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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 aim of this paper is to obtain a ring shaped region containing all the zeros of P(z). Our results not only generalize some known results but also a variety of interesting results can be deduced from them.

I.

Introduction and Statement of Results

The following beautiful result which is well-known in the theory of distribution of zeros of polynomials is due to Enestrom and Kakeya [9]. Theorem A. If P(z) =∑đ?‘›đ?‘— =0 aj zj is a polynomial of degree n such that

(1)

an ≼ an-1 ≼‌..≼ a1 ≼ a0 ≼ 0, then all the zeros of P(z) lie in đ?‘§ ≤ 1. In the literature ([2] ,[5] –[6], [8]-[11]) there exists some extensions and generalizations of this famous result. Aziz and Mohammad [1] provided the following generalization of Theorem A. Theorem B. Let P(z) =∑đ?‘›đ?‘— =0 aj zj is a polynomial of degree n with real positive coefficients. If t1 ≼ đ?‘Ą2 ≼ 0 can be found such that (2) đ?‘Žđ??˝ t1 t 2+ aj−1 t1 − t 2 a j−2 ≼ 0, j=1, 2, ‌ ;n+1, đ?‘Žâˆ’1 = đ?‘Žđ?‘› +1 = 0 , then all the zeros of P(z) lie in đ?‘§ ≤ đ?‘Ą1 . For đ?‘Ą1 = 1, đ?‘Ą2 = 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 special case. -----------------------------------------------------------------------------Mathematics, subject, classification (2002). 30C10,30C15, Keywords and Phrases , Enestrom-Kakeya Theorem, Zeros ,bounds. Theorem C. Let P(z) = ∑đ?‘›đ?‘—=0 đ?‘Žđ?‘— đ?‘§ đ?‘— be a polynomial of degree n. If for some t > 0. (3) đ?‘€đ?‘Žđ?‘Ľ đ?‘§ =đ?‘… đ?‘Ąđ?‘Ž0 đ?‘§ đ?‘› + đ?‘Ąđ?‘Ž1 − đ?‘Ž0 đ?‘§ đ?‘› −1 + â‹Ż + đ?‘Ąđ?‘Žđ?‘› − đ?‘Žđ?‘›âˆ’1 ≤ đ?‘€3 Where R is any positive real number, then all the zeros of P(z) lie in đ?‘€ 1 (4) đ?‘§ ≤ đ?‘€đ?‘Žđ?‘Ľ đ?‘Ž 3 , đ?‘… đ?‘›

The aim of this paper is to apply Schwarz Lemma to prove a more general result which includes Theorems A, B and C as special cases and yields a number of other interesting results for various choices of parameters đ?›ź, đ?‘&#x;, đ?‘Ą1 đ?‘Žđ?‘›đ?‘‘ đ?‘Ą2 . In fact we start by proving the following result. Theorem 1. Let P(z) = ∑đ?‘›đ?‘—=0 đ?‘Žđ?‘— đ?‘§ đ?‘— be a polynomial of degree n, If for some real numbers đ?‘Ą1 , đ?‘Ą2 with đ?’•đ?&#x;? ≠đ?&#x;Ž, đ?’•đ?&#x;? > đ?’•đ?&#x;? ≼ đ?&#x;Ž (5) đ?‘€đ?‘Žđ?‘Ľ đ?‘§ =đ?‘… đ?‘Žđ?‘› đ?‘Ą1 − đ?‘Ą2 + đ?›ź − đ?‘Žđ?‘›âˆ’1 + ∑đ?‘›đ?‘—=0 đ?‘Žđ?‘— đ?‘Ą1 đ?‘Ą2 + đ?‘Žđ?‘— −1 đ?‘Ą1 − đ?‘Ą2 − đ?‘Žđ?‘— −2 đ?‘§ đ?‘›âˆ’đ?‘— +1 ≤ đ?‘€1 (6) đ?‘€đ?‘–đ?‘› đ?‘§ =đ?‘… đ?‘Ž0 đ?‘Ą1 − đ?‘Ą2 + đ?‘Ž1 đ?‘Ą1 đ?‘Ą2 + đ?›˝ + ∑đ?‘›+2 đ?‘— =2 đ?‘Žđ?‘— đ?‘Ą1 đ?‘Ą2 + đ?‘Žđ?‘— −1 đ?‘Ą1 − đ?‘Ą2 −

(đ?‘Žđ?‘— −2 đ?‘§ đ?‘— ≤ đ?‘€2

Where R is a positive real number, then all the zeros of P(z) lie in the ring shaped region.

(7)

���

�1 �2 �0 � =R

đ?›˝ +đ?‘€2

, � ≤ � ≤ ���

� =R

đ?‘€1 + đ?‘Žđ?‘›

�

1

,đ?‘…

Taking t2 =0, we get the following generalization and refinement of Theorem C. Corollary 1. Let P(z) = ∑đ?‘›đ?‘—=0 đ?‘Žđ?‘— đ?‘§ đ?‘— be a polynomial of degree n. If for some t > 0. đ?‘€đ?‘Žđ?‘Ľ đ?‘§ =đ?‘… a n t + Îą − a n−1 + ∑nj=0 a j−1 t − a j−2 z n−j+2 ≤ M1 đ?‘€đ?‘–đ?‘› đ?‘§ =đ?‘… a 0 t + β + ∑n+2 a j−1 t − a j−2 z j ≤ M2 j=0 Where R is a real positive number then all the zeros of P(z) lie in www.ijmer.com

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International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4363-4372 ISSN: 2249-6645

�1 + �

z ≤ ��� � =�

đ?‘Žđ?‘›

1

,đ?‘…

In case we take t=1=R in corollary 1, we get the \following interesting result. Corollary 2. Let P(z) = ∑đ?‘›đ?‘—=0 đ?‘Žđ?‘— đ?‘§ đ?‘— be a polynomial of degree n, đ?‘€đ?‘Žđ?‘Ľ đ?‘§ =đ?‘… đ?‘Žđ?‘› + đ?›ź − đ?‘Žđ?‘›âˆ’1 + đ?‘Žđ?‘› −1 − đ?‘Žđ?‘› −2 đ?‘§ + â‹Ż + đ?‘Ž1 − đ?‘Ž0 đ?‘§ đ?‘›âˆ’1 + then all the zeros of P(z) lie in the circle z ≤ Max

M+ Îą an

đ?‘Žđ?‘œ đ?‘§ đ?‘›

≤ M,

,1

If for some Îą > 0, đ?›ź + đ?‘Žđ?‘› ≼ đ?‘Žđ?‘› −1 ≼ â‹Ż ≼ đ?‘Ž1 ≼ đ?‘Ž0 ≼ 0, then , đ?‘€ ≤ đ?‘Žđ?‘› + đ?›ź + đ?‘Žđ?‘› −1 + đ?‘Žđ?‘› −1 − đ?‘Žđ?‘›âˆ’2 + â‹Ż + đ?‘Ž1 − đ?‘Ž0 + đ?‘Ž0 = đ?‘Žđ?‘› + đ?›ź − đ?‘Žđ?‘› −1 + đ?‘Žđ?‘›âˆ’1 − đ?‘Žđ?‘›âˆ’2 + â‹Ż + đ?‘Ž1 − đ?‘Ž0 + đ?‘Ž0 = đ?‘Žđ?‘› + đ?›ź Using this observation in Corollary 2, we get the following generalization of Enestrom-Kakeya Theorem. Corollary 3. If, P(z)= đ?‘Žđ?‘› đ?‘§ đ?‘› + đ?‘Žđ?‘› −1 + đ?‘§ đ?‘› −1 + â‹Ż + đ?‘Ž1 đ?‘§ + đ?‘Ž0 , is a polynomial of degree n, such that for some đ?›ź ≼ 0, đ?›ź + đ?‘Žđ?‘› ≼ đ?‘Žâˆ’1 ≼ â‹Ż . ≼ đ?‘Ž0 ≼ 0, then all the zeros of P(z) lie in the circle 2đ?›ź

� ≤ 1+� , �

For đ?›ź = 0, đ?‘Ąâ„Žđ?‘–đ?‘ đ?‘&#x;đ?‘’đ?‘‘đ?‘˘đ?‘?đ?‘’đ?‘  đ?‘Ąđ?‘œ đ?‘‡â„Žđ?‘’đ?‘œđ?‘&#x;đ?‘’đ?‘š đ??´. đ??źđ?‘“ đ?‘¤đ?‘’ đ?‘Ąđ?‘Žđ?‘˜đ?‘’ đ?›ź = đ?‘Žđ?‘›âˆ’1 − đ?‘Žđ?‘› ≼ 0, then we get Corollary 2, of ([4] Aziz and Zarger). Next we present the following interesting result which includes Theorem A as a special case. Theorem 2. Let P(z) = ∑đ?‘›đ?‘—=0 đ?‘Žđ?‘— đ?‘§ đ?‘— be a polynomial of degree n, If for some real numbers with đ?’•đ?&#x;? ≠ đ?&#x;Ž, đ?’•đ?&#x;? > đ?’•đ?&#x;? ≼ đ?&#x;Ž (đ?&#x;–) đ?‘´đ?’‚đ?’™ đ?’› =đ?‘š ∑đ?’?đ?’‹=đ?&#x;Ž đ?’‚đ?’‹ đ?’•đ?&#x;? đ?’•đ?&#x;? + đ?’‚đ?’‹âˆ’đ?&#x;? đ?’•đ?&#x;? − đ?’•đ?&#x;? − đ?’‚đ?’‹âˆ’đ?&#x;? đ?’›đ?’?−đ?’‹ ≤ đ?‘´3 then all the zeros of P(z) lie in the region (9) đ?‘§ ≤ đ?‘&#x;1, Where , (10)

đ?‘&#x;1 =

2đ?‘€3 đ?‘Ž đ?‘› đ?‘Ą 1 −đ?‘Ą 2 đ?‘Ž đ?‘› −1 2 +4 đ?‘Ž đ?‘› đ?‘€3

1/2 − đ?‘Ž đ?‘Ą −đ?‘Ą − đ?‘Ž đ?‘› 1 2 đ?‘› −1

Taking đ?‘Ą2 = 0, we get the following result Corollary 4. Let P(z) = ∑đ?‘›đ?‘—=0 đ?‘Žđ?‘— đ?‘§ đ?‘— be a polynomial of degree n, If for some đ?‘Ą > 0, đ?‘´đ?’‚đ?’™ đ?’› =đ?‘š ∑đ?’?đ?’‹=đ?&#x;Ž đ?’‚đ?’‹âˆ’đ?&#x;? đ?’• − đ?’‚đ?’‹âˆ’đ?&#x;? đ?’›đ?’?−đ?’‹ ≤ đ?‘´3, then all the zeros of P(z) lie in the region đ?‘§ ≤ đ?‘&#x;1, Where , r= Remark.

2đ?‘€3 đ?‘Ž đ?‘› đ?‘Ąâˆ’ đ?‘Ž đ?‘› −1 2 +4 đ?‘Ž đ?‘› đ?‘€3

1/2 − đ?‘Ž đ?‘Ąâˆ’đ?‘Ž đ?‘› đ?‘› −1

Suppose polynomial P(z) = ∑đ?‘›đ?‘—=0 đ?‘Žđ?‘— đ?‘§ đ?‘— , satisfies the conditions of www.ijmer.com

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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 2đ?‘Žđ?‘› −1

� ≤

2

đ?‘Žđ?‘› − đ?‘Žđ?‘› −1

+ 4đ?‘Žđ?‘› đ?‘Žđ?‘›âˆ’1 − đ?‘Žđ?‘› − đ?‘Žđ?‘› −1

2đ?‘Ž

= 2đ?‘Ž đ?‘› −1 = 1, đ?‘› −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) = ∑đ?‘›đ?‘—=0 đ?‘Žđ?‘— đ?‘§ đ?‘— be a polynomial of degree n, Ifđ?›ź, đ?›˝ are complex numbers and with đ?‘Ą1 ≠0, đ?‘Ą2 đ?‘Žđ?‘&#x;đ?‘’ đ?‘&#x;đ?‘’đ?‘Žđ?‘™ đ?‘›đ?‘˘đ?‘šđ?‘?đ?‘’đ?‘&#x;đ?‘  đ?‘¤đ?‘–đ?‘Ąâ„Ž đ?‘Ą1 ≼ đ?‘Ą2 ≼ 0 and (11) đ?‘€đ?‘Žđ?‘Ľ đ?‘§ =đ?‘… đ?‘Žđ?‘› đ?‘Ą1 − đ?‘Ą2 + đ?›ź − đ?‘Žđ?‘› −1 đ?‘§ + ∑đ?‘›đ?‘—=0 đ?‘Žđ?‘— đ?‘Ą1 đ?‘Ą2 + đ?‘Žđ?‘— −1 đ?‘Ą1 −đ?‘Ą2 − đ?‘Žđ?‘— −2 đ?‘§ đ?‘› −đ?‘— −2 ≤ đ?‘€4 and, đ?‘— (12) đ?‘€đ?‘Žđ?‘Ľ đ?‘§ =đ?‘… đ?‘Žđ?‘› đ?‘Ą1 − đ?‘Ą2 + đ?›˝ − đ?‘Žđ?‘› −1 đ?‘§ + ∑đ?‘›+2 ≤ đ?‘€5 đ?‘— =2 đ?‘Žđ?‘— đ?‘Ą1 đ?‘Ą2 + đ?‘Žđ?‘— −1 đ?‘Ą1 −đ?‘Ą2 − đ?‘Žđ?‘— −2 đ?‘§

Where đ?‘Žâˆ’2= đ?‘Žâˆ’1 = 0 = đ?‘Žđ?‘›+1 = đ?‘Žđ?‘›+2 and R is any positive real number, then all the zeros of P(z) lie in the ring shaped region. 1

Min (r2 , R) ≤ đ?‘§ ≤ max đ?‘&#x;1 , đ?‘…

(13) Where đ?‘&#x;1 =

(14)

đ?›ź đ?‘… 2 đ?‘€4 +đ?‘… 2 đ?‘€4 − đ?‘Ž đ?‘›

2[ đ?‘Ž đ?‘… 2 đ?‘Ž đ?‘› đ?‘Ą 1− đ?‘Ą 2 −đ?‘Ž đ?‘› −1 +đ?›ź +đ?‘€42 ] đ?‘Ž đ?‘› đ?‘Ą 1− đ?‘Ą 2 +đ?›źâˆ’đ?‘Ž đ?‘› −1 2 +4 đ?›ź đ?‘… 2 đ?‘Ž đ?‘› đ?‘Ą 1− đ?‘Ą 2 +đ?›źâˆ’đ?‘Ž đ?‘› −1 +đ?‘€42

đ?›ź đ?‘…2 đ?‘€4 + đ?‘…2 đ?‘€4 − đ?‘Žđ?‘›

đ?‘Ž đ?‘› đ?‘… 2 đ?‘€4

1/2 −

đ?‘Žđ?‘› đ?‘Ą1− đ?‘Ą2 + đ?›ź − đ?‘Žđ?‘› −1

and (15)

1 đ?‘&#x;2

= đ?‘…2

2[đ?‘… 2 đ?›˝

đ?‘Ž 1 đ?‘Ą 1 đ?‘Ą 2 +đ?‘Ž 0 đ?‘Ą 1− đ?‘Ą 2 +đ?›˝ −đ?‘€52 ]

đ?‘Ž 1 đ?‘Ą 1 đ?‘Ą 2 +đ?‘Ž 0 đ?‘Ą 1− đ?‘Ą 2 +đ?›˝ ( đ?‘Ž 0 đ?‘Ą 1 đ?‘Ą 2 −đ?‘€5 )− đ?›˝ đ?‘€5 +

đ?‘…4

đ?‘Ž1 đ?‘Ą1 đ?‘Ą2 + đ?‘Ž0 đ?‘Ą1− đ?‘Ą2 + đ?›˝

+4 đ?‘Ž0

đ?‘Ž0 đ?‘Ą1 đ?‘Ą2 − đ?‘€5

2

đ?‘€5 đ?‘Ą1 đ?‘Ą2 (đ?‘… 2 đ?›˝ đ?‘Ž1 đ?‘Ą1 đ?‘Ą2 + đ?‘Ž0 đ?‘Ą1− đ?‘Ą2 ) + đ?›˝ + đ?‘€52

1/2

Taking đ?›ź = 0, đ?›˝ = 0, in Theorem 3, we get the following. Corollary 5. Let P(z) = ∑đ?‘›đ?‘—=0 đ?‘Žđ?‘— đ?‘§ đ?‘— be a polynomial of degree n, If ,đ?‘Ą1 ≠0 đ?‘Žđ?‘›đ?‘‘ đ?‘Ą2 are real numbers with đ?‘Ą1 ≼ đ?‘Ą2 ≼ 0, đ?‘›âˆ’đ?‘— +2 đ?‘€đ?‘Žđ?‘Ľ đ?‘§ =đ?‘… ∑đ?‘›+1 ≤ đ?‘€4 , đ?‘— =1 đ?‘Žđ?‘— đ?‘Ą1 đ?‘Ą2 + đ?‘Žđ?‘— −1 đ?‘Ą1− đ?‘Ą2 − đ?‘Žđ?‘— −2 đ?‘§ đ??Ł đ??Œđ??šđ??ą đ??ł =đ??‘ ∑đ??§+đ?&#x;? ≤ đ??Œđ?&#x;“ , đ??Ł=đ?&#x;? đ??šđ??Ł đ??­ đ?&#x;? đ??­ đ?&#x;? + đ??šđ??Łâˆ’đ?&#x;? đ??­ đ?&#x;?− đ??­ đ?&#x;? − đ??šđ??Łâˆ’đ?&#x;? đ??ł

Where R is any positive real number . then all the zeros of P(z) lie in the ring shaped region min (r2,R) ≤ đ?‘§ ≤ max đ?‘&#x;1 ,

1 đ?‘…

Where đ?‘&#x;1 =

đ?‘…4

đ?‘Ą 1− đ?‘Ą 2 ) đ?‘Ž đ?‘› −đ?‘Ž đ?‘› −1 2 đ?‘€4 − đ?‘Ž đ?‘›

2đ?‘€42 2 +4 đ?‘Ž đ?‘… 2 đ?‘€ 3 1/2 − đ?‘› 4

đ?‘Ą 1 −đ?‘Ą 2 )đ?‘Ž đ?‘› −đ?‘Ž đ?‘› −1 đ?‘€4 − đ?‘Ž đ?‘› đ?‘… 2

đ?‘&#x;2 =

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International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4363-4372 ISSN: 2249-6645 1 2𝑀52

𝑅4 𝑎1 𝑡1 𝑡2 + 𝑎0 𝑡1 − 𝑡2

2

𝑀5 − 𝑎0 𝑡1 𝑡2

2

+ 4𝑀53 𝑅 2 𝑎0 𝑡1 𝑡2

1 2

𝑅 2 𝑀5 − 𝑎0 𝑡1 𝑡2 𝑎1 𝑡1 𝑡2 + 𝑎0 𝑡1 − 𝑡2 The 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 and Schmesser [7]. LEMMA. 1. If 𝑓(z) is analytic in 𝑧 ≤ 1, 𝑓 𝑜 = 𝑎 𝑤ℎ𝑒𝑟𝑒 𝑎 < 1, 𝑓′ 𝑜 = 𝑏, 𝑓(𝑧) ≤ 1, 𝑜𝑛 𝑧 = 1, 𝑡ℎ𝑒𝑛 𝑓𝑜𝑟 𝑧 ≤ 1. 𝑓(𝑧) ≤

𝑧 2 + 𝑏 𝑧 + 𝑎 1− 𝑎

1− 𝑎

𝑧 2 + 𝑏 𝑧 + 1− 𝑎

𝑎 1− 𝑎

The example 𝑓 (z)=

𝑏 𝑧−𝑧 2 1+𝑎 𝑏 1− 𝑧−𝑎𝑧 2 1+𝑎

𝑎+

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 ′

0

= 𝑏 𝑎𝑛𝑑 𝑓(𝑧) ≤ 𝑀,

𝑓𝑜𝑟 𝑧 = 𝑅, 𝑡ℎ𝑒𝑛 𝑓(𝑧) ≤

𝑀 𝑧 𝑀 𝑧 +𝑅 2 𝑏 𝑅2

𝑀+ 𝑏 𝑧

for 𝑧 ≤ 𝑅

III. Proof of Theorem . 1 (16)

PROOFS OF THE THEOREMS.

Consider F(z) = 𝑡2 + 𝑧 𝑡1 − 𝑧 𝑃(𝑧) = −𝑎𝑛 𝑧 𝑛+2 + 𝑎𝑛 𝑡1 − 𝑡2 − 𝑎𝑛−1 𝑧 𝑛+1 + 𝑎𝑛 𝑡1 𝑡2 + 𝑎𝑛 −1 𝑡1 − 𝑡2 − 𝑎𝑛−2 𝑧 𝑛 +….+ 𝑎𝑛 𝑡1 𝑡2+ 𝑎0 𝑡1 − 𝑡2

𝑧 + 𝑎0 𝑡1 𝑡2

Let G(z) + 𝑧 𝑛+2 𝐹

1 𝑧

= −𝑎𝑛 + 𝑎𝑛 𝑡1 − 𝑡2 − 𝑎𝑛−1 𝑧 + 𝑎𝑛 𝑡1 𝑡2 + 𝑎𝑛 −1 𝑡1 − 𝑡2 − 𝑎𝑛 −2 𝑧 2 + ⋯ …+ 𝑎1 𝑡1 𝑡2 + 𝑎0 𝑡1 − 𝑡2 𝑧 𝑛 +1 + 𝑎0 𝑡1 𝑡2 𝑧 𝑛+2 = −𝑎𝑛 − 𝛼𝑧 + 𝑎𝑛 𝑡1 − 𝑡2 + 𝛼 − 𝑎𝑛−1 𝑧 + 𝑎𝑛 𝑡1 𝑡2 + 𝑎𝑛−1 𝑡1 − 𝑡2 − 𝑎𝑛 −2 𝑧 2 + ⋯ … + (𝑎1 𝑡1 𝑡2 + 𝑎0 𝑡1 − 𝑡2 𝑧 𝑛+1 + 𝑎0 𝑡1 𝑡2 ) 𝑧 𝑛+2 = − 𝑎𝑛 + 𝛼𝑧 + 𝐻(𝑧) Where, H(z)= 𝑧 𝑎𝑛 𝑡1 − 𝑡2 + 𝛼 − 𝑎𝑛 −1 + ∑𝑛𝑗=0 𝑎𝑗 𝑡1 𝑡2 + 𝑎𝑗 −1 𝑡1 − 𝑡2 − 𝑎𝑗 −2 𝑧 𝑛−𝑗 www.ijmer.com

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International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4363-4372 ISSN: 2249-6645 Clearly H(o)=0 and đ?&#x2018;&#x20AC;đ?&#x2018;&#x17D;đ?&#x2018;Ľ đ?&#x2018;§ =đ?&#x2018;&#x2026; đ??ť(đ?&#x2018;§) â&#x2030;¤ đ?&#x2018;&#x2026;đ?&#x2018;&#x20AC;1, We first assume that đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2030;¤ đ?&#x2018;&#x2026; đ?&#x2018;&#x20AC;1 + đ?&#x203A;ź Now for đ?&#x2018;§ â&#x2030;¤ đ?&#x2018;&#x2026;, by using Schwarz Lemma, we have (17)

G(z) = â&#x2C6;&#x2019; a n + Îąz + H(z) â&#x2030;Ľ a n â&#x2C6;&#x2019; đ?&#x203A;ź đ?&#x2018;§ â&#x2C6;&#x2019; đ??ť(đ?&#x2018;§) â&#x2030;Ľ a n â&#x2C6;&#x2019; đ?&#x203A;ź đ?&#x2018;§ â&#x2C6;&#x2019; |đ?&#x2018;&#x20AC;1 đ?&#x2018;§| = a n â&#x2C6;&#x2019; đ?&#x2018;&#x20AC;1 + đ?&#x203A;ź

đ?&#x2018;§|

> đ?&#x2018;&#x153;, đ?&#x2018;&#x2013;đ?&#x2018;&#x201C; đ?&#x2018;§ <đ?&#x2018;&#x20AC;

đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;

1+ đ?&#x203A;ź

â&#x2030;¤đ?&#x2018;&#x2026;

This shows that all the zeros of G(z) lie in đ?&#x2018;§ â&#x2030;Ľ 1

Replacing z by đ?&#x2018;§ and noting that F(z) = đ?&#x2018;§ đ?&#x2018;&#x203A;+2 đ??ş đ?&#x2018;§ â&#x2030;¤

đ?&#x2018;&#x20AC;1 + đ?&#x203A;ź đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;

đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; đ?&#x2018;&#x20AC;1 + đ?&#x203A;ź

1 đ?&#x2018;§

, it follows that all the zeros of F(z) lie in

if , đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2030;¤ đ?&#x2018;&#x2026; (đ?&#x2018;&#x20AC;1 + đ?&#x203A;ź )

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 đ?&#x2018;§ â&#x2030;¤đ?&#x2018;&#x20AC;

(18)

đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;

1+ đ?&#x203A;ź

Now assume đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; > đ?&#x2018;&#x2026; (đ?&#x2018;&#x20AC;1 + đ?&#x203A;ź ), then for đ?&#x2018;§ â&#x2030;¤ đ?&#x2018;&#x2026;, we have from (17). G z â&#x2030;Ľ a n â&#x2C6;&#x2019; đ?&#x203A;ź đ?&#x2018;§ â&#x2C6;&#x2019; đ??ť(đ?&#x2018;?) â&#x2030;Ľ đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019; đ?&#x203A;ź đ?&#x2018;&#x2026; â&#x2C6;&#x2019; đ?&#x2018;&#x20AC;1 đ?&#x2018;&#x2026; = đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019; đ?&#x203A;ź + đ?&#x2018;&#x20AC; đ?&#x2018;&#x2026; > 0, Thus G(z) â&#x2030; 0 đ?&#x2018;&#x201C;đ?&#x2018;&#x153;đ?&#x2018;&#x; đ?&#x2018;§ < đ?&#x2018;&#x2026;, đ?&#x2018;&#x201C;đ?&#x2018;&#x;đ?&#x2018;&#x153;đ?&#x2018;&#x161; đ?&#x2018;¤â&#x201E;&#x17D;đ?&#x2018;&#x2013;đ?&#x2018;?â&#x201E;&#x17D; it follows as before that all the zeros of F(z) and hence all the zeros of P(z) lie 1, in đ?&#x2018;§ â&#x2030;¤ đ?&#x2018;&#x2026; Combining this with (18), we infer that all the zeros of P(z) lie in đ??ł â&#x2030;¤ đ??Śđ??&#x161;đ??ą

19.

đ??&#x161;đ??§ đ??&#x152;đ?&#x;? + đ?&#x203A;&#x201A;

đ?&#x;?,

,đ??&#x2018;

Now to prove the second part of the Theorem, from (16) we can write F(z) as F(z)= t 2 + z t1 â&#x2C6;&#x2019; z P(z) = a 0 t1 t 2 + a1 t1 t 2 + a 0 t1 â&#x2C6;&#x2019; t 2 z + â&#x2039;Ż + a n t1 â&#x2C6;&#x2019; t 2 â&#x2C6;&#x2019; a nâ&#x2C6;&#x2019;1 z n+1 â&#x2C6;&#x2019; a n z n+2 =a 0 t1 t2 â&#x2C6;&#x2019; βz + a1 t1 t 2 + a 0 t1 â&#x2C6;&#x2019; t 2 + β z + â&#x2039;Ż + a n t1 â&#x2C6;&#x2019; t 2 â&#x2C6;&#x2019; a nâ&#x2C6;&#x2019;1 z n+1 â&#x2C6;&#x2019; a n z n+2 = a 0 t1 t 2 â&#x2C6;&#x2019; βz + T(z) Where đ?&#x2018;&#x2014; â&#x2C6;&#x2019;1 T(z) = z đ?&#x2018;&#x17D;1 đ?&#x2018;Ą1 đ?&#x2018;Ą2 + đ?&#x2018;&#x17D;0 đ?&#x2018;Ą1 â&#x2C6;&#x2019;đ?&#x2018;Ą2 + đ?&#x203A;˝ + â&#x2C6;&#x2018;n+2 j=2 đ?&#x2018;&#x17D;đ?&#x2018;&#x2014; đ?&#x2018;Ą1 đ?&#x2018;Ą2 + đ?&#x2018;&#x17D;đ?&#x2018;&#x2014; â&#x2C6;&#x2019;1 đ?&#x2018;Ą1 â&#x2C6;&#x2019;đ?&#x2018;Ą2 đ?&#x2018;&#x17D;đ?&#x2018;&#x2014; â&#x2C6;&#x2019;2 đ?&#x2018;§

Clearly T(o)=0, and

đ?&#x2018;&#x20AC;đ?&#x2018;&#x17D;đ?&#x2018;Ľ đ?&#x2018;§ =đ?&#x2018;&#x2026; đ?&#x2018;&#x2021;(đ?&#x2018;§) â&#x2030;¤ đ?&#x2018;&#x2026;đ?&#x2018;&#x20AC;2

We first assume that đ?&#x2018;&#x17D;0 đ?&#x2018;Ą1 đ?&#x2018;Ą2 â&#x2030;¤ β + M2 www.ijmer.com

4367 | Page


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 â&#x2030;¤ R, by using Schwasr lemma, we have F(z) â&#x2030;Ľ đ?&#x2018;&#x17D;0 đ?&#x2018;Ą1 đ?&#x2018;Ą2 â&#x2C6;&#x2019; β z â&#x2C6;&#x2019; T(z) â&#x2030;Ľ đ?&#x2018;&#x17D;0 đ?&#x2018;Ą1 đ?&#x2018;Ą2 â&#x2C6;&#x2019; β z â&#x2C6;&#x2019; M2 Z >0, if, z <

đ?&#x2018;&#x17D;0 đ?&#x2018;Ą1 đ?&#x2018;Ą2

â&#x2030;¤R

β +M 2

This shows that F(z) â&#x2030; 0, for z < z â&#x2030;Ľ

đ?&#x2018;&#x17D;0 đ?&#x2018;Ą1 đ?&#x2018;Ą2 β +M 2

, hence all the zeros of F(z) lie in

đ?&#x2018;&#x17D;0 đ?&#x2018;Ą1 đ?&#x2018;Ą2 β +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

z â&#x2030;Ľ

20.

đ?&#x2018;&#x17D;0 đ?&#x2018;Ą1 đ?&#x2018;Ą2 β +M 2

Now we assume đ?&#x2018;&#x17D;0 đ?&#x2018;Ą1 đ?&#x2018;Ą2 > β + M2 , then for z â&#x2030;¤ R, we have `

F(z) â&#x2030;Ľ đ?&#x2018;&#x17D;0 đ?&#x2018;Ą1 đ?&#x2018;Ą2 â&#x2C6;&#x2019; β z â&#x2C6;&#x2019; T(z) â&#x2030;Ľ đ?&#x2018;&#x17D;0 đ?&#x2018;Ą1 đ?&#x2018;Ą2 â&#x2C6;&#x2019; β R â&#x2C6;&#x2019; M2 R > đ?&#x2018;Ą1 đ?&#x2018;Ą2 đ?&#x2018;&#x17D;0 â&#x2C6;&#x2019;

β + M2 R

>0, This shows that all the zeros of F(z) and hence that of P(z) lie in z â&#x2030;ĽR Combining this with (20) , it follows that all the zeros of P(z) lie in z â&#x2030;Ľ min

21

đ?&#x2018;&#x17D;0 đ?&#x2018;Ą1 đ?&#x2018;Ą2 β +M 2

,R

Combining (19) and (20), the desired result follows. Proof of Theorem 2:-

Consider F(z) = đ?&#x2018;Ą2 + đ?&#x2018;§ đ?&#x2018;Ą1 â&#x2C6;&#x2019; đ?&#x2018;§ đ?&#x2018;&#x192;(đ?&#x2018;§)

22.

= đ?&#x2018;Ą2 + đ?&#x2018;§ đ?&#x2018;Ą1 â&#x2C6;&#x2019; đ?&#x2018;§

đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; đ?&#x2018;§ đ?&#x2018;&#x203A; + đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 đ?&#x2018;§ đ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 + â&#x2039;Ż + đ?&#x2018;&#x17D;1 đ?&#x2018;§ + đ?&#x2018;&#x17D;0

= â&#x2C6;&#x2019;đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; đ?&#x2018;§ đ?&#x2018;&#x203A;+2 + đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; đ?&#x2018;Ą1â&#x2C6;&#x2019; đ?&#x2018;Ą2

â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 đ?&#x2018;§ đ?&#x2018;&#x203A; +1 + â&#x2039;Ż + đ?&#x2018;&#x17D;0 đ?&#x2018;Ą1 đ?&#x2018;Ą2

Let 1

G(z) = đ?&#x2018;§ đ?&#x2018;&#x203A;+2 đ??š(đ?&#x2018;§ )

23.

= â&#x2C6;&#x2019;đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; + đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; đ?&#x2018;Ą1 â&#x2C6;&#x2019; đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 đ?&#x2018;§ + đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; đ?&#x2018;Ą1 đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 đ?&#x2018;Ą1 â&#x2C6;&#x2019; đ?&#x2018;Ą2 đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019;2 đ?&#x2018;§ 2 + đ??ť(đ?&#x2018;§) Where H(z) = đ?&#x2018;§ 2 đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; đ?&#x2018;Ą1 đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 đ?&#x2018;Ą1 â&#x2C6;&#x2019; đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019;2 + â&#x2039;Ż + đ?&#x2018;&#x17D;0 đ?&#x2018;Ą1 đ?&#x2018;Ą2 đ?&#x2018;§đ?&#x2018;&#x203A;

Clearly H(o)=0 = H(o), since H(z) â&#x2030;¤ M3 R2 for z = R www.ijmer.com

4368 | Page


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,

a n â&#x2030;¤ MR2 + R đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; đ?&#x2018;Ą1 â&#x2C6;&#x2019; đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1

24

Then by using Lemma 2. To H(z), it follows that M3 z M3 z

đ??ťz â&#x2030;¤

R2

M3

đ?&#x2018;&#x20AC;đ?&#x2018;&#x17D;đ?&#x2018;Ľ đ?&#x2018;§ =đ?&#x2018;&#x2026; H(z)

Hence from(23) we have G(z) â&#x2030;Ľ a n â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; đ?&#x2018;Ą1 â&#x2C6;&#x2019; đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 z â&#x2C6;&#x2019;

z 2M3 R2

đ?&#x2018;&#x2026;2

>0, if

M3 z

2

+ đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; đ?&#x2018;Ą1 â&#x2C6;&#x2019; đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 z â&#x2C6;&#x2019; a n < 0

That is if, z <

25.

1/2 đ?&#x2018;&#x17D; đ?&#x2018;&#x203A; đ?&#x2018;Ą 1 â&#x2C6;&#x2019;đ?&#x2018;Ą 2 â&#x2C6;&#x2019;đ?&#x2018;&#x17D; đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 2 +4 a n M â&#x2C6;&#x2019; đ?&#x2018;&#x17D; đ?&#x2018;&#x203A; đ?&#x2018;Ą 1 â&#x2C6;&#x2019;đ?&#x2018;Ą 2 â&#x2C6;&#x2019;đ?&#x2018;&#x17D; đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1

2M 3

1

=r â&#x2030;¤ R If đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; đ?&#x2018;Ą1 â&#x2C6;&#x2019; đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1

2

+ 4 a n M3 â&#x2030;¤ 2M3 R + đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; đ?&#x2018;Ą1 â&#x2C6;&#x2019; đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;â&#x2C6;&#x2019;1

2

Which implies a n â&#x2030;¤ M3 R2 + R đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; đ?&#x2018;Ą1 â&#x2C6;&#x2019; đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 Which is true by (24)..Hence all the zeros of G(z) lie in Z â&#x2030;Ľ 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 â&#x2030;¤ r. We now assume. a n â&#x2030;¤ M3 R2 + R đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; đ?&#x2018;Ą1 â&#x2C6;&#x2019; đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 ,

26.

then for z â&#x2030;¤ R, from (23) it follows that G(z) â&#x2030;Ľ a n â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; đ?&#x2018;Ą1 â&#x2C6;&#x2019; đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 z â&#x2C6;&#x2019; H(z) â&#x2030;Ľ đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019; đ?&#x2018;&#x2026; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; đ?&#x2018;Ą1 â&#x2C6;&#x2019; đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 â&#x2C6;&#x2019; R2 M3 >0, by (26). 1

1

This shows that all the zeros of G(z) lie in Z > đ?&#x2018;&#x2026;, and hence all the zeros of F(z) = z n+2 G(z ) lie in

z â&#x2030;¤ , but all the zeros of P(z) are also the zeros of R

F(z) , therefore it follows that all the zeros of P(z) lie in

1

Z â&#x2030;¤R 27.

From (25) amd (27) , we conclude that all the zeros of P(z) lie in 1

Z â&#x2030;¤ max(đ?&#x2018;&#x;, ) R

Which completes the proof of Theorem 2.

28. PROOF OF THEOREM 3. Consider the polynomial www.ijmer.com

4369 | Page


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) = đ?&#x2018;Ą2 + đ?&#x2018;§ đ?&#x2018;Ą1 â&#x2C6;&#x2019; đ?&#x2018;§ đ?&#x2018;&#x192;(đ?&#x2018;§) = đ?&#x2018;Ą1 đ?&#x2018;Ą2 + đ?&#x2018;Ą1 â&#x2C6;&#x2019; đ?&#x2018;Ą2 đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;§ 2 ) đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; đ?&#x2018;§ đ?&#x2018;&#x203A; + đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 đ?&#x2018;§ đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 + â&#x2039;Ż + đ?&#x2018;&#x17D;1 đ?&#x2018;§ + đ?&#x2018;&#x17D;0 = â&#x2C6;&#x2019;đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; đ?&#x2018;§ đ?&#x2018;&#x203A;+2 + đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; đ?&#x2018;Ą1â&#x2C6;&#x2019; đ?&#x2018;Ą2 + đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 đ?&#x2018;Ą1â&#x2C6;&#x2019; đ?&#x2018;Ą2

â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 đ?&#x2018;§ đ?&#x2018;&#x203A; +1 + â&#x2039;Ż + đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; đ?&#x2018;Ą1 đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 đ?&#x2018;§ đ?&#x2018;&#x203A;+1 + â&#x2039;Ż +

a 2 t1 t 2 + a1 đ?&#x2018;Ą1â&#x2C6;&#x2019; đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;0 z 2 + a1 t1 t 2 + a 0 đ?&#x2018;Ą1â&#x2C6;&#x2019; đ?&#x2018;Ą2 đ?&#x2018;§ + đ?&#x2018;&#x17D;0 t1 t 2 We have, 1 G(z) = đ?&#x2018;§ đ?&#x2018;&#x203A;+2 đ??š( ) đ?&#x2018;§ = đ?&#x2018;&#x;đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; + đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; đ?&#x2018;Ą1 â&#x2C6;&#x2019; đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 đ?&#x2018;§ + đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; đ?&#x2018;Ą1 đ?&#x2018;Ą2 + đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 đ?&#x2018;Ą1 â&#x2C6;&#x2019; đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019;2 đ?&#x2018;§ 2 +â&#x20AC;Ś+ đ?&#x2018;&#x17D;1 đ?&#x2018;Ą1 đ?&#x2018;Ą2 + đ?&#x2018;&#x17D;0 đ?&#x2018;Ą1â&#x2C6;&#x2019; đ?&#x2018;Ą2 đ?&#x2018;§ đ?&#x2018;&#x203A; +1 + đ?&#x2018;&#x17D;0 đ?&#x2018;Ą1 đ?&#x2018;Ą2 đ?&#x2018;§ đ?&#x2018;&#x203A;+2 = â&#x2C6;&#x2019;đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019; đ?&#x203A;źđ?&#x2018;§ + đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; đ?&#x2018;Ą1 â&#x2C6;&#x2019; đ?&#x2018;Ą2 + đ?&#x203A;ź â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 đ?&#x2018;§ + đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; đ?&#x2018;Ą1 đ?&#x2018;Ą2 + đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 đ?&#x2018;Ą1â&#x2C6;&#x2019; đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019;2 z 2 +. . + đ?&#x2018;&#x17D;1 đ?&#x2018;Ą1 đ?&#x2018;Ą2 + đ?&#x2018;&#x17D;0 đ?&#x2018;Ą1â&#x2C6;&#x2019; đ?&#x2018;Ą2 z n+1 + đ?&#x2018;&#x17D;0 đ?&#x2018;Ą1 đ?&#x2018;Ą2 đ?&#x2018;§ đ?&#x2018;&#x203A;+2 29. = â&#x2C6;&#x2019;đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019; đ?&#x203A;źđ?&#x2018;§ + đ??ť đ?&#x2018;§ Where H(z) = đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; đ?&#x2018;Ą1 â&#x2C6;&#x2019;đ?&#x2018;Ą2 + đ?&#x203A;ź â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 đ?&#x2018;§ + â&#x2039;Ż + đ?&#x2018;&#x17D;0 đ?&#x2018;Ą1 đ?&#x2018;Ą2 đ?&#x2018;§ đ?&#x2018;&#x203A;+2 We first assume that 30. a n â&#x2030;Ľ Îą R + M4 Then for z < đ?&#x2018;&#x2026;, we have 31. G(z) â&#x2030;Ľ a n â&#x2C6;&#x2019; Îą z â&#x2C6;&#x2019; H(z) Since H(z) â&#x2030;¤ M4 đ?&#x2018;&#x201C;đ?&#x2018;&#x153;đ?&#x2018;&#x; z â&#x2030;¤ R Therefore for đ?&#x2018;§ <R, from (31) with the help of (30), we have đ??ş(đ?&#x2018;§) > đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019; â&#x2C6;? đ?&#x2018;&#x2026; â&#x2C6;&#x2019; đ?&#x2018;&#x20AC;4 1 therefore , all the zeros of G(z) lie in z â&#x2030;Ľ R, in this case. Since F(z) = đ?&#x2018;§ đ?&#x2018;&#x203A; +2 G đ?&#x2018;§ therefore all the zeros of G(z) lie in 1

Z â&#x2030;¤ 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

32.

1

z â&#x2030;¤R Now we assume a n < Îą R + M4 , clearly H(o) = 0 and đ?&#x203A;ź â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 , Since by (11), H(z) â&#x2030;¤ M4 , for z = R, therefore it follows by Lemma2. that M4 z M4 z + R2 a n đ?&#x2018;Ą1 â&#x2C6;&#x2019;đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 + đ?&#x203A;ź H(z) â&#x2030;¤ R2 M4 + a n đ?&#x2018;Ą1 â&#x2C6;&#x2019;đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 + đ?&#x203A;ź z for z â&#x2030;¤ R Using this in (31), we get, M4 z M4 z + R2 a n đ?&#x2018;Ą1 â&#x2C6;&#x2019;đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 + đ?&#x203A;ź G(z) â&#x2030;Ľ đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019; đ?&#x203A;ź đ?&#x2018;§ â&#x2C6;&#x2019; R2 M4 + a n đ?&#x2018;Ą1 â&#x2C6;&#x2019;đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 + đ?&#x203A;ź z >0, if

H(o) = (đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; đ?&#x2018;Ą1 â&#x2C6;&#x2019;đ?&#x2018;Ą2 â&#x2C6;&#x2019;

đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; R2 M4 + a n đ?&#x2018;Ą1 â&#x2C6;&#x2019;đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 + đ?&#x203A;ź z â&#x2C6;&#x2019; z Îą R2 M4 + a n đ?&#x2018;Ą1 â&#x2C6;&#x2019;đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 + đ?&#x203A;ź z + M4 M4 z + R2 a n đ?&#x2018;Ą1 â&#x2C6;&#x2019;đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 + đ?&#x203A;ź > 0, this implies, Îą R2 a n đ?&#x2018;Ą1 â&#x2C6;&#x2019;đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 + đ?&#x203A;ź + M42 z 2 + Îą R2 M4 + R2 M4 a n t1 â&#x2C6;&#x2019; t 2 â&#x2C6;&#x2019; a nâ&#x2C6;&#x2019;1 +â&#x2C6;? â&#x20AC;Ś R2 M4 an đ?&#x2018;Ą1 â&#x2C6;&#x2019;đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 + đ?&#x203A;ź } z â&#x2C6;&#x2019; a n R2 M4 < 0, This gives, G(z) > 0, đ?&#x2018;&#x2013;đ?&#x2018;&#x201C; z < [ Îą R2 M4 + R2 M4 â&#x2C6;&#x2019; a n a n đ?&#x2018;Ą1 â&#x2C6;&#x2019;đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 + đ?&#x203A;ź 2 +.. â&#x20AC;Ś.4 Îą R2 a n đ?&#x2018;Ą1 â&#x2C6;&#x2019;đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 + đ?&#x203A;ź + M4 2 đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; R2 M4 ] â&#x2C6;&#x2019;

đ?&#x203A;ź R 2 M 4 +R 2 M 4 â&#x2C6;&#x2019; đ?&#x2018;&#x17D; đ?&#x2018;&#x203A;

a n đ?&#x2018;Ą 1 â&#x2C6;&#x2019;đ?&#x2018;Ą 2 â&#x2C6;&#x2019;đ?&#x2018;&#x17D; đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 +đ?&#x203A;ź

2 Îą R 2 a n đ?&#x2018;Ą 1 â&#x2C6;&#x2019;đ?&#x2018;Ą 2 â&#x2C6;&#x2019;đ?&#x2018;&#x17D; đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 +đ?&#x203A;ź +M 4 2 1

=r , 1

So, z < đ?&#x2018;&#x2026;, đ?&#x2018;&#x2013;đ?&#x2018;&#x201C; Îą R2 M4 + M4 â&#x2C6;&#x2019; a n R2 a n đ?&#x2018;Ą1 â&#x2C6;&#x2019;đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 + đ?&#x203A;ź www.ijmer.com

2

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International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4363-4372 ISSN: 2249-6645 4 Îą R2 a n đ?&#x2018;Ą1 â&#x2C6;&#x2019;đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 + đ?&#x203A;ź + M4 2

đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; R2 M4

< 2 Îą a n đ?&#x2018;Ą1 â&#x2C6;&#x2019;đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 + đ?&#x203A;ź R + Îą R2 M4 + 2RM4 2 + M4 â&#x2C6;&#x2019; a n R2 a n đ?&#x2018;Ą1 â&#x2C6;&#x2019;đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 + đ?&#x203A;ź

2

That is if, a n R2 M4 â&#x2030;¤ R2 Îą đ?&#x2018;&#x2026;2 an đ?&#x2018;Ą1 â&#x2C6;&#x2019;đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 + đ?&#x203A;ź + M4 2 + Îą R2 M4 + M4 â&#x2C6;&#x2019; a n R2 a n đ?&#x2018;Ą1 â&#x2C6;&#x2019;đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 + đ?&#x203A;ź R Which gives a n M4 â&#x2030;¤ M4 Îą R + M4 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; + đ?&#x2018;&#x2026; Îą R2 a n đ?&#x2018;Ą1 â&#x2C6;&#x2019;đ?&#x2018;Ą2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 + đ?&#x203A;ź + M4 2 Îą R2 + M4 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; Which is true because đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; < Îą R + M4 , 1

1

Consequently , all the zeros of G(z) lie in z â&#x2030;Ľ r , as F(z) = z n+1 G( z ), we conclude that all the zeros of 1

F(z) lie z â&#x2030;¤ đ?&#x2018;&#x;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 â&#x2030;¤ đ?&#x2018;&#x;1 Combining this with (32) it follows that all the zeros of P(z) lie in 1

34.

z â&#x2030;¤ đ?&#x2018;&#x161;đ?&#x2018;&#x17D;đ?&#x2018;Ľ đ?&#x2018;&#x;1 , đ?&#x2018;&#x2026; Again from (28) it follows that đ??š đ?&#x2018;§ = đ?&#x2018;&#x17D;0 t1 t 2 â&#x2C6;&#x2019; βz + a1 t1 t2 đ?&#x2018;Ą1â&#x2C6;&#x2019; đ?&#x2018;Ą2

+ đ?&#x203A;˝ z + â&#x2039;Ż + a n ( đ?&#x2018;Ą1 â&#x2C6;&#x2019;đ?&#x2018;Ą2

â&#x2C6;&#x2019;. đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 )z n+2 â&#x2C6;&#x2019; a n đ?&#x2018;§ đ?&#x2018;&#x203A;+2

35.

â&#x2030;Ľ đ?&#x2018;&#x17D;0 t1 t 2 â&#x2C6;&#x2019; đ?&#x203A;˝ đ?&#x2018;§ â&#x2C6;&#x2019; đ?&#x2018;&#x2021;(đ?&#x2018;§)

Where, T(z) = a n đ?&#x2018;§ đ?&#x2018;&#x203A;+2 + a n đ?&#x2018;Ą1 â&#x2C6;&#x2019;đ?&#x2018;Ą2 â&#x2C6;&#x2019;. đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1 đ?&#x2018;§ đ?&#x2018;&#x203A; +1 +. . + đ?&#x2018;&#x17D;0 t1 t 2 + a 0 đ?&#x2018;Ą1â&#x2C6;&#x2019; đ?&#x2018;Ą2 + đ?&#x203A;˝ z Clearly T(o)=0, and T(o) đ?&#x2018;&#x17D;1 t1 t 2 + a 0 đ?&#x2018;Ą1â&#x2C6;&#x2019; đ?&#x2018;Ą2 + đ?&#x203A;˝ Since by (12) , T(z) â&#x2030;¤ M5 for z = R using Lemma 2. To T(z), we have F(z) â&#x2030;Ľ đ?&#x2018;&#x17D;0 t1 t 2 â&#x2C6;&#x2019; β z â&#x2C6;&#x2019;

M5 z R2

M5 z + R2 đ?&#x2018;&#x17D;1 t1 t 2 + a 0 đ?&#x2018;Ą1â&#x2C6;&#x2019; đ?&#x2018;Ą2

= - R2 β đ?&#x2018;&#x17D;1 t1 t 2 + a 0 đ?&#x2018;Ą1â&#x2C6;&#x2019; đ?&#x2018;Ą2 đ?&#x2018;&#x17D;1 t1 t 2 + a 0 đ?&#x2018;Ą1â&#x2C6;&#x2019; đ?&#x2018;Ą2 R2 M5 + đ?&#x2018;&#x17D;1 t1 t 2 + a 0 đ?&#x2018;Ą1â&#x2C6;&#x2019; đ?&#x2018;Ą2

+đ?&#x203A;˝ z

+ đ?&#x203A;˝ + đ?&#x2018;&#x20AC;5 2 z 2 +

+ đ?&#x203A;˝ R2 đ?&#x2018;&#x17D;0 t1 t 2 â&#x2C6;&#x2019; R2 M5 â&#x2C6;&#x2019; β R2 M5 z + M5 R2 đ?&#x2018;&#x17D;1 t1 t2 +đ?&#x203A;˝ z

>0, if, R2 β đ?&#x2018;&#x17D;1 t1 t 2 + a 0 đ?&#x2018;Ą1â&#x2C6;&#x2019; đ?&#x2018;Ą2

+ đ?&#x203A;˝ + đ?&#x2018;&#x20AC;5 2 z 2 â&#x2C6;&#x2019; đ?&#x2018;&#x17D;1 t1 t 2 + a 0 đ?&#x2018;Ą1â&#x2C6;&#x2019; đ?&#x2018;Ą2

+đ?&#x203A;˝

R2 đ?&#x2018;&#x17D;0 t1 t2 â&#x2C6;&#x2019; R2 M5 â&#x2C6;&#x2019; β R2 M5 z â&#x2C6;&#x2019; R2 M5 đ?&#x2018;&#x17D;0 t1 t 2 < 0, Thus, F(z) > 0, đ?&#x2018;&#x2013;đ?&#x2018;&#x201C; z < â&#x2C6;&#x2019;R2 đ?&#x2018;&#x17D;1 t1 t2 + a 0 đ?&#x2018;Ą1â&#x2C6;&#x2019; đ?&#x2018;Ą2

+đ?&#x203A;˝

đ?&#x2018;&#x17D;0 t1 t 2 â&#x2C6;&#x2019; M5

β M5

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International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4363-4372 ISSN: 2249-6645 â&#x2C6;&#x2019; R4 đ?&#x2018;&#x17D;1 t1 t 2 + a 0 đ?&#x2018;Ą1â&#x2C6;&#x2019; đ?&#x2018;Ą2

+đ?&#x203A;˝

đ?&#x2018;&#x17D;0 t1 t 2 â&#x2C6;&#x2019; M5

β M5

2

+4 R2 β đ?&#x2018;&#x17D;1 t1 t 2 + a 0 đ?&#x2018;Ą1â&#x2C6;&#x2019; đ?&#x2018;Ą2 + đ?&#x203A;˝ + M5 2 R2 M5 đ?&#x2018;&#x17D;0 t1 t2 2 R2 β đ?&#x2018;&#x17D;1 t1 t 2 + a 0 đ?&#x2018;Ą1â&#x2C6;&#x2019; đ?&#x2018;Ą2 + đ?&#x203A;˝ + M5 2

1/2

đ?&#x2018;&#x;đ?&#x2018;§

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 â&#x2030;Ľ 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 â&#x20AC;&#x201C;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), 319329 [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

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