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Statement

$f(z)=\prod_{\nu=1}^n(z-z_\nu)$ and $E=\{|f(z)|\le1\}$ (p. 97). Quoted (p. 98): "Let the zeros $z_\nu$ of $f(z)$ belong to the disk $|z|\le r$. Erdös, Herzog and Piranian [2, Problem 7] raised the question whether there is always a component of $E$ with diameter at least $2-r$ ($r<2$). The answer is negative for $r>1$. To show this, let $f(z)=z^n-r^n$ ($r>1$). Then the set $E=\{|z^n-r^n|\le1\}$ has $n$ components."

The passage concludes, after the estimate recorded below: "Hence the (common) diameter of the components of $E$ tends to $0$ as $n\to\infty$." For any fixed $r$ with $1<r<2$ and $n$ large, no component of $E$ has diameter at least $2-r>0$. The case $0<r\le1$ is answered affirmatively by Theorem 3 (p. 99).

Source. Ch. Pommerenke, On metric properties of complex polynomials, Michigan Math. J. 8 (1961), no. 2, 97--115; the unnumbered passage on printed p. 98 (PDF p. 2 of the publisher's scan), read on the page image (the scan has no text layer). The copy read is identified in the source digest.

Read depth. Claims checked: the passage was read clause by clause on the page image on 2026-09-22, and its estimate (three lines) was read in full and followed. Nothing here is independently reviewed.

Proof pointer

Page 98. The zeros of $z^n-r^n$ are the $n$ points $r\zeta$, $\zeta^n=1$, and each component of $E$ contains a zero. For $z$ in the component containing the zero $r$, write $z^n-r^n=\omega$ with $|\omega|\le1$; then, as printed,

$$|z-r|=\bigl|(r^n+\omega)^{1/n}-r\bigr| =r\,\bigl|1+\omega n^{-1}r^{-n}+O(n^{-2}r^{-n})+\cdots-1\bigr| \le n^{-1}r^{-n+1}\bigl(1+O(n^{-1})\bigr)\to0 .$$

By symmetry the same holds at every zero, so the $n$ components are distinct for large $n$ and their common diameter tends to $0$. The paper does not spell out why $E$ has exactly $n$ components for every $n$; the estimate shows the components are eventually disjoint.

A filing observation, not a review verdict: Problem 1048 is posed for the open set $\{|f|<1\}$ and asks for a component of diameter strictly above $2-r$; every component of the open set lies in a component of $E$, so the example refutes that form too.

Dependencies

None beyond the binomial expansion of $(r^n+\omega)^{1/n}$.

Bears on

• Problem 1048: the negative answer for $1<r<2$, the range the site's DISPROVED (LEAN) label rests on; Theorem 3 answers the range $0<r\le1$ affirmatively for the closed set $E$; for the problem's open set and strict inequality this carries over for $0<r\le1/2$ (the argument is on the claim page), the theorem does not decide $1/2<r\le1$, and the degenerate case $r=0$ fails ($z^n$ gives the open unit disc, of diameter exactly $2$).