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# On The Zeros of Certain Class of Polynomials

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### On The Zeros of Certain Class of Polynomials

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) =β ππ=0 aj zj is a polynomial of degree n such that(1) an β₯ an-1 β₯β¦..β₯ a1 β₯ a0 β₯ 0,then all the zeros of P(z) lie in π§ β€ 1. In the literature ([2] ,[5] β[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) =β ππ=0 aj zj is a polynomial of degree n with real positive coefficients. If t1 β₯ π‘2 β₯ 0 canbe found such that(2) π π½ t1 t 2+ ajβ1 t1 β t 2 a jβ2 β₯ 0, j=1, 2, β¦ ;n+1, πβ1 = π π +1 = 0 ,then all the zeros of P(z) lie in π§ β€ π‘1 .For π‘1 = 1, π‘2 = 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) = β ππ=0 ππ π§ π be a polynomial of degree n. If for some t > 0.(3) πππ₯ π§ =π π‘π0 π§ π + π‘π1 β π0 π§ π β1 + β― + π‘π π β π πβ1 β€ π3Where R is any positive real number, then all the zeros of P(z) lie in π 1(4) π§ β€ πππ₯ π 3 , π π 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 πΌ, π, π‘1 πππ π‘2 . Infact we start by proving the following result. Theorem 1. Let P(z) = β ππ=0 ππ π§ π be a polynomial of degree n, If for some real numbers π‘1 , π‘2 with π π β  π, π π > ππ β₯ π(5) πππ₯ π§ =π π π π‘1 β π‘2 + πΌ β π πβ1 + β ππ=0 ππ π‘1 π‘2 + ππ β1 π‘1 β π‘2 β ππ β2 π§ πβπ +1 β€ π1(6) πππ π§ =π π0 π‘1 β π‘2 + π1 π‘1 π‘2 + π½ + β ππ+2 ππ π‘1 π‘2 + ππ β1 π‘1 β π‘2 β =2 (π π β2 π§ π β€ π2 Where R is a positive real number, then all the zeros of P(z) lie in the ring shaped region. π‘1 π‘2 π0 π1 + πΌ 1(7) πππ π§ =R π½ +π2 , π β€ π§ β€ πππ₯ π§ =R ππ , πTaking t2 =0, we get the following generalization and refinement of Theorem C. Corollary 1. Let P(z) = β ππ=0 ππ π§ π be a polynomial of degree n. If for some t > 0. πππ₯ π§ =π a n t + Ξ± β a nβ1 + βn j=0 a jβ1 t β a jβ2 z nβj+2 β€ M1 n+2 j πππ π§ =π a 0 t + Ξ² + βj=0 a jβ1 t β a jβ2 z β€ 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 π1 + πΌ 1 z β€ πππ₯ π§ =π ππ , π In case we take t=1=R in corollary 1, we get the following interesting result.Corollary 2. Let P(z) = β ππ=0 ππ π§ π be a polynomial of degree n, πππ₯ π§ =π π π + πΌ β π πβ1 + π π β1 β π π β2 π§ + β― + π1 β π0 π§ πβ1 + ππ π§ π β€ M, M+ Ξ±then all the zeros of P(z) lie in the circle z β€ Max an ,1If for some Ξ± > 0, πΌ + π π β₯ π π β1 β₯ β― β₯ π1 β₯ π0 β₯ 0, then , π β€ π π + πΌ + π π β1 + π π β1 β π πβ2 + β― + π1 β π0 + π0 = π π + πΌ β π π β1 + π πβ1 β π πβ2 + β― + π1 β π0 + π0 = ππ + πΌUsing this observation in Corollary 2, we get the following generalization of Enestrom-Kakeya Theorem. Corollary 3. If, P(z)= π π π§ π + π π β1 + π§ π β1 + β― + π1 π§ + π0 , is a polynomial of degree n, such that for someπΌ β₯ 0, πΌ + π π β₯ πβ1 β₯ β― . β₯ π0 β₯ 0,then all the zeros of P(z) lie in the circle 2πΌ π§ β€ 1+ , ππFor πΌ = 0, π‘βππ  ππππ’πππ  π‘π πβπππππ π΄. πΌπ π€π π‘πππ πΌ = π πβ1 β π π 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) = β ππ=0 ππ π§ π be a polynomial of degree n, If for some real numbers with π π β  π, π π > π π β₯ π π(π) π΄ππ π =πΉ β π=π π π π π π π + π πβπ π π β π π β π πβπ π πβπ β€ π΄3then all the zeros of P(z) lie in the region(9) π§ β€ π1,Where , 2π3(10) π1 = π π π‘ 1 βπ‘ 2 π π β1 2 +4 π π π3 1/2 β π π π‘ 1 βπ‘ 2 β π π β1Taking π‘2 = 0, we get the following result Corollary 4. Let P(z) = β ππ=0 ππ π§ π be a polynomial of degree n, If for some π‘ > 0, π β π=π π πβπ π β π πβπ π πβπ β€ π΄3, π΄ππ π =πΉthen all the zeros of P(z) lie in the region π§ β€ π1, Where , 2π3 r= π π π‘β π π β1 2 +4 π π π3 1/2 β π π‘βπ π π β1 Remark. Suppose polynomial P(z) = β ππ=0 ππ π§ π , 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 π β1 π§ β€ π π β π π β1 2 + 4π π π πβ1 β π π β π π β1 2π = 2π π β1 = 1, π β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) = β ππ=0 ππ π§ π be a polynomial of degree n, IfπΌ, π½ are complex numbers and with π‘1 β  0, π‘2 πππ ππππ ππ’πππππ  π€ππ‘β π‘1 β₯ π‘2 β₯ 0 and (11) πππ₯ π§ =π π π π‘1 β π‘2 + πΌ β π π β1 π§ + β ππ=0 ππ π‘1 π‘2 + ππ β1 π‘1 βπ‘2 β ππ β2 π§ π βπ β2 β€ π4 and, (12) πππ₯ π§ =π π π π‘1 β π‘2 + π½ β π π β1 π§ + β ππ+2 ππ π‘1 π‘2 + ππ β1 π‘1 βπ‘2 β ππ β2 π§ π β€ π5 =2 Where πβ2= πβ1 = 0 = π π+1 = π π+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) β€ π§ β€ max π1 , π Where 2[ π π 2 π π π‘ 1β π‘ 2 βπ π β1 +πΌ +π4 ] 2 (14) π1 = 2 πΌ π 2 π4 +π 2 π4 β π π π π π‘ 1β π‘ 2 +πΌβπ π β1 2 +4 πΌ π 2 π π π‘ 1β π‘ 2 +πΌβπ π β1 +π4 π π π 2 π4 1/2 β πΌ π2 π4 + π2 π4 β π π π π π‘1β π‘2 + πΌ β π π β1 and 1 2[π 2 π½ 2 π 1 π‘ 1 π‘ 2 +π 0 π‘ 1β π‘ 2 +π½ βπ5 ] (15) = π2 π2 π 1 π‘ 1 π‘ 2 +π 0 π‘ 1β π‘ 2 +π½ ( π 0 π‘ 1 π‘ 2 βπ5 )β π½ π5 + π4 π1 π‘1 π‘2 + π0 π‘1β π‘2 + π½ π0 π‘1 π‘2 β π5 2 2 +4 π0 π5 π‘1 π‘2 (π 2 π½ π1 π‘1 π‘2 + π0 π‘1β π‘2 ) + π½ + π5 1/2 Taking πΌ = 0, π½ = 0, in Theorem 3, we get the following. Corollary 5. Let P(z) = β ππ=0 ππ π§ π be a polynomial of degree n, If ,π‘1 β  0 πππ π‘2 are real numbers with π‘1 β₯ π‘2 β₯ 0, πππ₯ π§ =π β ππ+1 ππ π‘1 π‘2 + ππ β1 π‘1β π‘2 β ππ β2 π§ πβπ +2 β€ π4 , =1 π§+π β π£=π π π£ π­ π π­ π + π π£βπ π­ πβ π­ π β π π£βπ π³ π£ β€ π π , πππ± π³ =πWhere R is any positive real number . then all the zeros of P(z) lie in the ring shaped region 1 min (r2,R) β€ π§ β€ max π1 , π Where 2π42 π1 = 3 π4 π‘ 1β π‘ 2 ) π π βπ π β1 2 π4 β π π 2 +4 π π π 2 π4 1/2 β π‘ 1 βπ‘ 2 )π π βπ π β1 π4 β π π π2 π2 = 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 2π5 π4 π1 π‘1 π‘2 + π0 π‘1 β π‘2 2 π5 β π0 π‘1 π‘2 2 + 4π5 π 2 π0 π‘1 π‘2 2 β π 2 π5 β π0 π‘1 π‘2 π1 π‘1 π‘2 + π0 π‘1 β π‘2The 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 π(z) is analytic in π§ β€ 1, π π = π π€βπππ π < 1, πβ² π = π, π(π§) β€ 1, ππ π§ = 1, π‘βππ πππ π§ β€1. 1β π π§ 2 + π π§ + π 1β π π(π§) β€ π 1β π π§ 2 + π π§ + 1β πThe example π π+ π§βπ§ 2 1+π π (z)= π 1β π§βππ§ 2 1+π Shows that the estimate is sharp.form Lemma 1, one can easily deduce the following:LEMMA. 2. If π (z) is analytic in π§ β€ π, π 0 = 0, π β² 0 = π πππ π(π§) β€ π, πππ π§ = π, π‘βππ π π§ π π§ +π 2 π π(π§) β€ for π§ β€ π π2 π+ π π§ III. PROOFS OF THE THEOREMS.Proof of Theorem . 1 Consider(16) F(z) = π‘2 + π§ π‘1 β π§ π(π§) = βπ π π§ π+2 + π π π‘1 β π‘2 β π πβ1 π§ π+1 + π π π‘1 π‘2 + π π β1 π‘1 β π‘2 β π πβ2 π§π +β¦.+ π π π‘1 π‘2+ π0 π‘1 β π‘2 π§ + π0 π‘1 π‘2Let 1 G(z) + π§ π+2 πΉ π§= βπ π + π π π‘1 β π‘2 β π πβ1 π§ + π π π‘1 π‘2 + π π β1 π‘1 β π‘2 β π π β2 π§ 2 + β― β¦+ π1 π‘1 π‘2 + π0 π‘1 β π‘2 π§ π +1 + π0 π‘1 π‘2 π§ π+2 = βπ π β πΌπ§ + π π π‘1 β π‘2 + πΌ β π πβ1 π§ + π π π‘1 π‘2 + π πβ1 π‘1 β π‘2 β π π β2 π§ 2 + β― β¦ + (π1 π‘1 π‘2 + π0 π‘1 β π‘2 π§ π+1 + π0 π‘1 π‘2 ) π§ π+2 = β π π + πΌπ§ + π»(π§) Where,H(z)= π§ π π π‘1 β π‘2 + πΌ β π π β1 + β ππ=0 ππ π‘1 π‘2 + ππ β1 π‘1 β π‘2 β ππ β2 π§ πβ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π₯ π§ =π π»(π§) β€ ππ1, We first assume that π π β€ π π1 + πΌNow for π§ β€ π, by using Schwarz Lemma, we have(17) G(z) = β a n + Ξ±z + H(z) β₯ a n β πΌ π§ β π»(π§) β₯ a n β πΌ π§ β |π1 π§| = an β π1 + πΌ π§| > π, ππ ππ π§ < π1 + πΌ β€ π ππ This shows that all the zeros of G(z) lie in π§ β₯ π1 + πΌ 1 1 Replacing z by π§ and noting that F(z) = π§ π+2 πΊ π§ , it follows that all the zeros of F(z) lie in π1 + πΌ π§ β€ ππ if , π π β€ π (π1 + πΌ ) 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π(18) π§ β€ π1 + πΌ Now assume π π > π (π1 + πΌ ), then for π§ β€ π, we have from (17). G z β₯ a n β πΌ π§ β π»(π) β₯ π π β πΌ π β π1 π = ππ β πΌ + π π > 0, Thus G(z) β  0 πππ π§ < π, ππππ π€βππβ it follows as before that all the zeros of F(z) and hence all the zeros of P(z) lie 1, in π§ β€ π Combining this with (18), we infer that all the zeros of P(z) lie in ππ§ π,19. π³ β€ π¦ππ± π π+ π , π Now to prove the second part of the Theorem, from (16) we can write F(z) as F(z)= t 2 + z t1 β z P(z) = a 0 t1 t 2 + a1 t1 t 2 + a 0 t1 β t 2 z + β― + a n t1 β t 2 β a nβ1 z n+1 β a n z n+2=a 0 t1 t2 β Ξ²z + a1 t1 t 2 + a 0 t1 β t 2 + Ξ² z + β― + a n t1 β t 2 β a nβ1 z n+1 β a n z n+2 = a 0 t1 t 2 β Ξ²z + T(z)WhereT(z) = z π1 π‘1 π‘2 + π0 π‘1 βπ‘2 + π½ + βn+2 ππ π‘1 π‘2 + ππ β1 π‘1 βπ‘2 ππ β2 π§ π β1 j=2 Clearly T(o)=0, and πππ₯ π§ =π π(π§) β€ ππ2 We first assume that π0 π‘1 π‘2 β€ 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 β€ R, by using Schwasr lemma, we have F(z) β₯ π0 π‘1 π‘2 β Ξ² z β T(z) β₯ π0 π‘1 π‘2 β Ξ² z β M2 Z >0, if, π0 π‘1 π‘2 z < β€R Ξ² +M 2 π0 π‘1 π‘2 This shows that F(z) β 0, for z < , hence all the zeros of F(z) lie in Ξ² +M 2 π0 π‘1 π‘2 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0 π‘1 π‘220. z β₯ Ξ² +M 2 Now we assume π0 π‘1 π‘2 > Ξ² + M2 , then for z β€ R, we have ` F(z) β₯ π0 π‘1 π‘2 β Ξ² z β T(z) β₯ π0 π‘1 π‘2 β Ξ² R β M2 R > π‘1 π‘2 π0 β 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0 π‘1 π‘221 z β₯ min Ξ² +M 2 ,R Combining (19) and (20), the desired result follows.Proof of Theorem 2:- Consider22. F(z) = π‘2 + π§ π‘1 β π§ π(π§) = π‘2 + π§ π‘1 β π§ π π π§ π + π πβ1 π§ πβ1 + β― + π1 π§ + π0 = βπ π π§ π+2 + ππ π‘1β π‘2 β π πβ1 π§ π +1 + β― + π0 π‘1 π‘2Let 123. G(z) = π§ π+2 πΉ( π§ ) = βπ π + π π π‘1 β π‘2 β π πβ1 π§ + π π π‘1 π‘2 β π π β1 π‘1 β π‘2 π π β2 π§ 2 + π»(π§) Where H(z) = π§ 2 π π π‘1 π‘2 β π πβ1 π‘1 β π‘2 β π π β2 + β― + π0 π‘1 π‘2 π§ π Clearly H(o)=0 = H(o), since H(z) β€ 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 β€ MR2 + R π π π‘1 β π‘2 β π π β1 Then by using Lemma 2. To H(z), it follows that M3 z M3 z π»z β€ R2 M3 πππ₯ π§ =π H(z) Hence from(23) we have z 2M3 G(z) β₯ a n β π π π‘1 β π‘2 β π π β1 z β π2 R2 >0, if 2 M3 z + π π π‘1 β π‘2 β π πβ1 z β a n < 0That is if, 1/2 π π π‘ 1 βπ‘ 2 βπ π β1 2 +4 a n M β π π π‘ 1 βπ‘ 2 βπ π β125. z < 2M 3 1 =r β€ RIf 2 2 π π π‘1 β π‘2 β π π β1 + 4 a n M3 β€ 2M3 R + π π π‘1 β π‘2 β π πβ1Which implies a n β€ M3 R2 + R π π π‘1 β π‘2 β π π β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 β€ r. We now assume.26. a n β€ M3 R2 + R π π π‘1 β π‘2 β π π β1 ,then for z β€ R, from (23) it follows that G(z) β₯ a n β π π π‘1 β π‘2 β π π β1 z β H(z) β₯ π π β π π π π‘1 β π‘2 β π πβ1 β R2 M3 >0, by (26). 1 This shows that all the zeros of G(z) lie in Z > π, and hence all the zeros of F(z) = z n+2 G(z ) lie in 1 z β€ , 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 β€R27. From (25) amd (27) , we conclude that all the zeros of P(z) lie in 1 Z β€ max(π, ) 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) = π‘2 + π§ π‘1 β π§ π(π§) = π‘1 π‘2 + π‘1 β π‘2 π§ β π§ 2 ) π π π§ π + π πβ1 π§ π β1 + β― + π1 π§ + π0 = βπ π π§ π+2 + ππ π‘1β π‘2 β π πβ1 π§ π +1 + β― + π π π‘1 π‘2 + π πβ1 π‘1β π‘2 β π π β1 π§ π+1 + β― + a 2 t1 t 2 + a1 π‘1β π‘2 β π0 z 2 + a1 t1 t 2 + a 0 π‘1β π‘2 π§ + π0 t1 t 2 We have, 1 G(z) = π§ π+2 πΉ( ) π§ = ππ π + π π π‘1 β π‘2 β π π β1 π§ + π π π‘1 π‘2 + π π β1 π‘1 β π‘2 β π π β2 π§ 2 +β¦+ π1 π‘1 π‘2 + π0 π‘1β π‘2 π§ π +1 + π0 π‘1 π‘2 π§ π+2 = βπ π β πΌπ§ + π π π‘1 β π‘2 + πΌ β π π β1 π§ + π π π‘1 π‘2 + π π β1 π‘1β π‘2 β π π β2 z 2 +. . + π1 π‘1 π‘2 + π0 π‘1β π‘2 z n+1 + π0 π‘1 π‘2 π§ π+2 29. = βπ π β πΌπ§ + π» π§ Where H(z) = π π π‘1 βπ‘2 + πΌ β π π β1 π§ + β― + π0 π‘1 π‘2 π§ π+2 We first assume that 30. a n β₯ Ξ± R + M4 Then for z < π, we have 31. G(z) β₯ a n β Ξ± z β H(z) Since H(z) β€ M4 πππ z β€ RTherefore for π§ <R, from (31) with the help of (30), we have πΊ(π§) > π π β β π β π4 1therefore , all the zeros of G(z) lie in z β₯ R, in this case. Since F(z) = π§ π +2 G π§ therefore all the zeros of G(z) lie in 1Z β€ 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 β€R Now we assume a n < Ξ± R + M4 , clearly H(o) = 0 and H(o) = (π π π‘1 βπ‘2 β πΌ β π π β1 , Since by (11), H(z) β€ M4 , for z = R, therefore it follows by Lemma2. that M4 z M4 z + R2 a n π‘1 βπ‘2 β π π β1 + πΌ H(z) β€ R2 M4 + a n π‘1 βπ‘2 β π π β1 + πΌ z for z β€ R Using this in (31), we get, M4 z M4 z + R2 a n π‘1 βπ‘2 β π πβ1 + πΌ G(z) β₯ π π β πΌ π§ β 2 R M4 + a n π‘1 βπ‘2 β π π β1 + πΌ z >0, if π π R2 M4 + a n π‘1 βπ‘2 β π π β1 + πΌ z β z Ξ± R2 M4 + a n π‘1 βπ‘2 β π π β1 + πΌ z + M4 M4 z + R2 a n π‘1 βπ‘2 β π π β1 + πΌ > 0, this implies, Ξ± R2 a n π‘1 βπ‘2 β π πβ1 + πΌ + M4 z 2 + Ξ± R2 M4 + 2 2 2 R M4 a n t1 β t 2 β a nβ1 +β β¦ R M4 an π‘1 βπ‘2 β π π β1 + πΌ } z β a n R2 M4 < 0, This gives, G(z) > 0, ππ z < [ Ξ± R2 M4 + R2 M4 β a n a n π‘1 βπ‘2 β π π β1 + πΌ 2 +.. β¦.4 Ξ± R2 a n π‘1 βπ‘2 β π π β1 + πΌ + M4 2 π π R2 M4 ] πΌ R 2 M 4 +R 2 M 4 β π π a n π‘ 1 βπ‘ 2 βπ π β1 +πΌ β 2 Ξ± R 2 a n π‘ 1 βπ‘ 2 βπ π β1 +πΌ +M 4 2 1 =r , 1 So, z < π, ππ Ξ± R2 M4 + M4 β a n R2 a n π‘1 βπ‘2 β π π β1 + πΌ 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 π‘1 βπ‘2 β π πβ1 + πΌ + M4 2 π π R2 M4 < 2 Ξ± a n π‘1 βπ‘2 β π π β1 + πΌ R + Ξ± R2 M4 + 2RM4 2 + M4 β a n R2 a n π‘1 βπ‘2 β π π β1 + πΌ 2 That is if, a n R2 M4 β€ R2 Ξ± π2 an π‘1 βπ‘2 β π πβ1 + πΌ + M4 2 + Ξ± R2 M4 + M4 β a n R2 a n π‘1 βπ‘2 β π π β1 + πΌ R Which gives a n M4 β€ M4 Ξ± R + M4 β π π + π Ξ± R2 a n π‘1 βπ‘2 β π π β1 + πΌ + M4 2 Ξ± R2 + M4 β π π Which is true because π π < 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 β€ π1 , 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 β€ π1 Combining this with (32) it follows that all the zeros of P(z) lie in 1 34. z β€ πππ₯ π1 , π Again from (28) it follows that πΉ π§ = π0 t1 t 2 β Ξ²z + a1 t1 t2 π‘1β π‘2 + π½ z + β― + a n ( π‘1 βπ‘2 β. π πβ1 )z n+2 β a n π§ π+2 35. β₯ π0 t1 t 2 β π½ π§ β π(π§) Where, T(z) = a n π§ π+2 + a n π‘1 βπ‘2 β. π π β1 π§ π +1 +. . + π0 t1 t 2 + a 0 π‘1β π‘2 + π½ z Clearly T(o)=0, and T(o) π1 t1 t 2 + a 0 π‘1β π‘2 + π½ Since by (12) , T(z) β€ M5 for z = R using Lemma 2. To T(z), we have M5 z F(z) β₯ π0 t1 t 2 β Ξ² z β R2 M5 z + R2 π1 t1 t 2 + a 0 π‘1β π‘2 + π½ z = - R2 Ξ² π1 t1 t 2 + a 0 π‘1β π‘2 + π½ + π5 2 z 2 + π1 t1 t 2 + a 0 π‘1β π‘2 + π½ R2 π0 t1 t 2 β R2 M5 β Ξ² R2 M5 z + M5 R2 π1 t1 t2 R2 M5 + π1 t1 t 2 + a 0 π‘1β π‘2 + π½ z>0, if, R2 Ξ² π1 t1 t 2 + a 0 π‘1β π‘2 + π½ + π5 2 z 2 β π1 t1 t 2 + a 0 π‘1β π‘2 + π½ R2 π0 t1 t2 β R2 M5 β Ξ² R2 M5 z β R2 M5 π0 t1 t 2 < 0,Thus, F(z) > 0, ππ z < βR2 π1 t1 t2 + a 0 π‘1β π‘2 + π½ π0 t1 t 2 β 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 β R4 π1 t1 t 2 + a 0 π‘1β π‘2 + π½ π0 t1 t 2 β M5 Ξ² M5 2 +4 R2 Ξ² π1 t1 t 2 + a 0 π‘1β π‘2 + π½ + M5 2 R2 M5 π0 t1 t2 1/2 ππ§ 2 R2 Ξ² π1 t1 t 2 + a 0 π‘1β π‘2 + π½ + 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 β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