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

Containment of the two selected segments.

Let $f$ be monic of degree $n\ge 3$ with roots in the closed unit disc. The following estimate controls the segments from a critical point to its two nearest roots. To specify the constant, put

Here $t$ parametrises the segment. The product estimates below use $f^{\prime }(c)=0$ and the ordering of distances, not the root-disc hypothesis; the latter enters when bounding the combined segment length. To explain the other parameter and the coefficient $n(n-2)$, suppose $f(c)\ne 0$, order the roots $a_1,\dots ,a_n$ by distance from $c$, and set $v_j=(a_2-c)/(a_j-c)$. Then $v_2=1$, $|v_j|\le 1$ for $j\ge 2$, and $\sum _j{v}_j=0$ by $f^{\prime }(c)=0$. With $y=1-\mathrm{\Re }{v}_1$, this gives

Consequently

The factor $v_2=1$ contributes $(1-t{)}^2$. Applying AM–GM to the remaining $n-2$ squared factors in $f(c+t(a_2-c))/f(c)=\prod _j(1-t{v}_j)$ gives the defining expression for $M_n^2$. Thus the maximisation is an explicit bound for the second segment, not an additional optimisation hypothesis about the roots.

For the nearer segment, put $x_j=(a_1-c)/(a_j-c)$. Now $x_1=1$, $|x_j|\le 1$ for every $j$, and $\sum _j{x}_j=0$. Hence

so AM–GM gives

This bound is at most $M_n^2$ for every $n\ge 3$, not just for numerically tested degrees. Indeed, set $y=n/(n-1)$ in the defining maximum. This value is admissible because $n/(n-1)-(4-n)=(n-2{)}^2/(n-1)\ge 0$. The last factor then has base $(1-t{)}^2+2tn/(n-1)$, and the preceding factor has this same base plus $n(n-2)t^2\ge 0$. Their product therefore dominates the displayed nearer-segment bound. Thus $|f(z)|\le M_n|f(c)|$ on both segments, so $0<|f(c)|\le 1/M_n$ gives containment in $\{|f|\le 1\}$; the critical-point distance estimate gives total length at most $2$. If $f(c)=0$, two root occurrences already coincide at $c$. The quadratic case is the direct root segment treated earlier. For degree five the maximum evaluates to

The breakpoint comes from the constraint on $y$, not from a numerical search. For fixed $0<t<1$, write the two bases in the definition of $M_5$ as

They are positive for $-1\le y\le 5$, and ${\partial }_y\log (A{B}^3)=2t(B-A)/(AB)$. Thus the unconstrained maximum occurs at $A=B$, or $y=5/4-45t/8$. This reaches the lower endpoint $-1$ at $t=2/5$; thereafter the maximum is at $y=-1$. For $0\le t\le 2/5$ the resulting expression is $(1-t)(1+t/2+19t^2/4{)}^2$, whose derivative is

For $2/5\le t\le 1$, the square of the resulting expression is $(1-t{)}^2(1+t{)}^6(16t^2-4t+1)$, whose derivative is

The quadratic changes sign once on this interval, at $t_{\ast }$, so the second branch increases from the common breakpoint to $t_{\ast }$ and then decreases to zero. Since the first branch increases to that breakpoint, $t_{\ast }$ gives the global maximum. This proves the displayed value without sampling either optimisation variable. As an existence criterion this is weaker than the $13/25$ theorem. Its different content is the control of the two straight segments from the specified critical point, rather than a curve selected by area averaging. For $z^5-b$ it requires $|b|\le 1/M_5$, so it excludes part of a binomial family already covered by the trinomial theorem. No claim is made that every quintic has a critical value below this level.

A near-regular pentagon illustrates why a prescribed joining point is restrictive. For sufficiently small $ϵ>0$, set

All five roots of $F_{ϵ}$ lie in the open unit disc, approach a regular pentagon, and have escaping segments from the origin. Here is a proof of that assertion, including the effect of the radial contraction.

The roots of $P_{ϵ}$ are simple perturbations ${\zeta }_{ϵ}={\zeta }_0+O({ϵ}^3)$ of the fifth roots of unity. The identity

makes the roots invariant under conjugate inversion. Uniqueness of the root near each ${\zeta }_0$ therefore forces $|{\zeta }_{ϵ}|=1$. The roots $r{\zeta }_{ϵ}$ of $F_{ϵ}$ consequently have modulus $r<1$. The displacement estimate follows from the implicit function theorem at each simple root of $z^5-1$, since the first coefficient perturbations have order ${ϵ}^3$.

For ${\zeta }_0=1,e^{4\pi i/5},e^{-4\pi i/5}$, take $t=ϵ/4$. Substitution into the polynomial gives

The coefficient is positive, since $\mathrm{\Re }{\zeta }_0^2\ge (\sqrt{5}-1)/4$. For the remaining roots, with ${\zeta }_0=e^{2\pi i/5}$ or $e^{-2\pi i/5}$, take $t={ϵ}^2/4$ instead. Then

for sufficiently small $ϵ$. The two scales are needed because $\mathrm{\Re }{\zeta }_0^2$ is negative for this second pair; its positive linear contribution dominates at the smaller scale. Multiplication by $r^{10}=1+O({ϵ}^9)$ cannot remove either positive leading term. The remainders are uniform over the five root branches, so one sufficiently small $ϵ$ works for every segment. This is an ordinary asymptotic proof; it gives no numerical upper bound for the permitted $ϵ$.

The addendum [26] reports the stronger assertion that no critical point of this family supplies two contained straight arms of total length at most $2$. Its proof is not available. The origin calculation above does not prove it: $P_{ϵ}^{\prime }(0)=-{ϵ}^5$ shows that the origin is not a critical point. The stronger reported assertion is not used in any theorem here. Neither the proved failure at the origin nor the reported critical-point obstruction excludes curved inverse images of rays or a different joining point.

A different local question allows a freely chosen, noncritical vertex: for some $\delta >0$, does every monic quintic with open-disc roots within $\delta$ of a rotated regular pentagon admit two contained straight arms through some $h\in \mathbb{C}$ of total length below $2$? Neither the origin nor a critical point is prescribed in this question. The sufficient low-critical threshold preceding this paragraph is unchanged.