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# On the Zeros of Analytic Functions inside the Unit Disk

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### On the Zeros of Analytic Functions inside the Unit Disk

1. 1. International Journal of Engineering Inventionse-ISSN: 2278-7461, p-ISSN: 2319-6491Volume 2, Issue 5 (March 2013) PP: 54-60www.ijeijournal.com Page | 54On the Zeros of Analytic Functions inside the Unit DiskM. H. GulzarDepartment of Mathematics, University of Kashmir, Srinagar 190006Abstract: In this paper we find an upper bound for the number of zeros of an analytic function inside the unitdisk by restricting the coefficients of the function to certain conditions.AMS Mathematic Subject Classification 2010: 30C 10, 30C15Keywords and Phrases: Bound, Analytic Function, ZerosI. Introduction and Statement of ResultsA well-known result due to Enestrom and Kakeya [5] states that a polynomial njjj zazP0)( of degree nwith0...... 011   aaaa nnhas all its zeros in .1zQ. G. Mohammad [6] initiated the problem of finding an upper bound for the number of zeros of P(z) satisfyingthe above conditions in21z . Many generalizations and refinements were later given by researchers on thebounds for the number of zeros of P(z) in 10,  z (for reference see [1]),[2], [4]etc.).In this paper we consider the same problem for analytic functions and prove the following results:Theorem 1: Let 0)(0 jjj zazf be analytic for 1z , where njia jjj ,......,1,0,   . If forsome 0 ,......210   ,then the number of zeros of f(z) in 10,0 zMa, does not exceed0000 22log1log1ajj ,where100 22jjM Taking 0 , the following result immediately follows from Theorem 1:Corollary 1: Let 0)(0 jjj zazf be analytic for 1z , where njia jjj ,......,1,0,   . If......210   ,then the number of zeros of f(z) in 10,0 zMa, does not exceed
2. 2. On the Zeros of Analytic Functions inside the Unit Diskwww.ijeijournal.com Page | 550000 2log1log1ajj ,where100 2jjM  .If the coefficients ja are real i.e. njj ,....,1,0,0  , we get the following result from Theorem 1:Corollary 2 : Let 0)(0 jjj zazf be analytic for 1z , where......210  aaa ,then the number of zeros of f(z) in 10,0 zMa, does not exceed0002log1log1aaa ,where02 aM   .Taking 0)1(   k , 1k , we get the following result from Theorem 1:Corollary 3 : Let 0)(0 jjj zazf be analytic for 1z , wherenjia jjj ,......,1,0,   . If for some 1k ,......210  k ,then the number of zeros of f(z) in 10,0 zMa, does not exceed0000 2)12(log1log1akjj ,where100 2)12(jjkM Applying Theorem 1 to the function –if(z), we get the following result:Theorem 2 : Let 0)(0 jjj zazf be analytic for 1z , where njia jjj ,......,1,0,   . If forsome 0 ,......210   ,then the number of zeros of f(z) in 10,0 zMa, does not exceed
3. 3. On the Zeros of Analytic Functions inside the Unit Diskwww.ijeijournal.com Page | 560000 22log1log1ajj ,where100 22jjM  .Theorem 3 : Let 0)(0 jjj zazf be analytic for 1z . If for some 0 ,......210  aaa ,and for real some  ,nja j ,.......,1,0,2arg  ,then the number of zeros of f(z) in 10,0 zMa, does not exceed010 sin)1sin)(cos(log1log1aaajj ,where10 sin)sin)(cos(jjaM  .Taking 0 , Theorem 3 reduces to the following result:Corollary 4:Let 0)(0 jjj zazf be analytic for 1z . If,......210  aaa ,and for some real  ,nja j ,.......,1,0,2arg  ,then the number of zeros of f(z) in 10,0 zMa, does not exceed010 sin)1sin(coslog1log1aaajj ,where10 sin)sin(cosjjaaM  .Taking 0)1( ak  , 1k , we get the following result from Theorem 3:
4. 4. On the Zeros of Analytic Functions inside the Unit Diskwww.ijeijournal.com Page | 57Corollary 5:Let 0)(0 jjj zazf be analytic for 1z . If,......210  aaak ,and for some real  ,nja j ,.......,1,0,2arg  ,then the number of zeros of f(z) in 10,0 zMadoes not exceed010 sin)1sin(coslog1log1aaakjj ,where100 sin)sin(cos2jjaakM  .2. LemmasFor the proofs of the above results we need the following results:Lemma 1 : Let f(z) be analytic for 1z , 0)0( f and Mzf )( for 1z ,Then the number of zeros of f(z) in 10,  z , does not exceed)0(log1log1fM(see [7] , page 171 ).Lemma 2 : If for some t>0, 1 jj ata and ,...2,1,0,2arg  ja j , for some real  , then sin)(cos)( 111   jjjjjj ataataata .The proof of Lemma 2 follows from a lemma of Govil and Rahman [3].3. Proofs of Theorems:Proof of Theorem 1: Consider the function)()1()( zfzzF ......))(1( 2210  zazaaz......)()( 221100  zaazaaa......)()( 221100  zzz ......})(){( 21100  zzii  .For 1z ,03221100 ......)(  zF......2110  000 22jj .
5. 5. On the Zeros of Analytic Functions inside the Unit Diskwww.ijeijournal.com Page | 58Since F(z) is analytic for 1z , 0)0( 0  aF , it follows , by using Lemma 1, that the number of zeros ofF(z) in 10,  z , does not exceed0000 22log1log1ajj .On the other hand, consider)()1()( zfzzF ......))(1( 2210  zazaaz......)()( 221100  zaazaaa)(0 zqa  ,where......)()()( 22110  zaazaazq......)()( 22110  zzz ......})(){( 22110  zzi For 1z ,............)( 21102110  zqMjj  100 22  .Since q(z ) is analytic for 1z , q(0)=0, it follows , by Schwarz’s lemma , thatzMzq )( for 1z .Hence for 1z ,)()( 0 zqazF )(0 zqa zMa  00ifMaz0 .This shows that all the zeros of F(z) lie inMaz0 . Since the zeros of f(z) are also the zeros of F(z), itfollows that all the zeros of f(z) lie inMaz0 . Thus, the number of zeros of f(z) in10,0 zMa, does not exceed
6. 6. On the Zeros of Analytic Functions inside the Unit Diskwww.ijeijournal.com Page | 590000 22log1log1ajj .Proof of Theorem 2: Consider the function)()1()( zfzzF ......))(1( 2210  zazaaz......)()( 221100  zaazaaa......)()( 221100  zaazaaza For 1z , we have, by using the hypothesis and Lemma 2, sin)(cos)[()( 10100 aaaaazF ......sin)(cos)( 2121   aaaa10 sin)1sin)(cos(jjaa  .Since F(z) is analytic for 1z , 0)0( 0  aF , it follows , by using the lemma 1, that the number of zerosof F(z) in 10,  z , does not exceed010 sin)1sin)(cos(log1log1aaajj .Again , Consider the function)()1()( zfzzF ......))(1( 2210  zazaaz......)()( 221100  zaazaaa......)()( 221100  zaazaaza )(0 zqa  ,where......)()()( 22110  zaazaazq......)()( 22110  zaazaaz For 1z , by using lemma 2, we have, sin)(cos)()( 1010 aazq ......sin)(cos)( 2121   aaaaMaajj  10 sin)sin)(cos(  .Since q(z ) is analytic for 1z , q(0)=0, it follows , by Schwarz’s lemma , thatzMzq )( for 1z .Hence for 1z ,)()( 0 zqazF 
7. 7. On the Zeros of Analytic Functions inside the Unit Diskwww.ijeijournal.com Page | 60)(0 zqa zMa  00ifMaz0 .This shows that all the zeros of F(z) lie inMaz0 . Since the zeros of f(z) are also the zeros of F(z), itfollows that all the zeros of f(z) lie inMaz0 . Thus, the number of zeros of f(z) in10,0 zMa, does not exceed010 sin)1sin)(cos(log1log1aaajj .References[1] K. K. Dewan, Extremal Properties and Coefficient estimates for polynomials with restricted zeros and on location of zeros ofpolynomials, Ph.D Thesis,IIT Delhi, !980.[2] K. K.Dewan, Theory of Polynomials and Applications,Deep & Deep Publications Pvt. Ltd., New Delhi, 2007.[3] N. K. Govil and Q. I. Rahman, On the Enestrom-Kakeya Theorem, Tohoku Math. J .20(1968), 126-136.[4] M. H. Gulzar, On the number of zeros of a polynomial in a prescribed region, Research Journal of Pure Algebra-2(2), Feb. 2012, 35-46.[5] M. Marden, Geometry of Polynomials, IInd. Ed. Math. Surveys, No.3, Amer. Math. Soc. Providence, R.I. 1966.[6] Q. G. Mohammad, On the Zeros of Polynomials, Amer. Math. Monthly, 72 (1965), 631-633.[7] E. C Titchmarsh, The Theory of Functions, 2ndedition, Oxford University Press, London, 1939.