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\hypersetup{hidelinks, pdftitle={Sheil-Small's (1974) covering problem has a positive answer}, pdfauthor={Dave Goel}}

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\title[Sheil-Small's covering problem]{Sheil-Small's (1974) covering problem\\ has a positive answer}
\author{Dave Goel}
\thanks{Lean 4 proof: \href{https://github.com/david44hello/paper_sheil-small-1974-covering/tree/v3.20/lean}{\nolinkurl{github.com/david44hello/paper_sheil-small-1974-covering/tree/v3.20/lean}}\spacefactor3000\relax}
\email{david44hello@gmail.com}
\subjclass[2020]{Primary 30C10; Secondary 30C15, 42A05}
\keywords{self-inversive polynomial, covering, Schur--Cohn form, de Bruijn's theorem}
\date{November 2026}

\begin{document}

\begin{abstract}
T.~Sheil-Small asked whether every self-inversive polynomial maps the unit disc onto a domain that contains a disc whose
radius is the largest modulus of the polynomial's coefficients (Problem~4.24 in Hayman and Lingham's problem book). We
show that the answer is yes. More generally, every non-constant polynomial whose coefficients have the same moduli when
read in reverse order maps the unit disc onto a domain that contains an open disc of this radius. This class contains the
self-inversive polynomials and the polynomials whose
zeros, with multiplicity, are symmetric under taking reciprocals (the problem's printed definition of self-inversive).
The radius cannot be
increased. For a polynomial with all its zeros on the unit circle, the largest disc in the image has radius exactly half
the maximum modulus on the circle. For these polynomials the question is therefore equivalent to
the known bound: every coefficient has modulus at most half this maximum (O'Hara--Rodriguez, Saff--Sheil-Small, Kristiansen).
The new case is that of polynomials with zeros off the unit circle.
We treat this case with the
Schur--Cohn Hermitian form and that bound.
The main theorem is also proved in the Lean 4 proof assistant, with that bound
taken as given.
\end{abstract}

\maketitle

\section{Introduction}\label{sec:intro}

Let $\D=\{z\in\C:|z|<1\}$ be the open unit disc, and write $D(c,r)=\{w\in\C:|w-c|<r\}$ for the open disc with centre $c$ and
radius $r>0$. For a polynomial $P(z)=\sum_{k=0}^{n}a_kz^k$ of degree $n\ge1$ we put
\[
A(P)=\max_{0\le k\le n}|a_k|,\qquad M(P)=\max_{|z|=1}|P(z)|,\qquad
P^{*}(z)=z^{n}\,\overline{P(1/\bar z)}=\sum_{k=0}^{n}\overline{a_{n-k}}\,z^{k}.
\]
When $P$ is fixed we write $A=A(P)$ and $M=M(P)$. The polynomial $P$ is called \emph{self-inversive} if $P^{*}=\lambda P$ for some constant $\lambda$ with $|\lambda|=1$.
When $P(0)\ne0$, this says exactly that the zeros of $P$ are symmetric with respect to the unit circle: together with each
zero $\zeta$, the point $1/\bar\zeta$ is a zero of the same multiplicity (Lemma~\ref{lem:classes} below). This is the
definition used by Saff and Sheil-Small \cite[p.~21]{SS74}. Sheil-Small's book \cite[p.~228]{SS02} requires $\lambda=1$; every
self-inversive polynomial satisfies this after multiplication by a suitable constant of modulus $1$.
T.~Sheil-Small posed the following problem in the problem list of the 1973 Canterbury symposium on complex analysis,
published in 1974 \cite{Ha74}; it is Problem~4.24 in Hayman and Lingham's problem book \cite{HL19} (see its
Tables~1 and~2).

\begin{quote}
\textbf{Problem} (Sheil-Small). Let $P(z)=\sum_{0}^{n}a_kz^k$ be a self-inversive polynomial, i.e.\ if $\zeta$ is a zero of
$P(\zeta)$ with multiplicity $m$, then $1/\zeta$ is also a zero with multiplicity $m$. Is it true that $w=P(z)$ maps $\D$ onto
a domain containing a disc of radius $A=\max_{0\le k\le n}|a_k|$?
\end{quote}

The update in \cite{HL19} reads: ``No progress on this problem has been reported to us.'' We answer the question in the
affirmative. The table and Figure~\ref{fig:result} compare our results with what was known.

\begin{center}\small
\renewcommand{\arraystretch}{1.15}
\begin{tabular}{|p{0.45\textwidth}|p{0.45\textwidth}|}
\hline
\textbf{New result} & \textbf{Best result before this paper}\\
\hline
Every non-constant $P$ with $|a_k|=|a_{n-k}|$ for all $k$ (in particular every self-inversive $P$) maps $\D$ onto a
domain containing an open disc of radius $A$ (Theorem~\ref{thm:main}). &
Open: Problem~4.24 of \cite{HL19}, with no progress reported. \\
\hline
If all zeros of $P$ lie on the unit circle, the largest disc in $P(\D)$ has radius exactly $M/2$, so here the question
is equivalent to $M\ge2A$ (Theorem~\ref{thm:circle}). &
We have not found the disc of radius $M/2$ stated, but it follows from de Bruijn's theorem (1947; see Lemma~\ref{lem:cover}), applied at a point $\zeta$ of
maximum modulus, where $\zeta P'(\zeta)=\frac n2P(\zeta)$ (an identity implicit in Saff and Sheil-Small, 1974). The
inequality $M\ge2A$ (asked by Hayman, 1967) was proved by O'Hara--Rodriguez and Saff--Sheil-Small (1974) except for the
middle coefficient, and in full by Kristiansen (1979; a gap in one lemma was closed in the 1980 correction). \\
\hline
\end{tabular}
\end{center}

\begin{itemize}[leftmargin=1.5em]
\item \emph{Theorem~\ref{thm:main} (the covering theorem).} Let $n\ge1$, and let $P(z)=\sum_{k=0}^{n}a_kz^k$ be a polynomial
of degree at most $n$, not constant, with $|a_k|=|a_{n-k}|$ for $0\le k\le n$. Then $P(\D)$ contains an open disc of
radius $A=\max_k|a_k|$. This answers Sheil-Small's question \cite[Problem~4.24]{HL19}, under either reading of the
definition of self-inversive (the quotation has $1/\zeta$, our definition $1/\bar\zeta$; see below), and the radius $A$ cannot be increased.
\item \emph{Theorem~\ref{thm:circle} (all zeros on the unit circle).} If $P$ has degree $n\ge1$ and all its zeros on the
unit circle, and $|P(\zeta)|=M(P)$ with $|\zeta|=1$, then $P(\D)$ contains $D\bigl(P(\zeta)/2,M(P)/2\bigr)$ and no open
disc of larger radius.
\end{itemize}

\begin{figure}[ht]
\centering
\newcommand{\figonecurve}{plot[domain=-90:90, samples=181, variable=\t]
  ({(5+2*cos(\t)+2*cos(2*\t))*cos(2*\t)},{(5+2*cos(\t)+2*cos(2*\t))*sin(2*\t)}) -- cycle}
\newcommand{\figoneunitdisc}{%
  \draw[thick, fill=black!12] (-6.6,-4.2) circle (1);
  \node[align=center] at (-6.6,-4.2) {$\D$};
  \draw[->, thick] (-5.35,-4.2) -- node[above, inner sep=2pt] {$P$} (-3.0,-4.2);
  \node[anchor=east, inner sep=1pt] at (-2.4,5.2) {$P(\D)$};}
\begin{tikzpicture}[scale=0.3, every node/.style={font=\footnotesize, align=left}]
  \node[anchor=base, align=center] at (5.5,7.9) {\small known before this work};
  \fill[pattern=north east lines, pattern color=black!45] \figonecurve;
  \draw[thick] \figonecurve;
  \figoneunitdisc
  \node[fill=white, inner sep=2pt, font=\Large] at (3,0) {?};
  \node[anchor=west, text width=2.75cm] at (9.5,0) {Is a disc of radius $A$ always inside?\\[3pt] Open since 1974 when zeros lie off the unit circle, as here.};
\end{tikzpicture}
\hfill
\begin{tikzpicture}[scale=0.3, every node/.style={font=\footnotesize, align=left}]
  \node[anchor=base, align=center] at (5.5,7.9) {\small this work};
  \fill[green!12] \figonecurve;
  \figoneunitdisc
  \fill[green!55!black!45] (2,0) circle (5);
  \draw (2,0) circle (5);
  \draw[thick] \figonecurve;
  \fill (2,0) circle (5pt);
  \node[align=center] at (2,2.3) {disc of\\ radius $A$};
  \draw[thick] (2,0) -- node[below, inner sep=2pt] {$A=5$} (7,0);
  \node[anchor=west, text width=2.75cm] at (9.5,0) {Yes. Theorem~\ref{thm:main}: a disc of radius $A$ for every non-constant $P$ with $|a_k|=|a_{n-k}|$.};
\end{tikzpicture}
\par\smallskip
{\footnotesize $P(z)=1+z+5z^2+z^3+z^4$, with largest coefficient $A=5$\par}
\caption{The image of the unit disc $\D$ under the self-inversive polynomial $P(z)=1+z+5z^2+z^3+z^4$, drawn to scale.
Here $A=5$, and the zeros of $P$ lie off the unit circle. The class of Theorem~\ref{thm:main} contains the self-inversive polynomials and
the polynomials with $a_{n-k}=\pm a_k$ (the literal reading of Problem~4.24), and the polynomials with all
zeros on the unit circle are self-inversive (Lemma~\ref{lem:classes}). Left, known before this work: the question was listed as open, whichever reading is meant;
when all zeros lie on the unit circle, a disc of radius $A$ follows from results in print: the inequality $M\ge2A$
(1974, 1979) with de Bruijn's theorem. Right, this work: Theorem~\ref{thm:main} gives a disc of radius $A$ for every non-constant member of the class,
and $A$ is the best radius that holds for all of them (Section~\ref{sec:sharp}); Theorem~\ref{thm:circle} gives the exact largest radius $M/2$ when all zeros lie on the unit circle.}
\label{fig:result}
\end{figure}

We make two remarks on Theorem~\ref{thm:main}. First, the problem as printed in \cite{HL19} pairs a zero $\zeta$ with $1/\zeta$, not with
$1/\bar\zeta$. Read literally, it concerns the polynomials with $z^nP(1/z)=\pm P(z)$, that is, $a_{n-k}=\pm a_k$. With
complex coefficients this is a different class (for example $1+3iz+z^2$ is of this kind but is not self-inversive). Both
classes satisfy $|a_k|=|a_{n-k}|$ (Lemma~\ref{lem:classes}), so Theorem~\ref{thm:main} answers the question under either
reading. Second, the radius $A$ cannot be increased, and the disc must be taken open: for $P(z)=1+z^n$ we have $A=1$ and
$P(\D)=D(1,1)$. Section~\ref{sec:sharp} and Figure~\ref{fig:examples} give examples in which $A$ is large compared with
$|a_0|$ and the radius $A$ is still best possible.

Hayman asked in 1967 \cite[p.~24]{Ha67} whether every polynomial $P$ with all its zeros on the unit circle satisfies
$M(P)\ge2A(P)$; there the inequality is misprinted as $M\le2A$ \cite[Update~4.2]{HL19}. O'Hara and Rodriguez \cite[Theorem~3]{OR74}
and, independently, Saff and Sheil-Small \cite[Theorem~3]{SS74} proved $|a_k|\le M/2$ for every $k$ except the middle index
$k=n/2$ ($n$ even). Kristiansen \cite{Kr79,Kr80} proved the middle case; see \cite[Update~4.2]{HL19}, and \cite{RS02}
for an account of his proof. (Some later papers, for example \cite{JS16,Ru80}, still describe the middle case as
open.)

The disc of radius $M/2$ in Theorem~\ref{thm:circle} comes from a covering theorem of de Bruijn
\cite{dB47} (Lemma~\ref{lem:cover} is the form we use), applied at a point $\zeta$ of maximum modulus, where $\zeta P'(\zeta)=\frac n2P(\zeta)$. Each ingredient
appears in one of the two 1974 proofs that $\max_{|z|=1}|P'|=\frac n2M(P)$ for self-inversive $P$: Saff and Sheil-Small
\cite[Theorem~7]{SS74} work at a point of maximum modulus, and O'Hara and Rodriguez \cite[Theorem~1]{OR74} apply de Bruijn's
theorem at a boundary point. In Theorem~\ref{thm:circle}, that no open disc of radius larger than $M/2$ lies in $P(\D)$ is elementary. Thus the case of Sheil-Small's question in which all zeros
lie on the unit circle can be assembled from results in print since 1980, although we have not found it stated.
By Theorem~\ref{thm:circle}, a positive answer for self-inversive polynomials also implies the inequality $M\ge2A$ of
O'Hara--Rodriguez, Saff--Sheil-Small and Kristiansen for polynomials with all zeros on the unit circle, so any proof of this answer also proves that inequality (see
Remark~\ref{rem:where}). The new part of this paper is the case in which $P$ has zeros off the unit circle. For self-inversive $P$
the disc of radius $M/2$ still exists (Proposition~\ref{prop:half}), but $M/2$ may be smaller than $A$: for $1+4z+z^2$ we
have $M/2=3$ and $A=4$ (Figure~\ref{fig:examples}, right). We therefore reduce to the circle case by another route, which we now
sketch.

\subsection*{The idea of the proof}
Suppose that $w$ is a value omitted by $P$ on $\D$. Then $H=P-w$ has no zero in $\D$, and two classical facts about such a
polynomial do most of the work. First, on the unit circle $|zH'(z)|\le|nH(z)-zH'(z)|$. Second, write
$\bigl(H(z)\overline{H(\eta)}-H^{*}(z)\overline{H^{*}(\eta)}\bigr)/(1-z\bar\eta)=\sum_{i,j}K_{ij}z^{i}\bar\eta^{\,j}$; the Hermitian form of Schur and
Cohn, given by the matrix $(K_{ij})$, is positive semidefinite, and its entries are explicit in the coefficients of $H$ (Lemma~\ref{lem:omitted}). In the proof of Theorem~\ref{thm:circle}, the first fact, applied at a point $\zeta$ where $|P|$ is
largest, places every omitted value at distance at least $M/2$ from $P(\zeta)/2$. For Theorem~\ref{thm:main}, let $P$ have degree at most $n$ with $|a_k|=|a_{n-k}|$ for all $k$. If $a_0=0$, then $P(0)=0$, so every omitted value $w$ is non-zero, and a $2\times2$
minor of $(K_{ij})$ gives $|w|\ge A$; hence $D(0,A)\subset P(\D)$. If $a_0\ne0$, we normalize so that $a_0=a_n=1$. Then the
diagonal entries $K_{jj}$ all equal $|1-w|^2-1\ge0$, so $|1-w|\ge1$, which gives $D(1,1)\subset P(\D)$ and settles the case $A=1$. When
$A>1$, let $j$ be an index with $|a_j|=A$ and put
$E=a_{n-j}/\overline{a_j}$, so that $|E|=1$ and $E$ does not depend on $w$; the $2\times2$ minor on the indices $0$ and $j$ gives
$|w-(1+E)|\ge A$ unless $w=1-E$. Consequently every value in $D(1+E,A)$ is taken, except possibly the single value $w_*=1-E$ (Figure~\ref{fig:punctured}). If $w_*$ is indeed omitted,
then $P-w_*$ has all its zeros on the unit circle and the same $A$. Theorem~\ref{thm:circle}, together with the
inequality $M\ge2A$ (Theorem~\ref{thm:coef}), then applies to it.

Section~\ref{sec:prelim} relates the two readings to the condition $|a_k|=|a_{n-k}|$ and collects the facts about omitted
values. Section~\ref{sec:circle} proves
Theorem~\ref{thm:circle}, Section~\ref{sec:proof} proves Theorem~\ref{thm:main}, and Section~\ref{sec:sharp} discusses
sharpness.
An expanded version of this paper, with further results, is posted on Zenodo \cite{G}.

\section{Self-inversive polynomials and omitted values}\label{sec:prelim}

Throughout, $n\ge1$ is fixed, and for a polynomial $H(z)=\sum_{k=0}^{n}h_kz^k$ of degree \emph{at most} $n$ we write
$H^{*}(z)=z^{n}\,\overline{H(1/\bar z)}=\sum_{k=0}^{n}\overline{h_{n-k}}\,z^{k}$ (the coefficient $h_n$ may vanish). Then
$(H^{*})^{*}=H$, and $|H^{*}(z)|=|H(z)|$ when $|z|=1$. We first record how the two readings of ``self-inversive'' relate to
the condition $|a_k|=|a_{n-k}|$.

\begin{lemma}\label{lem:classes}
Let $P(z)=\sum_{k=0}^{n}a_kz^k$ be a polynomial of degree $n$ with $P(0)\ne0$.
\begin{enumerate}[label=\textup{(\alph*)},leftmargin=2em]
\item The zeros of $P$ are symmetric under $\zeta\mapsto1/\bar\zeta$, with multiplicities, if and only if $P^{*}=\lambda P$
for some $|\lambda|=1$. In that case $\overline{a_{n-k}}=\lambda a_k$ for every $k$.
\item The zeros of $P$ are symmetric under $\zeta\mapsto1/\zeta$, with multiplicities, if and only if there is
$\varepsilon\in\{1,-1\}$ such that $a_{n-k}=\varepsilon a_k$ for every $k$.
\end{enumerate}
In both cases $|a_k|=|a_{n-k}|$ for every $k$. Every polynomial of degree $n\ge1$ with all its zeros on the unit circle is
self-inversive.
\end{lemma}

\begin{proof}
(a) The zeros of $P^{*}$ are the points $1/\bar\zeta$, where $\zeta$ runs through the zeros of $P$, with the same
multiplicities, and $P^{*}$ has degree $n$ because its leading coefficient is $\overline{a_0}\ne0$. If the zeros are
symmetric, $P$ and $P^{*}$ have the same zeros and the same degree, so $P^{*}=\lambda P$ for a constant $\lambda\ne0$. Since
$P^{**}=P$, we get $P=\bar\lambda P^{*}=|\lambda|^2P$, so $|\lambda|=1$. The converse is clear, and comparing coefficients
in $P^{*}=\lambda P$ gives $\overline{a_{n-k}}=\lambda a_k$.
(b) The same argument with $\widetilde P(z)=z^{n}P(1/z)=\sum_k a_{n-k}z^k$ in place of $P^{*}$ gives
$\widetilde P=\varepsilon P$, and applying the map $P\mapsto\widetilde P$ twice gives $\varepsilon^2=1$.
If all zeros $\zeta_1,\dots,\zeta_n$ lie on the unit circle, then $1/\bar\zeta_l=\zeta_l$ and $P(0)\ne0$, so (a) applies.
\end{proof}

The next lemma is the source of all our inequalities. Part~(a) is a classical inequality (on $|z|=1$ it says $|H'(z)|\le|(H^{*})'(z)|$, the key step in the usual proof of the
Erd\H{o}s--Lax inequality \cite{La44}), and part~(b), the positivity of the Schur--Cohn Hermitian form, is the easy half of
the Schur--Cohn criterion for the location of zeros \cite{Sc17,Co22}; see also \cite{Ma66}. We include the short proofs.

\begin{lemma}\label{lem:omitted}
Let $H(z)=\sum_{k=0}^{n}h_kz^k$ be a polynomial of degree at most $n$ with no zero in $\D$. Then there are numbers
$\beta_1,\dots,\beta_n$ with $|\beta_l|\le1$ such that
\begin{equation}\label{eq:factor}
H(z)=h_0\prod_{l=1}^{n}(1-\bar\beta_lz),\qquad H^{*}(z)=\overline{h_0}\prod_{l=1}^{n}(z-\beta_l).
\end{equation}
Consequently:
\begin{enumerate}[label=\textup{(\alph*)},leftmargin=2em]
\item $|zH'(z)|\le|nH(z)-zH'(z)|$ for every $z$ with $|z|=1$;
\item the function
\[
K(z,\eta)=\frac{H(z)\overline{H(\eta)}-H^{*}(z)\overline{H^{*}(\eta)}}{1-z\bar\eta}
=\sum_{i,j=0}^{n-1}K_{ij}\,z^{i}\,\bar\eta^{\,j}
\]
is a polynomial in $z$ and $\bar\eta$, and its coefficient matrix $(K_{ij})_{0\le i,j\le n-1}$ is positive semidefinite;
\item for $0\le j\le n-1$,
\begin{equation}\label{eq:entries}
K_{jj}=\sum_{r=0}^{j}\bigl(|h_r|^2-|h_{n-r}|^2\bigr),\qquad
K_{0j}=h_0\overline{h_j}-\overline{h_n}\,h_{n-j}.
\end{equation}
\end{enumerate}
\end{lemma}

\begin{proof}
Since $0\in\D$, $h_0=H(0)\ne0$. Let $m\le n$ be the degree of $H$ and $z_1,\dots,z_m$ its zeros, repeated according to
multiplicity. By hypothesis $|z_l|\ge1$. Put $\beta_l=1/\bar z_l$ for $l\le m$ and $\beta_l=0$ for $m<l\le n$. Then
$H(z)=h_m\prod_{l\le m}(z-z_l)=h_0\prod_{l\le m}(1-z/z_l)$, which is the first formula in \eqref{eq:factor} because
$1/z_l=\bar\beta_l$. The second follows from the definition of $H^{*}$:
$H^{*}(z)=z^{n}\,\overline{h_0}\prod_{l=1}^{n}(1-\beta_l/z)=\overline{h_0}\prod_{l=1}^{n}(z-\beta_l)$.

(a) Let $|z|=1$ with $H(z)\ne0$. By \eqref{eq:factor}, $zH'(z)/H(z)=\sum_l\bigl(1-1/(1-\tau_l)\bigr)$ with $\tau_l=\bar\beta_lz$,
$|\tau_l|\le1$. For $|\tau|\le1$, $\tau\ne1$, we have $\operatorname{Re}\frac1{1-\tau}=\frac{1-\operatorname{Re}\tau}{|1-\tau|^2}\ge\frac12$,
because $2(1-\operatorname{Re}\tau)-|1-\tau|^2=1-|\tau|^2\ge0$. Hence $\operatorname{Re}\bigl(zH'(z)/H(z)\bigr)\le n/2$, and
\[
|nH(z)-zH'(z)|^2-|zH'(z)|^2=n|H(z)|^2\Bigl(n-2\operatorname{Re}\frac{zH'(z)}{H(z)}\Bigr)\ge0.
\]
At a zero of $H$ on the circle the inequality in (a) is an equality.

(b) For $|\beta|\le1$ and all $z,\eta\in\C$ one checks by multiplying out that
\begin{equation}\label{eq:one-factor}
(1-\bar\beta z)(1-\beta\bar\eta)-(z-\beta)(\bar\eta-\bar\beta)=(1-|\beta|^2)(1-z\bar\eta).
\end{equation}
Put $u_l=(1-\bar\beta_lz)(1-\beta_l\bar\eta)$ and $v_l=(z-\beta_l)(\bar\eta-\bar\beta_l)$. Then
$H(z)\overline{H(\eta)}=|h_0|^2\prod_lu_l$ and $H^{*}(z)\overline{H^{*}(\eta)}=|h_0|^2\prod_lv_l$, and the telescoping
identity $\prod_lu_l-\prod_lv_l=\sum_l\bigl(\prod_{r<l}v_r\bigr)(u_l-v_l)\bigl(\prod_{r>l}u_r\bigr)$ together with
\eqref{eq:one-factor} gives
\[
K(z,\eta)=|h_0|^2\sum_{l=1}^{n}(1-|\beta_l|^2)\,f_l(z)\,\overline{f_l(\eta)},\qquad
f_l(z)=\prod_{r<l}(z-\beta_r)\prod_{r>l}(1-\bar\beta_rz).
\]
Each $f_l$ is a polynomial of degree at most $n-1$. If $c_l\in\C^{n}$ is its vector of coefficients, then
$(K_{ij})=|h_0|^2\sum_l(1-|\beta_l|^2)\,c_l\,c_l^{\,*}$ (where $c_l^{\,*}$ is the conjugate transpose), a sum of positive
semidefinite matrices.

(c) Compare the coefficients of $z^{i}\bar\eta^{\,j}$ on both sides of
$(1-z\bar\eta)K(z,\eta)=H(z)\overline{H(\eta)}-H^{*}(z)\overline{H^{*}(\eta)}$. Since the $k$-th coefficient of $H^{*}$ is
$\overline{h_{n-k}}$, this gives $K_{ij}-K_{i-1,j-1}=h_i\overline{h_j}-\overline{h_{n-i}}\,h_{n-j}$ (with $K_{ij}=0$ when an
index is negative). Summing along the diagonal gives the formula for $K_{jj}$, and the case $i=0$ gives $K_{0j}$.
\end{proof}

\section{Polynomials with all zeros on the unit circle}\label{sec:circle}

In this section we prove the following theorem.

\begin{theorem}\label{thm:circle}
Let $P$ be a polynomial of degree $n\ge1$ with all its zeros on the unit circle, and let $|\zeta|=1$ with $|P(\zeta)|=M(P)$.
Then $P(\D)$ contains the disc $D\bigl(P(\zeta)/2,\,M(P)/2\bigr)$, and it contains no open disc of larger radius.
Consequently $P(\D)$ contains an open disc of radius $A(P)$ if and only if $M(P)\ge2A(P)$.
\end{theorem}

The first step is a covering lemma: a single boundary point $z_0$ forces a whole disc into the image. It is closely related
to a theorem of de Bruijn \cite{dB47}, which is quoted in \cite[p.~332]{OR74} for open and closed circular regions. To reach
the lemma from it, take $K=\D$ in the notation there, let $u$ run over $\D$, and let $z\in\D$ tend to $z_0$; this gives the lemma for polynomials of degree exactly
$n$. We give a short proof based on Lemma~\ref{lem:omitted}(a); it also covers polynomials of degree less than $n$.

\begin{lemma}\label{lem:cover}
Let $f$ be a polynomial of degree at most $n$, and let $|z_0|=1$ with $f'(z_0)\ne0$. Then $f(\D)$ contains the open disc with
centre $f(z_0)-z_0f'(z_0)/n$ and radius $|f'(z_0)|/n$.
\end{lemma}

\begin{proof}
Let $w\notin f(\D)$. Then $H=f-w$ has degree at most $n$ and no zero in $\D$, and $H'=f'$. Lemma~\ref{lem:omitted}(a) at
$z_0$ gives $|f'(z_0)|\le\bigl|n\bigl(f(z_0)-w\bigr)-z_0f'(z_0)\bigr|$, that is,
$\bigl|w-\bigl(f(z_0)-z_0f'(z_0)/n\bigr)\bigr|\ge|f'(z_0)|/n$. Therefore no omitted value lies in the disc.
\end{proof}

\begin{proposition}\label{prop:half}
Let $F$ be a polynomial of degree at most $n$, not constant, with $F^{*}=\lambda F$ for some $|\lambda|=1$. Let $M=M(F)$ and
let $|\zeta|=1$ with $|F(\zeta)|=M$. Then $\zeta F'(\zeta)=\frac n2F(\zeta)$, and $D\bigl(F(\zeta)/2,\,M/2\bigr)\subset F(\D)$.
\end{proposition}

\begin{proof}
For real $t$ put $g(t)=|F(e^{it})|^2$. Since $1/\bar z=z$ on the unit circle, $F^{*}=\lambda F$ gives
$\overline{F(e^{it})}=\lambda e^{-int}F(e^{it})$, so $g(t)=\lambda e^{-int}F(e^{it})^2$. Near $t_0$ with $e^{it_0}=\zeta$ we have $F(e^{it})\ne0$ (as $M>0$), and
logarithmic differentiation gives
\[
\frac{g'(t)}{g(t)}=-in+2i\,\frac{zF'(z)}{F(z)}\Big|_{z=e^{it}}.
\]
The left side is real, and it vanishes at $t_0$, where $g$ is largest. Hence $\zeta F'(\zeta)=\frac n2F(\zeta)$. In
particular $F'(\zeta)\ne0$, and Lemma~\ref{lem:cover} with $z_0=\zeta$ gives the disc with centre
$F(\zeta)-\frac12F(\zeta)=F(\zeta)/2$ and radius $\frac1n\cdot\frac n2|F(\zeta)|=M/2$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:circle}]
Write $M=M(P)$. By Lemma~\ref{lem:classes}, $P$ is self-inversive, so Proposition~\ref{prop:half} gives
$D(P(\zeta)/2,M/2)\subset P(\D)$. Now let $D(c,r)\subset P(\D)$; we show that $r\le M/2$. Since $P$ has no zero in $\D$, the
point $0$ is not in $D(c,r)$, so $|c|\ge r>0$. By the maximum principle $|P|<M$ in $\D$, so $D(c,r)\subset D(0,M)$. Since
$D(c,r)$ contains the points $c+tc/|c|$ with $0\le t<r$, we get $|c|+r\le M$. Together with $|c|\ge r$, this gives
$2r\le|c|+r\le M$. The largest radius is therefore exactly $M/2$, and a disc of radius $A$ lies in $P(\D)$ if and only if $A\le M/2$
(Figure~\ref{fig:examples}, left).
\end{proof}

To compare $A(P)$ with $M(P)$ we use the following theorem.

\begin{theorem}[\cite{Kr79,Kr80,OR74,SS74}]\label{thm:coef}
Let $P(z)=\sum_{k=0}^{n}a_kz^k$ have degree $n\ge1$ and all its zeros on the unit circle. Then $|a_k|\le M(P)/2$ for
$0\le k\le n$.
\end{theorem}

For $k\ne n/2$ this is \cite[Theorem~3]{OR74} and \cite[Theorem~3]{SS74}. For $k=n/2$ ($n$ even) it is due to Kristiansen \cite[Section~2]{Kr79},
who works with real trigonometric polynomials whose zeros are all real. To pass to his setting, put $m=n/2$. By
Lemma~\ref{lem:classes}, $P^{*}=\lambda P$ with $|\lambda|=1$, so $Q(\theta)=e^{-im\theta}P(e^{i\theta})$ satisfies
$\overline{Q}=\lambda Q$, and $T=cQ$ is real when $c^{2}=\lambda$. Then $T$ is a real trigonometric polynomial of exact
degree $m$ (its top coefficient is $c\,a_n\ne0$) whose $2m$ zeros in a period are all real, $\max|T|=M(P)$, and the
constant term of $T$ is $c\,a_{n/2}$. So the case $k=n/2$ says that the constant term $t_0$ of such a $T$ satisfies
$|t_0|\le\frac12\max|T|$; this is Conjecture~4 of \cite{SS74}, applied to $T$ and to $-T$. A gap in the proof of one of Kristiansen's lemmas was closed by T.~Sheil-Small in the correction \cite{Kr80}.

\begin{corollary}\label{cor:circle}
If $P$ has degree $n\ge1$ and all its zeros on the unit circle, then $P(\D)$ contains an open disc of radius $A(P)$.
\end{corollary}

\begin{proof}
By Theorem~\ref{thm:coef}, $A(P)\le M(P)/2$; now apply Theorem~\ref{thm:circle}.
\end{proof}

\section{The covering theorem}\label{sec:proof}

In this section we prove the main theorem, using Lemma~\ref{lem:omitted} for the omitted values and
Corollary~\ref{cor:circle} for the polynomials with all zeros on the unit circle.

\begin{theorem}\label{thm:main}
Let $n\ge1$, and let $P(z)=\sum_{k=0}^{n}a_kz^k$ be a polynomial of degree at most $n$ that is not constant. Suppose that
\begin{equation}\label{eq:mirror}
|a_k|=|a_{n-k}|\qquad(0\le k\le n).
\end{equation}
Then $P(\D)$ contains an open disc of radius $A=\max_{k}|a_k|$.
\end{theorem}

\begin{proof}

Since $P$ is not constant, $A>0$. We distinguish two cases.

\emph{Case 1: $a_0=0$.} Then $a_n=0$ as well, so the maximum $A$ is attained at an index $j$ with $1\le j\le n-1$ (in particular
$n\ge2$). Moreover $P(0)=0$.
Let $w\notin P(\D)$; then $w\ne0$. Apply Lemma~\ref{lem:omitted} to $H=P-w$, of degree at most $n$, whose coefficients are
$h_0=-w$, $h_n=0$ and $h_k=a_k$ for $0<k<n$. In the sum \eqref{eq:entries} for $K_{jj}$ the terms with $1\le r\le j$ vanish
by \eqref{eq:mirror}: for these $r$ both $r$ and $n-r$ lie between $1$ and $n-1$, so $h_r=a_r$ and $h_{n-r}=a_{n-r}$. Hence $K_{jj}=|h_0|^2-|h_n|^2=|w|^2$, and $K_{0j}=-w\,\overline{a_j}$.
The $2\times2$ principal minor of the positive semidefinite matrix $(K_{pq})$ on the indices $0$ and $j$ gives
$|K_{0j}|^2\le K_{00}K_{jj}$, that is, $|w|^2A^2\le|w|^4$. Hence $|w|\ge A$ for every omitted value $w$, and
$D(0,A)\subset P(\D)$.

\emph{Case 2: $a_0\ne0$.} Then $a_n\ne0$, so $P$ has degree $n$. Choose $\alpha$ with $|\alpha|=1$ and
$\alpha^{n}=a_0/a_n$, and replace $P$ by $P(\alpha z)/a_0$. This multiplies $P(\D)$ by $1/a_0$, hence every radius
by $1/|a_0|$; it multiplies $A$ by the same factor and preserves \eqref{eq:mirror}. We may therefore assume $a_0=a_n=1$, and then $A\ge1$.

Let $w\notin P(\D)$ and apply Lemma~\ref{lem:omitted} to $H=P-w$: now $h_0=1-w$, $h_n=1$ and $h_k=a_k$ for $0<k<n$. As
in Case~1, the terms with $1\le r\le j$ in \eqref{eq:entries} cancel, so all diagonal entries are equal:
\[
K_{jj}=|1-w|^2-1=:d\qquad(0\le j\le n-1).
\]
Positive semidefiniteness gives $d\ge0$, that is, $|w-1|\ge1$. If $A=1$, this shows that every value in $D(1,1)$ is taken,
and we are done.

Suppose $A>1$. Then $A$ is attained at some $j$ with $1\le j\le n-1$. Put $E=a_{n-j}/\overline{a_j}$ and $u=w-1$. By
\eqref{eq:mirror}, $|E|=1$. By \eqref{eq:entries}, $K_{0j}=(1-w)\overline{a_j}-a_{n-j}=-\overline{a_j}\,(u+E)$, and the
$2\times2$ minor on the indices $0$ and $j$ gives
\begin{equation}\label{eq:minor}
A\,|u+E|\le d=|u|^2-1.
\end{equation}
Since $|E|=1$, we also have $|u|^2-1=|u^2|-|E^2|\le|u^2-E^2|=|u+E|\,|u-E|$, so \eqref{eq:minor} gives
$A\,|u+E|\le|u+E|\,|u-E|$. If $u\ne-E$, dividing by $|u+E|$ gives $|u-E|\ge A$, that is, $|w-(1+E)|\ge A$. We have shown that every omitted value other than $w_*=1-E$ lies outside
$D(1+E,A)$ (Figure~\ref{fig:punctured}):
\begin{equation}\label{eq:punctured}
D(1+E,A)\setminus\{w_*\}\subset P(\D).
\end{equation}

If $w_*\in P(\D)$, then \eqref{eq:punctured} gives $D(1+E,A)\subset P(\D)$, and we are done. Suppose $w_*\notin P(\D)$, and
let $S=P-w_*$. Its constant term is $E$ and its leading coefficient is $1$. It has no zero in $\D$, so its $n$ zeros
have modulus at least $1$, while their product has modulus $|E|=1$. Hence all zeros of $S$ lie on the unit circle. The
coefficients of $S$ are $E$, $a_1,\dots,a_{n-1}$ and $1$, so $A(S)=A$. By Corollary~\ref{cor:circle}, $S(\D)$
contains an open disc of radius $A$, and so does $P(\D)=S(\D)+w_*$.
\end{proof}

\begin{remark}\label{rem:where}
Since $|w_*-(1+E)|=2$, the point $w_*=1-E$ lies in $D(1+E,A)$ only when $A>2$ (Figure~\ref{fig:punctured}, right), that is, only when $A>2|a_0|$ before the
normalization. Thus Theorem~\ref{thm:coef} is needed only when $A>2|a_0|$ and $w_*$
is omitted; otherwise \eqref{eq:punctured} already gives $D(1+E,A)\subset P(\D)$. The threshold is the same as in
\cite[Section~2, Remark~C]{OR74}, where the middle case of Theorem~\ref{thm:coef} is proved when $|a_{n/2}|\le2|a_n|$. Moreover, if $A$ is attained at some index $j\ne n/2$, then $A(S)=|a_j|$ with $j\ne n/2$, and
\cite[Theorem~3]{OR74} or \cite[Theorem~3]{SS74} suffices. Thus the middle case of Theorem~\ref{thm:coef}, due to Kristiansen, is used only when $n$ is even, $|a_{n/2}|$ is
strictly larger than every other $|a_k|$ and than $2|a_0|$, and $w_*$ is omitted. The middle case cannot be bypassed: by
Theorem~\ref{thm:circle}, for polynomials with all zeros on the unit circle Theorem~\ref{thm:main} is equivalent to
Theorem~\ref{thm:coef}, so every proof of Theorem~\ref{thm:main} also proves the middle case.
\end{remark}

\section{Sharpness}\label{sec:sharp}

We show that the radius $A$ in Theorem~\ref{thm:main} cannot be increased, and that this remains true when $A$ is large
compared with $|a_0|$.

\begin{figure}[htb]
\centering
\begin{tikzpicture}[scale=0.55]
\draw[->,gray] (-1,0)--(4.8,0); \draw[->,gray] (0,-2.8)--(0,3.1);
\draw[thick,domain=0:360,samples=181,smooth,variable=\t] plot ({(2+2*cos(\t))*cos(\t)},{(2+2*cos(\t))*sin(\t)});
\draw[dashed] (2,0) circle (2);
\fill (2,0) circle (1.5pt) node[below] {$2$};
\fill (0,0) circle (1.5pt); \node[anchor=north east, inner sep=1pt] at (-0.6,-0.05) {$0$};
\fill (4,0) circle (1.5pt) node[below right] {$4$};
\node at (2,-3.4) {$(1+z)^2$: $A=2$, $M=4$};
\end{tikzpicture}
\hspace{1.5em}
\begin{tikzpicture}[scale=0.55]
\draw[->,gray] (-2.8,0)--(6.8,0); \draw[->,gray] (0,-4.3)--(0,4.6);
\draw[thick,domain=0:360,samples=181,smooth,variable=\t] plot ({(4+2*cos(\t))*cos(\t)},{(4+2*cos(\t))*sin(\t)});
\draw[dashed] (2,0) circle (4);
\draw[dotted] (-2,-4.3)--(-2,4.6); \draw[dotted] (6,-4.3)--(6,4.6);
\draw[gray] (3,0) circle (3);
\node[anchor=north east, inner sep=1pt] at (-2.05,-0.1) {$-2$}; \node[anchor=north west, inner sep=1pt] at (6.05,-0.1) {$6$};
\fill (2,0) circle (1.5pt) node[below] {$2$};
\fill (0,0) circle (1.5pt) node[below left] {$0$};
\node at (2,-5) {$1+4z+z^2$: $A=4$, $M=6$};
\end{tikzpicture}
\caption{Two images of $\D$, in polar coordinates $w=\rho e^{i\psi}$. Left: the image under $(1+z)^2$ is bounded by the cardioid
$\rho=2+2\cos\psi$; all zeros lie on the unit
circle, and the largest disc inside it is $D(2,2)$, of radius $M/2=A$ (Theorem~\ref{thm:circle}). Right: the image under
$1+4z+z^2$ is bounded by the lima\c{c}on $\rho=4+2\cos\psi$; it lies in the strip $-2<\operatorname{Re}w<6$ (dotted), and
the dashed disc $D(2,4)$ of Theorem~\ref{thm:main} has the largest possible radius $A=4$ (Example~\ref{ex:limacon}); the grey disc is $D(3,3)$, of radius $M/2$.}
\label{fig:examples}
\end{figure}%
\begin{figure}[htb]
\centering
\begin{tikzpicture}[scale=0.7,baseline=(lab.base)]
\fill[gray!30] (1.5,0.8660) circle (1.5);
\draw (1.5,0.8660) circle (1.5);
\draw[->,gray] (-1.2,0)--(3.6,0); \draw[->,gray] (0,-1.5)--(0,2.8);
\draw[dashed] (1,0) circle (1);
\draw[densely dotted] (0.5,-0.8660)--(1.5,0.8660);
\fill (0,0) circle (1.5pt); \node[anchor=south east, inner sep=1pt] at (-0.05,0.05) {$0$};
\fill (1,0) circle (1.5pt); \node[anchor=north west, inner sep=1pt] at (1.05,-0.05) {$1$};
\fill (1.5,0.8660) circle (1.5pt); \node[anchor=south west, inner sep=1pt] at (1.36,0.97) {$1+E$};
\draw[fill=white] (0.5,-0.8660) circle (2.2pt); \node[anchor=north west, inner sep=1pt] at (0.55,-1.08) {$w_*$};
\node[anchor=south west] at (2.3,2.0) {$D(1+E,A)$};
\node (lab) at (1.2,-3.4) {$A=3/2$};
\end{tikzpicture}
\hspace{2.5em}
\begin{tikzpicture}[scale=0.7,baseline=(lab.base)]
\fill[gray!30] (1.5,0.8660) circle (3);
\draw (1.5,0.8660) circle (3);
\draw[->,gray] (-1.9,0)--(4.9,0); \draw[->,gray] (0,-2.4)--(0,4.2);
\draw[dashed] (1,0) circle (1);
\draw[densely dotted] (0.5,-0.8660)--(1.5,0.8660);
\fill (0,0) circle (1.5pt); \node[anchor=south east, inner sep=1pt] at (-0.05,0.05) {$0$};
\fill (1,0) circle (1.5pt); \node[anchor=north west, inner sep=1pt] at (1.05,-0.05) {$1$};
\fill (1.5,0.8660) circle (1.5pt); \node[anchor=south west, inner sep=1pt] at (1.36,0.97) {$1+E$};
\draw[fill=white] (0.5,-0.8660) circle (2.2pt); \node[anchor=north west, inner sep=1pt] at (0.55,-1.08) {$w_*$};
\node at (1.6,2.85) {$D(1+E,A)$};
\node (lab) at (1.5,-3.4) {$A=3$};
\end{tikzpicture}
\caption{The discs in Case~2 of the proof of Theorem~\ref{thm:main}, after the normalization $a_0=a_n=1$, drawn for
$E=e^{i\pi/3}$. The image $P(\D)$ itself is not drawn. The dashed circle is $|w-1|=1$, the boundary of $D(1,1)$. The grey
disc is $D(1+E,A)$, with its centre $1+E$, and the open dot is the point $w_*=1-E$, joined to $1+E$ by a dotted segment.
The left panel has $A=3/2$, the right panel $A=3$. See the proof of Theorem~\ref{thm:main} and Remark~\ref{rem:where}.}
\label{fig:punctured}
\end{figure}
\begin{example}\label{ex:one}
For $P(z)=1+z^n$ we have $A=1$ and $P(\D)=D(1,1)$, which contains no open disc of radius larger than $1$ and no closed
disc of radius $1$. Thus in Theorem~\ref{thm:main} the radius cannot be increased, and ``open'' cannot be replaced by ``closed''.
\end{example}

The next example shows that the radius stays sharp when $A$ is large compared with $|a_0|$.

\begin{example}\label{ex:limacon}
Let $a\ge4$, $m\ge1$, and $P(z)=1+az^{m}+z^{2m}$. With $z^{m}=e^{i\theta}$ we get $P=e^{i\theta}(a+2\cos\theta)$, so
\[
\operatorname{Re}P=a\cos\theta+2\cos^2\theta\qquad(|z|=1).
\]
As a function of $x=\cos\theta\in[-1,1]$, the right-hand side has derivative $a+4x\ge0$, so it takes values in $[2-a,2+a]$.
By the maximum principle for the harmonic function $\operatorname{Re}P$, the set $P(\D)$ lies in the strip
$2-a<\operatorname{Re}w<2+a$, of width $2a$. Every disc in $P(\D)$ therefore has radius at most $a=A$. By
Theorem~\ref{thm:main} a disc of radius $A$ does lie in $P(\D)$: here $E=1$, and $w_*=0$ is taken because
$P(z)=0$ when $z^m$ is the root of $t^2+at+1$ in $(-1,0)$ (the roots are negative with product $1$), so the disc of
the proof is $D(2,a)$ (Figure~\ref{fig:examples}, right, for $a=4$, $m=1$).
\end{example}



\section*{Acknowledgments}
This work was done overwhelmingly by AI models, steered by the author in a continuous critical dialogue, alternating between Anthropic's Claude and OpenAI's ChatGPT. The author takes full responsibility for the paper.

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