A problem of Erdős, Herzog, and Piranian [EHP58]. It is also listed as Problem 4.10 in [Ha74], where it is attributed to Erdős.

In [Va99] it is just aked whether the length is at most $2n+O(1)$. This is true, as a consequence of result of Tao [Ta25] below.

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 23 January 2026. View history