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Statement

$f(z)=\prod_{\nu=1}^n(z-z_\nu)$ and $E=\{|f(z)|\le1\}$ (p. 97). Since $|f(0)|=\prod|z_\nu|\le1$ when $|z_\nu|\le1$, the point $0$ lies in $E$ (Remark 1, p. 99).

Theorem 3 (p. 99). "Let $f(z)=\prod(z-z_\nu)$, $|z_\nu|\le r\le1$, and let $d_0$ be the diameter of the component $E_0$ of $E$ that contains $0$. Then

$$d_0\ge2\quad\text{for}\quad0\le r\le1/2,\qquad d_0>1/r\quad\text{for}\quad1/2<r\le(\sqrt5-1)/2,\qquad d_0>2-r^2\quad\text{for}\quad(\sqrt5-1)/2\le r\le1."$$

Remarks (p. 99, quoted in part): "Lemma 1 shows that the centroid $z_0$ lies in $E_0$ (compare Theorem 1 of [2])." "The inequality $d_0\ge2$ for $r\le1/2$ cannot be improved, as the example $f(z)\equiv z^n$ shows. Also, the polynomial $(z^n+1)(z-1)^2(z-e^{i\pi/n})^{-1}(z-e^{-i\pi/n})^{-1}$ has $d_0<1+\varepsilon$ for sufficiently large $n$ (see the proof of Theorem 7 in [2]). Hence the inequality $d_0>1$ is best possible, for $r=1$." "Since all three bounds $2$, $r^{-1}$, and $2-r^2$ are greater than or equal to $2-r$, Theorem 3 answers Problem 7 of Erdös, Herzog and Piranian affirmatively, for $0<r\le1$."

Source. Ch. Pommerenke, On metric properties of complex polynomials, Michigan Math. J. 8 (1961), no. 2, 97--115; Theorem 3 and its remarks on printed p. 99 (PDF p. 3 of the publisher's scan), Lemmas 1 and 2 on pp. 98--99 (PDF pp. 2--3), the proof on p. 100 (PDF p. 4), read on the page images (the scan has no text layer). The copy read is identified in the source digest.

Read depth. Claims checked: the statement, the three remarks and Lemmas 1 and 2 were read clause by clause on the page images; the proofs of the two lemmas (a few lines each) were followed. The proof of the theorem (p. 100, four numbered steps) was read for structure and not checked. Nothing here is independently reviewed.

Proof pointer

Page 100, in four steps. (1) Unless $f\equiv z^n$, $E_0$ contains a point with $|z|>1$: with a zero of modulus below 1 in $E_0$ one has $|f(0)|<1$, and the polynomial $g(z)=\prod(1-\bar z_\nu z)$ satisfies $|g|>|f|$ in $|z|<1$ and $|g|=|f|$ on $|z|=1$; if $E_0\subset\{|z|\le1\}$, the minimum principle applied to $g$, which has no zeros in $E_0$ and satisfies $|g|\ge1$ on its boundary, contradicts $g(0)=1$ at an interior point. (2) For $r\le1/2$, $|f|\le1$ on $|z|\le1/2$, so Lemma 2 makes $E$ connected, $E_0=E$ has capacity 1, and a continuum of capacity 1 has diameter at least 2. (3) For $1/2<r\le(\sqrt5-1)/2$, Lemma 1 puts the disk $|z-z_0|\le(1-r^2+|z_0|^2)^{1/2}$ in $E_0$; if $|z_0|\le(1-2r^2)/(2r)$ this disk contains $|z|\le r$, so $E$ is connected by Lemma 2 and $d_0\ge2>1/r$, and otherwise the disk has radius at least $1/(2r)$ and $E_0$, which is not that disk, has $d_0>1/r$. (4) For $(\sqrt5-1)/2\le r\le1$, the point of step (1) and the disk of Lemma 1 give $d_0>1+(1-r^2+|z_0|^2)^{1/2}-|z_0|\ge2-r^2$ when $|z_0|\le r^2/2$, and otherwise the disk has radius at least $1-r^2/2$ and $E_0$ properly contains it.

Dependencies

Within the paper: Lemma 1 (pp. 98--99; the disk $|z-z_0|\le(1-\sigma^2+|z_0|^2)^{1/2}$ about the centroid lies in $E$ when $\sigma^2-|z_0|^2\le1$, by the arithmetic-geometric mean inequality) and Lemma 2 (p. 99; a continuum in $E$ containing all the zeros makes $E$ connected, by the maximum principle). Outside it: the diameter of a continuum of capacity 1 is at least 2, and the 1958 paper's Theorem 7 polynomial for the sharpness remark (erdos_1958_metric_properties_polynomials).

Bears on

• Problem 1048: the affirmative answer for $0<r\le1$ to the question whether some component has diameter above $2-r$, complementing the negative example of p. 98 for $1<r<2$. A filing observation, not a review verdict: the theorem concerns the closed set $E$, and the problem is posed for the open set $\{|f|<1\}$ with a strict inequality; for $0<r\le1/2$ the bound carries over, since no critical point of $f$ lies on $|f|=1$, so the open set is connected and has the diameter of $E$ (the argument is on the claim page); for $1/2<r\le1$ the printed bounds, though strict and at least $2-r$, concern the closed component $E_0$, which may join several components of the open set at critical points on $|f|=1$, so the theorem does not decide the problem's question there.