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OPEN This is open, and cannot be resolved with a finite computation. - $250
If $p(z)\in\mathbb{C}[z]$ is a monic polynomial of degree $n$ then is the length of the curve $\{ z\in \mathbb{C} : \lvert p(z)\rvert=1\}$ maximised when $p(z)=z^n-1$?
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A problem of Erdős, Herzog, and Piranian [EHP58]. It is also listed as Problem 4.10 in [Ha74], where it is attributed to Erdős.

Let the maximal length of such a curve be denoted by $f(n)$.

  • The length of the curve when $p(z)=z^n-1$ is $2n+O(1)$, and hence the conjecture implies in particular that $f(n)=2n+O(1)$.

  • Dolzhenko [Do61] proved $f(n) \leq 4\pi n$, but few were aware of this work.

  • Pommerenke [Po61] proved $f(n)\ll n^2$.

  • Borwein [Bo95] proved $f(n)\ll n$ (Borwein was unaware of Dolzhenko's earlier work). The prize of \$250 is reported by Borwein [Bo95].

  • Eremenko and Hayman [ErHa99] proved the full conjecture when $n=2$, and $f(n)\leq 9.173n$ for all $n$.

  • Danchenko [Da07] proved $f(n)\leq 2\pi n$.

  • Fryntov and Nazarov [FrNa09] proved that $z^n-1$ is a local maximiser, and solved this problem asymptotically, proving that\[f(n)\leq 2n+O(n^{7/8}).\]

  • Tao [Ta25] has proved that $p(z)=z^n-1$ is the unique (up to rotation and translation) maximiser for all sufficiently large $n$.




Erdős, Herzog, and Piranian [EHP58] also ask whether the length is at least $2\pi$ if $\{ z: \lvert f(z)\rvert<1\}$ is connected (which $z^n$ shows is the best possible). This was proved by Pommerenke [Po59].

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This page was last edited 29 December 2025.

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Additional thanks to: Geoffrey Irving

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T. F. Bloom, Erdős Problem #114, https://www.erdosproblems.com/114, accessed 2026-01-16