Some extensions of Eneström-Kakeya Theorem
aa r X i v : . [ m a t h . C V ] N ov SOME EXTENSIONS OF ENESTR ¨OM-KAKEYA THEOREM
N.A. Rather , Suhail Gulzar , ∗ and S.H. Ahangar Department of Mathematics,University of Kashmir,Hazratbal Srinagar 190006, India Department of Computer Science & Engineering,Islamic University of Science & Technology,Awantipora, Kashmir 192122, India ∗ email:[email protected] Abstract.
In this paper we obtain some refinements of a well-knownresult of Enestr¨o-Kakeya concerning the bounds for the moduli of thezeros of polynomials with complex coefficients which improve upon someresults due to Aziz and Mohammad, Govil and Rahman and others. Introduction
A classical result due to Enestr¨om (see [5]) and Kakeya [7] concerningthe bounds for the moduli of the zeros of polynomials having postive andreal coefficients is often stated as in the following theorem (see [10, p.136]or [11, p.272]).
Theorem A. If P ( z ) = P nj =0 a j z j is a polynomial of degree n with realcoefficients satisfying a n ≥ a n − ≥ · · · ≥ a ≥ a > , then all its zeros lie in | z | ≤ . In literature [1, 2, 4, 6, 8–13] there exits several extensions of Enestr¨om-Kakeya theorem. A. Aziz and Q. G. Muhammad [3] used Schwarz’s Lemmaand proved among other things the following generalization of Enestr¨om-Kakeya Theorem.
Theorem B.
Let P ( z ) = P nj =0 a j z n be a polynomial of degree n with realcoefficients. If t > t ≥ can be found such that t t a r +( t − t ) a r − − a r − ≥ r = 1 , , · · · , n +1 ( a − = a n +1 = 0) , then all the zeros of P ( z ) lie in | z | ≤ t . N.K. Govil and Q. I. Rahman [6] extended the Enestr¨om-Kakeya theoremto the polynomials with complex coefficients by proving:
Theorem C.
Let P ( z ) = P nj =0 a j z n be a polynomial of degree n with com-plex coefficients such that | arg a ν − β | ≤ α ≤ π/ , ν = 0 , , , · · · , n, Keywords: Polynomials; zeros; Enestr¨om-Kakeya theorem for some real β and | a n | ≥ | a n − | ≥ · · · | a | ≥ | a | , then all the zeros of P ( z ) lie in | z | ≤ (cos α + sin α ) + 2 sin α | a n | n X ν =0 | a ν | . The following generalization of Theorem 1.3 is due to A. Aziz and Q. G.Mohammad [3],
Theorem D.
Let P ( z ) = P nj =0 a j z n be a polynomial of degree n with realcoefficients. If t > t ≥ can be found such that | arg a ν − β | ≤ α ≤ π/ , ν = 0 , , , · · · , n for some real β and for some t > ,t n | a n | ≤ t n − | a n − | ≤ · · · ≤ t k +1 | a k +1 | ≤ t k | a k | ,t k | a k | ≥ t k − | a k − | ≥ · · · ≥ t | a | ≥ | a | > , where ≤ k ≤ n, then P ( z ) has all its zeros in the circle | z | ≤ t (cid:26)(cid:18) t k | a k | t n | a n | − (cid:19) cos α + sin α (cid:27) + 2 sin α n X j =0 | a j || a n | t n − j − . Rather et al. [12] extended Theorem 1.2 to the polynomials with complexcoefficients and proved the following result.
Theorem E.
Let P ( z ) = P nj =0 a j z n be a polynomial of degree n with com-plex coefficients such that | arg a ν − β | ≤ α ≤ π/ , ν = 0 , , , · · · , n, for some real β . If t > t ≥ can be found such that t t | a r | + ( t − t ) | a r − | − | a r − | ≥ r = 1 , , · · · , k + 1 , and t t | a r | + ( t − t ) | a r − | − | a r − | ≤ r = k + 2 , · · · , n + 1 , ≤ k ≤ n, a − = a n +1 = 0 , then all the zeros of P ( z ) lie in | z | ≤ t ( | a k | + 2 t | a k +1 | t n − k | a n | − ! cos α + sin α ) + 2 sin α | a n | n − X j =0 | a j | t n − j − . (1.1)They [12] also obtained the following generalization of Theorem 1.2. Theorem F.
Let P ( z ) = P nj =0 a j z n be a polynomial of degree n with Re a j = α j and Im a j = β j , j = 0 , , · · · , n. If t > t ≥ can be foundsuch that t t α r + ( t − t ) α r − − α r − ≥ r = 1 , , · · · , k + 1 ,t t α r + ( t − t ) α r − − α r − ≤ r = k + 2 , · · · , n + 1 , and t t β r + ( t − t ) β r − − β r − ≥ r = 1 , , · · · , m + 1 , OME EXTENSIONS OF ENESTR ¨OM-KAKEYA THEOREM 3 t t β r + ( t − t ) β r − − β r − ≤ r = m + 2 , · · · , n + 1 , ≤ k ≤ n, ≤ m ≤ n, α − = β − = α n +1 = β n +1 = 0 , α n > , then all thezero of P ( z ) lie in | z | ≤ t | a n | n t k − n (cid:0) α k + t α k +1 (cid:1) + 2 t m − n (cid:0) β m + t β m +1 (cid:1) − (cid:0) α n + β n (cid:1)o . (1.2)In this paper, we shall first present the following result which among otherthings considerably improves the bound of Theorem 1.5 for 0 ≤ k ≤ n − . Theorem 1.1.
Let P ( z ) = P nj =0 a j z n be a polynomial of degree n ≥ withcomplex coefficients such that | arg a j − β | ≤ α ≤ π/ , j = 0 , , , · · · , n for some real β. If t ( = 0) and t with t ≥ t ≥ can be found such that t t | a r | + ( t − t ) | a r − | − | a r − | ≥ r = 1 , , · · · , k + 1 , and t t | a r | + ( t − t ) | a r − | − | a r − | ≤ r = k + 2 , · · · , n + 1 , ≤ k ≤ n − , a − = a n +1 = 0 , then all the zeros of P ( z ) lie in (cid:12)(cid:12)(cid:12)(cid:12) z + a n − a n − ( t − t ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ t | a k +1 | + 2 | a k || a n | t n − k − cos α + 2 sin α | a n | n − X ν =0 | a ν | t n − ν − + (cid:18) t + (cid:12)(cid:12)(cid:12)(cid:12) a n − a n (cid:12)(cid:12)(cid:12)(cid:12)(cid:19) (sin α − cos α ) . (1.3) Remark 1.2.
In general Theorem 1.7 gives much better result than The-orem 1.5 for 0 ≤ k ≤ n − . To see this, we show that the circle definedby (1.3) is contained in the circle defined by (1.1). Let z = w be any pointbelonging to the circle defined by (1.3), then (cid:12)(cid:12)(cid:12)(cid:12) w + a n − a n − ( t − t ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:18) t + (cid:12)(cid:12)(cid:12)(cid:12) a n − a n (cid:12)(cid:12)(cid:12)(cid:12)(cid:19) (sin α − cos α )+ 2 t | a k +1 | + 2 | a k || a n | t n − k − cos α + 2 sin α | a n | n − X ν =0 | a ν | t n − ν − . This implies, | w | = (cid:12)(cid:12)(cid:12)(cid:12) w + a n − a n − ( t − t ) + ( t − t ) − a n − a n (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12) w + a n − a n − ( t − t ) (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) ( t − t ) − a n − a n (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:18) t + (cid:12)(cid:12)(cid:12)(cid:12) a n − a n (cid:12)(cid:12)(cid:12)(cid:12)(cid:19) (sin α − cos α ) + 2 t | a k +1 | + 2 | a k || a n | t n − k − cos α + 2 sin α | a n | n − X ν =0 | a ν | t n − ν − + (cid:12)(cid:12)(cid:12)(cid:12) ( t − t ) − a n − a n (cid:12)(cid:12)(cid:12)(cid:12) . SOME EXTENSIONS OF ENESTR ¨OM-KAKEYA THEOREM
Using Lemma 2.1 with j = n + 1 and noting that a n +1 = 0 , we get | w | ≤ (cid:18) t + (cid:12)(cid:12)(cid:12)(cid:12) a n − a n (cid:12)(cid:12)(cid:12)(cid:12)(cid:19) (sin α − cos α ) + 2 t | a k +1 | + 2 | a k || a n | t n − k − cos α + 2 sin α | a n | n − X ν =0 | a ν | t n − ν − + (cid:18)(cid:12)(cid:12)(cid:12)(cid:12) a n − a n (cid:12)(cid:12)(cid:12)(cid:12) − ( t − t ) (cid:19) cos α + (cid:18) t − t + (cid:12)(cid:12)(cid:12)(cid:12) a n − a n (cid:12)(cid:12)(cid:12)(cid:12)(cid:19) sin α =2 t | a k +1 | + | a k | t n − k − | a n | ! cos α + 2 sin α | a n | n − X ν =0 | a ν | t n − ν − + 2 sin α | a n − || a n | + t (sin α − cos α )= t ( | a k | + 2 t | a k +1 | t n − k | a n | − ! cos α + sin α ) + 2 sin α | a n | n − X j =0 | a j | t n − j − . This shows that the point z = w also belongs to the circle defined by (1.1).Hence the circle defined by (1.3) is contained in the circle defined by (1.1).For t = 0 , in Theorem 1.7, we obtain the following result which consid-erably improves the bound of Theorem 1.4 for 0 ≤ k ≤ n − . Corollary 1.3.
Let P ( z ) = P nj =0 a j z n be a polynomial of degree n ≥ withcomplex coefficients such that | arg a j − β | ≤ α ≤ π/ , j = 0 , , , · · · , n for some real β. If t > can be found such that t n | a n | ≤ t n − | a n − | ≤ · · · ≤ t k +1 | a k +1 | ≤ t k | a k | ,t k | a k | ≥ t k − | a k − | ≥ · · · ≥ t | a | ≥ | a | > ≤ k ≤ n − , a − = a n +1 = 0 , then all the zeros of P ( z ) lie in (cid:12)(cid:12)(cid:12)(cid:12) z + a n − a n − t (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12) a n − a n (cid:12)(cid:12)(cid:12)(cid:12) (sin α − cos α )+ 2 | a k || a n | t n − k − cos α + 2 sin α | a n | n − X ν =0 | a ν | t n − ν − . (1.4)Next, as a generalization of Theorem 1.6, we prove the following result. Theorem 1.4.
Let P ( z ) = P nj =0 a j z n be a polynomial of degree n ≥ with Re a j = α j and Im a j = β j , j = 0 , , · · · , n. If t = 0 , t ≥ t ≥ can befound such that t t α r + ( t − t ) α r − − α r − ≥ r = 1 , , · · · , k + 1 ,t t α r + ( t − t ) α r − − α r − ≤ r = k + 2 , · · · , n + 1 , and t t β r + ( t − t ) β r − − β r − ≥ r = 1 , , · · · , m + 1 ,t t β r + ( t − t ) β r − − β r − ≤ r = m + 2 , · · · , n + 1 , OME EXTENSIONS OF ENESTR ¨OM-KAKEYA THEOREM 5 ≤ k ≤ n − , ≤ m − ≤ n, α − = β − = α n +1 = β n +1 = 0 , α n > , thenall the zeros of P ( z ) lie in (cid:12)(cid:12)(cid:12)(cid:12) z + α n − − ( t − t ) α n a n (cid:12)(cid:12)(cid:12)(cid:12) ≤ α k +1 t + α k ) t k +11 | a n | t n + 2( β m +1 t + β m ) t m +11 | a n | t n − t α n + t β n + α n − | a n | . (1.5) Remark 1.5.
In general Theorem 1.10 also gives much better result thanTheorem 1.6 for 0 ≤ k ≤ n − . For this, we show that the circle definedby (1.5) is contained in the circle defined by (1.2). Let z = w be any pointbelonging to the circle defined by (1.5) then (cid:12)(cid:12)(cid:12)(cid:12) w + α n − − ( t − t ) α n a n (cid:12)(cid:12)(cid:12)(cid:12) ≤ α k +1 t + α k ) t k +11 | a n | t n + 2( β m +1 t + β m ) t m +11 | a n | t n − t α n + t β n + α n − | a n | . This implies | w | = (cid:12)(cid:12)(cid:12)(cid:12) w + α n − − ( t − t ) α n a n − α n − − ( t − t ) α n a n (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12) w + α n − − ( t − t ) α n a n (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) α n − − ( t − t ) α n a n (cid:12)(cid:12)(cid:12)(cid:12) ≤ α k +1 t + α k ) t k +11 | a n | t n + 2( β m +1 t + β m ) t m +11 | a n | t n − t α n + t β n + α n − | a n | + α n − − ( t − t ) α n | a n | = t | a n | n t k − n (cid:0) α k + t α k +1 (cid:1) + 2 t m − n (cid:0) β m + t β m +1 (cid:1) − (cid:0) α n + β n (cid:1)o . Hence the point z = w belongs to the circle defined by (1.2) and therefore,the circle defined by (1.5) is contained in the circle defined (1.2).For β j = 0 , j = 0 , , · · · , n in Theorem 1.10, we obtain the following result. Corollary 1.6.
Let P ( z ) = P nj =0 a j z n be a polynomial of degree n ≥ withreal and positive coefficients. If t ( = 0) ≥ t ≥ can be found such that t t a r + ( t − t ) a r − − a r − ≥ r = 1 , , · · · , k + 1 , and t t a r + ( t − t ) a r − − a r − ≤ r = k + 2 , · · · , n + 1 , ≤ k ≤ n − , a − = a n +1 = 0 , then all the zeros of P ( z ) lie in (cid:12)(cid:12)(cid:12)(cid:12) z + a n − a n − ( t − t ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ t + a n − a n + 2 t a k +1 + 2 a k a n t n − k −
11 SOME EXTENSIONS OF ENESTR ¨OM-KAKEYA THEOREM Preliminaries
Lemma 2.1. ( [12]) If | arg a j − β | ≤ α ≤ π/ , j = 0 , , , · · · , n and β real, then for t > t ≥ , | t t a j + ( t − t ) a j − − a j − |≤ | t t | a j | + ( t − t ) | a j − | − | a j − || cos α + ( t t | a j | + ( t − t ) | a j − | + | a j − | ) sin α. Proof of theorems
Proof of Theorem 1.7.
Consider the polynomial F ( z ) =( t − z )( t + z ) P ( z )= − a n z n +2 + ( a n ( t − t ) − a n − ) z n +1 + n X ν =2 ( a ν t t + a ν − ( t − t ) − a ν − ) z ν + ( a t t + a ( t − t )) z + a t t = − a n z n +2 + ( a n ( t − t ) − a n − ) z n +1 + n X ν =0 ( a ν t t + a ν − ( t − t ) − a ν − ) z ν ( a − = a − = 0)Let | z | > t , then | F ( z ) | ≥| a n || z | n +1 " (cid:12)(cid:12)(cid:12)(cid:12) z + a n − a n − ( t − t ) (cid:12)(cid:12)(cid:12)(cid:12) − n X ν =0 (cid:12)(cid:12)(cid:12)(cid:12) a ν t t + a ν − ( t − t ) − a ν − a n (cid:12)(cid:12)(cid:12)(cid:12) | z | n − ν +1 > | a n || z | n +1 " (cid:12)(cid:12)(cid:12)(cid:12) z + a n − a n − ( t − t ) (cid:12)(cid:12)(cid:12)(cid:12) − | a n | n X ν =0 | a ν t t + a ν − ( t − t ) − a ν − | t n − ν +11 (3.1) OME EXTENSIONS OF ENESTR ¨OM-KAKEYA THEOREM 7
By Lemma 2.1, we have n X ν =0 | a ν t t + a ν − ( t − t ) − a ν − | t n − ν +11 ≤ n X ν =0 || a ν | t t + | a ν − | ( t − t ) − | a ν − || t n − ν +11 cos α + n X ν =0 || a ν | t t + | a ν − | ( t − t ) + | a ν − || t n − ν +11 sin α = k +1 X ν =0 | a ν | t t + | a ν − | ( t − t ) − | a ν − | t n − ν +11 cos α − n X ν = k +2 | a ν | t t + | a ν − | ( t − t ) − | a ν − | t n − ν +11 cos α + n X ν =0 | a ν | t t + | a ν − | ( t − t ) + | a ν − | t n − ν +11 sin α = ( t | a n | + | a n − | )(sin α − cos α )+ 2 t | a k +1 | + 2 | a k | t n − k − cos α + 2 sin α n − X ν =0 | a ν | t n − ν − Using this in (3.1), we obtain | F ( z ) | > | a n || z | n +1 ( (cid:12)(cid:12)(cid:12)(cid:12) z + a n − a n − ( t − t ) (cid:12)(cid:12)(cid:12)(cid:12) − (cid:18) t + (cid:12)(cid:12)(cid:12)(cid:12) a n − a n (cid:12)(cid:12)(cid:12)(cid:12)(cid:19) (sin α − cos α ) − t | a k +1 | + 2 | a k || a n | t n − k − cos α − α n − X ν =0 (cid:12)(cid:12)(cid:12)(cid:12) a ν a n (cid:12)(cid:12)(cid:12)(cid:12) t n − ν − ) > , (3.2)whenever (cid:12)(cid:12)(cid:12)(cid:12) z + a n − a n − ( t − t ) (cid:12)(cid:12)(cid:12)(cid:12) > (cid:18) t + (cid:12)(cid:12)(cid:12)(cid:12) a n − a n (cid:12)(cid:12)(cid:12)(cid:12)(cid:19) (sin α − cos α )+ 2 t | a k +1 | + 2 | a k || a n | t n − k − cos α + 2 sin α n − X ν =0 (cid:12)(cid:12)(cid:12)(cid:12) a ν a n (cid:12)(cid:12)(cid:12)(cid:12) t n − ν − . Hence, all the zeros of F ( z ) whose modulus is greater than t lie in the circle (cid:12)(cid:12)(cid:12)(cid:12) z + a n − a n − ( t − t ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:18) t + (cid:12)(cid:12)(cid:12)(cid:12) a n − a n (cid:12)(cid:12)(cid:12)(cid:12)(cid:19) (sin α − cos α )+ 2 t | a k +1 | + 2 | a k || a n | t n − k − cos α + 2 sin α | a n | n − X ν =0 | a ν | t n − ν − . We now show that all those zeros of F ( z ) whose modulus is less than orequal to t also satisfy (1.3). Let | z | ≤ t , then by using Lemma 2.1, we SOME EXTENSIONS OF ENESTR ¨OM-KAKEYA THEOREM have (cid:12)(cid:12)(cid:12)(cid:12) z + a n − a n − ( t − t ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ t + (cid:12)(cid:12)(cid:12)(cid:12) a n − a n − ( t − t ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ t + (cid:18)(cid:12)(cid:12)(cid:12)(cid:12) a n − a n (cid:12)(cid:12)(cid:12)(cid:12) − ( t − t ) (cid:19) cos α + (cid:18) t − t + (cid:12)(cid:12)(cid:12)(cid:12) a n − a n (cid:12)(cid:12)(cid:12)(cid:12)(cid:19) sin α ≤ (cid:18) t + (cid:12)(cid:12)(cid:12)(cid:12) a n − a n (cid:12)(cid:12)(cid:12)(cid:12)(cid:19) (sin α − cos α )+ 2 t | a k +1 | + 2 | a k || a n | t n − k − cos α + 2 sin α n − X ν =0 (cid:12)(cid:12)(cid:12)(cid:12) a ν a n (cid:12)(cid:12)(cid:12)(cid:12) t n − ν − if t (1 − cos α ) + 2 (cid:18) t + (cid:12)(cid:12)(cid:12)(cid:12) a n − a n (cid:12)(cid:12)(cid:12)(cid:12)(cid:19) cos α + t sin α ≤ t | a k +1 | + 2 | a k || a n | t n − k − cos α + 2 sin α n − X ν =0 (cid:12)(cid:12)(cid:12)(cid:12) a ν a n (cid:12)(cid:12)(cid:12)(cid:12) t n − ν − (3.3)Now, by given hypothesis, n X ν = k +2 | a ν | t t + | a ν − | ( t − t ) − | a ν − | t n − ν +11 ≥ ≤ k ≤ n − , that is, 2 (cid:18) t + (cid:12)(cid:12)(cid:12)(cid:12) a n − a n (cid:12)(cid:12)(cid:12)(cid:12)(cid:19) ≤ t | a k +1 | + 2 | a k || a n | t n − k − . (3.4)Therefore, (3.3) holds, if t (1 − cos α ) + t sin α ≤ α n − X ν =0 (cid:12)(cid:12)(cid:12)(cid:12) a ν a n (cid:12)(cid:12)(cid:12)(cid:12) t n − ν − (3.5)Again, by hypothesis, t ≤ t + | a n − || a n | (3.6)Using (3.6) in (3.4), we obtain t ≤ t + | a n − || a n | ≤ t | a k +1 | + | a k || a n | t n − k − ≤ t | a k +1 | + | a k || a n | t n − k − = | a k +1 || a n | t n − k − + | a k || a n | t n − k − ≤ n − X ν =0 (cid:12)(cid:12)(cid:12)(cid:12) a ν a n (cid:12)(cid:12)(cid:12)(cid:12) t n − ν − (0 ≤ k ≤ n − t (1 − cos α ) + t sin α ≤ t sin α, OME EXTENSIONS OF ENESTR ¨OM-KAKEYA THEOREM 9 or if, cos α + sin α ≥ , where 0 ≤ α ≤ π/ . (3.7)Note that when 0 ≤ α ≤ π/ , cos α + sin α = √ α + π/ ≥ √ . √ | z | ≤ t, then for0 ≤ k ≤ n − (cid:12)(cid:12)(cid:12)(cid:12) z + a n − a n − ( t − t ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:18) t + (cid:12)(cid:12)(cid:12)(cid:12) a n − a n (cid:12)(cid:12)(cid:12)(cid:12)(cid:19) (sin α − cos α )+ 2 t | a k +1 | + 2 | a k || a n | t n − k − cos α + 2 sin α | a n | n − X ν =0 | a ν | t n − ν − . Hence all the zeros of F ( z ) lie in the circle defined by (1.3). But all thezeros of P ( z ) are also the zeros of F ( z ) , we conclude that all the zeros of P ( z ) lie in the circle defined by (1.3). This proves theorem 1.7. (cid:3) Proof of Theorem 1.10.
Consider the polynomial, G ( z ) =( t − z )( t + z ) P ( z )= − a n z n +2 + ( a n ( t − t ) − a n − ) z n +1 + n X ν =2 ( a ν t t + a ν − ( t − t ) − a ν − ) z ν + ( a t t + a ( t − t )) z + a t t = − a n z n +2 + ( a n ( t − t ) − a n − ) z n +1 + n X ν =0 ( a ν t t + a ν − ( t − t ) − a ν − ) z ν ( a − = a − = 0)Let | z | > t , then | G ( z ) | ≥| z | n +1 " | a n z + a n − − ( t − t ) a n |− n X ν =0 | a ν t t + a ν − ( t − t ) − a ν − | | z | n − ν +1 > | z | n +1 " | a n z + α n − − ( t − t ) α n | − | β n − | − ( t − t ) | β n |− n X ν =0 | a ν t t + a ν − ( t − t ) − a ν − | t n − ν +11 (3.8) Now by hypothesis, n X ν =0 | a ν t t + a ν − ( t − t ) − a ν − | t ν ≤ n X ν =0 | α ν t t + α ν − ( t − t ) − α ν − | t ν + n X ν =0 | β ν t t + β ν − ( t − t ) − β ν − | t ν ≤ k +1 X ν =0 | α ν t t + α ν − ( t − t ) − α ν − | t ν + n X ν = k +2 | α ν t t + α ν − ( t − t ) − α ν − | t ν + m +1 X ν =0 | β ν t t + β ν − ( t − t ) − β ν − | t ν + n X ν = m +2 | β ν t t + β ν − ( t − t ) − β ν − | t ν = 2( α k +1 t + α k ) t k +21 − ( α n t + α n − ) t n +11 + 2( β m +1 t + β m ) t m +21 − ( β n t + β n − ) t n +11 . Using this in (3.8), we obtain | G ( z ) | ≥| z | n +1 (cid:26) | a n z + α n − − ( t − t ) α n | − ( β n − − ( t − t ) β n ) − α k +1 t + α k ) 1 t n − k − + ( α n t + α n − ) − β m +1 t + β m ) 1 t n − m − + ( β n t + β n − ) (cid:27) = | z | n +1 (cid:26) | a n z + α n − − ( t − t ) α n | + t β n − α k +1 t + α k ) 1 t n − k − + ( α n t + α n − ) − β m +1 t + β m ) 1 t n − m − (cid:27) > , if (cid:12)(cid:12) a n z + α n − − ( t − t ) α n (cid:12)(cid:12) > α k +1 t + α k ) t k +11 t n + 2( β m +1 t + β m ) t m +11 t n − ( t α n + t β n + α n − ) . OME EXTENSIONS OF ENESTR ¨OM-KAKEYA THEOREM 11
Hence all the zeros of G ( z ) whose modulus is greater than t lie in the circle (cid:12)(cid:12)(cid:12)(cid:12) z + α n − − ( t − t ) α n a n (cid:12)(cid:12)(cid:12)(cid:12) ≤ α k +1 t + α k ) t k +11 | a n | t n + 2( β m +1 t + β m ) t m +11 | a n | t n − t α n + t β n + α n − | a n | . (3.9)Now, we show that all the zeros of G ( z ) whose modulus is less than equalto t also lie in the circle defined by (1.5). Let | z | ≤ t , then we have (cid:12)(cid:12) a n z + α n − − ( t − t ) α n (cid:12)(cid:12) ≤ | a n | t + | α n − − ( t − t ) α n |≤ t α n + t β n + α n − − ( t − t ) α n = 2 t β n + 2( t α n + α n − ) − ( t α n + 2 t β n + α n − ) . By hypothesis, n X ν = k +2 α ν t t + α ν − ( t − t ) − α ν − t n − ν +11 ≤ , ≤ k ≤ n − , this gives 2 ( t α n + α n − ) ≤ t α k +1 + 2 α k t n − k − (3.10)Similarly for 0 ≤ m ≤ n − , t β n + β n − ) ≤ t β m +1 + 2 β m t n − m − . (3.11)Also, we have t β n ≤ t β n + β n − (3.12)Combining (3.11) and (3.12), we obtain2 t β n ≤ t β m +1 + 2 β m t n − m − (3.13)Using (3.10) and (3.13) in (3.9), we have (cid:12)(cid:12) a n z + α n − − ( t − t ) α n (cid:12)(cid:12) ≤ α k +1 t + α k ) t k +11 t n + 2( β m +1 t + β m ) t m +11 t n − ( t α n + t β n + α n − ) . This shows that all the zeros of G ( z ) whose modulus is less than or equalto t also satisfy the inequality (1.5). Thus we conclude that all the zerosof G ( z ) hence that of P ( z ) lie in the circle defined by (1.5). This completesthe proof of Theorem 1.10. (cid:3) References [1] N. Anderson, E. B. Saff, R. S. Verga, On the Enestr¨om-Kakeya theorem and its sharpness,
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