Paths in Polynomial Lemniscates: A Degree-Seven Counterexample and Radial Connections

Unformalised remark: a bounded-radius concyclic class

When the common root circle is small enough, its self-inversive polynomial gives a contained adjacent-root chord. This ordinary argument has no complete formal counterpart in the supplied record. Let $f$ be monic of degree $n\ge 3$ with distinct zeros on a circle of radius $\rho$. If $2{\rho }^n\le 1$, two adjacent zeros are joined by their straight chord, whose length is at most $2\rho \sin (\pi /n)<2$, and the chord lies in $\{|f|<1\}$. (If a zero is repeated, the short-connection conclusion is immediate; the distinct case is the substantive one.)

For each fixed degree $n$, the radius condition is $\rho \le 2^{-1/n}$. It allows arbitrary spacing on that circle, but does not cover every radius $\rho <1$. No restriction on the circle’s centre is imposed.

After translating and scaling, let $g$ be the monic polynomial with roots $w_1,\dots ,w_n$ on the unit circle. Such a polynomial is self-inversive: $z^n\overline{g(1/\stackrel{―}{z})}$ is a constant multiple of $g(z)$, with the constant of modulus one. Equivalently, $e^{-in\theta /2}g(e^{i\theta })$ is real up to a fixed phase. Choose

Then $g$ and $q$ have the same leading and constant coefficients, and their boundary values have a common real phase. If every root gap contained a point with $|g|>|q|$, the real boundary function of $g-q$ would have alternating signs at these points. Advancing the angle by $2\pi$ multiplies that function by $(-1{)}^n$, so the alternation also holds across the final gap. The intermediate value theorem would give $n$ distinct unit-circle zeros of $g-q$. This is impossible: its degree is at most $n-1$, and $g=q$ cannot satisfy the assumed strict inequalities. Thus some adjacent-root arc satisfies $|g|\le |q|\le 2$. The interior of that arc contains no zero of $q$, since $g$ is nonzero there. As consecutive zeros of $q$ are separated by angle $2\pi /n$, the selected arc has angular width at most $2\pi /n$. This explains why the same selected pair also has the required short chord. To pass from this arc to its chord $[a,b]$, let $\nu$ be the unit normal pointing into the circular segment bounded by them. On the open chord, logarithmic differentiation gives

The endpoint terms vanish because their directions are tangent to the chord. Every remaining numerator is positive because the other roots lie strictly on the far side of its line; $n\ge 3$ ensures at least one such term. The harmonic function $\log |g|$ has no interior maximum and tends to $-\mathrm{\infty }$ at $a,b$. A maximum cannot lie on the open chord either, since moving into the circular segment increases the function. It is therefore attained on the arc, and the maximum on the chord is strictly smaller. Scaling back gives $|f|<2{\rho }^n\le 1$ throughout the chord, with $f=0$ at its endpoints. These are the arguments of Theorem C and Lemma S, Sections 3–4 of the concyclic note.

The argument is an ordinary proof outside Lean. The exact-rational checker checks finitely many identities used in the proof and configurations, while the numerical checker is regression and stress-test evidence. The arc constant $2$ is attained by the regular $n$-gon. Thus the bound obtained by passing through the arc is $\rho \le 2^{-1/n}$, although this cutoff tends to $1$ as $n\to \mathrm{\infty }$. This does not assert optimality of the chord bound or exclude another contained path at a larger radius. Nor does arc sharpness make the pair of radial segments shortest. For example, for $g(z)=z^3-1/8$ the chord between $1/2$ and $e^{2\pi i/3}/2$ has length $\sqrt{3}/2<1$, whereas the two radii have total length $1$. The binomial calculation above gives a maximum of $9/64<1$ on that chord. The length-$2$ equality at $z^n-1$, proved below, concerns the uncontracted closed-level problem.