Problem Statement
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 $p(z)=z^n-1$?
Categories:
Polynomials Analysis
Progress
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].
Source: erdosproblems.com/114 | Last verified: January 13, 2026