{"title":"On Zero Free Regions for Derivatives of a Polynomial","authors":"Mohammad Hedayetullah Mir, I. Nazir, I. A. Wani","doi":"10.46793/kgjmat2303.403m","DOIUrl":null,"url":null,"abstract":". Let P n denote the set of polynomials of the form p ( z ) = ( z − a ) m n − m Y k =1 ( z − z k ) , with | a | ≤ 1 and | z k | ≥ 1 for 1 ≤ k ≤ n − m. For the polynomials of the form p ( z ) = z Q n − 1 k =1 ( z − z k ) , with | z k | ≥ 1, where 1 ≤ k ≤ n − 1, Brown [2] stated the problem “Find the best constant C n such that p 0 ( z ) does not vanish in | z | < C n ”. He also conjectured in the same paper that C n = 1 n . This problem was solved by Aziz and Zarger [1]. In this paper, we obtain the results which generalizes the results of Aziz and Zarger.","PeriodicalId":44902,"journal":{"name":"Kragujevac Journal of Mathematics","volume":"1 1","pages":""},"PeriodicalIF":1.0000,"publicationDate":"2023-01-01","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"","citationCount":"1","resultStr":null,"platform":"Semanticscholar","paperid":null,"PeriodicalName":"Kragujevac Journal of Mathematics","FirstCategoryId":"1085","ListUrlMain":"https://doi.org/10.46793/kgjmat2303.403m","RegionNum":0,"RegionCategory":null,"ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":null,"EPubDate":"","PubModel":"","JCR":"Q1","JCRName":"MATHEMATICS","Score":null,"Total":0}
引用次数: 1
Abstract
. Let P n denote the set of polynomials of the form p ( z ) = ( z − a ) m n − m Y k =1 ( z − z k ) , with | a | ≤ 1 and | z k | ≥ 1 for 1 ≤ k ≤ n − m. For the polynomials of the form p ( z ) = z Q n − 1 k =1 ( z − z k ) , with | z k | ≥ 1, where 1 ≤ k ≤ n − 1, Brown [2] stated the problem “Find the best constant C n such that p 0 ( z ) does not vanish in | z | < C n ”. He also conjectured in the same paper that C n = 1 n . This problem was solved by Aziz and Zarger [1]. In this paper, we obtain the results which generalizes the results of Aziz and Zarger.
. set of polynomials》让P n denote表格P (z) = z(−a) k = 1 Y m n−−z z (k)里,用| a | k≤1和z | |≥1为≤k≤n−m . polynomials》为P (z) = z表格Q n−1 k = k(−z z z)里,用| k |≥1,哪里≤k≤n−1,布朗[2]stated《百康斯坦”问题找到C n P 0 (z)确实如此,那不是其为消失在| z | < C n”。他还把它和C = 1的纸放在同一个纸上。这个问题已经解决了,阿齐兹和扎格[1]。在这篇文章中,我们向阿齐兹和扎格尔的继任者报告了结果。