---

Source. Theorem 2, p. 349, proof pp. 349--351, of P. Erdős, F. Herzog and G. Piranian, Polynomials whose zeros lie on the unit circle, Duke Math. J. 22 (1955), 347--351, DOI 10.1215/S0012-7094-55-02237-7, the edition named on the source card.

Statement

Theorem 2 (p. 349, quoted). "Let $P(z)=\prod_{r=1}^{n}(z-z_r)$, with $\lvert z_r\rvert=1$. If $n\le4$, there exist two values $\theta'$ and $\theta''$ such that $\lvert P(re^{i\theta'})\rvert\le\lvert1-r^n\rvert$ and $\lvert P(re^{i\theta''})\rvert\ge1+r^n$ for $0\le r<\infty$."

The print uses $r$ both as the product index and as the modulus. In words: for every monic polynomial of degree $n\le4$ whose zeros all lie on the unit circle there are two half-lines from the origin, of directions $\theta'$ and $\theta''$, such that along the whole of the first $\lvert P\rvert\le\lvert1-r^n\rvert$ and along the whole of the second $\lvert P\rvert\ge1+r^n$, where $r$ is the distance from the origin. The bounds are those attained by $z^n-1$ and $z^n+1$.

Remark (p. 351). For $n=4$ the inequality $\lvert P\rvert\ge1$ need not hold everywhere on the bisector of the greatest of the four angles between consecutive zeros: the paper's example has zeros $e^{i\pi/3}$, $-1$ and $e^{-i\pi/3}$ (double), with $\lvert P(1/2)\rvert=(9/16)3^{1/2}<1$, and by continuity the same holds for some configuration whose four angles are distinct and positive.

Read depth. Claims checked: the statement, the case split and the closing remark were read clause by clause on the page images of pp. 349--351. The inequalities of the proof were followed but not rechecked in detail. Nothing here is independently reviewed.

Proof pointer

Pages 349--351, written here in outline. The cases $n=1,2$ are called trivial and omitted. For $n=3$ and $n=4$ the zeros are described by the angles $\alpha,\beta,\gamma$ (and $\delta$) that consecutive radii to them form at the origin, labelled as convenient. For $\theta'$ the paper rotates one zero to $1$ and takes the positive real axis: for $n=3$ it shows $(1-r^3)^2-\lvert P(r)\rvert^2\ge0$ by trigonometric estimates on the angles, and for $n=4$ one factor is $\lvert1-r\rvert$, one is at most $1+r$ and the remaining pair has product at most $1+r^2$. For $\theta''$ it takes the bisector of an angle $\alpha$ chosen by the ordering (the largest angle for $n=3$), or for $n=4$ with $\alpha<\pi/2$ the direction perpendicular to it, and bounds the product of distances below by $1+r^n$, comparing distances to the zeros with distances to their negatives.

Dependencies

None.

Bears on

The paper links this theorem to no Erdős problem in the corpus; its open question on the greatest admissible degree is on its own page.