International Journal of Modern Engineering Research (IJMER)                www.ijmer.com         Vol.2, Issue.6, Nov-Dec....
International Journal of Modern Engineering Research (IJMER)                  www.ijmer.com         Vol.2, Issue.6, Nov-De...
International Journal of Modern Engineering Research (IJMER)                      www.ijmer.com         Vol.2, Issue.6, No...
International Journal of Modern Engineering Research (IJMER)                www.ijmer.com         Vol.2, Issue.6, Nov-Dec....
International Journal of Modern Engineering Research (IJMER)                 www.ijmer.com         Vol.2, Issue.6, Nov-Dec...
International Journal of Modern Engineering Research (IJMER)             www.ijmer.com         Vol.2, Issue.6, Nov-Dec. 20...
International Journal of Modern Engineering Research (IJMER)                   www.ijmer.com         Vol.2, Issue.6, Nov-D...
International Journal of Modern Engineering Research (IJMER)               www.ijmer.com         Vol.2, Issue.6, Nov-Dec. ...
International Journal of Modern Engineering Research (IJMER)                 www.ijmer.com         Vol.2, Issue.6, Nov-Dec...
International Journal of Modern Engineering Research (IJMER)                  www.ijmer.com         Vol.2, Issue.6, Nov-De...
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On The Zeros of Certain Class of Polynomials

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  1. 1. International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4363-4372 ISSN: 2249-6645 On The Zeros of Certain Class of Polynomials B.A. Zargar Post-Graduate Department of Mathematics, U.K, Srinagar.Abstract: Let P(z) be a polynomial of degree n with real or complex coefficients . The aimof this paper is to obtain a ring shaped region containing all the zeros of P(z). Our results not only generalize some knownresults but also a variety of interesting results can be deduced from them. I. Introduction and Statement of ResultsThe following beautiful result which is well-known in the theory of distribution of zeros of polynomials is due to Enestromand Kakeya [9]. Theorem A. If P(z) =βˆ‘ 𝑗𝑛=0 aj zj is a polynomial of degree n such that(1) an β‰₯ an-1 β‰₯…..β‰₯ a1 β‰₯ a0 β‰₯ 0,then all the zeros of P(z) lie in 𝑧 ≀ 1. In the literature ([2] ,[5] –[6], [8]-[11]) there exists some extensions and generalizations of this famous result. Azizand Mohammad [1] provided the following generalization of Theorem A. Theorem B. Let P(z) =βˆ‘ 𝑗𝑛=0 aj zj is a polynomial of degree n with real positive coefficients. If t1 β‰₯ 𝑑2 β‰₯ 0 canbe found such that(2) π‘Ž 𝐽 t1 t 2+ ajβˆ’1 t1 βˆ’ t 2 a jβˆ’2 β‰₯ 0, j=1, 2, … ;n+1, π‘Žβˆ’1 = π‘Ž 𝑛 +1 = 0 ,then all the zeros of P(z) lie in 𝑧 ≀ 𝑑1 .For 𝑑1 = 1, 𝑑2 = 0, this reduces to Enestrom-Kakeya Theorem (Theorem A). Recently Aziz and Shah [3] have proved the following more general result which includes Theorem A as a specialcase. ------------------------------------------------------------------------------Mathematics, subject, classification (2002). 30C10,30C15, Keywords and Phrases , Enestrom-Kakeya Theorem, Zeros ,bounds. Theorem C. Let P(z) = βˆ‘ 𝑗𝑛=0 π‘Žπ‘— 𝑧 𝑗 be a polynomial of degree n. If for some t > 0.(3) π‘€π‘Žπ‘₯ 𝑧 =𝑅 π‘‘π‘Ž0 𝑧 𝑛 + π‘‘π‘Ž1 βˆ’ π‘Ž0 𝑧 𝑛 βˆ’1 + β‹― + π‘‘π‘Ž 𝑛 βˆ’ π‘Ž π‘›βˆ’1 ≀ 𝑀3Where R is any positive real number, then all the zeros of P(z) lie in 𝑀 1(4) 𝑧 ≀ π‘€π‘Žπ‘₯ π‘Ž 3 , 𝑅 𝑛 The aim of this paper is to apply Schwarz Lemma to prove a more general result which includes Theorems A, Band C as special cases and yields a number of other interesting results for various choices of parameters 𝛼, π‘Ÿ, 𝑑1 π‘Žπ‘›π‘‘ 𝑑2 . Infact we start by proving the following result. Theorem 1. Let P(z) = βˆ‘ 𝑗𝑛=0 π‘Žπ‘— 𝑧 𝑗 be a polynomial of degree n, If for some real numbers 𝑑1 , 𝑑2 with 𝒕 𝟏 β‰  𝟎, 𝒕 𝟏 > π’•πŸ β‰₯ 𝟎(5) π‘€π‘Žπ‘₯ 𝑧 =𝑅 π‘Ž 𝑛 𝑑1 βˆ’ 𝑑2 + 𝛼 βˆ’ π‘Ž π‘›βˆ’1 + βˆ‘ 𝑗𝑛=0 π‘Žπ‘— 𝑑1 𝑑2 + π‘Žπ‘— βˆ’1 𝑑1 βˆ’ 𝑑2 βˆ’ π‘Žπ‘— βˆ’2 𝑧 π‘›βˆ’π‘— +1 ≀ 𝑀1(6) 𝑀𝑖𝑛 𝑧 =𝑅 π‘Ž0 𝑑1 βˆ’ 𝑑2 + π‘Ž1 𝑑1 𝑑2 + 𝛽 + βˆ‘ 𝑗𝑛+2 π‘Žπ‘— 𝑑1 𝑑2 + π‘Žπ‘— βˆ’1 𝑑1 βˆ’ 𝑑2 βˆ’ =2 (π‘Ž 𝑗 βˆ’2 𝑧 𝑗 ≀ 𝑀2 Where R is a positive real number, then all the zeros of P(z) lie in the ring shaped region. 𝑑1 𝑑2 π‘Ž0 𝑀1 + 𝛼 1(7) 𝑀𝑖𝑛 𝑧 =R 𝛽 +𝑀2 , 𝑅 ≀ 𝑧 ≀ π‘€π‘Žπ‘₯ 𝑧 =R π‘Žπ‘› , 𝑅Taking t2 =0, we get the following generalization and refinement of Theorem C. Corollary 1. Let P(z) = βˆ‘ 𝑗𝑛=0 π‘Žπ‘— 𝑧 𝑗 be a polynomial of degree n. If for some t > 0. π‘€π‘Žπ‘₯ 𝑧 =𝑅 a n t + Ξ± βˆ’ a nβˆ’1 + βˆ‘n j=0 a jβˆ’1 t βˆ’ a jβˆ’2 z nβˆ’j+2 ≀ M1 n+2 j 𝑀𝑖𝑛 𝑧 =𝑅 a 0 t + Ξ² + βˆ‘j=0 a jβˆ’1 t βˆ’ a jβˆ’2 z ≀ M2Where R is a real positive number then all the zeros of P(z) lie in www.ijmer.com 4363 | Page
  2. 2. International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4363-4372 ISSN: 2249-6645 𝑀1 + 𝛼 1 z ≀ π‘€π‘Žπ‘₯ 𝑧 =𝑅 π‘Žπ‘› , 𝑅 In case we take t=1=R in corollary 1, we get the following interesting result.Corollary 2. Let P(z) = βˆ‘ 𝑗𝑛=0 π‘Žπ‘— 𝑧 𝑗 be a polynomial of degree n, π‘€π‘Žπ‘₯ 𝑧 =𝑅 π‘Ž 𝑛 + 𝛼 βˆ’ π‘Ž π‘›βˆ’1 + π‘Ž 𝑛 βˆ’1 βˆ’ π‘Ž 𝑛 βˆ’2 𝑧 + β‹― + π‘Ž1 βˆ’ π‘Ž0 𝑧 π‘›βˆ’1 + π‘Žπ‘œ 𝑧 𝑛 ≀ M, M+ Ξ±then all the zeros of P(z) lie in the circle z ≀ Max an ,1If for some Ξ± > 0, 𝛼 + π‘Ž 𝑛 β‰₯ π‘Ž 𝑛 βˆ’1 β‰₯ β‹― β‰₯ π‘Ž1 β‰₯ π‘Ž0 β‰₯ 0, then , 𝑀 ≀ π‘Ž 𝑛 + 𝛼 + π‘Ž 𝑛 βˆ’1 + π‘Ž 𝑛 βˆ’1 βˆ’ π‘Ž π‘›βˆ’2 + β‹― + π‘Ž1 βˆ’ π‘Ž0 + π‘Ž0 = π‘Ž 𝑛 + 𝛼 βˆ’ π‘Ž 𝑛 βˆ’1 + π‘Ž π‘›βˆ’1 βˆ’ π‘Ž π‘›βˆ’2 + β‹― + π‘Ž1 βˆ’ π‘Ž0 + π‘Ž0 = π‘Žπ‘› + 𝛼Using this observation in Corollary 2, we get the following generalization of Enestrom-Kakeya Theorem. Corollary 3. If, P(z)= π‘Ž 𝑛 𝑧 𝑛 + π‘Ž 𝑛 βˆ’1 + 𝑧 𝑛 βˆ’1 + β‹― + π‘Ž1 𝑧 + π‘Ž0 , is a polynomial of degree n, such that for some𝛼 β‰₯ 0, 𝛼 + π‘Ž 𝑛 β‰₯ π‘Žβˆ’1 β‰₯ β‹― . β‰₯ π‘Ž0 β‰₯ 0,then all the zeros of P(z) lie in the circle 2𝛼 𝑧 ≀ 1+ , π‘Žπ‘›For 𝛼 = 0, π‘‘β„Žπ‘–π‘  π‘Ÿπ‘’π‘‘π‘’π‘π‘’π‘  π‘‘π‘œ π‘‡β„Žπ‘’π‘œπ‘Ÿπ‘’π‘š 𝐴. 𝐼𝑓 𝑀𝑒 π‘‘π‘Žπ‘˜π‘’ 𝛼 = π‘Ž π‘›βˆ’1 βˆ’ π‘Ž 𝑛 β‰₯ 0, then we get Corollary 2, of ([4] Aziz andZarger). Next we present the following interesting result which includes Theorem A as a special case. Theorem 2. Let P(z) = βˆ‘ 𝑗𝑛=0 π‘Žπ‘— 𝑧 𝑗 be a polynomial of degree n, If for some real numbers with 𝒕 𝟏 β‰  𝟎, 𝒕 𝟏 > 𝒕 𝟐 β‰₯ 𝟎 𝒏(πŸ–) 𝑴𝒂𝒙 𝒛 =𝑹 βˆ‘ 𝒋=𝟎 𝒂 𝒋 𝒕 𝟏 𝒕 𝟐 + 𝒂 π’‹βˆ’πŸ 𝒕 𝟏 βˆ’ 𝒕 𝟐 βˆ’ 𝒂 π’‹βˆ’πŸ 𝒛 π’βˆ’π’‹ ≀ 𝑴3then all the zeros of P(z) lie in the region(9) 𝑧 ≀ π‘Ÿ1,Where , 2𝑀3(10) π‘Ÿ1 = π‘Ž 𝑛 𝑑 1 βˆ’π‘‘ 2 π‘Ž 𝑛 βˆ’1 2 +4 π‘Ž 𝑛 𝑀3 1/2 βˆ’ π‘Ž 𝑛 𝑑 1 βˆ’π‘‘ 2 βˆ’ π‘Ž 𝑛 βˆ’1Taking 𝑑2 = 0, we get the following result Corollary 4. Let P(z) = βˆ‘ 𝑗𝑛=0 π‘Žπ‘— 𝑧 𝑗 be a polynomial of degree n, If for some 𝑑 > 0, 𝒏 βˆ‘ 𝒋=𝟎 𝒂 π’‹βˆ’πŸ 𝒕 βˆ’ 𝒂 π’‹βˆ’πŸ 𝒛 π’βˆ’π’‹ ≀ 𝑴3, 𝑴𝒂𝒙 𝒛 =𝑹then all the zeros of P(z) lie in the region 𝑧 ≀ π‘Ÿ1, Where , 2𝑀3 r= π‘Ž 𝑛 π‘‘βˆ’ π‘Ž 𝑛 βˆ’1 2 +4 π‘Ž 𝑛 𝑀3 1/2 βˆ’ π‘Ž π‘‘βˆ’π‘Ž 𝑛 𝑛 βˆ’1 Remark. Suppose polynomial P(z) = βˆ‘ 𝑗𝑛=0 π‘Žπ‘— 𝑧 𝑗 , satisfies the conditions of www.ijmer.com 4364 | Page
  3. 3. International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4363-4372 ISSN: 2249-6645 Theorem A, then it can be easily verified that from Corollary 4. M3 =an-1 , then all the zeros of P(z) lie in 2π‘Ž 𝑛 βˆ’1 𝑧 ≀ π‘Ž 𝑛 βˆ’ π‘Ž 𝑛 βˆ’1 2 + 4π‘Ž 𝑛 π‘Ž π‘›βˆ’1 βˆ’ π‘Ž 𝑛 βˆ’ π‘Ž 𝑛 βˆ’1 2π‘Ž = 2π‘Ž 𝑛 βˆ’1 = 1, 𝑛 βˆ’1 which is the conclusion of Enestrom-Kekaya Theorem. Finally , we prove the following generalization of Theorem 1 of ([2], Aziz and Shah). Theorem 3. Let P(z) = βˆ‘ 𝑗𝑛=0 π‘Žπ‘— 𝑧 𝑗 be a polynomial of degree n, If𝛼, 𝛽 are complex numbers and with 𝑑1 β‰  0, 𝑑2 π‘Žπ‘Ÿπ‘’ π‘Ÿπ‘’π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿπ‘  π‘€π‘–π‘‘β„Ž 𝑑1 β‰₯ 𝑑2 β‰₯ 0 and (11) π‘€π‘Žπ‘₯ 𝑧 =𝑅 π‘Ž 𝑛 𝑑1 βˆ’ 𝑑2 + 𝛼 βˆ’ π‘Ž 𝑛 βˆ’1 𝑧 + βˆ‘ 𝑗𝑛=0 π‘Žπ‘— 𝑑1 𝑑2 + π‘Žπ‘— βˆ’1 𝑑1 βˆ’π‘‘2 βˆ’ π‘Žπ‘— βˆ’2 𝑧 𝑛 βˆ’π‘— βˆ’2 ≀ 𝑀4 and, (12) π‘€π‘Žπ‘₯ 𝑧 =𝑅 π‘Ž 𝑛 𝑑1 βˆ’ 𝑑2 + 𝛽 βˆ’ π‘Ž 𝑛 βˆ’1 𝑧 + βˆ‘ 𝑗𝑛+2 π‘Žπ‘— 𝑑1 𝑑2 + π‘Žπ‘— βˆ’1 𝑑1 βˆ’π‘‘2 βˆ’ π‘Žπ‘— βˆ’2 𝑧 𝑗 ≀ 𝑀5 =2 Where π‘Žβˆ’2= π‘Žβˆ’1 = 0 = π‘Ž 𝑛+1 = π‘Ž 𝑛+2 and R is any positive real number, then all the zeros of P(z) lie in the ring shaped region. 1 (13) Min (r2 , R) ≀ 𝑧 ≀ max π‘Ÿ1 , 𝑅 Where 2[ π‘Ž 𝑅 2 π‘Ž 𝑛 𝑑 1βˆ’ 𝑑 2 βˆ’π‘Ž 𝑛 βˆ’1 +𝛼 +𝑀4 ] 2 (14) π‘Ÿ1 = 2 𝛼 𝑅 2 𝑀4 +𝑅 2 𝑀4 βˆ’ π‘Ž 𝑛 π‘Ž 𝑛 𝑑 1βˆ’ 𝑑 2 +π›Όβˆ’π‘Ž 𝑛 βˆ’1 2 +4 𝛼 𝑅 2 π‘Ž 𝑛 𝑑 1βˆ’ 𝑑 2 +π›Όβˆ’π‘Ž 𝑛 βˆ’1 +𝑀4 π‘Ž 𝑛 𝑅 2 𝑀4 1/2 βˆ’ 𝛼 𝑅2 𝑀4 + 𝑅2 𝑀4 βˆ’ π‘Ž 𝑛 π‘Ž 𝑛 𝑑1βˆ’ 𝑑2 + 𝛼 βˆ’ π‘Ž 𝑛 βˆ’1 and 1 2[𝑅 2 𝛽 2 π‘Ž 1 𝑑 1 𝑑 2 +π‘Ž 0 𝑑 1βˆ’ 𝑑 2 +𝛽 βˆ’π‘€5 ] (15) = π‘Ÿ2 𝑅2 π‘Ž 1 𝑑 1 𝑑 2 +π‘Ž 0 𝑑 1βˆ’ 𝑑 2 +𝛽 ( π‘Ž 0 𝑑 1 𝑑 2 βˆ’π‘€5 )βˆ’ 𝛽 𝑀5 + 𝑅4 π‘Ž1 𝑑1 𝑑2 + π‘Ž0 𝑑1βˆ’ 𝑑2 + 𝛽 π‘Ž0 𝑑1 𝑑2 βˆ’ 𝑀5 2 2 +4 π‘Ž0 𝑀5 𝑑1 𝑑2 (𝑅 2 𝛽 π‘Ž1 𝑑1 𝑑2 + π‘Ž0 𝑑1βˆ’ 𝑑2 ) + 𝛽 + 𝑀5 1/2 Taking 𝛼 = 0, 𝛽 = 0, in Theorem 3, we get the following. Corollary 5. Let P(z) = βˆ‘ 𝑗𝑛=0 π‘Žπ‘— 𝑧 𝑗 be a polynomial of degree n, If ,𝑑1 β‰  0 π‘Žπ‘›π‘‘ 𝑑2 are real numbers with 𝑑1 β‰₯ 𝑑2 β‰₯ 0, π‘€π‘Žπ‘₯ 𝑧 =𝑅 βˆ‘ 𝑗𝑛+1 π‘Žπ‘— 𝑑1 𝑑2 + π‘Žπ‘— βˆ’1 𝑑1βˆ’ 𝑑2 βˆ’ π‘Žπ‘— βˆ’2 𝑧 π‘›βˆ’π‘— +2 ≀ 𝑀4 , =1 𝐧+𝟐 βˆ‘ 𝐣=𝟏 𝐚 𝐣 𝐭 𝟏 𝐭 𝟐 + 𝐚 π£βˆ’πŸ 𝐭 πŸβˆ’ 𝐭 𝟐 βˆ’ 𝐚 π£βˆ’πŸ 𝐳 𝐣 ≀ 𝐌 πŸ“ , 𝐌𝐚𝐱 𝐳 =𝐑Where R is any positive real number . then all the zeros of P(z) lie in the ring shaped region 1 min (r2,R) ≀ 𝑧 ≀ max π‘Ÿ1 , 𝑅 Where 2𝑀42 π‘Ÿ1 = 3 𝑅4 𝑑 1βˆ’ 𝑑 2 ) π‘Ž 𝑛 βˆ’π‘Ž 𝑛 βˆ’1 2 𝑀4 βˆ’ π‘Ž 𝑛 2 +4 π‘Ž 𝑛 𝑅 2 𝑀4 1/2 βˆ’ 𝑑 1 βˆ’π‘‘ 2 )π‘Ž 𝑛 βˆ’π‘Ž 𝑛 βˆ’1 𝑀4 βˆ’ π‘Ž 𝑛 𝑅2 π‘Ÿ2 = www.ijmer.com 4365 | Page
  4. 4. International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4363-4372 ISSN: 2249-6645 1 1 3 2 2𝑀5 𝑅4 π‘Ž1 𝑑1 𝑑2 + π‘Ž0 𝑑1 βˆ’ 𝑑2 2 𝑀5 βˆ’ π‘Ž0 𝑑1 𝑑2 2 + 4𝑀5 𝑅 2 π‘Ž0 𝑑1 𝑑2 2 βˆ’ 𝑅 2 𝑀5 βˆ’ π‘Ž0 𝑑1 𝑑2 π‘Ž1 𝑑1 𝑑2 + π‘Ž0 𝑑1 βˆ’ 𝑑2The result was also proved by Shah and Liman [12]. II. LEMMAS For the proof of these Theorems, we need the following Lemmas. The first Lemma is due to Govil, Rahman andSchmesser [7].LEMMA. 1. If 𝑓(z) is analytic in 𝑧 ≀ 1, 𝑓 π‘œ = π‘Ž π‘€β„Žπ‘’π‘Ÿπ‘’ π‘Ž < 1, 𝑓′ π‘œ = 𝑏, 𝑓(𝑧) ≀ 1, π‘œπ‘› 𝑧 = 1, π‘‘β„Žπ‘’π‘› π‘“π‘œπ‘Ÿ 𝑧 ≀1. 1βˆ’ π‘Ž 𝑧 2 + 𝑏 𝑧 + π‘Ž 1βˆ’ π‘Ž 𝑓(𝑧) ≀ π‘Ž 1βˆ’ π‘Ž 𝑧 2 + 𝑏 𝑧 + 1βˆ’ π‘ŽThe example 𝑏 π‘Ž+ π‘§βˆ’π‘§ 2 1+π‘Ž 𝑓 (z)= 𝑏 1βˆ’ π‘§βˆ’π‘Žπ‘§ 2 1+π‘Ž Shows that the estimate is sharp.form Lemma 1, one can easily deduce the following:LEMMA. 2. If 𝑓 (z) is analytic in 𝑧 ≀ 𝑅, 𝑓 0 = 0, 𝑓 β€² 0 = 𝑏 π‘Žπ‘›π‘‘ 𝑓(𝑧) ≀ 𝑀, π‘“π‘œπ‘Ÿ 𝑧 = 𝑅, π‘‘β„Žπ‘’π‘› 𝑀 𝑧 𝑀 𝑧 +𝑅 2 𝑏 𝑓(𝑧) ≀ for 𝑧 ≀ 𝑅 𝑅2 𝑀+ 𝑏 𝑧 III. PROOFS OF THE THEOREMS.Proof of Theorem . 1 Consider(16) F(z) = 𝑑2 + 𝑧 𝑑1 βˆ’ 𝑧 𝑃(𝑧) = βˆ’π‘Ž 𝑛 𝑧 𝑛+2 + π‘Ž 𝑛 𝑑1 βˆ’ 𝑑2 βˆ’ π‘Ž π‘›βˆ’1 𝑧 𝑛+1 + π‘Ž 𝑛 𝑑1 𝑑2 + π‘Ž 𝑛 βˆ’1 𝑑1 βˆ’ 𝑑2 βˆ’ π‘Ž π‘›βˆ’2 𝑧𝑛 +….+ π‘Ž 𝑛 𝑑1 𝑑2+ π‘Ž0 𝑑1 βˆ’ 𝑑2 𝑧 + π‘Ž0 𝑑1 𝑑2Let 1 G(z) + 𝑧 𝑛+2 𝐹 𝑧= βˆ’π‘Ž 𝑛 + π‘Ž 𝑛 𝑑1 βˆ’ 𝑑2 βˆ’ π‘Ž π‘›βˆ’1 𝑧 + π‘Ž 𝑛 𝑑1 𝑑2 + π‘Ž 𝑛 βˆ’1 𝑑1 βˆ’ 𝑑2 βˆ’ π‘Ž 𝑛 βˆ’2 𝑧 2 + β‹― …+ π‘Ž1 𝑑1 𝑑2 + π‘Ž0 𝑑1 βˆ’ 𝑑2 𝑧 𝑛 +1 + π‘Ž0 𝑑1 𝑑2 𝑧 𝑛+2 = βˆ’π‘Ž 𝑛 βˆ’ 𝛼𝑧 + π‘Ž 𝑛 𝑑1 βˆ’ 𝑑2 + 𝛼 βˆ’ π‘Ž π‘›βˆ’1 𝑧 + π‘Ž 𝑛 𝑑1 𝑑2 + π‘Ž π‘›βˆ’1 𝑑1 βˆ’ 𝑑2 βˆ’ π‘Ž 𝑛 βˆ’2 𝑧 2 + β‹― … + (π‘Ž1 𝑑1 𝑑2 + π‘Ž0 𝑑1 βˆ’ 𝑑2 𝑧 𝑛+1 + π‘Ž0 𝑑1 𝑑2 ) 𝑧 𝑛+2 = βˆ’ π‘Ž 𝑛 + 𝛼𝑧 + 𝐻(𝑧) Where,H(z)= 𝑧 π‘Ž 𝑛 𝑑1 βˆ’ 𝑑2 + 𝛼 βˆ’ π‘Ž 𝑛 βˆ’1 + βˆ‘ 𝑗𝑛=0 π‘Žπ‘— 𝑑1 𝑑2 + π‘Žπ‘— βˆ’1 𝑑1 βˆ’ 𝑑2 βˆ’ π‘Žπ‘— βˆ’2 𝑧 π‘›βˆ’π‘— www.ijmer.com 4366 | Page
  5. 5. International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4363-4372 ISSN: 2249-6645Clearly H(o)=0 and π‘€π‘Žπ‘₯ 𝑧 =𝑅 𝐻(𝑧) ≀ 𝑅𝑀1, We first assume that π‘Ž 𝑛 ≀ 𝑅 𝑀1 + 𝛼Now for 𝑧 ≀ 𝑅, by using Schwarz Lemma, we have(17) G(z) = βˆ’ a n + Ξ±z + H(z) β‰₯ a n βˆ’ 𝛼 𝑧 βˆ’ 𝐻(𝑧) β‰₯ a n βˆ’ 𝛼 𝑧 βˆ’ |𝑀1 𝑧| = an βˆ’ 𝑀1 + 𝛼 𝑧| > π‘œ, 𝑖𝑓 π‘Žπ‘› 𝑧 < 𝑀1 + 𝛼 ≀ 𝑅 π‘Žπ‘› This shows that all the zeros of G(z) lie in 𝑧 β‰₯ 𝑀1 + 𝛼 1 1 Replacing z by 𝑧 and noting that F(z) = 𝑧 𝑛+2 𝐺 𝑧 , it follows that all the zeros of F(z) lie in 𝑀1 + 𝛼 𝑧 ≀ π‘Žπ‘› if , π‘Ž 𝑛 ≀ 𝑅 (𝑀1 + 𝛼 ) Since all the zeros of P(z) are also the zeros of F(z), we conclude that all the zeros of P(z) lie in π‘Žπ‘›(18) 𝑧 ≀ 𝑀1 + 𝛼 Now assume π‘Ž 𝑛 > 𝑅 (𝑀1 + 𝛼 ), then for 𝑧 ≀ 𝑅, we have from (17). G z β‰₯ a n βˆ’ 𝛼 𝑧 βˆ’ 𝐻(𝑍) β‰₯ π‘Ž 𝑛 βˆ’ 𝛼 𝑅 βˆ’ 𝑀1 𝑅 = π‘Žπ‘› βˆ’ 𝛼 + 𝑀 𝑅 > 0, Thus G(z) β‰  0 π‘“π‘œπ‘Ÿ 𝑧 < 𝑅, π‘“π‘Ÿπ‘œπ‘š π‘€β„Žπ‘–π‘β„Ž it follows as before that all the zeros of F(z) and hence all the zeros of P(z) lie 1, in 𝑧 ≀ 𝑅 Combining this with (18), we infer that all the zeros of P(z) lie in 𝐚𝐧 𝟏,19. 𝐳 ≀ 𝐦𝐚𝐱 𝐌 𝟏+ 𝛂 , 𝐑 Now to prove the second part of the Theorem, from (16) we can write F(z) as F(z)= t 2 + z t1 βˆ’ z P(z) = a 0 t1 t 2 + a1 t1 t 2 + a 0 t1 βˆ’ t 2 z + β‹― + a n t1 βˆ’ t 2 βˆ’ a nβˆ’1 z n+1 βˆ’ a n z n+2=a 0 t1 t2 βˆ’ Ξ²z + a1 t1 t 2 + a 0 t1 βˆ’ t 2 + Ξ² z + β‹― + a n t1 βˆ’ t 2 βˆ’ a nβˆ’1 z n+1 βˆ’ a n z n+2 = a 0 t1 t 2 βˆ’ Ξ²z + T(z)WhereT(z) = z π‘Ž1 𝑑1 𝑑2 + π‘Ž0 𝑑1 βˆ’π‘‘2 + 𝛽 + βˆ‘n+2 π‘Žπ‘— 𝑑1 𝑑2 + π‘Žπ‘— βˆ’1 𝑑1 βˆ’π‘‘2 π‘Žπ‘— βˆ’2 𝑧 𝑗 βˆ’1 j=2 Clearly T(o)=0, and π‘€π‘Žπ‘₯ 𝑧 =𝑅 𝑇(𝑧) ≀ 𝑅𝑀2 We first assume that π‘Ž0 𝑑1 𝑑2 ≀ Ξ² + M2 www.ijmer.com 4367 | Page
  6. 6. International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4363-4372 ISSN: 2249-6645 Now for z ≀ R, by using Schwasr lemma, we have F(z) β‰₯ π‘Ž0 𝑑1 𝑑2 βˆ’ Ξ² z βˆ’ T(z) β‰₯ π‘Ž0 𝑑1 𝑑2 βˆ’ Ξ² z βˆ’ M2 Z >0, if, π‘Ž0 𝑑1 𝑑2 z < ≀R Ξ² +M 2 π‘Ž0 𝑑1 𝑑2 This shows that F(z) β‰ 0, for z < , hence all the zeros of F(z) lie in Ξ² +M 2 π‘Ž0 𝑑1 𝑑2 z β‰₯ Ξ² +M 2 But all the zeros of P(z) are also the zeros of F(z), we conclude that all the zeros of P(z) lie in π‘Ž0 𝑑1 𝑑220. z β‰₯ Ξ² +M 2 Now we assume π‘Ž0 𝑑1 𝑑2 > Ξ² + M2 , then for z ≀ R, we have ` F(z) β‰₯ π‘Ž0 𝑑1 𝑑2 βˆ’ Ξ² z βˆ’ T(z) β‰₯ π‘Ž0 𝑑1 𝑑2 βˆ’ Ξ² R βˆ’ M2 R > 𝑑1 𝑑2 π‘Ž0 βˆ’ Ξ² + M2 R >0, This shows that all the zeros of F(z) and hence that of P(z) lie in z β‰₯R Combining this with (20) , it follows that all the zeros of P(z) lie in π‘Ž0 𝑑1 𝑑221 z β‰₯ min Ξ² +M 2 ,R Combining (19) and (20), the desired result follows.Proof of Theorem 2:- Consider22. F(z) = 𝑑2 + 𝑧 𝑑1 βˆ’ 𝑧 𝑃(𝑧) = 𝑑2 + 𝑧 𝑑1 βˆ’ 𝑧 π‘Ž 𝑛 𝑧 𝑛 + π‘Ž π‘›βˆ’1 𝑧 π‘›βˆ’1 + β‹― + π‘Ž1 𝑧 + π‘Ž0 = βˆ’π‘Ž 𝑛 𝑧 𝑛+2 + π‘Žπ‘› 𝑑1βˆ’ 𝑑2 βˆ’ π‘Ž π‘›βˆ’1 𝑧 𝑛 +1 + β‹― + π‘Ž0 𝑑1 𝑑2Let 123. G(z) = 𝑧 𝑛+2 𝐹( 𝑧 ) = βˆ’π‘Ž 𝑛 + π‘Ž 𝑛 𝑑1 βˆ’ 𝑑2 βˆ’ π‘Ž π‘›βˆ’1 𝑧 + π‘Ž 𝑛 𝑑1 𝑑2 βˆ’ π‘Ž 𝑛 βˆ’1 𝑑1 βˆ’ 𝑑2 π‘Ž 𝑛 βˆ’2 𝑧 2 + 𝐻(𝑧) Where H(z) = 𝑧 2 π‘Ž 𝑛 𝑑1 𝑑2 βˆ’ π‘Ž π‘›βˆ’1 𝑑1 βˆ’ 𝑑2 βˆ’ π‘Ž 𝑛 βˆ’2 + β‹― + π‘Ž0 𝑑1 𝑑2 𝑧 𝑛 Clearly H(o)=0 = H(o), since H(z) ≀ M3 R2 for z = R www.ijmer.com 4368 | Page
  7. 7. International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4363-4372 ISSN: 2249-6645 We first assume,24 a n ≀ MR2 + R π‘Ž 𝑛 𝑑1 βˆ’ 𝑑2 βˆ’ π‘Ž 𝑛 βˆ’1 Then by using Lemma 2. To H(z), it follows that M3 z M3 z 𝐻z ≀ R2 M3 π‘€π‘Žπ‘₯ 𝑧 =𝑅 H(z) Hence from(23) we have z 2M3 G(z) β‰₯ a n βˆ’ π‘Ž 𝑛 𝑑1 βˆ’ 𝑑2 βˆ’ π‘Ž 𝑛 βˆ’1 z βˆ’ 𝑅2 R2 >0, if 2 M3 z + π‘Ž 𝑛 𝑑1 βˆ’ 𝑑2 βˆ’ π‘Ž π‘›βˆ’1 z βˆ’ a n < 0That is if, 1/2 π‘Ž 𝑛 𝑑 1 βˆ’π‘‘ 2 βˆ’π‘Ž 𝑛 βˆ’1 2 +4 a n M βˆ’ π‘Ž 𝑛 𝑑 1 βˆ’π‘‘ 2 βˆ’π‘Ž 𝑛 βˆ’125. z < 2M 3 1 =r ≀ RIf 2 2 π‘Ž 𝑛 𝑑1 βˆ’ 𝑑2 βˆ’ π‘Ž 𝑛 βˆ’1 + 4 a n M3 ≀ 2M3 R + π‘Ž 𝑛 𝑑1 βˆ’ 𝑑2 βˆ’ π‘Ž π‘›βˆ’1Which implies a n ≀ M3 R2 + R π‘Ž 𝑛 𝑑1 βˆ’ 𝑑2 βˆ’ π‘Ž 𝑛 βˆ’1 Which is true by (24)..Hence all the zeros of G(z) lie in Z β‰₯ r. since F z = z n+2 G z , It follows that all the zeros of F(z) and hence that of P(z) lie in z ≀ r. We now assume.26. a n ≀ M3 R2 + R π‘Ž 𝑛 𝑑1 βˆ’ 𝑑2 βˆ’ π‘Ž 𝑛 βˆ’1 ,then for z ≀ R, from (23) it follows that G(z) β‰₯ a n βˆ’ π‘Ž 𝑛 𝑑1 βˆ’ 𝑑2 βˆ’ π‘Ž 𝑛 βˆ’1 z βˆ’ H(z) β‰₯ π‘Ž 𝑛 βˆ’ 𝑅 π‘Ž 𝑛 𝑑1 βˆ’ 𝑑2 βˆ’ π‘Ž π‘›βˆ’1 βˆ’ R2 M3 >0, by (26). 1 This shows that all the zeros of G(z) lie in Z > 𝑅, and hence all the zeros of F(z) = z n+2 G(z ) lie in 1 z ≀ , but all the zeros of P(z) are also the zeros of F(z) , therefore it follows that all the zeros of P(z) lie in R 1 Z ≀R27. From (25) amd (27) , we conclude that all the zeros of P(z) lie in 1 Z ≀ max(π‘Ÿ, ) R Which completes the proof of Theorem 2.28. PROOF OF THEOREM 3. Consider the polynomial www.ijmer.com 4369 | Page
  8. 8. International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4363-4372 ISSN: 2249-6645 F(z) = 𝑑2 + 𝑧 𝑑1 βˆ’ 𝑧 𝑃(𝑧) = 𝑑1 𝑑2 + 𝑑1 βˆ’ 𝑑2 𝑧 βˆ’ 𝑧 2 ) π‘Ž 𝑛 𝑧 𝑛 + π‘Ž π‘›βˆ’1 𝑧 𝑛 βˆ’1 + β‹― + π‘Ž1 𝑧 + π‘Ž0 = βˆ’π‘Ž 𝑛 𝑧 𝑛+2 + π‘Žπ‘› 𝑑1βˆ’ 𝑑2 βˆ’ π‘Ž π‘›βˆ’1 𝑧 𝑛 +1 + β‹― + π‘Ž 𝑛 𝑑1 𝑑2 + π‘Ž π‘›βˆ’1 𝑑1βˆ’ 𝑑2 βˆ’ π‘Ž 𝑛 βˆ’1 𝑧 𝑛+1 + β‹― + a 2 t1 t 2 + a1 𝑑1βˆ’ 𝑑2 βˆ’ π‘Ž0 z 2 + a1 t1 t 2 + a 0 𝑑1βˆ’ 𝑑2 𝑧 + π‘Ž0 t1 t 2 We have, 1 G(z) = 𝑧 𝑛+2 𝐹( ) 𝑧 = π‘Ÿπ‘Ž 𝑛 + π‘Ž 𝑛 𝑑1 βˆ’ 𝑑2 βˆ’ π‘Ž 𝑛 βˆ’1 𝑧 + π‘Ž 𝑛 𝑑1 𝑑2 + π‘Ž 𝑛 βˆ’1 𝑑1 βˆ’ 𝑑2 βˆ’ π‘Ž 𝑛 βˆ’2 𝑧 2 +…+ π‘Ž1 𝑑1 𝑑2 + π‘Ž0 𝑑1βˆ’ 𝑑2 𝑧 𝑛 +1 + π‘Ž0 𝑑1 𝑑2 𝑧 𝑛+2 = βˆ’π‘Ž 𝑛 βˆ’ 𝛼𝑧 + π‘Ž 𝑛 𝑑1 βˆ’ 𝑑2 + 𝛼 βˆ’ π‘Ž 𝑛 βˆ’1 𝑧 + π‘Ž 𝑛 𝑑1 𝑑2 + π‘Ž 𝑛 βˆ’1 𝑑1βˆ’ 𝑑2 βˆ’ π‘Ž 𝑛 βˆ’2 z 2 +. . + π‘Ž1 𝑑1 𝑑2 + π‘Ž0 𝑑1βˆ’ 𝑑2 z n+1 + π‘Ž0 𝑑1 𝑑2 𝑧 𝑛+2 29. = βˆ’π‘Ž 𝑛 βˆ’ 𝛼𝑧 + 𝐻 𝑧 Where H(z) = π‘Ž 𝑛 𝑑1 βˆ’π‘‘2 + 𝛼 βˆ’ π‘Ž 𝑛 βˆ’1 𝑧 + β‹― + π‘Ž0 𝑑1 𝑑2 𝑧 𝑛+2 We first assume that 30. a n β‰₯ Ξ± R + M4 Then for z < 𝑅, we have 31. G(z) β‰₯ a n βˆ’ Ξ± z βˆ’ H(z) Since H(z) ≀ M4 π‘“π‘œπ‘Ÿ z ≀ RTherefore for 𝑧 <R, from (31) with the help of (30), we have 𝐺(𝑧) > π‘Ž 𝑛 βˆ’ ∝ 𝑅 βˆ’ 𝑀4 1therefore , all the zeros of G(z) lie in z β‰₯ R, in this case. Since F(z) = 𝑧 𝑛 +2 G 𝑧 therefore all the zeros of G(z) lie in 1Z ≀ R . As all the zeros of P(z) are the zeros of F(z), it follows that all the zeros of P(z) lie in 1 32. z ≀R Now we assume a n < Ξ± R + M4 , clearly H(o) = 0 and H(o) = (π‘Ž 𝑛 𝑑1 βˆ’π‘‘2 βˆ’ 𝛼 βˆ’ π‘Ž 𝑛 βˆ’1 , Since by (11), H(z) ≀ M4 , for z = R, therefore it follows by Lemma2. that M4 z M4 z + R2 a n 𝑑1 βˆ’π‘‘2 βˆ’ π‘Ž 𝑛 βˆ’1 + 𝛼 H(z) ≀ R2 M4 + a n 𝑑1 βˆ’π‘‘2 βˆ’ π‘Ž 𝑛 βˆ’1 + 𝛼 z for z ≀ R Using this in (31), we get, M4 z M4 z + R2 a n 𝑑1 βˆ’π‘‘2 βˆ’ π‘Ž π‘›βˆ’1 + 𝛼 G(z) β‰₯ π‘Ž 𝑛 βˆ’ 𝛼 𝑧 βˆ’ 2 R M4 + a n 𝑑1 βˆ’π‘‘2 βˆ’ π‘Ž 𝑛 βˆ’1 + 𝛼 z >0, if π‘Ž 𝑛 R2 M4 + a n 𝑑1 βˆ’π‘‘2 βˆ’ π‘Ž 𝑛 βˆ’1 + 𝛼 z βˆ’ z Ξ± R2 M4 + a n 𝑑1 βˆ’π‘‘2 βˆ’ π‘Ž 𝑛 βˆ’1 + 𝛼 z + M4 M4 z + R2 a n 𝑑1 βˆ’π‘‘2 βˆ’ π‘Ž 𝑛 βˆ’1 + 𝛼 > 0, this implies, Ξ± R2 a n 𝑑1 βˆ’π‘‘2 βˆ’ π‘Ž π‘›βˆ’1 + 𝛼 + M4 z 2 + Ξ± R2 M4 + 2 2 2 R M4 a n t1 βˆ’ t 2 βˆ’ a nβˆ’1 +∝ … R M4 an 𝑑1 βˆ’π‘‘2 βˆ’ π‘Ž 𝑛 βˆ’1 + 𝛼 } z βˆ’ a n R2 M4 < 0, This gives, G(z) > 0, 𝑖𝑓 z < [ Ξ± R2 M4 + R2 M4 βˆ’ a n a n 𝑑1 βˆ’π‘‘2 βˆ’ π‘Ž 𝑛 βˆ’1 + 𝛼 2 +.. ….4 Ξ± R2 a n 𝑑1 βˆ’π‘‘2 βˆ’ π‘Ž 𝑛 βˆ’1 + 𝛼 + M4 2 π‘Ž 𝑛 R2 M4 ] 𝛼 R 2 M 4 +R 2 M 4 βˆ’ π‘Ž 𝑛 a n 𝑑 1 βˆ’π‘‘ 2 βˆ’π‘Ž 𝑛 βˆ’1 +𝛼 βˆ’ 2 Ξ± R 2 a n 𝑑 1 βˆ’π‘‘ 2 βˆ’π‘Ž 𝑛 βˆ’1 +𝛼 +M 4 2 1 =r , 1 So, z < 𝑅, 𝑖𝑓 Ξ± R2 M4 + M4 βˆ’ a n R2 a n 𝑑1 βˆ’π‘‘2 βˆ’ π‘Ž 𝑛 βˆ’1 + 𝛼 2 + www.ijmer.com 4370 | Page
  9. 9. International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4363-4372 ISSN: 2249-6645 4 Ξ± R2 a n 𝑑1 βˆ’π‘‘2 βˆ’ π‘Ž π‘›βˆ’1 + 𝛼 + M4 2 π‘Ž 𝑛 R2 M4 < 2 Ξ± a n 𝑑1 βˆ’π‘‘2 βˆ’ π‘Ž 𝑛 βˆ’1 + 𝛼 R + Ξ± R2 M4 + 2RM4 2 + M4 βˆ’ a n R2 a n 𝑑1 βˆ’π‘‘2 βˆ’ π‘Ž 𝑛 βˆ’1 + 𝛼 2 That is if, a n R2 M4 ≀ R2 Ξ± 𝑅2 an 𝑑1 βˆ’π‘‘2 βˆ’ π‘Ž π‘›βˆ’1 + 𝛼 + M4 2 + Ξ± R2 M4 + M4 βˆ’ a n R2 a n 𝑑1 βˆ’π‘‘2 βˆ’ π‘Ž 𝑛 βˆ’1 + 𝛼 R Which gives a n M4 ≀ M4 Ξ± R + M4 βˆ’ π‘Ž 𝑛 + 𝑅 Ξ± R2 a n 𝑑1 βˆ’π‘‘2 βˆ’ π‘Ž 𝑛 βˆ’1 + 𝛼 + M4 2 Ξ± R2 + M4 βˆ’ π‘Ž 𝑛 Which is true because π‘Ž 𝑛 < Ξ± R + M4 , 1 1 Consequently , all the zeros of G(z) lie in z β‰₯ r , as F(z) = z n+1 G( z ), we conclude that all the zeros of 1 F(z) lie z ≀ π‘Ÿ1 , since every zero of P(z) is also a zero of F(z), it follows that all the zeros of P(z) lie in 33. z ≀ π‘Ÿ1 Combining this with (32) it follows that all the zeros of P(z) lie in 1 34. z ≀ π‘šπ‘Žπ‘₯ π‘Ÿ1 , 𝑅 Again from (28) it follows that 𝐹 𝑧 = π‘Ž0 t1 t 2 βˆ’ Ξ²z + a1 t1 t2 𝑑1βˆ’ 𝑑2 + 𝛽 z + β‹― + a n ( 𝑑1 βˆ’π‘‘2 βˆ’. π‘Ž π‘›βˆ’1 )z n+2 βˆ’ a n 𝑧 𝑛+2 35. β‰₯ π‘Ž0 t1 t 2 βˆ’ 𝛽 𝑧 βˆ’ 𝑇(𝑧) Where, T(z) = a n 𝑧 𝑛+2 + a n 𝑑1 βˆ’π‘‘2 βˆ’. π‘Ž 𝑛 βˆ’1 𝑧 𝑛 +1 +. . + π‘Ž0 t1 t 2 + a 0 𝑑1βˆ’ 𝑑2 + 𝛽 z Clearly T(o)=0, and T(o) π‘Ž1 t1 t 2 + a 0 𝑑1βˆ’ 𝑑2 + 𝛽 Since by (12) , T(z) ≀ M5 for z = R using Lemma 2. To T(z), we have M5 z F(z) β‰₯ π‘Ž0 t1 t 2 βˆ’ Ξ² z βˆ’ R2 M5 z + R2 π‘Ž1 t1 t 2 + a 0 𝑑1βˆ’ 𝑑2 + 𝛽 z = - R2 Ξ² π‘Ž1 t1 t 2 + a 0 𝑑1βˆ’ 𝑑2 + 𝛽 + 𝑀5 2 z 2 + π‘Ž1 t1 t 2 + a 0 𝑑1βˆ’ 𝑑2 + 𝛽 R2 π‘Ž0 t1 t 2 βˆ’ R2 M5 βˆ’ Ξ² R2 M5 z + M5 R2 π‘Ž1 t1 t2 R2 M5 + π‘Ž1 t1 t 2 + a 0 𝑑1βˆ’ 𝑑2 + 𝛽 z>0, if, R2 Ξ² π‘Ž1 t1 t 2 + a 0 𝑑1βˆ’ 𝑑2 + 𝛽 + 𝑀5 2 z 2 βˆ’ π‘Ž1 t1 t 2 + a 0 𝑑1βˆ’ 𝑑2 + 𝛽 R2 π‘Ž0 t1 t2 βˆ’ R2 M5 βˆ’ Ξ² R2 M5 z βˆ’ R2 M5 π‘Ž0 t1 t 2 < 0,Thus, F(z) > 0, 𝑖𝑓 z < βˆ’R2 π‘Ž1 t1 t2 + a 0 𝑑1βˆ’ 𝑑2 + 𝛽 π‘Ž0 t1 t 2 βˆ’ M5 Ξ² M5 www.ijmer.com 4371 | Page
  10. 10. International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4363-4372 ISSN: 2249-6645 βˆ’ R4 π‘Ž1 t1 t 2 + a 0 𝑑1βˆ’ 𝑑2 + 𝛽 π‘Ž0 t1 t 2 βˆ’ M5 Ξ² M5 2 +4 R2 Ξ² π‘Ž1 t1 t 2 + a 0 𝑑1βˆ’ 𝑑2 + 𝛽 + M5 2 R2 M5 π‘Ž0 t1 t2 1/2 π‘Ÿπ‘§ 2 R2 Ξ² π‘Ž1 t1 t 2 + a 0 𝑑1βˆ’ 𝑑2 + 𝛽 + M5 2 Thus it follows by the same reasoning as in Theorem 1, that all the zeros of F(z) and hence that of P(z) lie in 36. z β‰₯ min(r2, R) Combining (34) and (36) , the desired result follows. REFERENCES[1] A. Aziz and Q.G Mohammad, On thr zeros of certain class of polynomials and related analytic functions, J. Math. Anal Appl. 75(1980), 495-502[2] A.Aziz and W.M Shah, On the location of zeros of polynomials and related analytic functions, Non-linear studies, 6(1999),97-104[3] A.Aziz and W.M Shah, On the zeros of polynomials and related analytic functions, Glasnik, Matematicke, 33 (1998), 173-184;[4] A.Aziz and B.A. Zarger, Some extensions of Enestrem-Kakeya Theorem, Glasnik, Matematicke, 31 (1996), 239-244[5] K.K.Dewan and M. Biakham,, On the Enestrom –Kakeya Theorem, J.Math.Anal.Appl.180 (1993), 29-36[6] N.K, Govil and Q.I. Rahman, On the Enestrem-Kakeya Theorem, Tohoku Math.J. 20 (1968), 126-136.[7] N.K, Govil , Q.I. Rahman and G. Schmeisser, on the derivative of polynomials, Illinois Math. Jour. 23 (1979), 319- 329[8] P.V. Krishnalah, On Kakeya Theorem , J.London. Math. Soc. 20 (1955), 314-319[9] M. Marden, Geometry of polynomials, Mathematicial surveys No.3. Amer. Math. Soc. Providance , R.I. 1966[10] G.V. Milovanovic, D.S. Mitrinovic, Th.M. Rassias Topics in polynomials, Extremal problems Inequalitics, Zeros(Singapore, World Scientific)1994.[11] Q.I. Rahman and G.Schmessier, Analytic Theory of polynomials, (2002), Clarantone, Press Oxford.[12] W.M. Shah and A. Liman, On the zeros of a class of polynomials, Mathematical inequalioties and Applications, 10(2007), 793-799 www.ijmer.com 4372 | Page

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