\documentclass[11pt,reqno]{amsart} \usepackage[T1]{fontenc} \usepackage{lmodern,indentfirst} \usepackage[colorlinks=true,linkcolor=blue,citecolor=blue,urlcolor=blue]{hyperref} \addtolength{\textheight}{\baselineskip} \newtheorem{theorem}{Theorem}[section] \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \DeclareMathOperator{\conv}{conv} \DeclareMathOperator{\dist}{dist} \renewcommand{\Re}{\operatorname{Re}} \renewcommand{\Im}{\operatorname{Im}} \title{Schmeisser's conjecture in large degree} \author{Qiyue Tang} \date{November 7, 2026} \keywords{Schmeisser's conjecture, critical points, convex hulls} \subjclass[2020]{Primary 30C15; Secondary 30C10} \begin{document} \raggedbottom \begin{abstract} We prove Schmeisser's conjecture for all sufficiently large degrees. \end{abstract} \maketitle \section{Introduction} Let \(P\) be a complex polynomial of degree \(n\ge2\), with zeros \(z_1,\ldots,z_n\) counted with multiplicity. Its critical points are the zeros of \(P'\). The Gauss--Lucas theorem places them in the convex hull of \(z_1,\ldots,z_n\), the set of all sums \(\sum_{j=1}^n t_jz_j\) with real \(t_j\ge0\) and \(\sum_{j=1}^n t_j=1\). Sendov's conjecture, proposed in 1958 \cite{schmeisser2004} and recorded under Ilieff's name as Problem 4.5 in Hayman's book \cite{hayman1967,schmeisser1971}, asks whether each zero is within distance one of a critical point whenever all zeros lie in the closed unit disk \(\overline{\mathbb D}=\{z\in\mathbb C:|z|\le1\}\). The constant one is attained by \(P(z)=z^n-1\), whose zeros have modulus one and whose only critical point is zero. Early work established the estimate in small degree and for restricted zero configurations. Schmeisser treated hulls with every vertex on the unit circle \cite{schmeisser1969}, then studied polynomials with gaps in their powers \cite{schmeisser1971}. His 1977 paper treated triangular hulls and asked whether, for zeros in the closed unit disk, the distance bound holds at every point of their hull. He recorded this convex-hull bound for degrees at most five, for hulls with all vertices on the circle, and for triangular hulls \cite[\S3]{schmeisser1977}. The general question is Problem 4.30 in the anniversary edition of Hayman's book \cite{hayman}. Paw{\l}owski obtained a degree-dependent bound for the distance from the centroid \(n^{-1}\sum_{j=1}^n z_j\) to the nearest critical point \cite{pawlowski}. Zhang sharpened it to \(\sqrt{(n-2)/(n-1)}\) by a second-moment estimate \cite{zhang}. For Sendov's conjecture, D\'egot proved the unit-distance bound in sufficiently large degree for a zero at a prescribed modulus in \((0,1)\) \cite{degot}. Tao obtained a single degree threshold valid for all zeros \cite{tao}. Mazur later gave an all-degree proof, with a further exposition by Tao \cite{mazur,tao2026}. In that exposition the convex-hull problem is Conjecture 18. \begin{theorem}\label{thm:main} There exists an integer \(N\ge 2\) such that the following holds for every complex polynomial \(P\) of degree \(n\ge N\). If all zeros of \(P\) lie in the closed unit disk, then for every point \(a\) in their convex hull there is a zero \(\xi\) of \(P'\) with \(|\xi-a|\le 1\). \end{theorem} A point violating the distance bound can be moved to an edge of the convex hull. Near the unit circle, the two endpoint zeros give equal-value relations for the polynomial. At intermediate moduli, a reciprocal-polynomial comparison yields a degree bound. Near the origin, points where the zeros meet their smallest enclosing circle give a lower bound on the scale at a rightmost zero. Estimates for the logarithmic derivative of \(P'\) then rule out both bounded and unbounded values at that scale. \section{Normalization and polynomial fibers}\label{sec:normalization} Write \(\conv E\) for the convex hull of a set \(E\subset\mathbb C\). List the critical points of \(P\) as \(\xi_1,\ldots,\xi_{n-1}\), with multiplicity. A strict counterexample consists of \(P\) with all zeros in \(\overline{\mathbb D}\) and \(a\in\conv\{z_1,\ldots,z_n\}\) such that every \(|\xi_j-a|>1\). In a strict counterexample the smallest disk containing the zeros has positive radius \(R\le1\), since a polynomial with just one distinct zero has a critical point there. If the disk has center \(c\), replace \(P(z)\) by a monic multiple of \(P(c+Rz)\) and the witness by \((a-c)/R\). Critical distances are multiplied by \(1/R\ge1\), and affine maps preserve convex combinations. By Gauss--Lucas the new witness is nonzero. Rotate the configuration so that this witness, still denoted by \(a\), is positive real. The inequalities \(|\xi_j|\le1\) and \(|\xi_j-a|>1\) imply \(\Re\xi_j0\). Replacing \(a\) by the last point of the root hull on the ray \([a,\infty)\) preserves the strict inequalities. For sufficiently large \(n\), Tao's theorem \cite[Theorem 1.2]{tao} gives a critical point within distance one of every zero, so \(a\) cannot be a vertex. It lies in the relative interior of an edge with distinct zeros as endpoints. If the hull is a segment, use the segment itself. In either case \(a<1\), since a convex combination of points in the disk can have real part one only if all points used equal one. If Theorem~\ref{thm:main} fails, these reductions give a sequence of normalized strict counterexamples with \(n\to\infty\). We use \(P,a,z_j,\xi_j\) for the data at degree \(n\), suppressing the sequence index. Unless specified otherwise, asymptotic estimates refer to \(n\to\infty\). Passing to a subsequence, we may assume that \(a\) converges in \([0,1]\). \begin{lemma}[{\cite[Theorems 2 and 3]{degot}}]\label{lem:fibers} Let \(P\) be a complex polynomial of degree \(n\ge2\) and let \(P'(b)\ne0\). For each \(w\in\mathbb C\), there is a point \(\beta\) with \(P(\beta)=w\) and \[ |\beta-b|\le n\,\frac{|w-P(b)|}{|P'(b)|}. \] If \(P(\beta)=P(\gamma)\) and \(\beta\ne\gamma\), the perpendicular bisector of \([\beta,\gamma]\) intersects the convex hull of the critical points. \end{lemma} \begin{proof} The case \(w=P(b)\) follows by taking \(\beta=b\). Otherwise let \(\beta_1,\ldots,\beta_n\) be the zeros of \(P-w\) with multiplicity. Then \(P'(b)/(P(b)-w)=\sum_j(b-\beta_j)^{-1}\). At least one summand has modulus at least \(n^{-1}|P'(b)/(P(b)-w)|\), which gives the radius bound. For the bisector assertion, take the two distinct points to \(-1\) and \(1\) by a complex affine map. If their bisector misses the convex hull of the critical points, we can interchange the points so that the critical points all have positive real part. The equality \(P(-1)=P(1)\) gives \[ \int_{-1}^{1}\prod_{j=1}^{n-1}(t-\xi_j)\,dt=0 . \] As a polynomial in the \(\xi_j\), this integral is symmetric and affine in each variable separately. Choose \(\varepsilon>0\) with \(\Re\xi_j\ge\varepsilon\) for all \(j\). For sufficiently large real \(R\) the disk with center \(R\) and radius \(R-\varepsilon/2\) contains all \(\xi_j\), since \(|\xi_j-R|^2\le R^2-2R\varepsilon+|\xi_j|^2\). This disk lies in \(\Re z\ge\varepsilon/2\). The Grace--Walsh--Szeg\H{o} coincidence theorem \cite[Theorem 1]{degot} gives a point \(\xi\) in the disk with \(\int_{-1}^1(t-\xi)^{n-1}\,dt=0\). Integration gives \((1-\xi)^n=(-1-\xi)^n\), hence \(|1-\xi|=|-1-\xi|\) and \(\Re\xi=0\), a contradiction. \end{proof} We use the root and critical point probability measures \[ \mu_n=\frac1n\sum_{j=1}^n\delta_{z_j}, \qquad \nu_n=\frac1{n-1}\sum_{j=1}^{n-1}\delta_{\xi_j}, \] where \(\delta_z\) is unit mass at \(z\). For a compactly supported measure \(\sigma\), its logarithmic potential off its support is \(U_\sigma(w)=\int\log|w-z|\,d\sigma(z)\). Weak convergence means convergence of integrals against continuous functions on the common compact disk. \begin{lemma}\label{lem:exterior} If \(\mu_n\) and \(\nu_n\) converge weakly to \(\mu\) and \(\nu\), respectively, then \(U_\mu=U_\nu\) on \(|w|>1\). The means of \(\mu_n\) and \(\nu_n\) agree for each \(n\). \end{lemma} \begin{proof} The coefficient of \(z^{n-2}\) in \(P'/n\) is both \(-(n-1)\sum_j z_j/n\) and \(-\sum_j \xi_j\). Thus the means agree. For \(|w|>1\), \[ \frac{wP'(w)}{nP(w)} =\int\frac{1}{1-z/w}\,d\mu_n(z). \] The integrand has real part at least \(1/2\) and modulus at most \(1/(1-1/|w|)\). The product formulas give \[ (n-1)U_{\nu_n}(w)-nU_{\mu_n}(w) =\log\left|\frac{wP'(w)}{nP(w)}\right|-\log|w|. \] The right side is bounded for fixed \(w\). Divide by \(n\) and take the limit. \end{proof} \section{Witnesses approaching the unit circle}\label{sec:boundary} Let \(\sigma\) be normalized arc length on the unit circle. \begin{lemma}\label{lem:robust} Let \(P=P_n\) be monic of degree \(n\) along a sequence with \(n\to\infty\), with all roots in the closed unit disk. Suppose \(b_n\in[0,1]\) is a root, \(b_n\to b>0\), and \(|\xi_j-b_n|\ge r_n\) for all \(j\), where \(r_n\to1\). Then \(b=1\), \(\mu_n\to\sigma\), and \(\nu_n\to\delta_0\) weakly. Also \(|P(0)|^{1/n}\to1\). \end{lemma} \begin{proof} For large \(n\), \(r_n>0\) and \(b_n\) is simple. Fix \(d\in(0,b)\). By Gauss--Lucas, \(|\xi_j|\le1\), so the exclusion implies \(2\operatorname{Re}\xi_j\le(1+b_n^2-r_n^2)/b_n=b+o(1)\). In particular, \[ |\xi_j-b_n|^2-|\xi_j-d|^2 =(b_n-d)(b_n+d-2\operatorname{Re}\xi_j) \ge (b-d)d+o(1) \] uniformly in \(j\). There is a fixed \(\eta>0\), chosen also with \(\eta<(b-d)/2\), such that this difference stays positive upon replacing \(b_n\) by any \(\gamma\) with \(|\gamma-b_n|<\eta\). The fiber estimate in Lemma~\ref{lem:fibers} gives a point \(\gamma\) with \(P(\gamma)=P(d)\) and \(|\gamma-b_n|\le n|P(d)|/|P'(b_n)|\). If the bound were smaller than \(\eta\), the critical points would all lie strictly on the \(d\)-side of the perpendicular bisector, contrary to the second part of Lemma~\ref{lem:fibers}. Therefore \[ |P(d)|\ge \frac{\eta}{n}|P'(b_n)|\ge\eta r_n^{n-1}. \] On the other hand, \[ |P(d)|^{2/n}\le\frac1n\sum_{j=1}^n|z_j-d|^2 \le 1+d^2-2d\operatorname{Re}\int z\,d\mu_n(z). \] First letting \(n\to\infty\), then \(d\downarrow0\), bounds the limit superior of the real part of the mean by zero. By Lemma~\ref{lem:exterior}, the critical-point mean therefore has real part at most \(b/8\) for large \(n\). If the proportion of critical points with real part at most \(b/4\) is \(p_n\), that mean has real part at least \((1-p_n)b/4-p_n\). Hence \(p_n\ge b/(8+2b)>0\). Uniformly for \(0\le d\le b/2\), the squared-distance comparison gives \(|d-\xi_j|/|b_n-\xi_j|\le1+o(1)\) for every \(j\) and \(|d-\xi_j|/|b_n-\xi_j|\le1-\kappa\) on the positive proportion just identified, for some fixed \(\kappa>0\). The latter assertion uses \(b_n-d\ge b/3\) and \(b_n+d-2\operatorname{Re}\xi_j\ge b/3\) for those \(j\) and all large \(n\). Moving \(d=b/4\) to a point \(z\) within distance \(\varepsilon r_n\) increases each ratio by at most \(\varepsilon\). Choose \(\varepsilon<\kappa/2\) small enough that \((1-\kappa/2)^{b/(8+2b)}(1+2\varepsilon)<1\). Since \(r_n\to1\), this yields a fixed disk about \(b/4\) on which the product of the ratios decays exponentially, as it does on the segment. Thus \[ |P'(z)|\le e^{-c n}|P'(b_n)| \] on the segment and the disk, for some \(c>0\). Integrate from \(d=b/4\) to zero and use the lower bound on \(|P(d)|\) to obtain \[ 1\ge |P(0)|\ge \frac{\eta}{2n}|P'(b_n)| \ge \frac{\eta}{2}r_n^{n-1}. \] In particular, \(|P'(b_n)|=O(n)\), and integration also gives \(P(z)/P(0)=1+O(ne^{-cn})\) uniformly on the small disk about \(b/4\). Any subsequential weak limit of \(\mu_n\) is supported on the unit circle. In fact, for \(00\), dividing this sum by \((n-1)\log((1+\varepsilon)/r_n)\) bounds the fraction of critical points with \(|b_n-\xi_j|\ge1+\varepsilon\) by a quantity tending to zero. For any weak limit \(\nu\), its Cauchy transform \(\int(z-w)^{-1}\,d\nu(w)\) agrees with \(1/z\) outside the closed unit disk by differentiation of the potentials in Lemma~\ref{lem:exterior}. The complement of the proper circular arc \(A_b\) remains connected after removing the origin. Continuation would give a pole at the origin if \(b<1\), which is impossible because the origin would lie off the support. Thus \(b=1\). By continuation the logarithmic potential agrees with \(\log|z|\) off \(A_1\), since their real gradients agree there and both have the same behavior at infinity. The arc has planar measure zero, and the distributional Laplacian of \(\log|z-w|\) is \(2\pi\delta_w\). Consequently the limit is \(\delta_0\). \end{proof} \begin{proposition}\label{prop:boundary} There is no strict counterexample sequence of degrees \(n\to\infty\) whose supporting-edge witnesses \(a\in(0,1)\) tend to \(1\). \end{proposition} We use the primitive estimate and Fourier comparison from Tao's boundary analysis \cite[\S5]{tao}. The fiber relations at the two endpoints of the edge give the needed bound at \(a\) by convexity. \begin{proof} Select a root \(b_n\) with \(\Re b_n\ge a\), which is possible because \(a\) lies in the convex hull. Then \(|b_n-a|^2\le1-a^2\). Rotate by \(-\arg b_n\) to apply Lemma~\ref{lem:robust} with \(r_n=1-|b_n-a|\). The rotation tends to the identity, so in the original coordinates \(\mu_n\to\sigma\), \(\nu_n\to\delta_0\), and \(|P(0)|^{1/n}\to1\). The convergence \(\mu_n\to\sigma\) implies that every point of the unit circle is at distance \(o(1)\) from some root, uniformly on the circle. This follows by a finite cover by disks of arbitrarily small radius, each with positive limiting mass. Hence the convex hull contains a centered disk of radius \(1-o(1)\). Indeed, in every unit direction \(e\), a root at distance \(o(1)\) from \(e\) gives \(\max_j\Re(z_j\bar e)\ge1-o(1)\) uniformly in \(e\). The intersection of the corresponding supporting half-planes contains that disk. For the outward unit normal \(e\) of the edge through \(a\), this maximum is \(\Re(a\bar e)\). Hence \(e\to1\), and both endpoints \(u,v\) of the edge have real parts tending to one and moduli at most one. Thus \(u,v\to1\), and the unit edge direction \(d=(v-u)/|v-u|\) satisfies \(d^2\to-1\). Write \(\mathbb E\) for averaging with respect to \(\nu_n\), and set \begin{gather*} m=\mathbb E\xi,\qquad V=\mathbb E|\xi-m|^2,\qquad Q=\mathbb E|\xi|^2,\\ L(z)=\mathbb E\log|z-\xi|,\qquad c_n=\frac{\log|P(0)|}{n}. \end{gather*} Here \(\xi\) is the integration variable, \(Q=V+|m|^2\to0\), and \(c_n\le0\). The potential \(L\) is finite off the critical points, and its logarithmic singularities on the unit circle are integrable. Constants in the estimates can depend on the fixed contours or compact sets used in the proof. We first derive a primitive estimate on suitable paths. Suppose \(z\) is joined to a point of modulus \(1/4\) by a piecewise smooth path of uniformly bounded length whose distance from every critical point is bounded below by a positive constant. In the outward orientation and with arc-length parametrization \(w(s)\), assume \(d\log|w(s)|/ds\ge c>0\) on each smooth piece. The paths used below satisfy \(1/4\le |w|\le1\). The identity \[ \frac1{w-\xi}=\frac1{w-m}+\frac{\xi-m}{(w-m)^2} +\frac{(\xi-m)^2}{(w-m)^2(w-\xi)} \] gives \(\mathbb E(1/(w-\xi))=1/(w-m)+O(V)\). For \(B(w)=P'(w)(w-m)/n\), the product formula for \(P'\) gives \[ B'(w)=P'(w)\left(\frac1n+\frac{n-1}{n}(w-m)\mathbb E\frac1{w-\xi}\right) =P'(w)(1+O(V)). \] Also, \[ \frac{d}{ds}\log|P'(w(s))| =(n-1)\Re\left(w'(s)\mathbb E\frac1{w(s)-\xi}\right). \] Here \(\mathbb E(1/(w-\xi))=1/w+o(1)\) uniformly on the path and \(|w'(s)|=1\). The derivative is therefore at least \(c(n-1)/2\) for large \(n\). If \(z\) is the endpoint at arc length \(s_0\), this gives \(|P'(w(s))|\le |P'(z)|e^{-c(n-1)(s_0-s)/2}\). Its integral is at most \(2|P'(z)|/(c(n-1))\). On \(|w|\le1/4\), the weak convergence \(\nu_n\to\delta_0\) and the product formula give \(\sup |P'(w)|=O(3^{-n})\): all but \(o(n)\) factors have modulus at most \(1/4+1/32\), and the remaining ones are at most \(5/4\). In \(P(z)-P(0)-B(z)\), integrate \(P'-B'\) along \(w(s)\) and add \(P(w(0))-P(0)-B(w(0))\). The integral is \(O(V|P'(z)|/n)\), and the added term is \(O(3^{-n})\) by integration along the segment from \(0\) to \(w(0)\). Since \(|z-m|\ge1/8\) for large \(n\), this gives \begin{equation}\label{eq:boundary-primitive} P(z)=P(0)+(1+O(V))\frac{P'(z)(z-m)}n+O(3^{-n}). \end{equation} For a root \(\lambda\) to which this estimate applies, use \(|P(0)|\ge (2/3)^n\) for large \(n\) to obtain \[ \frac{|P'(\lambda)|}{n}|\lambda-m| =|P(0)|(1+O(V)+O(2^{-n})). \] Taylor expansion of the logarithmic kernel about \(m\) gives \(L(\lambda)=\log|\lambda-m|+O(V)\). The remainder bound is uniform even when some \(\xi\) are far from \(m\). Expand for \(|\xi-m|\) less than half the fixed lower bound for \(|\lambda-m|\), and on the complementary region use boundedness of the kernel and the lower bound on \(|\xi-m|\) to dominate the remainder by \(C|\xi-m|^2\). Taking logarithms of the last displayed estimate and using \(\log(|P'(\lambda)|/n)=(n-1)L(\lambda)\) gives \begin{equation}\label{eq:boundary-fiber} L(\lambda)=c_n+O(V/n)+O(2^{-n}). \end{equation} For \(z\) near \(1\), with \(|z|\le1\), a radial segment to \(z/(4|z|)\) satisfies the path conditions. In fact these segments lie in a fixed compact subset of \(\{|w-1|<1\}\), separated from the critical points by \(|\xi_j-a|>1\) and \(a\to1\). We can thus apply \eqref{eq:boundary-fiber} at \(u\) and \(v\). For real \(x\) with \(a+xd\in[u,v]\), \[ \frac{d^2}{dx^2}L(a+xd) =-\operatorname{Re}\mathbb E\frac{d^2}{(a+xd-\xi)^2}=1+o(1), \] where the convergence is uniform by the separation near \(1\), \(\nu_n\to\delta_0\), and \(d^2\to-1\). By convexity \(L(a)\) is at most the convex combination of \(L(u)\) and \(L(v)\) with the weights that express \(a\) in terms of \(u,v\). Since \(L(a)>0\) and \(c_n\le0\), \begin{equation}\label{eq:boundary-interpolation} |c_n|+L(a)\le C V/n+O(2^{-n}). \end{equation} In particular \(|P(0)|\to1\). Also the nonnegative sum \(\sum_j\log|a-\xi_j|=(n-1)L(a)\) tends to zero, so the distance of every critical point from \(A_1\) tends to zero uniformly. We next apply \eqref{eq:boundary-fiber} to roots near any compact subarc of the unit circle disjoint from \(A_1\). From a nearby point with radius \(r_0\) and argument \(\theta_0\), vary the argument at unit speed toward \(0\) if \(\cos\theta_0>1/2\), or toward the nearer of \(\pi,-\pi\) if \(\cos\theta_0<1/2\). After an angular change of size \(s\), use radius \(r=r_0-\kappa s\), with fixed small \(\kappa>0\). Then follow the corresponding real axis to radius \(1/4\). At radius one the angular sweep stays uniformly away from \(A_1\), whose intersections with the unit circle have arguments \(\pm\pi/3\). The separation persists for small \(\kappa\) and \(r_0\) close to one. The positive axis segment stays at least \(1/4\) from the circle \(|w-1|=1\). The negative one stays at least \(1/4\) from \(A_1\subset\{\Re w\ge0\}\). The outward derivative of \(\log r\) per unit arc length on the angular part is \(\kappa/(r\sqrt{\kappa^2+r^2})\), and on the axis it is \(1/r\). These bounds are uniform on a neighborhood of the chosen subarc, and all critical points are uniformly close to \(A_1\). From \eqref{eq:boundary-interpolation} and \eqref{eq:boundary-fiber}, all these roots satisfy \begin{equation}\label{eq:boundary-root} L(\lambda)=O(V/n)+O(2^{-n}). \end{equation} For \(z\) uniformly separated from the critical points and bounded away from zero, expansion about \(\xi=0\) gives \[ L(z)=\log|z|-\operatorname{Re}(m/z)+O(Q). \] As in the earlier kernel expansion, the remainder is bounded by \(C|\xi|^2\) by using Taylor expansion for small \(\xi\) and bounded kernels for larger \(\xi\). Choose roots approaching \(1,i,-1,-i\). They satisfy \eqref{eq:boundary-root}. Since \(\log|\lambda|\le0\), the four bounds on \(\Re(m/\lambda)\) bound both signs of \(\Re m\) and \(\Im m\) by \(O(Q)+o(|m|)+O(2^{-n})\). Using \(Q=V+|m|^2\) and \(m\to0\), we obtain \begin{equation}\label{eq:boundary-mean} |m|=O(V)+O(2^{-n}),\qquad 1-a=O(V)+O(2^{-n}). \end{equation} The second bound follows by averaging \(|a-\xi_j|^2>1\), which gives \(1-a^21/2\), and is at most \(2\). Cauchy--Schwarz now gives the estimate. The Fourier coefficients of \(\log|e^{i\theta}-\xi|\) follow from the series of \(\log(1-\xi e^{-i\theta})\), first for \(|\xi|<1\), then for \(|\xi|=1\) by dominated convergence in the rotated coordinate. The constant term is zero, and only \(-\Re(\xi^2e^{-2i\theta})/2\) contributes to the cosine integral at frequency two. Averaging over the critical points gives \[ \int L(e^{i\theta})\,d\sigma(\theta)=0,\qquad \operatorname{Re}\mathbb E\xi^2 =-4\int L(e^{i\theta})\cos(2\theta)\,d\sigma(\theta). \] Omit fixed neighborhoods of the two angles \(\pm\pi/3\) with total measure \(\delta\). On the omitted set the integral of \(|L|\) is at most \(C\sqrt\delta Q+\delta|m|\). On the remaining arcs use \eqref{eq:boundary-circle} for the negative part. Since the integral of \(|L|\) is twice that of its negative part, we get \begin{equation}\label{eq:boundary-fourier} |\operatorname{Re}\mathbb E\xi^2| \le (C\sqrt\delta+o(1))V+O(2^{-n}) \end{equation} for every fixed sufficiently small \(\delta>0\). The \(o(1)\) and exponential constant here may depend on \(\delta\), while \(C\) is absolute. For each critical point set \(\ell=\log|a-\xi|\ge0\). The identity \[ \operatorname{Re}\xi =\frac{|\xi|^2-(1-a^2)-(e^{2\ell}-1)}{2a} \] expresses its numerator as a difference of nonnegative quantities. As \((x-y)^2\le x^2+y^2\) for \(x,y\ge0\), and \(0\le\ell\le\log 2\), we obtain, for large \(n\), \[ \operatorname{Re}(\xi^2)=2(\operatorname{Re}\xi)^2-|\xi|^2 \le-\tfrac14|\xi|^2+C((1-a)^2+\ell^2). \] Here the coefficient of \(|\xi|^2\) before weakening is at most \(1/(2a^2)-1\), using \(|\xi|\le1\). By \eqref{eq:boundary-interpolation}, \(\mathbb E\ell^2\le(\log 2)L(a)=o(V)+O(2^{-n})\), and \((1-a)^2=o(V)+O(2^{-n})\) by \eqref{eq:boundary-mean}. Averaging and applying \eqref{eq:boundary-fourier} gives \(Q/4\le (C\sqrt\delta+o(1))V+O(2^{-n})\). Choose \(\delta\) so that \(C\sqrt\delta<1/8\), then use \(V\le Q\) to conclude \(Q=O(2^{-n})\). Set \(\tau=\max_j|\xi_j|\le\sqrt{(n-1)Q}\). The product formula for \(P'\) and a telescoping difference of products imply \[ \sup_{|z|\le1+n^{-3}}|P'(z)-nz^{n-1}| \le n(n-1)\tau(1+n^{-3}+\tau)^{n-2} =O(n^2\tau). \] Integration gives the same bound for \(P(z)-(z^n+P(0))\) there. Put \(R=|P(0)|^{1/n}\to1\). At each root \(\omega\) of \(z^n+P(0)\), on the circle \(|z-\omega|=n^{-3}\), expansion of \((1+(z-\omega)/\omega)^n-1\) gives a modulus at least \(n^{-2}/(2R)\) for large \(n\), since the terms of order at least two have total modulus at most \(\sum_{k\ge2}(n^{-2}/R)^k/k!=O(n^{-4})\). Multiplication by \(|\omega|^n=|P(0)|\to1\) shows that \(z^n+P(0)\) dominates the error. By Rouch\'e's theorem each disk contains one root of \(P\), and any two distinct roots of \(P\) are separated by at least \(2R\sin(\pi/n)-2n^{-3}\ge c/n\) for a fixed \(c>0\). By Tao's theorem \cite[Theorem 1.2]{tao}, for large \(n\) each endpoint \(u,v\) is within distance \(1\) of critical points \(c_u,c_v\), respectively. Set \(D=|c_u-c_v|\le2\tau=o(n^{-2})\). At the midpoint \(w=(u+v)/2\), \[ |w-c_u|^2 =\frac{|u-c_u|^2+|v-c_u|^2}{2}-\frac{|u-v|^2}{4} \le1+D+D^2/2-\frac{|u-v|^2}{4}\le1 \] for large \(n\). Interchanging \(u,v\) gives the same for \(c_v\). The two closed unit disks centered at \(c_u,c_v\) cover the two halves of the edge by convexity, contradicting the strict witness. \end{proof} \section{Witnesses with an interior radial limit} \begin{proposition}\label{prop:interior} A normalized strict counterexample sequence cannot have \(a\to a_*\in(0,1)\). \end{proposition} \begin{proof} Choose a zero \(b\) of maximal real part. Then \(\Re b\ge a\) and \(|b-a|^2\le1-a^2<1\), so \(P'(b)\ne0\). Pass to a subsequence with \(b\to b_*\). If \(b_*=a_*\), then \(|\xi_j-b|\ge1-|b-a|\) for every \(j\). Rotate each polynomial to place \(b\) on the positive real axis. Lemma~\ref{lem:robust} then forces \(a_*=1\). Suppose \(b_*\ne a_*\). Choose a fixed \(t\in(0,1)\) so close to one that \(t-1+|b_*-ta_*|^2>0\). Put \(\gamma=b-ta\) and \(C=ta(2\Re b-(1+t)a)\). Then \(C\ge t(1-t)a^2\), \(|\gamma|^2\le1-t(2-t)a^2<1\), and \(t\ge1-|\gamma|^2\) eventually. Direct expansion gives \begin{align*} |b-\xi_j|^2-|1-\overline\gamma\xi_j|^2 ={}&t(|\xi_j-a|^2-1)+C\\ &+(t-1+|\gamma|^2)(1-|\xi_j|^2) \ge C . \end{align*} Since \(|1-\overline\gamma\xi_j|\le2\), this gives \(|b-\xi_j|\ge\sqrt{1+C/4}\,|1-\overline\gamma\xi_j|\). The identity \[ |1-\overline\gamma\xi_j|^2-|\gamma-\xi_j|^2 =(1-|\gamma|^2)(1-|\xi_j|^2) \] also gives \(|b-\xi_j|^2-|\gamma-\xi_j|^2\ge C\). Set \(\eta=\min(1,C/10)\). For \(|\beta-b|\le\eta\), replacing \(b\) by \(\beta\) reduces the difference by at most \(2|\xi_j-b|\eta+\eta^2\le4\eta+\eta^2\le C/2\). Hence \(\beta\ne\gamma\) and all critical points are strictly closer to \(\gamma\) than to \(\beta\). The bisector assertion in Lemma~\ref{lem:fibers} then excludes \(P(\beta)=P(\gamma)\) in this disk. The radius bound in the same lemma gives \[ |P(\gamma)|\ge\frac{\eta}{n}|P'(b)|. \] For \(F(z)=\sum_{k=0}^d f_kz^k\) of degree \(d\), define the reciprocal polynomial \(F^*(z)=\sum_{k=0}^d\overline{f_{d-k}}z^k\). For \(q=P'/n\), the coefficient of \(z^k\) in \(q^*\) is \(1-k/n\) times the coefficient of \(z^k\) in \(P^*\), for \(0\le k\le n-1\). The same formula gives zero for \(k=n\), so \(q^*=P^*-z(P^*)'/n\). Taking the logarithmic derivative of the product for \(P^*\) gives \[ P^*(\gamma)=\prod_{j=1}^n(1-\gamma\overline{z_j})\ne0, \qquad \frac{q^*(\gamma)}{P^*(\gamma)} =\frac1n\sum_{j=1}^n\frac{1}{1-\gamma\overline{z_j}} . \] Each summand has real part at least \(1/2\). Also \(|P(\gamma)|\le|P^*(\gamma)|\), by the squared-distance identity above applied to each \(z_j\). Multiplying the bounds on \(|b-\xi_j|\) yields \[ |P'(b)|\ge n(1+C/4)^{(n-1)/2}|q^*(\gamma)| \ge\frac n2(1+C/4)^{(n-1)/2}|P^*(\gamma)|. \] The inequalities for \(|P(\gamma)|\) give \(\eta(1+C/4)^{(n-1)/2}\le2\). Since \(C\ge t(1-t)a^2\) stays bounded away from zero, this contradicts \(n\to\infty\). \end{proof} \section{The logarithmic derivative at a rightmost zero} \label{sec:unbounded-field} For a polynomial $F$ of degree at least two and a simple zero $b$, the value $F''(b)/F'(b)$ is the sum of the reciprocals $1/(b-c)$ over its critical points $c$, counted with multiplicity. The next estimate bounds the corresponding sum of squared moduli when $b$ has largest real part among the zeros. \begin{lemma}\label{lem:field-bound} There is an absolute constant $C$ with the following property. Let $b$ be a simple zero of a complex polynomial $F$ of degree at least two. Suppose that every zero $u$ of $F$ satisfies $\Re u\leq\Re b$. With $c$ running over the critical points of $F$, counted with multiplicity, set \[ r=\min_c|b-c|,\qquad S=\sum_c\frac1{b-c},\qquad Q=\sum_c\frac1{|b-c|^2}. \] Then $Qr^2\leq C(1+r|S|)$. \end{lemma} \begin{proof} Replacing \(F(z)\) by \(F(b+rz)\) puts the chosen zero at zero and the nearest critical point at distance one. The two sums become \(rS\) and \(r^2Q\). If the estimate fails, there is a sequence in the normalized coordinates with \(Q/(1+|S|)\to\infty\). All limits in this proof refer to that sequence. Put \(v_c=-1/c\), \(h=|S|\), and \(B=-\sum_c v_c^2\). By Gauss--Lucas, \(\Re v_c\ge0\), and by normalization \(|v_c|\le1\). Thus \(\sum_c(\Re v_c)^2\le\sum_c\Re v_c=\Re S\le h\), and Cauchy--Schwarz gives \[ 0\leq Q-\Re B=2\sum_c(\Re v_c)^2\leq2h,\qquad |\Im B|\leq 2\sqrt{hQ}. \] In particular $B/Q\to1$. Also, for $\Re z>0$, the product over the zeros of $F$ gives \begin{equation}\label{eq:rightmost-log-derivative} \Re\left(\frac{F'(z)}{F(z)}-\frac1z\right) =\sum_{u\ne 0}\Re\frac1{z-u}\geq0, \end{equation} where the zeros \(u\) are counted with multiplicity. Both cases below use, for \(|z|\le1/2\), \[ \frac{F'(z)}{F'(0)} =\prod_c(1+v_c z) =\exp\bigl(Sz+Bz^2/2+O(Q|z|^3)\bigr). \] Indeed, the power series for \(\log(1+w)\) at zero has remainder bounded by \(2|w|^3\) after its quadratic term for \(|w|\le1/2\), and \(\sum_c|v_c|^3\le Q\). Suppose first that \(h/\sqrt Q\to\infty\) along a subsequence. Then \(e=Q/h^2\to0\). For \(z=\zeta/S\) with \(\zeta\) in a fixed bounded set, the quadratic term is \(O(e)\) and the remainder is \(O(e/h)\). Expanding the exponential to first order in the quadratic term gives an error \(O(e/h+e^2)=o(e)\). Since \(F(0)=0\), integration gives \begin{gather*} \frac{S F(\zeta/S)}{F'(0)}=e^\zeta-1+\frac{B}{S^2}A(\zeta)+o(e),\\ A(\zeta)=\int_0^\zeta e^t\frac{t^2}{2}\,dt =\frac{e^\zeta(\zeta^2-2\zeta+2)-2}{2}, \end{gather*} For \(\zeta_0=\pm2\pi i\), Rouch\'e's theorem on a small fixed circle about \(\zeta_0\) gives a zero \(\zeta\to\zeta_0\). Since \(e^\zeta-1=(\zeta-\zeta_0)+O(|\zeta-\zeta_0|^2)\), the equation at the zero first gives \(\zeta-\zeta_0=O(e)\), then \(\zeta-\zeta_0=-(B/S^2)A(\zeta_0)+o(e)\). Evaluating \(A(\zeta_0)=\zeta_0^2/2-\zeta_0\) and dividing by \(S\) gives zeros of \(F\) at \[ t_\pm=\frac{\pm2\pi i}{S} +\frac{B}{S^3}(2\pi^2\pm2\pi i)+o(e/h). \] The real parts of the leading terms are \(\pm2\pi\Im S/h^2\). Since \(\Re t_\pm\le0\) and the corrections are \(O(e/h)\), we have \(|\Im S|/h=O(e)\), and \(\Re S\ge0\) gives \(S/h\to1\). Adding the expansions and using \(B/Q\to1\) gives \[ \Re(t_++t_-)=\bigl(4\pi^2+o(1)\bigr)Q/h^3>0, \] a contradiction. Otherwise \(h/\sqrt Q\) is bounded, and we pass to a subsequence with \(S/\sqrt Q\to\eta\in\mathbb C\). At \(z/\sqrt Q\) the product expansion has exponent \[ \frac{S}{\sqrt Q}z+\frac{B}{2Q}z^2+O(|z|^3/\sqrt Q). \] Integrating from zero shows that \(\sqrt Q\,F(z/\sqrt Q)/F'(0)\) and its derivative converge locally uniformly to \(G(z)\) and \(G'(z)\), where \[ G(z)=\int_0^z e^{\eta t+t^2/2}\,dt . \] Applying \eqref{eq:rightmost-log-derivative} at \(z/\sqrt Q\) and dividing by \(\sqrt Q\) gives \(\Re(G'(z)/G(z))\ge\Re(1/z)\) in the limit whenever \(\Re z>0\) and \(G(z)\ne0\). Along the rays of arguments $\pm\pi/3$, the function $G$ has limits $L_+$ and $L_-$, respectively. We may rotate the integration rays to arguments $\pm\pi/2$: on the connecting arcs of radius $R$, the integrand is bounded by $e^{-R^2/4+|\eta|R}$, and the arc integrals tend to zero. It follows that \[ L_+-L_-=i\int_{-\infty}^{\infty}e^{-u^2/2+i\eta u}\,du =i\sqrt{2\pi}\,e^{-\eta^2/2}. \] For the last equality, differentiating the real-line integral in $\eta$ and integrating by parts shows that its derivative is $-\eta$ times the integral. Its value at $\eta=0$ is the Gaussian integral. Hence at least one of the two limits is nonzero. On that ray, at $z=R e^{\pm\pi i/3}$, we have $G(z)$ bounded away from zero for large $R$ and $|G'(z)|\leq e^{-R^2/4+|\eta|R}$. This contradicts $\Re(G'(z)/G(z))\geq\Re(1/z)=1/(2R)$. \end{proof} \begin{lemma}\label{lem:unbounded-field} Let $(P_j)$ be a sequence of polynomials of degree at least two with all zeros in the closed unit disk. Suppose $a_j>0$ and every critical point $c$ of $P_j$ satisfies $|c-a_j|>1$. Let $b_j=x_j+iy_j$ be a zero of $P_j$ of largest real part, and suppose $x_j\geq a_j$ and $b_j\to i$. With $\rho_j=\max(x_j,1-y_j)$, the sequence \[ \rho_j\left|\frac{P_j''(b_j)}{P_j'(b_j)}\right| \] is bounded. \end{lemma} \begin{proof} Since $|b_j-a_j|^2\leq 1-2a_jx_j+a_j^2\leq1-a_j^2$, each $b_j$ is a simple zero. Suppose that the displayed sequence tends to infinity along a subsequence. Suppress the index $j$ and set \[ \begin{gathered} F(u)=P(i+\rho u),\qquad \beta=(b-i)/\rho,\\ X=x/\rho,\qquad A=a/\rho. \end{gathered} \] Thus $01$ gives $\Re v1$ implies $|b-c|\geq\rho/2$, so $|\beta-v|\geq1/2$ in this case. Lemma~\ref{lem:field-bound} applied to $F$ at $\beta$ gives $Q\leq C_1 h$ for all sufficiently large $j$, with a fixed $C_1\geq1$. Let \(z\) be a zero with \(|z-\beta|\le1/16\) and \(|K(z)|\ge h/2\). Since \(|\beta-v|\ge1/4\), we have \(|z-v|\ge|\beta-v|/2\ge1/8\) and \(\sum_v|z-v|^{-2}\le4Q\). In particular \(z\) is simple. For \(|\zeta|\le8\), apply the series for \(\log(1+w)\) to each factor \(1+\zeta/(K(z)(z-v))\) in the quotient of derivatives. The linear terms sum to \(\zeta\), giving \[ \frac{F'(z+\zeta/K(z))}{F'(z)}=\exp\bigl(\zeta+O(h^{-1})\bigr). \] For large \(h\), each \(|\zeta/(K(z)(z-v))|\le1/2\). The error in the logarithm is then bounded by \(|\zeta/K(z)|^2\sum_v|z-v|^{-2}\). Integration from \(\zeta=0\) gives \[ \frac{K(z)F(z+\zeta/K(z))}{F'(z)}=e^\zeta-1+O(h^{-1}). \] The errors are uniform in \(z\). On either circle of radius \(M/h\) about \(\pm2\pi i\), \(|e^\zeta-1|\ge M/(2h)\) for large \(h\) by Taylor expansion. Choose the fixed \(M\) so that this lower bound exceeds the error in the last display. Rouch\'e's theorem gives zeros \(z+t_+\) and \(z+t_-\) with \begin{equation}\label{eq:two-nearby-roots} t_\pm=\frac{\pm2\pi i}{K(z)}+O(h^{-2}). \end{equation} At \(z=\beta\), the real parts in \eqref{eq:two-nearby-roots} are \(\pm2\pi\Im S/h^2+O(h^{-2})\). Both are nonpositive, so \(|\Im S|=O(1)\). Each \((\beta-v)^{-1}\) has positive real part, giving \(\Re S>0\) and \(S/h\to1\). Fix \(0<\epsilon\le\min(1/16,1/(16C_1))\). For \(|z-\beta|\le\epsilon\), subtracting the reciprocals term by term gives \(|K(z)-S|\le 2\epsilon Q\le h/8\), since \(|z-v|\ge|\beta-v|/2\). Therefore \(\Re K(z)\ge3h/4\) and \(|K(z)|\le5h/4\) for large \(j\). At any zero in the disk, the leading term of \(t_-\) in \eqref{eq:two-nearby-roots} has modulus at most \(8\pi/(3h)\) and imaginary part \(-2\pi\Re K(z)/|K(z)|^2\le-24\pi/(25h)\). Including the error gives \(|t_-|\le10/h\) and \(\Im t_-\le-2/h\). Take \(\lfloor\epsilon h/20\rfloor\) such steps from \(\beta\). At each step the total distance traveled so far is at most \(\epsilon/2\), so the next step is available until the last one. The imaginary parts strictly decrease. For large \(h\) there are at least \(\epsilon h/80\) steps in the latter half, and each has index at least \(\epsilon h/80\). Their endpoints satisfy \(|u|\le|\beta|+\epsilon/2<2\) and \(\Im u\le-\epsilon/40\). We bound the number of zeros in this region by a circle identity. In the following sums, $u$ runs over the zeros of $F$ and $v$ over the critical points, each with multiplicity. At the scaled point $2$, put \[ \begin{gathered} L=F'(2)/F(2),\qquad D=K(2)-L,\\ R_0=\sum_u\frac1{|2-u|^2},\qquad R_1=\sum_v\frac1{|2-v|^2},\\ E_0=\sum_u\frac{1-|i+\rho u|^2}{2\rho|2-u|^2},\qquad E_1=\sum_v\frac{1-|i+\rho v|^2}{2\rho|2-v|^2}. \end{gathered} \] The scaled roots satisfy $\Re u\leq1$, so $\Re L\geq R_0$ and $|D|\leq1$ by $K(2)-L=(F'/F)'(2)/L$ and $|(F'/F)'(2)|\leq R_0$. The bound $\Re v1\) and \(|z|\le1\), expand \(\log|w-z|=\log|w|-\Re\sum_{k\ge1}z^k/(kw^k)\). The integral of this series against \(\mu-\nu\) has zero real part and tends to zero at infinity, so all its coefficients vanish. Thus \(\int z^k\,d\mu(z)=\int z^k\,d\nu(z)\) for integers \(k\ge1\), and the same holds for \(k=0\) because both measures have mass one. The Poisson extension to the disk of the function \(z\mapsto\dist(z,\Gamma)\) on the circle is continuous on the closed disk and harmonic inside. It is strictly positive off \(\Gamma\) by the Poisson formula. Approximation of the boundary function by trigonometric polynomials and the maximum principle approximate this extension uniformly by harmonic polynomials. These have equal integrals for \(\mu\) and \(\nu\) by the moment identities. Since the integral against \(\nu\) is zero, \(\mu\) is supported on \(\Gamma\) as well. Trigonometric polynomial approximation on the circle now gives \(\mu=\nu\). Put \[ r_n(w)=\frac{P(w)}{P'(w)},\quad g_n(w)=\frac{P''(w)}{nP'(w)},\quad g(w)=\int\frac{d\mu(z)}{w-z}. \] On each compact subset of \(\mathbb C\setminus\Gamma\), all critical points stay a positive distance away for large \(n\). Thus \(r_n,g_n\) are holomorphic there and \(g_n\to g\) locally uniformly by the product for \(P'\) and weak convergence of \(\nu_n\). For the same reason \(n^{-1}\log|P'/n|\to U_\mu\) uniformly there. For \(P\), take \(\log0=-\infty\) and truncate \(\log|w-z|\) below at a fixed constant smaller than its minimum for \(w\) in the compact set and \(z\in\Gamma\). The resulting continuous kernels converge uniformly after integration against \(\mu_n\), by weak convergence and compactness of the family parametrized by \(w\). Their limiting integral is \(U_\mu(w)\). Thus \(n^{-1}\log|P|\le U_\mu+o(1)\) and \(|r_n|\le\exp(o(n))\) locally uniformly. The equation \(r_n'=1-ng_nr_n\) determines the limit where \(g\ne0\). Fix \(w_0\) with \(g(w_0)\ne0\) and take primitives \(\phi_n'=g_n\) with \(\phi_n(w_0)=0\). On a sufficiently small closed disk of radius \(r\) about \(w_0\), \(|g_n(w)-g(w_0)|\le |g(w_0)|/2\) for large \(n\). Integrating along segments proves injectivity, since \(\phi_n(w)-\phi_n(w')\) differs from \(g(w_0)(w-w')\) by at most half its modulus. On the boundary the same estimate with \(w'=w_0\) shows by Rouch\'e's theorem that the image contains the disk of radius \(|g(w_0)|r/2\) about zero. Choose \(\delta=|g(w_0)|r/8\). For \(w\) in a fixed small neighborhood of \(w_0\), \(s=\phi_n(w)\) satisfies \(|s|<\delta\). The chain rule gives \[ \frac{d}{ds}(r_n\circ\phi_n^{-1}) =\frac{1}{g_n\circ\phi_n^{-1}}-n(r_n\circ\phi_n^{-1}). \] Multiply by \(e^{ns}\) and integrate along the segment from \(s-\delta\) to \(s\): \[ r_n(w)=e^{-n\delta}r_n(\phi_n^{-1}(s-\delta)) +\int_0^\delta\frac{e^{-nt}}{g_n(\phi_n^{-1}(s-t))}\,dt . \] The first term is at most \(\exp(-n\delta+o(n))\) in modulus. The functions \(1/(g_n\circ\phi_n^{-1})\) have uniformly bounded derivatives on \(|s|\le2\delta\) by Cauchy's estimate. Replacing the reciprocal in the integrand by \(1/g_n(w)\) gives \((1-e^{-n\delta})/(ng_n(w))\), with error \(O(\int_0^\delta te^{-nt}\,dt)=O(n^{-2})\). Hence \(nr_n\to1/g\) locally uniformly wherever \(g\ne0\). The complement of \(\Gamma\) is connected and \(wg(w)\to1\) at infinity, so \(g\) is not identically zero there. If \(g\) had a zero of order \(k\ge1\), take a small circle surrounding that zero alone and lying off \(\Gamma\). Uniform convergence on the circle would give \(nr_n\) the winding number \(-k\) of \(1/g\) there. This is impossible because \(r_n\) is holomorphic throughout the enclosed disk for large \(n\). Therefore \(g\) never vanishes off \(\Gamma\), and the convergence to \(1/g\) excludes all zeros of \(P\) from every compact set disjoint from \(\Gamma\). The argument applies to every weakly convergent subsequence, proving the uniform assertion. \end{proof} \section{Witnesses approaching the origin}\label{sec:origin} We consider monic polynomials \(P\) of degree \(n\) whose roots have minimum enclosing disk \(\{|z|\le1\}\). Let \(a>0\) be real, in the convex hull of the roots, and assume that \(|\xi-a|>1\) for every critical point \(\xi\) of \(P\). Sums over roots and critical points include their algebraic multiplicities. By Gauss--Lucas, \begin{equation}\label{eq:origin-crescent} |\xi|\le1,\qquad |\xi|\ge1-a,\qquad \Re\xi1/100\). \end{lemma} \begin{proof} Suppose \(nx\le1/100\). We have \(a\le x\le1/200\). Call a root on the unit circle a contact root. The convex hull of these roots contains zero. Otherwise there is a unit direction \(e\) with \(\Re(r\bar e)>0\) for every contact root \(r\). For sufficiently small \(\varepsilon>0\), each such root has \(|r-\varepsilon e|^2=1-2\varepsilon\Re(r\bar e)+\varepsilon^2<1\), and the finitely many other roots stay strictly inside the disk after the center moves to \(\varepsilon e\). This contradicts minimality. Thus the angular gap between successive contact roots is at most \(\pi\). The gap containing \(1\) has endpoints \(r_\pm\) with arguments \(\theta_-<0<\theta_+\) satisfying \[ \arccos x\le\theta_+\le\pi-\arccos x,\qquad -\pi+\arccos x\le\theta_-\le-\arccos x . \] The inner bounds follow from \(\Re r_\pm\le x\), and the outer bounds follow from the gap length. Put \(t=2x\), \(s=\sqrt{1-t^2}\), \(\gamma=\arccos t\), \(p=t-is\), and \(q=t+is\). The angular distances from \(r_-\) to \(p\) and from \(r_+\) to \(q\) are at most \(\arcsin x+\arcsin(2x)\le4x\), since \((\arcsin y)'\le4/3\) for \(0\le y\le1/2\). The Euclidean distances have the same bound. For \(|z|<1\) let \(\omega(z)\) be the harmonic measure of the open right-hand arc from \(p\) to \(q\), given by \[ \omega(z)=\frac1{2\pi}\int_{-\gamma}^{\gamma} \frac{1-|z|^2}{|e^{i\theta}-z|^2}\,d\theta . \] Set \(\omega=0\) on the complementary arc away from \(p,q\). The imaginary part of \((q-z)/(p-z)\) is \(2s(t-\Re z)/|p-z|^2\). When \(\Re z=t\) in the open disk, the ratio is negative real, since \(-s<\Im z0\), and \(\Re(q/(q-r))\ge1/2\). If \(\Omega_q\ge1/2\), this gives \(\arg H(q)+\pi\Omega_q\ge\pi/2-1/2\). For \(|q-r|\le1/2\), the angle \(\arg(p-r)+\pi/2\) lies between zero and \(\arcsin(1/(4s))<1/3\), because \(p-r\) lies within \(1/2\) of \(-2is\) and has positive real part. Hence \[ \arg\frac{q}{q-r} =\frac{\pi}{2}-\pi\omega(r)-\bigl(\arg(p-r)+\pi/2\bigr). \] If \(\Omega_q<1/2\), all these near terms lie in the sector with arguments between \(\pi/2-\pi\Omega_q-1/3\) and \(\pi/2\). Its width is less than \(2\pi/3\), so projection onto its bisector bounds the modulus of their sum below by half the sum of their moduli. Since \(|q-r_+|\le4x<1/2\), the near sum includes the term for \(r_+\), of modulus at least \(1/(4x)\). The remaining, far terms have total modulus at most \(2n\). Their ratio to the near sum in modulus is at most \(16nx\le4/25\), so they change its argument by at most \(\arcsin(4/25)<1/6\). Again \(\arg H(q)+\pi\Omega_q\ge\pi/2-1/2\). Applying the same bound to the polynomial with conjugate zeros gives \(-\arg H(p)+\pi\Omega_p\ge\pi/2-1/2\), where \(\Omega_p\) is defined with \(p\) in place of \(q\). The two near sets are disjoint, and all harmonic measures are nonnegative. Thus \eqref{eq:origin-phase} gives \[ \pi\sum_\xi\omega(\xi)\ge\pi-\gamma-1\ge\pi/2-1>1/2 . \] The vectors \(p-\xi,q-\xi\) have first coordinate \(t-\Re\xi\) and second coordinates \(-s-\Im\xi,s-\Im\xi\). Their dot product is \(|\xi|^2-2t\Re\xi+2t^2-1\), and their determinant is \(2s(t-\Re\xi)>0\). Dividing and subtracting \((t-a)/s\) gives \[ \cot\Theta(\xi)-\frac{t-a}{s} =\frac{|\xi-a|^2-1+2at-a^2}{2s(t-\Re\xi)}>0. \] The cotangent decreases on \((0,\pi)\), so \(\Theta(\xi)<\arctan(s/(t-a))\), where \(t>a\). Subtracting \(\gamma=\arctan(s/t)\) gives \[ \pi\omega(\xi) < \arctan\frac{as}{1-at}\le2a. \] This yields \(\pi\sum_\xi\omega(\xi)<2(n-1)x\le1/50\), a contradiction. \end{proof} \begin{lemma}\label{lem:quadrant-product} Let polynomials \(f_j\) converge locally uniformly on \(\mathbb C\) to a nonconstant entire function \(f\). Suppose that the roots of \(f_j\) have real part at most \(1\) and imaginary part at most \(0\), and that every zero of \(f'\) is real. Then \[ f(z)=e^{\alpha z}g(z),\qquad f'(z)=e^{\alpha z}h(z), \] where \(\Re\alpha\ge0\), \(\Im\alpha\le0\), and \(g,h\) are genus-zero products of minimal exponential type. Such a product has the form \(Cz^m\prod_{\lambda\ne0}(1-z/\lambda)\), where \(C\ne0\), \(m\ge0\) is the multiplicity at zero, and the nonzero zeros are counted with multiplicity and satisfy \(\sum_{\lambda\ne0}|\lambda|^{-1}<\infty\). Minimal exponential type means that for each \(\varepsilon>0\) there is \(C_\varepsilon>0\) such that the modulus is at most \(C_\varepsilon e^{\varepsilon |z|}\) for all \(z\). If \(\Im\alpha<0\), then \(f\) has no zeros. \end{lemma} \begin{proof} Let \(z_0=3+i\). By Hurwitz \(f(z_0)\ne0\). The reciprocals \((z_0-\lambda)^{-1}\) for the zeros \(\lambda\) of \(f_j\) lie in the closed fourth quadrant, so \[ \sum_{\lambda}\frac1{|z_0-\lambda|} \le\sqrt2\,|f_j'(z_0)/f_j(z_0)|=O(1). \] This follows by projecting onto the direction \(1-i\). On disks whose boundaries avoid zeros of \(f\), the zeros converge with multiplicities by Rouch\'e's theorem, so \(\sum_{\lambda:f(\lambda)=0}|z_0-\lambda|^{-1}<\infty\). For zeros of \(f_j\) with \(|z_0-\lambda|>R>2|z-z_0|\), the logarithm of the normalized factor \(1+(z-z_0)/(z_0-\lambda)\) differs from \((z-z_0)/(z_0-\lambda)\) by at most \(|z-z_0|^2/|z_0-\lambda|^2\). The sum of these errors is \(O(|z-z_0|^2/R)\) by the reciprocal sum bound. For fixed \(R\) with the cutoff circle avoiding the zeros of \(f\), the sum of the tail reciprocals tends to \(f'(z_0)/f(z_0)-\sum_{|z_0-\lambda|1\), put \(L_j(s)=f_j'(s)/f_j(s)\) and \(K_j(s)=f_j''(s)/f_j'(s)\). The root factorization gives \[ |L_j'(s)|\le\sum_\lambda |s-\lambda|^{-2} \le\frac{\Re L_j(s)}{s-1},\qquad |K_j(s)-L_j(s)|=\left|\frac{L_j'(s)}{L_j(s)}\right| \le\frac1{s-1}. \] Pass to the limit in \(j\) by Hurwitz and local uniform convergence of the functions and derivatives. Then let \(s\to+\infty\). The logarithmic derivatives of \(g,h\) tend to zero there by their absolutely summable inverse distances to \(z_0\). More explicitly, \(|z_0-\lambda|/|s-\lambda|\le 1+|z_0-3|/|s-\lambda|\) for \(s\ge3\), since \(\Re\lambda\le1\). Thus \(\alpha'=\alpha\). Suppose \(\tau=-\Im\alpha>0\). Multiply \(f\) by a constant of modulus one so that \(h\) is real entire, meaning \(h(\bar z)=\overline{h(z)}\). This is possible because all its zeros are real and its canonical factors are real. With \(g^*(z)=\overline{g(\bar z)}\), write \(g=u+iv\) for the real entire functions \(u=(g+g^*)/2\) and \(v=(g-g^*)/(2i)\). The imaginary part of \(h=g'+\alpha g\) on the real axis gives \(v'+(\Re\alpha)v=\tau u\). In particular \(v\) is not identically zero, since otherwise \(u\equiv0\). On the real axis away from the zeros of \(g\), \[ \Im\frac{g'(s)}{g(s)} =\sum_{\lambda:g(\lambda)=0}\Im\frac1{s-\lambda}\le0. \] The left side equals \((uv'-u'v)/|g|^2\). Hence \(u'v-uv'\ge0\) on the real axis by continuity, and substituting \(v'+(\Re\alpha)v=\tau u\) gives \(vv''-(v')^2=\tau(u'v-uv')\ge0\). If \(v\) had a real zero \(s_0\) of order \(m\ge1\) with leading coefficient \(c\ne0\), then \(vv''-(v')^2\) would have leading term \(-mc^2(s-s_0)^{2m-2}\), contradicting nonnegativity near \(s_0\). Thus \((\log|v|)''=(vv''-(v')^2)/v^2\ge0\) on \(\mathbb R\). The minimal exponential type of \(g\) and \(g^*\) gives \(\log|v(s)|\le \log C_\varepsilon+\varepsilon|s|\) for each \(\varepsilon>0\). By convexity, every secant slope lies between \(-\varepsilon\) and \(\varepsilon\), as follows by comparing it with secants whose left or right endpoint tends to infinity in absolute value. Letting \(\varepsilon\downarrow0\) shows that \(|v|\) is constant on the real axis. The function \(v\) is real and nonvanishing there, so it is constant by continuity and the identity theorem. The relation with \(u\) makes \(u\) and \(g\) constant. \end{proof} \begin{proposition}\label{prop:origin} There is no sequence of the normalized counterexamples with \(n\to\infty\) and \(a\to0\). \end{proposition} \begin{proof} Select a root \(b=x+iy\) with maximal real part \(x\ge a\). Lemma~\ref{lem:semicircle} implies that any limit of \(b\) is \(i\) or \(-i\). Pass to a subsequence and conjugate if necessary so that \(b\to i\). The inequality \(|b-a|^2\le1-2ax+a^2<1\) implies that \(b\) is simple. Use the scale and coordinates \begin{gather*} \rho=\max(x,1-y),\qquad \beta=(b-i)/\rho,\\ X=x/\rho,\qquad A=a/\rho,\qquad v_\xi=(\xi-i)/\rho . \end{gather*} Here \(01/100\), so we can pass to a subsequence with \((n-1)\rho\to N_0>0\). The polynomials \[ f_n(u)=\frac{P(i+\rho u)}{\rho P'(b)} \] satisfy \(f_n(\beta)=0\) and \(f_n'(\beta)=1\). By the product formula, \(f_n'(u)=\prod_\xi(1+(u-\beta)/(\beta-v_\xi))\). Applying \(|1+z|^2\le\exp(2\Re z+|z|^2)\) to each factor gives \[ |f_n'(u)|\le \exp\left(\Re((u-\beta)S)+\frac{|u-\beta|^2Q}{2}\right). \] Since \(\beta\) is bounded, the derivatives are locally uniformly bounded, and integrating from \(\beta\) gives the same boundedness for \(f_n\). By Montel's theorem pass to a subsequence with \(f_n\to f\) locally uniformly and \(\beta\to\beta_\infty\). Then \(f(\beta_\infty)=0\) and \(f'(\beta_\infty)=1\). Every scaled critical point satisfies \[ a\Re v_\xi-\frac{\rho}{2}|v_\xi|^2-\frac{a^2}{2\rho} <\Im v_\xi\le-\frac{\rho}{2}|v_\xi|^2, \] by the two disk inequalities for \(\xi\). As \(A\le1\), every bounded critical-point limit is real and has real part at most \(1/2\). By Hurwitz the zeros of \(f'\) have the same property. We may apply Lemma~\ref{lem:quadrant-product}. For real \(s>1\), define \[ K_n(s)=\sum_\xi(s-v_\xi)^{-1},\quad R_n(s)=\sum_\xi|s-v_\xi|^{-2},\quad E_n(s)=\sum_\xi\frac{1-|\xi|^2}{2\rho |s-v_\xi|^2}. \] All zeros of \(f'\) have real part at most \(1/2\), so \(f'(s)\ne0\). Hence \(K_n(s)=f_n''(s)/f_n'(s)\to f''(s)/f'(s)\), and \(R_n(s)\le \Re K_n(s)/(s-1/2)\) is bounded. Expanding \(|\xi|^2=|i+\rho v_\xi|^2\) and using \[ \frac{|v_\xi|^2}{|s-v_\xi|^2} =1-2s\Re\frac1{s-v_\xi}+\frac{s^2}{|s-v_\xi|^2} \] gives \[ \Im K_n(s)=-\frac{(n-1)\rho}{2} +\rho s\Re K_n(s)-\frac{\rho s^2R_n(s)}2-E_n(s). \] From \(|\xi-a|>1\) we have \(0\le 1-|\xi|^2