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International Journal of Engineering Research and Development is an international premier peer reviewed open access engineering and technology journal promoting the discovery, innovation, advancement …

International Journal of Engineering Research and Development is an international premier peer reviewed open access engineering and technology journal promoting the discovery, innovation, advancement and dissemination of basic and transitional knowledge in engineering, technology and related disciplines.

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  • 1. International Journal of Engineering Research and Development e-ISSN: 2278-067X, p-ISSN: 2278-800X, www.ijerd.com Volume 9, Issue 2 (November 2013), PP. 41-52 Location of Zero-free Regions of Polynomials M. H. Gulzar Department of Mathematics University of Kashmir, Srinagar 190006 Abstract: In this paper we locate zero-free regions of polynomials when their coefficients are restricted to certain conditions. Mathematics Subject Classification: 30C10, 30C15 Keywords and phrases: Coefficient, Polynomial, Zero. I. INTRODUCTION AND STATEMENT OF RESULTS The problem of finding the regions containing all or some or no zero of a polynomial is very important in the theory of polynomials. In this connection, a lot of papers have been published by various researchers (e.g. see 1,2,3,4,5,6). Recently, M. H. Gulzar [5] proved the following results: Theorem A: Let Let P ( z )   a z j 0 j j be a polynomial o f degree n such that Re( a j )   j , Im( a j )   j and for some positive integers  ,  and for some real numbers k1 , k 2 ,  1 ,  2 ;0  k1  1,0  k 2  1,0   1  1,0   2  1, k1 n   n1  ......   1       1  ......  1   1 0 k 2  n   n1  ......   1       1  ...... 1   1  0 . z Then the number of zeros of P(z) in R ( R  0, c  1) does not exceed c M 1 log 1 , where log c a0 M 1  an R n1  a0  R n [  n   n  k1 (  n   n )  k 2 (  n   n )       ]  R  [    1 (  0   0 )   0 ]  R  [    2 (  0   0 )   0 ] for R  1 and M 1  an R n1  a0  R n [  n   n  k1 (  n   n )  k 2 (  n   n )       ]  R[       1 (  0   0 )   2 (  0   0 )   0   0 ] for R  1 . Theorem B: Let Let P ( z )   a z j 0 j j be a polynomial o f degree n such that Re( a j )   j , Im( a j )   j and for some positive integers  ,  ,  ,  and for some real numbers k1 , k 2 , k 3 , k 4 ; 1 ,  2 ,  3 ,  4 ;0  k1  1,0  k 2  1,0  k 3  1,0  k 4  1 0   1  1,0   2  1,0   3  1,0   4  1 , k1 k 2 k3  k2  n 2[ ] 2  ......  2  2   2   ......  2   1 0 n 2[ ]1 2 n 2[ ] 2  ......  2  1   2  1  ......  3   2 1  ......  2   2   2   ......  2   3  0 n 2[ ]1 2  ......  2 1   2 1  ......  3   4  1 . . 41
  • 2. Location of Zero-free Regions of Polynomials Then, for even n, the number of zeros of P(z) in in z M 1 log 2 , log c a0 R ( R  0, c  1) does not exceed c where M 2  an R n1  a0  R n [  n   n  k1 (  n   n )  k 2 (  n   n )       ]  R  [    1 (  0   0 )   0 ]  R  [    2 (  0   0 )   0 ] for R  1 and M 2  an R n1  a0  R n [  n   n  k1 (  n   n )  k 2 (  n   n )       ]  R[       1 (  0   0 )   2 (  0   0 )   0   0 ] for R  1 . R If n is odd, the number of zeros of P(z) in z  ( R  0, c  1) does not exceed c  M 2 is same as M 2 except that k1 , k 2 , k3 , k 4 are respectively replaced by k 2 , k1 , k 4 , k 3 and Theorem C: Let Let P ( z )   1 , 2 , 3 , 4  M2 1 log , where log c a0 are respectively replaced by  2 , 1 , 4 , 3 .  a z j 0 j j be a polynomial o f degree n such that  ,  , and for some real numbers k1 , k 2 , 1 , 2 , 0  k1  1,0  k 2  1,0   1  1,0   2  1, for some positive integers k1 a n 2[ ] 2 k2 a  ...... a 2   2  a 2   ...... a 2   1 a 0 n 2[ ]1 2  ...... a 2  1  a 2  1  ...... a 3   2 a1 . Then for even n the number of zeros of P(z) in z R ( R  0, c  1) does not exceed c M 1 log 3 , log c a0 R  1, M 3  an R n2  an1 R n1  a0  R n [ a1  (1  k1 ) an  (1  k 2 ) an1 where for  cos  (2 a 2   2 a 2   k1 a n  k 2 a n 1   2 a1   1 a 0 ) n2  sin  (k1 a n  k 2 a n 1   2 a1   1 a 0  2 a j )] j 2 R  1, M 3  an R n2  an1 R n1  a0  R[ a1  (1  k1 ) an  (1  k 2 ) an1 and for  cos  (2 a 2   2 a 2   k1 a n  k 2 a n 1   2 a1   1 a 0 ) n2  sin  (k1 a n  k 2 a n 1   2 a1   1 a 0  2 a j )]. j 2 If n is odd, then the number of zeros of P(z) in z R ( R  0, c  1) does not exceed c where M 4 is same as M 3 except that k 1 , k 2 are respectively replaced by k 2 , k1 and M 1 log 4 , log c a0  1 , 2 are respectively replaced by  2 ,  1 . In this paper we find regions containing no zero of the polynomials in theorems A,B,C and prove the following results: 42
  • 3. Location of Zero-free Regions of Polynomials Theorem 1: Let Let P ( z )   a z j 0 j be a polynomial o f degree n such that j Re( a j )   j , Im( a j )   j and for some positive integers  ,  and for some real numbers k1 , k 2 ,  1 ,  2 ;0  k1  1,0  k 2  1,0   1  1,0   2  1, k1 n   n1  ......   1       1  ......  1   1 0 k 2  n   n1  ......   1       1  ...... 1   1  0 . Then that P(z) has no zero in z  M* a0 R  1 and no zero in z  a0 for R  1, where M **  an R n1  R n [  n   n  k1 (  n   n )  k 2 (  n   n )       ] , for * M  R  [    1 (  0   0 )   0 ]  R  [    2 (  0   0 )   0 ] and M **  an R n1  R n [  n   n  k1 (  n   n )  k 2 (  n   n )       ]  R[       1 (  0   0 )   2 (  0   0 )   0   0 ] . Combining Theorem 1 and Theorem A, we get the following result: Corollary 1: Let Let P ( z )   a z j 0 j be a polynomial o f degree n such that j Re( a j )   j , Im( a j )   j and for some positive integers  ,  and for some real numbers k1 , k 2 ,  1 ,  2 ;0  k1  1,0  k 2  1,0   1  1,0   2  1, k1 n   n1  ......   1       1  ......  1   1 0 k 2  n   n1  ......   1       1  ...... 1   1  0 . a0 Then the number of zeros of P(z) in M * R  1 and the number of zeros of P(z) in for  z  a0 M ** R ( R  0, c  1) does not exceed c  z  M 1 log 1 for log c a0 R ( R  0, c  1) does not exceed c M 1 log 1 log c a0 R  1 , where M 1 , M * , M ** are given as in Theorem 1 and Theorem A. Theorem 2: Let Let P ( z )   a z j 0 j j be a polynomial o f degree n such that Re( a j )   j , Im( a j )   j and for some positive integers  ,  ,  ,  and for some real numbers k1 , k 2 , k 3 , k 4 ; 1 ,  2 ,  3 ,  4 ;0  k1  1,0  k 2  1,0  k 3  1,0  k 4  1 0   1  1,0   2  1,0   3  1,0   4  1 , k1 k 2 k3  k2  n 2[ ] 2  ......  2  2   2   ......  2   1 0 n 2[ ]1 2 n 2[ ] 2  ......  2  1   2  1  ......  3   2 1  ......  2   2   2   ......  2   3  0 n 2[ ]1 2  ......  2 1   2 1  ......  3   4  1 . . 43
  • 4. Location of Zero-free Regions of Polynomials Then, for even n, P(z) has no zero in z  a0 M1 * , for R  1 and no zero in z  a0 ** for R  1, where for R  1, where M1 M 1  an R n2  an1 R n1  R n [  n   n   n1   n1 *  2( 2   2  1   2    2 1 )  (k1  n  k 3  n )  (k 2  n 1  k 4  n1 )  ( 2 1   4 1 )  ( 1 0   3  0 )  (1   2 )  1  (1   1 )  0  (1   3 ) 1 (1   4 )  0  a1 ] and M 1  an R n2  an1 R n1  R[  n   n   n1   n1 **  2( 2   2  1   2    2 1 )  (k1  n  k 3  n )  (k 2  n1  k 4  n1 )  ( 2 1   4 1 )  ( 1 0   3  0 )  (1   2 )  1  (1   1 )  0  (1   3 ) 1 (1   4 )  0  a1 ]. z  If n is odd, then P(z) has no zero in * ** * a0 M2 * R  1 and no zero in z  , for a0 M2 ** ** M 2 and M 2 are same as M 1 and M 1 except that k1 , k 2 , k3 , k 4 are respectively replaced by k 2 , k1 , k 4 , k 3 and  1 , 2 , 3 , 4 are respectively replaced by  2 , 1 , 4 , 3 . Combining Theorem 2 and Theorem B, we get the following result:  j Corollary 2: Let Let P ( z )   a j z be a polynomial o f degree n such that j 0 Re( a j )   j , Im( a j )   j and for some positive integers  ,  ,  ,  and for some real numbers k1 , k 2 , k 3 , k 4 ; 1 ,  2 ,  3 ,  4 ;0  k1  1,0  k 2  1,0  k 3  1,0  k 4  1 0   1  1,0   2  1,0   3  1,0   4  1 , k1 k 2 k3  k2  n 2[ ] 2  ......  2  2   2   ......  2   1 0 n 2[ ]1 2 n 2[ ] 2  ......  2  1   2  1  ......  3   2 1  ......  2   2   2   ......  2   3  0 n 2[ ]1 2  ......  2 1   2 1  ......  3   4  1 . . Then, for even n, the number of zeros of P(z) in a0 M3 *  z  R ( R  0, c  1) does not exceed c a0 M R 1  z  ( R  0, c  1) does not log 2 for R  1 and the number of zeros of P(z) in ** c log c a0 M3 M 1 * ** exceed log 2 for R  1, where M 2 , M 3 , M 3 are as given in Theorem 2 and Theorem B. log c a0 44
  • 5. Location of Zero-free Regions of Polynomials a0 If n is odd, the number of zeros of P(z) in M4 *  z  R ( R  0, c  1) does not exceed c  a0 M2 1 R log for R  1 and the number of zeros of P(z) in  z  ( R  0, c  1) does not ** log c a0 c M4  M2 1 * **  * ** log exceed , where M 2 , M 4 , M 4 are same as M 2 , M 3 , M 3 except that k1 , k 2 , k3 , k 4 log c a0 are respectively replaced by k 2 , k1 , k 4 , k 3 and Theorem 3: Let Let P ( z )   1 , 2 , 3 , 4 are respectively replaced by  2 , 1 , 4 , 3 .  a z j be a polynomial o f degree n such that j j 0  ,  , and for some real numbers k1 , k 2 , 1 , 2 , 0  k1  1,0  k 2  1,0   1  1,0   2  1, for some positive integers k1 a n 2[ ] 2 k2 a  ...... a 2   2  a 2   ...... a 2   1 a 0 n 2[ ]1 2  ...... a 2  1  a 2  1  ...... a 3   2 a1 . Then for even n P(z) has no zero in z  a0 M5 * , for a0 R  1 and no zero in z  M5 ** for R  1, where M 5  an R n2  an1 R n1  R n [ a1  (1  k1 ) an  (1  k 2 ) an1 *  cos  (2 a 2   2 a 2   k1 a n  k 2 a n 1   2 a1   1 a 0 ) n2  sin  (k1 a n  k 2 a n 1   2 a1   1 a 0  2 a j )] j 2 and M 5  an R n2  an1 R n1  R[ a1  (1  k1 ) an  (1  k 2 ) an1 **  cos  (2 a 2   2 a 2   k1 a n  k 2 a n 1   2 a1   1 a 0 ) n2  sin  (k1 a n  k 2 a n 1   2 a1   1 a 0  2 a j )] j 2 and for odd n P(z) has no zero in z  a0 M6 * , for R  1 and no zero in z  a0 M6 ** for R  1, where * ** * ** M 6 and M 6 are same as M 5 and M 5 except that k1 , k 2 are respectively replaced by k 2 , k1 and  1 , 2 are respectively replaced by  2 ,  1 . Combining Theorem 3 and Theorem C, we get the following result: Corollary 3: Let Let P ( z )   a z j 0 j j be a polynomial o f degree n such that  ,  , and for some real numbers k1 , k 2 , 1 , 2 , 0  k1  1,0  k 2  1,0   1  1,0   2  1, for some positive integers 45
  • 6. Location of Zero-free Regions of Polynomials k1 a n 2[ ] 2 k2 a  ...... a 2   2  a 2   ...... a 2   1 a 0 n 2[ ]1 2  ...... a 2  1  a 2  1  ...... a 3   2 a1 . Then for even n the number of zeros of P(z) in a0 M5 *  z  R ( R  0, c  1) does not exceed c a0 M R 1  z  ( R  0, c  1) does not log 3 for R  1 and the number of zeros of P(z) in ** c log c a0 M5 M 1 * ** exceed log 3 for R  1,where M 3 , M 5 , M 5 are as given in Theorem 3 and Theorem C. log c a0 If n is odd, then the number of zeros of P(z) in a0 M6 *  z  R ( R  0, c  1) does not exceed c a0 M R 1  z  ( R  0, c  1) does not log 4 for R  1 and the number of zeros of P(z) in ** c log c a0 M6 M 1 * ** * ** exceed log 4 for R  1,where M 4 , M 6 , M 6 are same as M 3 , M 5 , M 5 except that log c a0 k1 , k 2 are respectively replaced by k 2 , k1 and  1 , 2 are respectively replaced by  2 ,  1 . For different values of the parameters, we get many interesting results from the above theorems. II. PROOFS OF THEOREMS Proof of Theorem 1: Consider the polynomial F ( z )  (1  z ) P( z )  (1  z )( a n z n  a n 1 z n 1  ......  a1 z  a 0 )  a n z n 1  (a n  a n 1 ) z n  ......  (a1  a 0 ) z  a 0  a n z n 1  a 0  (k1  1) n z n  (k1 n   n 1 ) z n  ( n 1   n  2 ) z n 1  ...... (  1    ) z  1  (     1 ) z   ....... (1   1 0 ) z  ( 1  1) 0 z  i{(k 2  1)  n z n  (k 2  n   n1 ) z n  (  n1   n2 ) z n1  ......  (   1    ) z  1  (      1 ) z   ...... ( 1   2  0 ) z  ( 2  1)  0 z}  a 0  G ( z ) ,where G ( z )  a n z n 1  (k1  1) n z n  (k1 n   n 1 ) z n  ( n 1   n  2 ) z n 1  ...... (  1    ) z  1  (     1 ) z   ....... (1   1 0 ) z  ( 1  1) 0 z  i{(k 2  1)  n z n  (k 2  n   n1 ) z n  (  n1   n2 ) z n1  ......  (   1    ) z  1  (      1 ) z   ...... ( 1   2  0 ) z  ( 2  1)  0 z} For z  R , we have by using the hypothesis G( z)  an R n1  (1  k1 )  n R n  ( n1  k1 n )R n  ( n2   n1 )R n1  ...... 46
  • 7. Location of Zero-free Regions of Polynomials  (     1 ) R  1  (     1 ) R   ....... ( 1   1 0 ) R  (1   1 )  0 R  (1  k 2 )  n R n  (  n 1  k 2  n ) R n  (  n  2   n 1 ) R n 1  ......  (      1 ) R  1  (      1 ) R   ...... (  1   2  0 ) R  (1   2 )  0 R R  1, G( z)  an R n1  R n [(1  k1 )  n   n1  k1 n   n2   n1  ...... For     2    1       1  (1  k 2 )  n   n1  k 2  n   n 2   n 1  ......     2    1       1 ]  R  [     1    1    2  ......   1   1 0  (1   1 )  0 ]  R  [      1    1    2  ......  1   2  0  (1   2 )  0 ]  an R n1  R n [  n   n  k1 (  n   n )  k 2 (  n   n )       ]  R  [    1 (  0   0 )   0 ]  R  [    2 (  0   0 )   0 ]  M* For R  1, G( z)  an R n1  R n [  n   n  k1 (  n   n )  k 2 (  n   n )       ]  R[       1 (  0   0 )   2 (  0   0 )   0   0 ]  M ** Since G(z) is analytic for z  R and G(0)=0, it follows by Schwarz Lemma that G( z)  M * z for R  1 and G( z)  M ** z for R  1. Hence, for R  1 , F ( z)  a0  G( z)  a0  G ( z )  a0  M * z if z  0 a0 . M* And for R  1, F ( z)  a0  G( z)  a0  G( z)  a 0  M ** z if z  0 a0 M ** . This shows that F(z) has no zero in z  for R  1. a0 M * , for R  1 and no zero in z  47 a0 M **
  • 8. Location of Zero-free Regions of Polynomials Since the zeros of P(z) are also the zeros of F(z) , it follows that that P(z) has no zero in z  and no zero in z  a0 M ** for a0 M* R  1 , thereby proving Theorem 1. Proof of Theorem 2: Let n be even. Consider the polynomial F ( z )  (1  z 2 ) P ( z )  (1  z 2 )( a n z n  a n 1 z n 1  ......  a1 z  a 0 )  az n  2  a n 1 z n 1  (a n  a n  2 ) z n  (a n 1  a n 3 ) z n 1  ......  (a 3  a1 ) z 3  ( a 2  a 0 ) z 2  a1 z  a 0  a n z n  2  a n 1 z n 1  (k1  1) n z n  (k1 n   n  2 ) z n  (k 2  1) z n 1  (k 2 n1   n3 ) z n1  ( n2   n4 ) z n2  ( n3   n5 ) z n3  ......  ( 3   2 1 ) z 3  ( 2 1   1 ) z 3  ( 2   1 0 ) z 2  ( 1 0   0 ) z 2  a1 z  a0  i{(k 3  1)  n z n  (k 3  n   n2 ) z n  (k 4  1)  n1 z n1  (k 4  n1   n3 ) z n1  (  n2   n4 ) z n2  (  n3   n5 ) z n3  ......  (  3   4 1 ) z 3  ( 4 1  1 ) z 3  (  2   3  0 ) z 2  ( 3  0   0 ) z 2 .  a 0  G ( z ) , where G ( z )  a n z n  2  a n 1 z n 1  (k1  1) n z n  (k1 n   n  2 ) z n  (k 2  1) z n 1  (k 2 n 1   n 3 ) z n 1  ( n 2   n  4 ) z n 2  ( n 3   n 5 ) z n 3  ......  ( 3   2 1 ) z 3  ( 2 1   1 ) z 3  ( 2   1 0 ) z 2  ( 1 0   0 ) z 2  a1 z  i{(k 3  1)  n z n  (k 3  n   n  2 ) z n  (k 4  1)  n 1 z n 1  (k 4  n 1   n 3 ) z n 1  (  n 2   n 4 ) z n 2  (  n 3   n 5 ) z n 3  ......  (  3   4 1 ) z 3  ( 4 1  1 ) z 3  (  2   3  0 ) z 2  ( 3  0   0 ) z 2 . For z  R , we have by using the hypothesis G( z)  an R n2  an1 R n1  (1  k1 )  n R n  ( n2  k1 n )R n  (1  k 2 )  n1 R n1  ( n 3  k 2 n 1 ) R n 1  ( n  4   n  2 ) R n  2  ( n 5   n 3 ) R n 3  ......  ( 2   2  2 ) R 2  2  ( 2   2  2 ) R 2 ...... ( 2  1   2  1 ) R 2  1  ( 2  1   2  3 ) R 2  1  ...... ( 3   2 1 ) R 3  (1   2 )  1 R 3  ( 2   1 0 ) R 2  (1   1 )  0 R 2  a1 R  (1  k 3 )  n R n  (  n  2  k 3  n ) R n  ......  (  2    2   2 ) R 2   2  (  2    2   2 ) R 2   ...... (  2 1   2 1 ) R 2 1  (  2 1   2 3 ) R 2 1  (  3   4  1 ) R 3  (1   4 )  1 R 3  (  2   3  0 ) R 2  (1   3 )  0 R 2 R  1, G( z)  an R n2  an1 R n1  R n [(1  k1 )  n   n2  k1 n  (1  k 2 )  n1 For 48 , for R 1
  • 9. Location of Zero-free Regions of Polynomials   n 3  k 2 n 1   n  4   n  2   n 5   n 3  ......   2   2  2   2   2  2 ......  2  1   2  1   2  1   2  3  ......  3   2 1  (1   2 )  1  ( 2   1 0 )  (1   1 )  0  a 0  (1  k 3 )  n   n  2  k 3  n  ......  2    2   2   2    2   2  ......  2 1   2 1   2 1   2 3   3   4  1  (1   4 )  1  (  2   3  0 )  (1   3 )  0  a1 ]  an R n2  an1 R n1  R n [  n   n   n1   n1  2( 2   2  1   2    2 1 )  (k1  n  k 3  n )  (k 2  n 1  k 4  n1 )  ( 2 1   4 1 )  ( 1 0   3  0 )  (1   2 )  1  (1   1 )  0  (1   3 ) 1 (1   4 )  0  a1 ]  M1 . For R  1, G( z)  an R n2  an1 R n1  R[  n   n   n1   n1 *  2( 2   2  1   2    2 1 )  (k1  n  k 3  n )  (k 2  n 1  k 4  n1 )  ( 2 1   4 1 )  ( 1 0   3  0 )  (1   2 )  1  (1   1 )  0  (1   3 ) 1 (1   4 )  0  a1 ]  M1 . ** Since G(z) is analytic for z  R and G(0)=0, it follows by Schwarz Lemma G( z)  M 1 z for R  1 and G( z )  M 1 z for R  1. Hence, for R  1 , F ( z)  a0  G( z) * that **  a0  G ( z )  a0  M * z if z  0 a0 * . M1 And for R  1, F ( z)  a0  G( z)  a0  G( z)  a 0  M ** z if z  0 a0 M1 ** . This shows that F(z) has no zero in z  a0 M1 * , for R  1 and no zero in z  49 a0 M1 **
  • 10. Location of Zero-free Regions of Polynomials for R  1. Since the zeros of P(z) are also the zeros of F(z) , it follows that that P(z) has no zero in and no zero in z  a0 M1 ** for z  a0 M1 R  1 , thereby proving Theorem 1 for even n. The proof for odd n is similar and is omitted. Proof of Theorem 3: Suppose that n is even and the coefficient conditions hold i.e. k1 a n  ......  a 2   2  a 2   ......  a 2   1 a 0 k 2 a n 1  ......  a 2  1  a 2  1  ......  a 3   2 a1 . Consider the polynomial F ( z )  (1  z 2 ) P ( z )  (1  z 2 )( a n z n  a n 1 z n 1  ......  a1 z  a 0 )  az n  2  a n 1 z n 1  (a n  a n  2 ) z n  (a n 1  a n 3 ) z n 1  ......  (a 3  a1 ) z 3  ( a 2  a 0 ) z 2  a1 z  a 0  a n z n  2  a n 1 z n 1  (k1  1)a n z n  (k1 a n  a n  2 ) z n  (k 2  1)a n z n 1  (k 2 a n 1  a n 3 ) z n 1  (a n  2  a n  4 ) z n  2  (a n 3  a n 5 ) z n 3  ......  (a 2   2  a 2  ) z 2   2  (a 2  a 2   2 ) z 2   ...... (a 2  1  a 2  1 ) z 2  1  (a 2  1  a 2  3 ) z 2  1  ...... (a 3   2 a1 ) z 3  ( 2 a1  a1 ) z 3  (a 2   1 a 0 ) z 2  ( 1 a 0  a 0 ) z 2  a1 z  a 0  a 0  G ( z ) , where G ( z )  a n z n  2  a n 1 z n 1  (k1  1)a n z n  (k1 a n  a n  2 ) z n  (k 2  1)a n z n 1  (k 2 a n 1  a n 3 ) z n 1  (a n  2  a n  4 ) z n  2  (a n 3  a n 5 ) z n 3  ......  (a 2   2  a 2  ) z 2   2  (a 2   a 2   2 ) z 2   ...... (a 2  1  a 2  1 ) z 2  1  (a 2  1  a 2  3 ) z 2  1  ...... (a3   2 a1 ) z 3  ( 2 a1  a1 ) z 3  (a 2   1 a 0 ) z 2  ( 1 a 0  a 0 ) z 2  a1 z For z  R , we have by using the hypothesis and Lemma 3 G( z)  an R n2  an1 R n1  (1  k1 ) an R n  an2  k1an R n  (1  k 2 ) an1 R n1  a n 3  k 2 a n 1 R n 1  a n  4  a n  2 R n  2  a n 5  a n 3 R n 3  ......  a 2  a 2  2 R 2  2  a 2  a 2   2 R 2 ...... a 2  1  a 2  1 R 2  1  a 2  1  a 2  3 R 2  1  ...... a3   2 a1 R 3  (1   2 ) a1 R 3  a 2   1 a 0 R 2  (1   1 )  0 R 2  a1 R  an R n2  an1 R n1  R n [(1  k1 ) an  (1  k 2 ) an1 50 * , for R 1
  • 11. Location of Zero-free Regions of Polynomials  ( a n  2  k1 a n ) cos  ( a n  2  k1 a n ) sin   ( a n  4  a n  2 ) cos  ( a n  4  a n  2 ) sin   ...... ( a 2   a 2   2 ) cos  ( a 2   a 2   2 ) sin   ( a 2   a 2   2 ) cos  ( a 2   a 2   2 ) sin   ...... ( a 2  1  a 2  1 ) cos  ( a 2  1  a 2  1 ) sin   ( a 2  1  a 2  3 ) cos  ( a 2  1  a 2  3 ) sin   ...... ( a 3   2 a1 ) cos  ( a 3   2 a1 ) sin   (1   2 ) a1  (1   1 ) a 0  ( a 2   1 a 0 ) cos  ( a 2   1 a 0 ) sin   a1 ]  an R n2  an1 R n1  R n [ a1  (1  k1 ) an  (1  k 2 ) an1  cos  (2 a 2   2 a 2   k1 a n  k 2 a n 1   2 a1   1 a 0 ) n2  sin  (k1 a n  k 2 a n 1   2 a1   1 a 0  2 a j )] j 2 for R  1 .  M5 For R  1 , G( z)  an R n2  an1 R n1  R[ a1  (1  k1 ) an  (1  k 2 ) an1 *  cos  (2 a 2   2 a 2   k1 a n  k 2 a n 1   2 a1   1 a 0 ) n2  sin  (k1 a n  k 2 a n 1   2 a1   1 a 0  2 a j )] j 2  M5 ** . Since G(z) is analytic for z  R and G(0)=0, it follows by Schwarz Lemma G( z)  M 5 z for R  1 and G( z)  M 5 z for R  1. Hence, for R  1 , F ( z)  a0  G( z) * that **  a0  G ( z )  a0  M 5 z * if z  0 a0 * . M5 And for R  1, F ( z)  a0  G( z)  a0  G( z )  a0  M 5 z ** if z  0 a0 M5 ** . This shows that F(z) has no zero in for R  1. z  a0 M5 * , for R  1 and no zero in z  51 a0 M5 **
  • 12. Location of Zero-free Regions of Polynomials Since the zeros of P(z) are also the zeros of F(z) , it follows that that P(z) has no zero in and no zero in z  a0 M5 ** for z  a0 M5 * , for R 1 R  1 , thereby proving Theorem 3 for even n. For odd n the proof is similar and is omitted. REFERENCES [1]. [2]. [3]. [4]. [5]. [6]. K. K. Dewan, Extremal Properties and Coefficient estimates for Polynomials with restricted Zeros and on location of Zeros of Polynomials, Ph.D. Thesis, IIT Delhi, 1980. N. K. Govil and Q. I. Rahman, On the Enestrom- Kakeya Theorem, Tohoku Math. J. 20(1968), 126136. M. H. Gulzar, On the Location of Zeros of Polynomials with Restricted Coefficients, Int. Journal of Advanced Scientific Research and Technology, Vol.3, Issue 2, June 2012, 546-556. M. H. Gulzar, On the Zeros of Complex Polynomials in a Given Circle, International Journal of Engineering Research and Development, Vol. 8, Issue 11 (October 2013), 54-61. M. Marden, Geometry of Polynomials,Math. Surveys No. 3, Amer. Math. Society (R.I. Providence) (1996). Q. G. Mohammad, On the Zeros of Polynomials, Amer. Math. Monthly 72(1965), 631-633. 52