Journal of Inequalities in Pure and Applied Mathematics

Size: px

Start display at page:

Download "Journal of Inequalities in Pure and Applied Mathematics"

Ginger Kennedy
5 years ago
Views:

Journal of Inequalities in Pure and Applied Mathematics ON THE LOCATION OF ZEROS OF COMPLEX POLYNOMIALS MATTHIAS DEHMER Technische Universität Darmstadt Department of Computer Science Hochschulstraße

1 Journal of Inequalities in Pure and Applied Mathematics ON THE LOCATION OF ZEROS OF COMPLEX POLYNOMIALS MATTHIAS DEHMER Technische Universität Darmstadt Department of Computer Science Hochschulstraße Darmstadt, Germany volume 7, issue 1, article 26, 2006 Received 05 April, 2004; accepted 12 December, 2005 Communicated by: D Stefanescu matthias@dehmerorg Abstract Home Page c 2000 Victoria University ISSN (electronic):

2 Abstract This paper proves bounds for the zeros of complex valued polynomials The assertions stated in this work have been specialized in the area of the location of zeros for complex polynomials in terms of two foci: (i) finding bounds for complex valued polynomials with special conditions for the coefficients and (ii) locating zeros of complex valued polynomials without special conditions for the coefficients especially we are searching for bounds, which again are positive roots of concomitant polynomials As a result we obtain new zero bounds for univariate polynomials with complex coefficients 2000 Mathematics Subject Classification: 30C15 Key words: Complex polynomials, zeros, inequalities I thank the referee of the present paper for fruitful comments Furthermore I thank Frank Emmert-Streib (Stowers Institute, USA) and Ralf Tandetzky (University of Jena, Germany) for helpful discussions and comments concerning this work 1 Introduction 3 2 Bounds for the Zeros of with Special Conditions for the Coefficients 4 3 Further Bounds for the Zeros of 11 4 Evaluation of Zero Bounds 21 5 Conclusions 25 References Page 2 of 27 J Ineq Pure and Appl Math 7(1) Art 26,

3 1 Introduction The problems in the analytic theory of polynomials concerning locating zeros of complex polynomials have been frequently investigated Over many decades, a large number of research papers, eg, [1, 2, 3, 5, 6, 8, 9, 12, 13, 14] and monographs [7, 10, 11] have been published The new theorems in this paper provide closed disks in the complex plane K(z 0, r) := {z C z z 0 r}, z 0 C, r R +, containing all zeros of a complex valued polynomial The steps to achieve this are as follows: Let n f n : C C; f n (z) = a i z i, a i C, n 1, be a complex polynomial Construct a bound S = S(a 0, a 1,, a n ) in such a way that all zeros of f(z) are situated in the closed disk K(z 0, S(a 0, a 1,, a n )) := {z C z z 0 S(a 0, a 1,, a n )} Without loss of generality we set z 0 = 0 A special case is the existence of certain conditions, thus resulting in special bounds This paper is organized as follows: Section 2 provides such bounds on the basis of certain conditions, concerning the polynomials coefficients In Section 3 mainly zero bounds are proved which are positive roots of algebraic equations The main result of Section 3 is a theorem which is an extension of a classical result of Cauchy Section 4 shows applications of the bounds in such a way that the bounds will be evaluated with certain polynomials The paper finishes in Section 5 with conclusions Page 3 of 27 J Ineq Pure and Appl Math 7(1) Art 26,

4 2 Bounds for the Zeros of with Special Conditions for the Coefficients Definition 21 Let be a complex polynomial f n : C C; f n (z) = n a i z i, n 1, (21) f n (z) = a n z n + f n 1 (z), f 0 (z) := a 0 denotes the recursive description of f n (z) Theorem 21 Let P (z) be a complex polynomial, such that P (z) is reducible in C[z], with P (z) := f n1 (z)g n2 (z) = (b n1 z n 1 + f n1 1(z))(c n2 z n 2 + g n2 1(z)) b n1 > b i, 0 i n 1 1, c n2 > c i, 0 i n 2 1 If n 1 + n 2 > 1, then all zeros of the polynomial P (z) lie in the closed disk (22) K(0, δ), where δ > 1 is the positive root of the equation (23) z n 1+n z n 1+n z n 1+n 2 + z n z n = 0 It holds for 1 < δ < Page 4 of 27 J Ineq Pure and Appl Math 7(1) Art 26,

5 Proof Expanding f n1 1(z), we conclude analogously Assuming f n1 1(z) = g n2 1(z) = n 1 1 n 2 1 b i z i, c i z i (24) b n1 > b i, 0 i n 1 1, c n2 > c i, 0 i n 2 1, it follows that f n1 1(z) b n1 = n 1 1 b i z i b n1 < n 1 1 n 1 1 b i b n1 z i With the inequalities (24) above, we find that P (z) = (b n1 z n 1 + f n1 1(z))(c n2 z n 2 + g n2 1(z)) b n1 c n2 z n 1+n 2 z i = z n 1 1 z 1 { b n1 z n 1 g n2 1(z) + c n2 z n 2 f n1 1(z) + f n1 1(z) g n2 1(z) } Page 5 of 27 J Ineq Pure and Appl Math 7(1) Art 26,

6 { [ = b n1 c n2 z n 1+n 2 z n g 1 n 2 1(z) c n2 + f n 1 1(z) g ]} n 2 1(z) > b n1 c n2 b n1 { z n 1+n 2 c n2 + ( z n 1 1)( z n 2 1) ( z 1) 2 [ z n 1 z n z n 2 f n 1 1(z) b n1 z 1 + z n 2 z n1 1 z 1 ]} = b n 1 c n2 { z n 1+n z n 1+n z n 1+n 2 + z n2+1 + z n1+1 1}, ( z 1) }{{ 2 } >0 Let H( z ) := z n 1+n z n 1+n z n 1+n 2 + z n z n 1+1 1, and we assume n 1 + n 2 > 1 Hence P (z) > 0 if H( z ) > 0 Applying Descartes Rule of Signs [4] to H( z ), we can conclude that H( z ) has either one or three positive zeros δ 1, δ 2, and δ 3 Now we examine the positive zeros of H( z ) At first Dividing H( z ) by z 1 yields H(1) = 0 ( z n 1+n z n 1+n z n 1+n 2 + z n z n 1+1 1) : ( z 1) Page 6 of 27 J Ineq Pure and Appl Math 7(1) Art 26,

7 = z n 1+n z n 1+n 2 n 1 +n 2 1 j=n 1 +1 n 2 z j + z j j=0 Further, let R( z ) := z n 1+n z n 1+n 2 n 1 +n 2 1 j=n 1 +1 n 2 z j + z j j=0 We see that R(1) = 0 and infer H( z ) = ( z 1)R( z ) = ( z 1) 2 Q( z ), deg(q( z )) = n 1 + n 2, hence δ 1 = δ 2 = 1 is a zero with multiplicity two Altogether, H has exactly three positive zeros Now, we note that sign{h(0)} = 1, sign{h( )} = 1, and we choose K > 1 such that H(K) > 0 On the other hand, it holds that H(1) = H (1) = 0 and H (1) < 0 Therefore, there exists a θ > 0 such that H(z) < 0, 1 z 1 + θ Now, we obtain H(1 + θ) < 0 < H(K) and therefore δ 3 := δ (1 + θ, K), hence δ > 1 Thus, P (z) > 0, if H( z ) > 0, z > δ According to this, all zeros of P (z) lie in the closed disk K(0, δ) In order to examine the value of δ > 1, we replace z with δ in equation (23) and examine the equation R(δ) = δ n 1+n δ n 1+n 2 n 1 +n 2 1 j=n 1 +1 n 2 δ j + δ j = 0 j=0 Page 7 of 27 J Ineq Pure and Appl Math 7(1) Art 26,

8 The formulas 1 + δ + δ δ n 1 + δ n δ n 1+n 2 1 = δn 1+n 2 1 δ 1 yield the result n 1 +n 2 1 j=n 1 +1 n 1 +n 2 1 j=n 1 +1 δ j = δn 1+n 2 1 δ 1 δ j = δn 1+n 2 1 δ 1 δ n 1+n δ n 1+n 2 δn 1+n 2 δ n 1+1 From this equation follows the inequality and, furthermore, δ 1 (1 + δ + δ δ n 1 ) δn δ 1 δ n 1+n δ n 1+n 2 δn 1+n 2 δ n 1+1 δ 1 = δn 1+n 2 δ n 1+1 δ 1 + δn = 0 } δ {{ 1 } >0 (25) δ n 1+1 (δ n 2+1 4δ n 2 + 2δ n ) < 0 < 0, Inequality (25) implies δ n 2 ( δ ) < 1 δ Page 8 of 27 J Ineq Pure and Appl Math 7(1) Art 26,

9 and we finally conclude (26) δ 2 4δ + 2 < 0 To solve inequality (26) take a closer look at ξ 2 4ξ + 2 = 0, and see that ξ 1,2 = 2 ± 2 With the fact that δ > 1, the inequality holds for 1 < δ < This completes the proof of Theorem 21 Theorem 21 can be strengthened by Theorem (22) Theorem 22 Let P (z) := a n z n + a n 1 z n a 0, a n 0, n 1, i = 0, 1,, n be a complex polynomial such that a n > a i, i = 0, 1,, n 1 Then all zeros of P (z) lie in the closed disk K(0, 2) Proof For z 1 the conclusion of Theorem 22 is evident If we assume that z > 1 we obtain immediately P (z) = a n z n + a n 1 z n a 0 a n z n { a n 1 z n a 0 } [ = a n z {1 n an 1 1 a n z + + a 0 a n 1 z n ]} Page 9 of 27 J Ineq Pure and Appl Math 7(1) Art 26,

10 Now if a n > a i, i = 0, 1,, n 1 is assumed, we conclude [ P (z) a n z {1 n an 1 1 a n z + + a ]} 0 1 a n z } n n > a n z {1 n 1 z j j=1 } > a n z {1 n 1 z j j=1 { } = a n z n z = a n z n { z 2} z 1 }{{} >0 Hence, we get P (z) > 0 if z > 2 Page 10 of 27 J Ineq Pure and Appl Math 7(1) Art 26,

11 3 Further Bounds for the Zeros of Complex Polynomials Now we prove Theorem 32, which is similar to Theorem 31 Theorem 31 is a classic result for the location of zeros found by Cauchy [7] The zero bound of the new Theorem 32, as well as Theorem 31, depends on an algebraic equation s positive root Theorem 31 (Cauchy) Let f(z) = a n z n + a n 1 z n a 0, a n 0, k = 0, 1,, n be a complex polynomial All zeros of f(z) lie in the closed disk K(0, ρ C ), where ρ C denotes the positive zero of Theorem 32 Let H C (z) := a 0 + a 1 z + + a n 1 z n 1 a n z n f(z) = a n z n + a n 1 z n a 0, a n 0, k = 0, 1,, n be a complex polynomial All zeros of f(z) lie in the closed disk K(0, max(1, δ)), where M := max a j a n and δ 1 denotes the positive root of the equation 0 j n 1 Proof On the basis of the inequality z n+1 (1 + M)z n + M = 0 (31) f(z) a n [ z n M{ z n } ], Page 11 of 27 J Ineq Pure and Appl Math 7(1) Art 26,

12 we get the modulus of f(z) f(z) a n [ z n M{ z n } ] [ ] = a n z n M z n 1 z 1 [ ] z n+1 z n (1 + M) + M = a n z 1 Define F (z) := z n+1 z n (1 + M) + M Using the Descartes Rule of Signs we have that F (z) has exactly two positive zeros δ 1 and δ 2, and F (δ 1 = 1) = 0 holds With sign{f (0)} = 1 and from the fact that F (z) has exactly two positive zeros, we finally conclude that f(z) > 0 for z > max(1, δ) Hence, all zeros of f(z) lie in K(0, max(1, δ)) Now, we express a further theorem, the proof of which is similar to that of Theorem 32 Theorem 33 Let Page 12 of 27 f(z) = a n z n + a n 1 z n a 0, a n 0, k = 0, 1,, n J Ineq Pure and Appl Math 7(1) Art 26,

13 be a complex polynomial All zeros of f(z) lie in the closed disk K(0, max(1, δ)), where M := max a n j a n j 1 0 j n a n, a 1 := 0 and δ 1 denotes the positive root of the equation z n+2 (1 + M)z n+1 + M = 0 Furthermore we can state immediately a new Theorem 34 which is very similar to a well known theorem due to Cauchy [7] It provides an upper bound for all zeros of a complex polynomial f(z) In many cases the bound of Theorem 34 is sharper than the bound of Cauchy [7] (K(0, 1 + M)) Theorem 34 Let f(z) = a n z n + a n 1 z n a 0, a n 0, k = 0, 1,, n be a complex polynomial All zeros of f(z) lie in the closed disk K(0, 1 + M) Proof For the zeros with z 1, we have nothing to prove Assuming z > 1 and define P (z) := (1 z)f(z) = a n z n+1 + (a n a n 1 )z n + + (a 1 a 0 )z + a 0 Now, we obtain the estimation P (z) a n z n+1 { a n a n 1 z n + a n 1 a n 2 z n a 1 a 0 z + a 0 } Page 13 of 27 J Ineq Pure and Appl Math 7(1) Art 26,

14 { [ = a n z n+1 a n a n 1 a n z n + }] + a 1 a 0 a n z + a 0 0 a n { } a n z n+1 M n z i { } = a n z n+1 M z n+1 1 z 1 } > a n { z n+1 z n M z 1 = a { n z n+2 z n+1 (1 + z 1 M) } }{{} >0 Finally, we conclude a n 1 a n 2 a n z n+2 z n+1 (1 + M) = z n+1 [ z (1 + M)] > 0 z n 1 + We infer P (z) > 0 if z > 1+ M Hence, all zeros of P (z) lie in K(0, 1+ M) Because of the fact that all zeros of f(z) are zeros of P (z), Theorem 34 holds also for f(z) Now, we prove Theorem 35, which provides a zero bound for complex polynomials that are in a more general form than the previous ones Page 14 of 27 J Ineq Pure and Appl Math 7(1) Art 26,

15 Theorem 35 Let f(z) = a n z n + a n 1 z n a 0, a n 0, k = 0, 1,, n be a complex polynomial and H(z) := (f(z)) σ, N σ 1 All zeros of H(z) lie in the closed disk K(0, ρ), where ρ is the positive zero of (32) H(z) := an σ z nσ {( n 1 ) σ a i z i ( σ 1 n 1 ) σ j ( j) σ + a n j z nj a i z i j=1 Proof Define H(z) := (f(z)) σ, this yields the estimation ( n ) σ ( n 1 ) σ H(z) = a i z i = a i z i + a n z n ( σ n 1 ) σ j ( j) = σ (a n z n ) j a i z i j=0 ( = (a σ 1 n 1 ) σ j ( j) σ nz n ) σ + (a n z n ) j a i z i j=0 ( σ 1 n 1 ) σ j ( j) a n z n σ σ (a n z n ) j a i z i j=0 Page 15 of 27 J Ineq Pure and Appl Math 7(1) Art 26,

16 ( n 1 ) σ ( σ 1 n 1 ) σ j ( j) = a n z n σ σ a i z i + (a n z n ) j a i z i j=1 ( n 1 ) σ a n z n σ ( σ 1 n 1 ) σ j ( j) a i z i + σ (a n z n ) j a i z i a n σ z nσ := H( z ) {( n 1 ) σ a i z i j=1 ( σ 1 n 1 ) σ j ( j) σ + a n z n j a i z i j=1 H(z) is a polynomial with only real and positive coefficients Therefore we apply the Descartes Rule of Signs and obtain that H(z) has exactly one positive zero ρ Furthermore, we determine easily H(0) = a 0 σ < 0 With ( σ 1 n 1 ) σ j ( j) σ H 1 (z) := (a n z n ) σ and H2 (z) := (a n z n ) j a i z i, we infer j=0 deg( H 1 (z)) > deg( H 2 (z)) Finally, it follows that all zeros of H(z) lie in the closed disk K(0, ρ) Page 16 of 27 J Ineq Pure and Appl Math 7(1) Art 26,

17 The following corollary states that the bound of Theorem 35 is a generalization of Cauchy s bound concerning Theorem 31 Corollary 36 σ = 1 in Equation (32) leads to the bound of Cauchy, Theorem 31 Proof If we set σ = 1 in equation (32), we obtain n 1 H(z) = a n z n a i z i Therefore, we have H(z) = H C (z) and H(z) posseses the same positive zero as H C (z) Thus we have the bound of Cauchy, Theorem 31 The last theorem in this paper expresses another bound for the zeros of H(z) Here, the belonging concomitant polynomial is easier to solve that the concomitant polynomial of Theorem 35 Theorem 37 Let f(z) = a n z n + a n 1 z n a 0, a n 0, k = 0, 1,, n be a complex polynomial and H(z) := (f(z)) σ, N σ 1 Furthermore let B := max 0 i n 1 a i and K := max 0 j σ 1 a n j B σ j( σ j) All zeros of H(z) lie in the closed disk K(0, ρ), where ρ > 1 is the largest positive zero of the equation a n σ ( z 1) σ K z σ B σ z + B σ Page 17 of 27 J Ineq Pure and Appl Math 7(1) Art 26,

18 Proof Starting from H( z ), obtained in the proof of Theorem 35 and the inequality n 1 n 1 a i z i B z i, we have H(z) a n σ z nσ { ( z B σ n 1 z 1 [ σ 1 + j=1 ) σ a n j z nj B σ j ( σ j ) ( ) ]} z n σ j 1 z 1 If we now assume z > 1 we obtain the inequalities { ( ) z H(z) > a n σ z nσ B σ n σ 1 σ 1 ( ) } z + K z nj n σ j 1 z 1 z 1 j=1 { } > a n σ z nσ B σ z nσ σ 1 ( z 1) + K z nσ σ ( z 1) σ j j=1 [ ]} = z { a nσ n σ B σ ( z 1) + K σ 1 ( z 1) j σ ( z 1) σ On the basis of the relation σ 1 σ 1 ( z 1) j < z j, j=1 j=0 j=1 Page 18 of 27 J Ineq Pure and Appl Math 7(1) Art 26,

19 we conclude furthermore that [ ]} H(z) > z { a nσ n σ B σ ( z 1) + K σ ( z 1) z σ 1 σ z 1 [ ]} > z { a nσ n σ B σ ( z 1) + K z σ σ ( z 1) σ+1 z nσ { = an σ ( z 1) σ+1 K z σ B σ z + B σ} ( z 1) }{{ σ+1 } >0 Define P (z) := a n σ (z 1) σ+1 Kz σ B σ z + B σ We see immediately that deg(p (z)) = deg ( a n σ (z 1) σ+1 Kz σ B σ z + B σ) ( σ+1 ( ) σ + 1 = deg a n σ ( 1) )z j σ+1 j Kz σ B σ z + B σ j hence we infer = σ + 1, j=0 (33) lim z P (z) = Now, it holds P (0) = { an σ + B σ : σ is odd a n σ + B σ : σ is even Furthermore we obtain P (1) = K < 0 In the first case P (0) = a n σ + B σ > 0 it follows that P (z) has at least two positive zeros α 1 < α 2 and α 2 > 1 The Page 19 of 27 J Ineq Pure and Appl Math 7(1) Art 26,

20 case P (0) = a n σ + B σ > 0 B σ > a n σ leads us to the same situation Now, we assume P (0) = a n σ + B σ < 0 B σ < a n σ and it holds that P (1) < 0 With equation (33) we conclude that P (z) has at least one positive zero α > 1 Let α 1, α 2,, α τ, 1 τ deg(p (z)) be the positive zeros of P (z) and ρ := max (α 1, α 2,, α τ ) 1 τ deg( H(x)) Finally, we infer that P (z) > 0 if z > ρ > 1 This inequality completes the proof of Theorem 37 Page 20 of 27 J Ineq Pure and Appl Math 7(1) Art 26,

21 4 Evaluation of Zero Bounds Let f(z) = n a iz i, a n 0, n 1, be a complex polynomial and z 1, z 2,, z n be the zeros of f(z) Define ξ := max ( z 1, z 2,, z n ) and S 1 := max(1, δ), Theorem(21) S 2 := δ, Theorem(32) S 3 := δ, Theorem(33) S 4 := 1 + M, Theorem(34) S 5 := ρ, Theorem(37) Now, we consider the closed disks z S i, i = 1, 2,, 4 for each complex polynomial f j (z), j = 1, 2,, 6 1 f 1 (z) := 360 z 6 194z 5 79z z z 2 12z 5 therefore it holds Zeros of f 1 (z): = (18z 2 7z 5) (20z 4 3z 3 + z + 1), b 2 > b i, 0 i 1, c 4 > c i, 0 i 3 z 1 = 0, , 2775i, z 2 = 0, , 2775i, z 3 = 0, , 3864i, z 4 = 0, , 3864i, Page 21 of 27 J Ineq Pure and Appl Math 7(1) Art 26,

22 z 5 = 0, 3673, z 6 = 0, f 2 (z) := z 5 iz 4 + iz 2 z + i Zeros of f 2 (z): z 1 = 1, , 1223i, z 2 = 0, 8766i, z 3 = 0, 5950i, z 4 = 1, 5262i, z 5 = 1, , 1223i 3 f 3 (z) := z z z Zeros of f 3 (z): z 1 = 0, 5810, z 2 = 2, 5866, z 3 = 997, f 4 (z) := 10z 4 + z 3 + (3 + 2i)z 2 + z( 2 + i) + i = (5z 2 2z + i) (2z 2 + z + 1), Page 22 of 27 J Ineq Pure and Appl Math 7(1) Art 26,

23 therefore it holds Zeros of f 4 (z): b 2 > b i, 0 i 1, c 2 > c i, 0 i 1 5 f 5 (z) := z 7 z + 1 Zeros of f 5 (z): z 1 = 0, 25 0, 6614i, z 2 = 0, , 6614i, z 3 = 0, , 2863i, z 4 = 0, , 2863i z 1 = 1, 1127, z 2 = 0, , 9008i, z 3 = 0, , 9008i, z 4 = 0, , 9525i, z 5 = 0, , 9525i, z 6 = 0, , 2628i, z 7 = 0, , 2628i 6 f 6 (z) := (2z 4 z 3 + z 2 + 2z 1) 2, σ = 2 Page 23 of 27 J Ineq Pure and Appl Math 7(1) Art 26,

24 f i S 1 S 2 S 3 S 4 S 5 ξ f 1 3,2918 1,4895 2,5366 2,5388 2,1757 (σ = 1) 0,756 f 2-1,9659 2,4069 2,4142-1,5262 f ,0 3001,0 3001,0-997,9914 f 4 3,1746 1,7897 1,8595 1, (σ = 1) 1,25 f 5-1,9919 1,2301 3,0-1,1127 f 6-2,7494 4,2499 4,25 3,850 (σ = 2) 0,4184 Figure 1: Comparison of zero bounds Zeros of f 6 (z): z 1 = 0, 8935, z 2 = 0, 4184, z 3 = 0, , 0485i, z 4 = 0, , 04857i, z 5 = 0, 8935, z 6 = 0, 4184, z 7 = 0, , 04857i, z 8 = 0, , 04857i Page 24 of 27 J Ineq Pure and Appl Math 7(1) Art 26,

25 5 Conclusions Table 1 shows the overall results of the comparison of the new zero bounds In general, the comparison between zero bounds is very difficult Hence, it is not possible to obtain general assertions for describing the quality of zero bounds Table 1 shows that the quality of the proven bounds depends on the polynomial under consideration We observe that the bounds S 3, S 4 are very large for f 3 (z) This is due to the fact that the number M := max a j 0 j n 1 a n is very large In the case of a polynomial H(z) := (f(z)) σ, the bound of Theorem 35 is directly applicable (without expanding H(z)) in order to determine the value of the bound The new bounds of the present paper can be used in many applications The characteristic property of our new bounds is that we can compute these zero bounds more effectively than the classic bound of Theorem 31 Page 25 of 27 J Ineq Pure and Appl Math 7(1) Art 26,

26 References [1] H ALZER, Bounds for the zeros of polynomials, Complex Variables, Theory Appl, 8(2) (2001), [2] M DEHMER, Schranken für die Nullstellen komplexwertiger Polynome, Die Wurzel, 8 (2000), [3] F HEIGL, Neuere Entwicklungen in der Geometrie der Polynome, Jber Deutsch Math-Verein, 65 (1962/63), [4] P HENRICI, Applied and computational complex analysis Volume I, Wiley Classics Library Edition, (1988) [5] F KITTANEH, Bounds for the zeros of polynomials from matrix inequalities, Arch Math, 81(5) (2003), [6] S LIPKA, Über die Abgrenzung der Wurzeln von algebraischen Gleichungen, Acta Litt Sci (Szeged), 3 (1927), [7] M MARDEN, Geometry of polynomials, Mathematical Surveys Amer Math Soc, Rhode Island, 3 (1966) [8] QG MOHAMMAD, Location of the zeros of polynomials, Amer Math Monthly, 74 (1962), [9] QG MOHAMMAD, On the zeros of polynomials, Amer Math Monthly, 72 (1966), Page 26 of 27 J Ineq Pure and Appl Math 7(1) Art 26,

27 [10] N OBRESCHKOFF, Verteilung und Berechnung der Nullstellen reeller Polynome, Hochschulbücher für Mathematik 55 Berlin, VEB Deutscher Verlag der Wissenschaften, (1963) [11] QI RAHMAN AND G SCHMEISSER, Analytic Theory of Polynomials Critical Points, Zeros and Extremal Properties, Clarendon Press, (2002) [12] DM SIMEUNOVIĆ, On the location of the zeros of polynomials, Math Morav, 2 (1998), [13] W SPECHT, Die Lage der Nullstellen eines Polynoms, Mathematische Nachrichten, 15 (1956), [14] EC WESTERFIELD, A new bound for the zeros of polynomials, Amer Math Monthly, 40 (1933), Page 27 of 27 J Ineq Pure and Appl Math 7(1) Art 26,

POLYNOMIALS WITH A SHARP CAUCHY BOUND AND THEIR ZEROS OF MAXIMAL MODULUS

M athematical Inequalities & Applications Volume 18, Number 4 (2015), 1387 1392 doi:10.7153/mia-18-108 POLYNOMIALS WITH A SHARP CAUCHY BOUND AND THEIR ZEROS OF MAXIMAL MODULUS HARALD K. WIMMER (Communicated