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

Unformalised remark: a coefficient condition for four-term polynomials

One more monomial introduces a partial sum that need not lie in the unit disc. Bounding that sum gives the following ordinary coefficient criterion. Let

assume all roots of $g$ lie in the open unit disk, and let $w_1,w_2$ be roots of the two smallest moduli. If

then the complete radial spokes from $w_1$ and $w_2$ to the origin lie in $\{|g|<1\}$, and their broken line has length strictly below $2$. The simpler coefficient condition

makes every root-to-origin segment contained; the coefficient $a$ has no additional restriction beyond the assumed root locations. This coefficient condition includes $b=0$ and $|c|\le 1$, but is not automatic for all four-term polynomials with roots in the disc.

The mechanism is visible in one identity. At a root $w$ and for $0\le u\le 1$,

Here the root equation has eliminated $a{w}^r$ before taking absolute values. The three weights are nonnegative and sum to $1-u^m$. The controlled terms are $c$, $c+b{w}^s$ and $-w^m$; their moduli are strictly below one under the root-dependent condition. The same identity explains why no bound on $a$ is needed. Lean checks the factorization and the resulting strict radial-segment theorem under the exact weak exponent hypotheses $1\le s\le r\le m$; it also checks the coefficient-only corollary above. Its weak bound $|b|+|c|\le 1$ still gives strict containment: the root-disc hypothesis implies $|c|<1$, and, if $b\ne 0$, $|w{|}^s<1$ gives $|c|+|b||w{|}^s<|c|+|b|\le 1$. If $b=0$, the required bound is simply $|c|<1$.

An alternative criterion uses a complex power sum to select two indices. Let $S$ index a finite family of roots, $N=|S|\ge 2$, and

The squared-modulus identity gives

Consequently, if every $|w_i|<1$, $|c|<1$, and

then two distinct indices $i,j\in S$ satisfy $|c+b{w}_i^s|<1$ and $|c+b{w}_j^s|<1$. Indeed, the displayed identity bounds the sum of squared moduli strictly below $N-1$. If at most one index had modulus below one, the other $N-1$ terms would each contribute at least one. This explains both the threshold $N-1$ and why the criterion selects two indices. Applying the resulting contained-segment theorem to these inequalities proves that both root segments lie strictly in $\{|g|<1\}$. Retaining the signed cross term permits the inequality to hold for configurations outside the coefficient-only triangle $|b|+|c|\le 1$. The formal hypotheses do not require $i\mapsto w_i$ to be injective, so distinct indices need not denote distinct root values. For the polynomial consequence, the family must be a submultiset of the zeros, respecting their algebraic multiplicities. A repeated zero then already gives a constant path between two occurrences. Repeating a simple zero in an arbitrary indexed family cannot supply a second occurrence. The two sufficient conditions are not ordered by strength. For example, $g(w)=w^5+19/20$ has all roots in the open unit disc and satisfies the coefficient condition with $b=0$. For its full five-root list, however, the left side of the power-sum condition is $5(19/20{)}^2=361/80>4=N-1$, so that condition fails. Its benefit is the use of cancellation in the cross term, not inclusion of every case covered by the coefficient condition. Conversely,

has all roots of modulus $1/\sqrt{2}$ or $(4/5{)}^{1/3}$, hence below one, but $|b|+|c|=6/5>1$. Its full root list has $M=\sum _i{w}_i^2=1$: the two quadratic roots contribute $1$, and the three cubic roots contribute $0$. The power-sum left side is $5(16/25+4/25)-16/25=84/25<4$. Thus this criterion does cover root-disc polynomials excluded by the coefficient condition.

For $q\ge 2$ and $f(z)=g((z-h{)}^q)$, assume also that all roots of $f$ lie in the open unit disc. If $y^q=w$ and $\omega =e^{2\pi i/q}$, the roots $h+{\omega }^{j}y$ satisfy

Thus every quotient root has $|w|=|y{|}^q<1$, which supplies the hypothesis on the zeros used above, and every lifted segment has length $|y|<1$. Two selected spokes join through $h$ with total length below $2$; repeated root occurrences instead permit the constant path. The Lean theorem assumes the finite root family, its complex power sum $M$, and the displayed inequality. It neither derives that inequality from arbitrary coefficients nor verifies the family’s polynomial multiplicities, the cyclic lifting, or the final path construction. Completeness of the root list is not required for the selection argument: a submultiset respecting multiplicity suffices. Without one of the power-sum, root-dependent, or coefficient-only inequalities, the tetranomial case is not covered by this argument; this family does not solve unrestricted Erdős #1041.