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

Unformalised remark: roots close to a regular polygon

Let $f(z)=\prod _{i=1}^n(z-a_i)$, $n\ge 2$, with all $|a_i|\le 1$. Put ${\rho }_i=|a_i|$ and $D=|\mathrm{disc}(f)|/n^n$. The Vandermonde determinant estimate says that, when $D\ge 1-\eta$ and $0\le \eta \le 1/(80n^2)$,

where the second inequality is for $i\ne j$. Moreover, the roots lie within $7\sqrt{\eta }$ of a rotated regular $n$-gon, after a bijection. These estimates come from near equality in Hadamard’s determinant inequality, as follows.

Take the Vandermonde matrix $V_{im}=a_i^m$, $0\le m<n$, and let $G=V{V}^{\ast }$, $s_i=G_{ii}=\sum _{m=0}^{n-1}{\rho }_i^{2m}$ and $H=\mathrm{diag}(s_i^{-1/2})G\mathrm{diag}(s_i^{-1/2})$. The normalisation makes $H$ positive semidefinite with diagonal entries one, and

Every factor is at most one by Hadamard’s inequality, so $D\ge 1-\eta$ forces each factor to be at least $1-\eta$. In particular,

Hadamard–Fischer applied to the $\{i,j\}$ principal block gives $detH\le 1-|H_{ij}{|}^2$, hence $|G_{ij}|\le n\sqrt{\eta }$. For $t=a_i\overline{a_j}$, the finite geometric sum then gives

Together with $1-t=1-{\rho }_j^2+\overline{a_j}(a_j-a_i)$, this proves the displayed separation bound. These two bounds need only $0\le \eta <1$; the smaller range is used for the matching with a regular polygon.

For that matching, put $\kappa =n^2\eta /(n-1)\le 1/80$ and $u_i=a_i/{\rho }_i$. The radial bound gives ${\rho }_i^{2n}\ge 1-\kappa$, while $(1-t)G_{ij}=1-t^n$ gives $|1-a_i^n\overline{a_j^n}|\le 2n\sqrt{\eta }$ for $i\ne j$. Using $|1-b{e}^{i\theta }{|}^2=(1-b{)}^2+b|1-e^{i\theta }{|}^2$ with $b={\rho }_i^n{\rho }_1^n\ge 1-\kappa$, we obtain

Choose the nearest $n$th-root phase to $u_i$ among $u_1{e}^{2\pi ij/n}$. The inequality $|e^{i\theta }-1|\ge 2|\theta |/\pi$ for $|\theta |\le \pi$ shows that its distance from $a_i$ is at most

The separation bound exceeds $14\sqrt{\eta }$ in the stated range, so two roots cannot be assigned the same phase. There are $n$ roots and $n$ phases, giving the required bijection. This ordinary argument uses standard determinant inequalities. It has no new Lean declaration, and no priority is claimed for the stability estimate.

The discriminant hypothesis $D\ge 1-\eta$ with $\eta \le 1/(80n^2)$ is stronger than closeness to the unit circle alone. The regular $n$-gon has $D=1$, whereas a configuration with a repeated root has $D=0$. In particular, radial information alone cannot supply this discriminant bound.

We next compare moduli pointwise on a specified segment. No ordering of complex polynomial values is involved. Write $a_k={\rho }_k{u}_k$, with $|u_k|=1$. Under $D\ge 1-\eta$ and $0\le \eta \le 1/(10n^4)$, the same source proves

The elementary step is

Here is where the discriminant hypothesis enters. The two bounds just proved imply

For $k=i$, the required inequality is $2s\le 1+{\rho }_i$, which follows from $s\le {\rho }_i\le 1$. For $k\ne i$, it is immediate if $\mathrm{\Re }(u_i\overline{u_k})\le 0$. Otherwise $s\le {\rho }_i$ gives

Multiplying the factorwise inequalities proves the comparison. This argument uses the two elementary bounds, not the polygon-matching range; it therefore also covers $n=2$, when $1/(10n^4)>1/(80n^2)$. Moving the roots inward reduces this modulus bound, but does not select the angular directions of two contained segments.

For exact regular-polygon directions, write $a_i={\rho }_i\omega {\zeta }^i$, with $|\omega |=1$ and $\zeta =e^{2\pi i/n}$. In any degree $n\ge 2$, the sufficient condition

makes the same factorwise comparison hold. Indeed, for $k\ne i$ with positive cosine, $s\le 1$ gives $2s\mathrm{\Re }({\zeta }^{i-k})\le 2\cos (2\pi /n)\le 1+{\rho }_k$; nonpositive cosines and $k=i$ are handled as above. Since $\prod _k(z-\omega {\zeta }^k)=z^n-{\omega }^n$, we obtain

Every root-to-origin segment is therefore contained in $\{|f|\le 1\}$. If all ${\rho }_i<1$, containment is strict: the displayed bound is below $1$ for $s>0$, and $|f(0)|=\prod _i{\rho }_i<1$. Any two roots can then be joined through the origin with length ${\rho }_i+{\rho }_j<2$. For $2\le n\le 6$ the lower bound on ${\rho }_k$ is nonpositive, so every choice of positive radii at most one is allowed. For $n\ge 7$ the argument instead requires each radius to be at least $2\cos (2\pi /n)-1$. This is a sufficient condition for the factorwise comparison, not a claimed necessary condition for a contained path. The favourable index in the general unit-circle averaging identity may move with $s$, however; an existence statement at one level does not supply a fixed pair of roots.