This is Problem 4.9 in [Ha74], where it is attributed to Erdős.

A problem of Erdős, Herzog, and Piranien [EHP58], who ask more generally whether\[\sum_C \mathrm{diam}(C) \leq n2^{1/n}\]and\[\sum_{C}\max(0, \mathrm{diam}(C)-1)\ll 1\]for all such $f$, where $C$ ranges over the connected components of the set in question. The example $f(z)=z^n-1$ has $\sum_C \mathrm{diam}(C)=(1+o(1))n2^{1/n}$.

They also asked whether, if the roots of $f$ are all in the disc $\{\lvert z\rvert\leq 1\}$, the total number of connected components with diameter $>1$ is absolutely bounded, but noted in an addendum that the answer is no: consider $z^n+1$ and move the zeros $e^{i\pi/n}$ and $e^{-i\pi/n}$ a short distance along the circle towards $1$. The set $\{z: \lvert f(z)\rvert \leq 1\}$ for $f(z)=z^n+1$ looks like $n$ 'leaves' joined at $0$, and this moving of two roots makes $\approx n/2$ of the leaves become disconnected at $0$. Each of the resulting components has diameter $>2^{1/n}-\epsilon$, and hence there are $\gg n$ components of diameter $>1$.

In [Er61] Erdős asks the weaker question given here, with the definition of the set altered so that $<1$ is replaced by $\leq 1$.

Pommerenke [Po61] proved that the answer is no to most of these questions, by showing that, for any $0<d<4$ and $k\geq 1$ there exist monic polynomials $f\in \mathbb{C}[x]$ such that $\{z: \lvert f(z)\rvert\leq 1\}$ has at least $k$ connected components of diameter $\geq d$.

This was independently proved by Huang [Hu25] (unaware of the previous work of Pommerenke). Pólya [Po28] showed that $4$ is the best possible here, in that no connected component can have diameter $>4$.

A picture of the set in question for $z^5-1$ is here.

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This page was last edited 29 December 2025.