2 \equiv 0 \pmod{2\pi}$ (where $\Phi_2 = 180^\circ$ and $P_1 = (-\frac{1}{\sqrt{2}}, 0)$), $P_1$ whips around $34\times$ faster ($\frac{\ln 2}{3-2\sqrt{2}} \approx 4.0399\text{ rad/unit }b$). Consequently, $P_1$ spends over $80\%$ of its parameter trajectory lingering in the right half-plane ($\operatorname{Re}(P_1) > 0$). Across all first five zeta zeros $b_1, \dots, b_5$, $\Phi_2(b_n)$ clusters tightly within $[-12.50^\circ, +20.60^\circ]$ of the positive real axis (Table \ref{tab:five_zeros_dyadic}).
    
    \item \textbf{Opposing-Hemisphere Alignment of Leading Inner Circles ($P_3, P_5, P_7$):}
    Because $P_1$ has the largest radius $\mathcal{R}_1 \approx 0.7071$ and clusters near $0^\circ$ in the right half-plane, cancellation toward $(0,0)$ requires the largest inner circles $P_3$ ($\mathcal{R}_3 \approx 0.4082$), $P_5$ ($\mathcal{R}_5 \approx 0.3162$), and $P_7$ ($\mathcal{R}_7 \approx 0.2673$) to swing into the \textbf{opposing left hemisphere} where $\cos(b\ln m) < 0$ (Table \ref{tab:b1_5circles}). Any interval of $b$ in which $\cos(b\ln 3) > 0$ and $\cos(b\ln 5) > 0$ simultaneously reinforces $P_1$, driving the cumulative sum away from the origin ($|\Sigma_3| > 1.0$) and ruling out low-lying zeros.
    
    \item \textbf{Stationary-Phase Cutoff Circle $M(b) \approx \sqrt{b / 2\pi}$ and Finite Exclusion:}
    The angular step between consecutive odd circles $P_m$ and $P_{m+2}$ is $\Delta\Psi \approx -\frac{2b}{m}$. Once $m$ exceeds the Riemann-Siegel stationary-phase cutoff $M(b) = \lfloor\sqrt{b/2\pi}\rfloor$, successive inner circles form cancelling Cornu-type spirals bounded by an explicit remainder $E_M(b) = \mathcal{O}(b^{-1/4})$ \cite{Titchmarsh1986,Edwards1974}. Thus, if the finite sum of the first $M(b)$ concentric circles exceeds $E_M(b)$ in magnitude, $b$ is guaranteed not to be a zero.
    
    \item \textbf{1D Collinear Locking via Functional Equation Symmetry:}
    Unlike $a \neq \frac{1}{2}$ (where off-center circles wander in 2D), the conjugate symmetry $1 - s = \bar{s}$ on $a = \frac{1}{2}$ locks $\zeta(\frac{1}{2}+bi) = e^{-i\theta(b)}Z(b)$ ($Z(b) \in \mathbb{R}$) onto a 1D line through the origin \cite{Edwards1974}. Whenever the leading inner circles $P_3, P_5, \dots$ sweep across the opposing hemisphere and cause the signed projection $Z(b)$ to change sign between $b_a$ and $b_b$, the concentric circle sum is topologically forced to pass directly through $(0,0)$.
\end{enumerate}

\begin{table}[H]
\centering
\caption{Evaluation of the first five concentric dyadic circles $P_m(\frac{1}{2}+i b_1)$ ($m = 1, 3, 5, 7, 9$) at the first Riemann zeta zero $b_1 \approx 14.134725$, showing monotonic contraction of the cumulative sum $\Sigma_k = \sum_{j=1}^k P_{2j-1}$ toward $(0,0)$.}
\label{tab:b1_5circles}
\small
\begin{tabular}{@{}llllll@{}}
\toprule
\textbf{Circle ($m$)} & \textbf{Radius $\mathcal{R}_m$} & \textbf{Angle $\Psi_m(b_1)$} & \textbf{Point $P_m(\frac{1}{2}+ib_1)$} & \textbf{Cumulative Sum $\Sigma_k$} & \textbf{Distance $|\Sigma_k|$} \\ \midrule
$P_1 = t_1 + t_2$ & $0.7071$ & $+3.71^\circ$ (Q1) & $(+0.7056, \, +0.0457)$ & $+0.7056 + 0.0457i$ & $\mathbf{0.7071}$ \\
$P_3 = t_3 + t_6$ & $0.4082$ & $-166.02^\circ$ (Q3) & $(-0.3962, \, -0.0986)$ & $+0.3095 - 0.0530i$ & $\mathbf{0.3140}$ \\
$P_5 = t_5 + t_{10}$ & $0.3162$ & $+140.29^\circ$ (Q2) & $(-0.2433, \, +0.2021)$ & $+0.0662 + 0.1491i$ & $\mathbf{0.1632}$ \\
$P_7 = t_7 + t_{14}$ & $0.2673$ & $-132.21^\circ$ (Q3) & $(-0.1796, \, -0.1980)$ & $-0.1133 - 0.0489i$ & $\mathbf{0.1234}$ \\
$P_9 = t_9 + t_{18}$ & $0.2357$ & $+24.26^\circ$ (Q1) & $(+0.2149, \, +0.0968)$ & $+0.1016 + 0.0480i$ & $\mathbf{0.1123}$ \\ \bottomrule
\end{tabular}
\end{table}

\begin{table}[H]
\centering
\caption{Comparison of polar angles $\Psi_m(b_n)$ and 5-circle residual distance $|\Sigma_5|$ across the first five non-trivial Riemann zeta zeros ($b_1$ through $b_5$).}
\label{tab:five_zeros_dyadic}
\small
\begin{tabular}{@{}llllllll@{}}
\toprule
\textbf{Zeta Zero $b_n$} & \textbf{$\Psi_1(b_n)$} & \textbf{$\Psi_3(b_n)$} & \textbf{$\Psi_5(b_n)$} & \textbf{$\Psi_7(b_n)$} & \textbf{$\Psi_9(b_n)$} & \textbf{Initial $|\Sigma_1|$} & \textbf{5-Circle $|\Sigma_5|$} \\ \midrule
$b_1 \approx 14.1347$ & $+3.71^\circ$ & $-166.02^\circ$ & $+140.29^\circ$ & $-132.21^\circ$ & $+24.26^\circ$ & $0.7071$ & $\mathbf{0.1123}$ \\
$b_2 \approx 21.0220$ & $-12.50^\circ$ & $+104.24^\circ$ & $-151.03^\circ$ & $+163.70^\circ$ & $-139.01^\circ$ & $0.7071$ & $\mathbf{0.1216}$ \\
$b_3 \approx 25.0109$ & $+20.60^\circ$ & $-113.73^\circ$ & $-125.75^\circ$ & $+112.08^\circ$ & $+111.94^\circ$ & $0.7071$ & $\mathbf{0.1504}$ \\
$b_4 \approx 30.4249$ & $-9.50^\circ$ & $-124.62^\circ$ & $+64.90^\circ$ & $-161.65^\circ$ & $+120.26^\circ$ & $0.7071$ & $\mathbf{0.2320}$ \\
$b_5 \approx 32.9351$ & $+8.74^\circ$ & $+95.61^\circ$ & $-148.34^\circ$ & $-63.27^\circ$ & $-177.51^\circ$ & $0.7071$ & $\mathbf{0.2918}$ \\ \bottomrule
\end{tabular}
\end{table}

\section{\texorpdfstring{Removing Odd-Even Dyadic Pairs: The Multiples-of-4 Subseries}{Removing Odd-Even Dyadic Pairs: The Multiples-of-4 Subseries}}

When we remove all odd-even dyadic pairs $P_m(s) = t_m(s) + t_{2m}(s)$ ($m = 1, 3, 5, \dots$) from the full series $\sum_{n=1}^\infty t_n(s)$, we strip out all odd indices ($n \equiv 1, 3 \pmod 4$) and all singly-even indices ($n = 2m \equiv 2 \pmod 4$). The remaining terms of the series are strictly the \textbf{multiples of 4}:
\begin{equation}
n = 4k \in \{4, 8, 12, 16, 20, \dots\}, \qquad k \in \mathbb{N}.
\end{equation}

\begin{proposition}[Factorization of the Multiples-of-4 Subseries]
\label{prop:mult4}
Every multiple-of-$4$ term $t_{4k}(s)$ has non-alternating sign $(-1)^{4k-1} = -1$ and factors as $t_{4k}(s) = k^{-s} t_4(s)$, where $t_4(s) = \frac{-4^{-s}}{1 - 2^{1-s}}$. Consequently, the $K$-term partial sum $\Sigma_K^{(4)}(s)$ and analytic continuation of the multiples-of-$4$ subseries satisfy:
\begin{equation}
\Sigma_K^{(4)}(s) = \sum_{k=1}^K t_{4k}(s) = t_4(s) \sum_{k=1}^K \frac{1}{k^s}, \qquad \sum_{k=1}^\infty t_{4k}(s) = t_4(s)\,\zeta(s).
\end{equation}
\end{proposition}

\begin{theorem}[Exclusive Origin-Convergence of the Multiples-of-4 Subseries at Zeta Zeros]
\label{thm:mult4_zero}
On the critical line $s = \frac{1}{2} + ib$, the $K$-term multiples-of-$4$ partial sum obeys the Euler-Maclaurin spiral decomposition:
\begin{equation}
\label{eq:em_spiral}
\Sigma_K^{(4)}\left(\frac{1}{2} + ib\right) = \underbrace{t_4\left(\frac{1}{2} + ib\right)\zeta\left(\frac{1}{2} + ib\right)}_{\text{Target Center } C_{\mathrm{target}}(b)} + \frac{t_4\left(\frac{1}{2} + ib\right)}{\frac{1}{2} - ib} K^{\frac{1}{2} - ib} + \mathcal{O}\left(K^{-1/2}\right).
\end{equation}
The target center $C_{\mathrm{target}}(b)$ of the multiples-of-$4$ subseries lies at the origin $(0,0)$ \textbf{if and only if} $b = b_n$ is a zero of $\zeta(\frac{1}{2}+ib)$. For all non-zero choices of $b$, the subseries spirals around a non-zero point $C_{\mathrm{target}}(b) \neq (0,0)$.
\end{theorem}

\begin{proof}
On the critical line $a = \frac{1}{2}$, the conic radius of the anchor term $t_4(\frac{1}{2}+ib)$ is:
\begin{equation}
R_4(b) = \left|t_4\left(\frac{1}{2} + ib\right)\right| = \frac{1}{2\sqrt{3 - 2\sqrt{2}\cos(b\ln 2)}} \in \left[\frac{\sqrt{2}-1}{2} \approx 0.2071, \;\; \frac{\sqrt{2}+1}{2} \approx 1.2071\right].
\end{equation}
Because $R_4(b) \ge 0.2071 > 0$ for all $b \in \mathbb{R}$, the factor $t_4(\frac{1}{2}+ib)$ never vanishes. Applying the Euler-Maclaurin summation formula \cite{Edwards1974,Apostol1976} to $\sum_{k=1}^K k^{-s}$ and multiplying by $t_4(s)$ yields \eqref{eq:em_spiral}. Thus, $C_{\mathrm{target}}(b) = t_4(\frac{1}{2}+ib)\zeta(\frac{1}{2}+ib) = (0,0)$ if and only if $\zeta(\frac{1}{2}+ib) = 0$.
\end{proof}

Table \ref{tab:mult4_comparison} illustrates Theorem \ref{thm:mult4_zero}: at true zeta zeros ($b_1, b_2, b_3, b_4, b_5$), the anchor conic $t_4(b_n)$ is compressed near its periapsis minimum ($0.2071$) and the first five terms $t_4, t_8, t_{12}, t_{16}, t_{20}$ contract strongly toward $C_{\mathrm{target}}(b_n) = (0,0)$. Conversely, at non-zero values of $b$ (such as $b = 0$, $b = \frac{\pi}{\ln 2} \approx 4.5324$, $b = 10.0$, and the Gram point $b = 17.8456$), adding $t_8, \dots, t_{20}$ reinforces $t_4$ toward the off-origin center $C_{\mathrm{target}}(b) \neq (0,0)$. Finally, applying the dyadic pairing theorem recursively to the multiples of $4$ by pairing $(t_{4m} + t_{8m})$, $(t_{16m} + t_{32m}), \dots$ restores alternating signs and sums the complete binary tower of each odd $m$ to $\mathcal{T}_m(s) = \sum_{r=0}^\infty t_{2^r m}(s) = (1 - 2^{-s})^{-1}m^{-s}$.

\begin{table}[H]
\centering
\caption{Comparison of the multiples-of-$4$ subseries $\Sigma_5^{(4)}(b) = \sum_{k=1}^5 t_{4k}(\frac{1}{2}+ib)$ at non-zero values of $b$ versus the first five Riemann zeta zeros ($b_1$ through $b_5$).}
\label{tab:mult4_comparison}
\small
\begin{tabular}{@{}llllll@{}}
\toprule
\textbf{Input $b$} & \textbf{Category} & \textbf{Radius $R_4(b)$} & \textbf{Target $|C_{\mathrm{target}}(b)|$} & \textbf{5-Term $|\Sigma_5^{(4)}(b)|$} & \textbf{Ratio $\frac{|\Sigma_5^{(4)}|}{R_4(b)}$} \\ \midrule
$b = 0.0000$ & Non-Zero (Real Axis) & $1.2071$ & $1.7620 \neq 0$ & $3.9010$ & $3.232$ (Diverges) \\
$b = \frac{\pi}{\ln 2} \approx 4.5324$ & Non-Zero (Half-Period) & $0.2071$ & $0.1398 \neq 0$ & $0.2477$ & $1.196$ (Off-Origin) \\
$b = 10.0000$ & Non-Zero & $0.5791$ & $0.8972 \neq 0$ & $0.7469$ & $1.290$ (Off-Origin) \\
$b = 17.8456$ & Non-Zero (Gram Point $g_1$) & $1.0514$ & $2.4606 \neq 0$ & $2.2878$ & $2.176$ (Off-Origin) \\ \midrule
$\mathbf{b_1 \approx 14.1347}$ & \textbf{1st Zeta Zero} & $0.2107$ & $\mathbf{0.0000}$ & $\mathbf{0.0467}$ & $\mathbf{0.222}$ ($77.8\%$ cancelled) \\
$\mathbf{b_2 \approx 21.0220}$ & \textbf{2nd Zeta Zero} & $0.2443$ & $\mathbf{0.0000}$ & $\mathbf{0.0536}$ & $\mathbf{0.219}$ ($78.1\%$ cancelled) \\
$\mathbf{b_3 \approx 25.0109}$ & \textbf{3rd Zeta Zero} & $0.2968$ & $\mathbf{0.0000}$ & $\mathbf{0.0752}$ & $\mathbf{0.253}$ ($74.7\%$ cancelled) \\
$\mathbf{b_4 \approx 30.4249}$ & \textbf{4th Zeta Zero} & $0.2293$ & $\mathbf{0.0000}$ & $\mathbf{0.0887}$ & $\mathbf{0.387}$ ($61.3\%$ cancelled) \\
$\mathbf{b_5 \approx 32.9351}$ & \textbf{5th Zeta Zero} & $0.2260$ & $\mathbf{0.0000}$ & $\mathbf{0.0996}$ & $\mathbf{0.441}$ ($55.9\%$ cancelled) \\ \bottomrule
\end{tabular}
\end{table}

\section{Conclusion}

By unifying the Dirichlet prefactor with individual terms, we have shown that the critical line $a = \frac{1}{2}$ is distinguished not only algebraically by functional equation symmetry, but geometrically: it is the \textbf{unique real axis slice where the base-2 epicyclic anchor $S_2(s)$ and all its dyadic descendents $P_m(s)$ are perfectly centered at the origin $(0,0)$} \cite{BohrCourant1914}. Furthermore, the phase distribution across the concentric circles $P_m(\frac{1}{2}+ib)$ and the exclusive origin-convergence of the complementary multiples-of-$4$ subseries $\sum_{k=1}^\infty t_{4k}(s) = t_4(s)\zeta(s)$ provide complementary geometric criteria that separate Riemann zeta zeros from all other points on the critical line.

\section*{Acknowledgments}

The author gratefully acknowledges \url{https://tacoyaci.github.io/} for her informal notes and helpful feedback on an earlier version of this manuscript.

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\end{document}