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Univalent polynomials and Koebe’s one-quarter theorem

Univalent polynomials and Koebe’s one-quarter theorem The famous Koebe $$\frac{1}{4}$$ 1 4 theorem deals with univalent (i.e., injective) analytic functions f on the unit disk $${\mathbb {D}}$$ D . It states that if f is normalized so that $$f(0)=0$$ f ( 0 ) = 0 and $$f'(0)=1$$ f ′ ( 0 ) = 1 , then the image $$f({\mathbb {D}})$$ f ( D ) contains the disk of radius $$\frac{1}{4}$$ 1 4 about the origin, the value $$\frac{1}{4}$$ 1 4 being best possible. Now suppose f is only allowed to range over the univalent polynomials of some fixed degree. What is the optimal radius in the Koebe-type theorem that arises? And for which polynomials is it attained? A plausible conjecture is stated, and the case of small degrees is settled. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Analysis and Mathematical Physics Springer Journals

Univalent polynomials and Koebe’s one-quarter theorem

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Publisher
Springer Journals
Copyright
Copyright © 2019 by Springer Nature Switzerland AG
Subject
Mathematics; Analysis; Mathematical Methods in Physics
ISSN
1664-2368
eISSN
1664-235X
DOI
10.1007/s13324-019-00305-x
Publisher site
See Article on Publisher Site

Abstract

The famous Koebe $$\frac{1}{4}$$ 1 4 theorem deals with univalent (i.e., injective) analytic functions f on the unit disk $${\mathbb {D}}$$ D . It states that if f is normalized so that $$f(0)=0$$ f ( 0 ) = 0 and $$f'(0)=1$$ f ′ ( 0 ) = 1 , then the image $$f({\mathbb {D}})$$ f ( D ) contains the disk of radius $$\frac{1}{4}$$ 1 4 about the origin, the value $$\frac{1}{4}$$ 1 4 being best possible. Now suppose f is only allowed to range over the univalent polynomials of some fixed degree. What is the optimal radius in the Koebe-type theorem that arises? And for which polynomials is it attained? A plausible conjecture is stated, and the case of small degrees is settled.

Journal

Analysis and Mathematical PhysicsSpringer Journals

Published: Apr 25, 2019

References