---

Statement

Lengths here are spherical, on the Riemann sphere.

Theorem 2 (p. 2, quoted). "Let $f$ be a rational function of degree $d$. Then the spherical length of the preimage under $f$ of any circle $C$ is at most $d$ times the length of a great circle."

The paper notes that the bound is best possible, as $f(z)=z^d$ with $C$ the real line shows (p. 2), and calls the rational problem much easier than the polynomial one (p. 1).

Source. Alexandre Eremenko and Walter Hayman, On the length of lemniscates, Michigan Math. J. 46 (1999), no. 2, 409--415, DOI 10.1307/mmj/1030132418; page numbers are those of the authors' corrected preprint (pp. 1--9) named on the source card, not the journal's pagination.

Read depth. Claims checked: the statement and its proof (p. 8) were read on the print. Nothing here is independently reviewed.

Proof pointer

P. 8, following Borwein. With great circles of length $2\pi$, the Poincaré integral-geometric formula gives the spherical length of a curve as one quarter of the integral, over the sphere, of the number of its intersections with the great circle centred at each point. By Lemma 1 the preimage meets every great circle, apart from finitely many, at most $2d$ times, so its spherical length is at most $2\pi d$.

Dependencies

• Lemma 1.
• The Poincaré integral-geometric formula on the sphere (Santaló).

Bears on

• #114: background only. It is the rational, spherical analogue of the problem's question, solved completely; it does not bound the Euclidean length of a polynomial lemniscate or bear on whether $z^n-1$ is the maximizer.