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1) The document presents theorems on the number and location of zeros of polynomials and analytic functions. 2) Theorem 1 shows that if a polynomial's coefficients satisfy certain conditions, it will have at least p zeros within a given region. 3) Theorem 2 generalizes this to the case where the coefficients are complex numbers. 4) Theorem 3 proves a similar result for the zeros of analytic functions within a disk, under conditions on the function's Taylor coefficients.

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International Journal of Computational Engineering Research||Vol, 03||Issue, 4|| www.ijceronline.com ||April||2013|| Page 172 On The Zeros of Polynomials and Analytic Functions M. H. Gulzar Department of Mathematics University of Kashmir, Srinagar 190006 I. INTRODUCTION AND STATEMENT OF RESULTS Regarding the zeros of a polynomial , Jain [2] proved the following results: Theorem A: Let n n p p p p zazazazaazP    ............)( 1 110 be a polynomial of degree n such that pp aa 1 for some  np ,......,2,1 , , 1 1 n n pj jjp aaaMM     nn aMnp  ),1( , 1 1 1     p j jjp aamm .0),2( 1  mnp Then P(z) has at least p zeros in 1 )( 1     p aa M p z pp , provided 1 1 )( 1     p aa M p pp and                              11 11 0 p aa M p p aa M p ma pp p pp . Theorem B: Let n n p p p p zazazazaazP    ............)( 1 110 be a polynomial of degree n such that pp aa 1 for some  np ,......,2,1 , ,,.......,1,0, 2 arg nja j    for some real  and  and 011 ...... aaaa nn   . Then P(z) has at least p zeros in 1 1     p aa L p z pp , where     n pj jjpnnp aaaaaLL 1 1 sin)(cos)(  and Abstract In this paper we obtain some results on the zeros of polynomials and related analytic functions, which generalize and improve upon the earlier well-known results. Mathematics Subject Classification: 30C10, 30C15 Key-words and phrases: Polynomial, Analytic Function , Zero.
On The Zeros Of Polynomials... www.ijceronline.com ||April||2013|| Page 173      1 1 101 sin)(cos)( p j jjpp aaaall  0),12( 1  lnp , provided 1 11 0 11                       p pp p pp p aa L p p aa L p la . In this paper, we prove the following results: Theorem 1: Let n n p p p p zazazazaazP    ............)( 1 110 be a polynomial of degree n such that pp aa 1 for some  1,......,2,1  np and ,0 0111 ............ aaaaaa ppnn    . Then P(z) has at least p zeros in 1 )( )2( 1 0 0      p aa M p Kz aKaaK a pp n n n  , where pnn aaaM  2 , provided   1 1 01 1 0 11                                 p pp p p pp p aa M p aa p aa M p a and K<1. Remark 1: Taking 0 in Theorem 1, we get the following result: Corollary 1: Let n n p p p p zazazazaazP    ............)( 1 110 be a polynomial of degree n such that pp aa 1 for some  1,......,2,1  np 0111 ............ aaaaaa ppnn   . Then P(z) has at least p zeros in 1 )( 2 1 1 1 10 0      p aa M p Kz aKaa a pp n n n  , where pnn aaaM 1 , provided   1 1 1 01 1 1 0 11                                   p pp p p pp p aa M p aa p aa M p a and 11 K . This result was earlier proved by Roshan Lal et al [4] . If the coefficients are positive in Theorem 1, we have the following result: Corollary 2: Let n n p p p p zazazazaazP    ............)( 1 110 be a polynomial of degree n such that pp aa 1 for some  1,......,2,1  np and ,0 .0............ 0111   aaaaaa ppnn  Then P(z) has at least p zeros in 1 )( )})1(2{ 1 2 2 202 0      p aa M p Kz aKaK a pp n n  , where pn aaM  )(22  , provided
On The Zeros Of Polynomials... www.ijceronline.com ||April||2013|| Page 174   1 1 2 01 1 2 0 11                                   p pp p p pp p aa M p aa p aa M p a and 12 K . If the coefficients of the polynomial P(z) are complex, we prove the following result: Theorem 2: Let n n p p p p zazazazaazP    ............)( 1 110 be a polynomial of degree n such that pp aa 1 for some  1,......,2,1  np and ,0 0111 ............ aaaaaa ppnn    and for some real  and  , .,......,1,0, 2 arg nja j    Then P(z) has at least p zeros in             1 1 3 3 p aa M p Kz pp , where      1 1 13 sin)(cos)1sin)(cos( n pj jjpn aaaaM  , provided 1 1 3 1 3 0 11                                 p pp p pp p aa M p m p aa M p a Remark 2: Taking 0 in Theorem 2, it reduces to Theorem B. Remark 3: It is easy to see that 3 K <1. Next, we prove the following result on the zeros of analytic functions : Theorem 3: Let 0)( 0    j j j zazf be analytic in 4 Kz  and for some natural number p with pa a p p 1 2 1   , ............ 11210   ppp aaaaaa , for some 0 ,and   1 1 10 1 0 1 2 1                                   p pp p p p pp p p aa a p aa p aa a p a  . Then f(z) has at least p zeros in             p pp a aa p p Kz 1 4 1 . Remark 4: Taking 0 , Theorem 3 reduces to the following result: Corollary 3: Let 0)( 0    j j j zazf be analytic in 4 Kz  and for some natural number p with pa a p p 1 2 1   ,
On The Zeros Of Polynomials... www.ijceronline.com ||April||2013|| Page 175 ............ 11210   ppp aaaaaa , for some 0 ,and   1 1 10 1 0 11                                   p pp p p p pp p p aa a p aa p aa a p a . Then f(z) has at least p zeros in             p pp a aa p p Kz 1 4 1 . Cor.3 was earlier proved by Roshan Lal et al [4]. II. LEMMA For the proofs of the above results , we need the following lemma due to Govil and Rahman [1]: Lemma : If 1 a and 2 a are complex numbers such that ,2,1, 2 arg  ja j   for some real numbers  and  , then  sin)(cos)( 212121 aaaaaa  . III. PROOFS OF THE THEOREMS 3.1 Proof of Theorem 1: Consider the polynomial )()1()( zPzzF  j n pj jj p pp p j j jj n n p p p p zaazaazaaa zazazazaaz           1 11 1 1 10 1 110 )()()( )............)(1( 1  n n za )()( zz   , where )( z      1 1 10 )( p j j jj zaaa and 1 1 11 )()()(      n n n pj jj p pp zaaazaaz . For Kz  (<1), we have, by using the hypothesis,              1 1 11 )( n n pj n j jj p pp KaKaaKaaz                   1 1 )1( 1 1 1 1 1 )( n pj pn jj pn nn pn n pp pp KaaKaaKaKKaa  pnnnn pp pp aaaaaKKaa     11 1 1 )(   pnnnn pp pp aaaaaKKaa     11 1 1 )(   pnn pp pp aaaKKaa    2)( 1 1
On The Zeros Of Polynomials... www.ijceronline.com ||April||2013|| Page 176 p pnn pp pp aaa aap aa              2 )( )( 1 1  pnn p pnn pp aaa aaa aap                 2 2 )( 1 1 1 1 12                     p pp p pnn p aa aaa p  1 1 1                  p pp p p aa M p (1) Also for Kz  (<1),      1 1 10 )( p j j jj Kaaaz      1 1 10 )( p j jj aaKa )( 010 aaKa p                         12 )( 011 0 p aa aaa aap a p pnn pp   01 1 0 1 aa p aa M p a p pp               (2) Since, by hypothesis   1 1 01 1 0 11                                 p pp p p pp p aa M p aa p aa M p a , it follows from (1) and (2) that )()( zz   for Kz  . Hence, by Rouche’s theorem, )( z and )()( zz   i.e. F(z) have the same number of zeros in Kz  Since the zeros of P(z) are also the zeros of F(z) and since )( z has at least p zeros in Kz  ,it follows that P(z) has at least p zeros in Kz  .That prove the first part of Theorem 1. To prove the second part, we show that P(z) has no zero in n nn Kaaa a z   0 0 2  . Let )()1()( zPzzF  1 1 10 1 110 )( )............)(1(         n n j n j jj n n p p p p zazaaa zazazazaaz )(0 zga  , where
On The Zeros Of Polynomials... www.ijceronline.com ||April||2013|| Page 177 1 1 1 1 )()(      n n n j j jj zazaazg . For Kz  (<1), we have, by using the hypothesis, 1 1 1 )(      n n j n j jj zazaazg 1 1 1      n n j n j jj KaKaa            n n n j jjnn KaaaaaK 1 1 11 )(  n nnnn KaaaaaK   011   n nnnn KaaaaaK   011   n nn KaaaK  0 2  . Since g(z) is analytic for Kz  , g(0)=0, we have, by Schwarz’s lemma,  zKaaaKzg n nn  0 2)(  for Kz  . Hence, for Kz  , )()( 0 zgazF   zKaaaKa zga n nn   00 0 2 )(  0 if )2( 0 0 n nn KaaaK a z    . This shows that F(z) and therefore P(z) has no zero in )2( 0 0 n nn KaaaK a z    . That proves Theorem 1 completely. Proof of Theorem 2: Consider the polynomial )()1()( zPzzF  j n pj jj p pp p j j jj n n p p p p zaazaazaaa zazazazaaz           1 11 1 1 10 1 110 )()()( )............)(1( 1  n n za )()( zz   , where )( z      1 1 10 )( p j j jj zaaa and
On The Zeros Of Polynomials... www.ijceronline.com ||April||2013|| Page 178 1 1 11 )()()(      n n n pj jj p pp zaaazaaz .. For 3 Kz  (<1), we have, by using the hypothesis,              1 3 1 3131 )( n n pj n j jj p pp KaKaaKaaz                   1 1 )1( 31 1 313 1 331 n pj pn jj pn nn pn n pp pp KaaKaaKaKKaa                1 1 11 1 331 n pj jjnnn pp pp aaaaaKKaa                 1 1 11 1 331 n pj jjnnn pp pp aaaaaKKaa        cos}){( 1 1 331 nnn pp pp aaaKKaa   sin)(cos)(...... sin)(cos)(sin}){( 2121 111     nnnn ppppnn aaaa aaaaaa   cos)1sin)(cos( 1 331 pn pp pp aaKKaa         1 1 1 sin)( n pj jj aa  3 1 331 MKKaa pp pp    3 1 1 3 1 3 1 11 M p aa M p p aa M p aa p pp p pp pp                          1 1 3 1                      p pp p p aa M p m p aa M p a pp              1 1 3 0 mKa  30 (3) Also , for 3 Kz  , we have, by using the lemma and the hypothesis, jp j jj zaaaz      1 1 10 )(           1 1 130 1 1 310 p j jj p j j jj aaKa Kaaa
On The Zeros Of Polynomials... www.ijceronline.com ||April||2013|| Page 179  .......sin)(cos)( 010130   aaaaKa      sin)(cos)( 2121   pppp aaaa      sin)(cos)( 10130 jjp aaaaKa mKa  30 (4) Thus, for 3 Kz  , we have from (3) and (4), )()( zz   . Since )( z and )( z are analytic for 3 Kz  , it follows by Rouche’s theorem that )( z and )()( zz   i.e. F(z) have the same number of zeros in 3 Kz  . But the zeros of P(z) are also the zeros of F(z). Therefore, we conclude that P(z) has at least p zeros in Kz  , as the same is true of )( z . That proves Theorem 2. Proof of Theorem 3: Consider the function )()1()( zfzzF           1 11 1 1 10 2 210 )()()( ......))(1( pj j jj p pp p j j jj zaazaazaaa zazaaz ),()( zz   where      1 1 10 )()( p j j jj zaaaz ,      1 11 )()()( pj j jj p pp zaazaaz . For 4 Kz  ( 4 K <1, by hypothesis for pa a p p 1 2 1   ), we have                 )1( 4 1 1 1 441 )( pj pj jj pp pp KaaKKaaz                1 1 1 441 pj jj pp pp aaKKaa p pp pp aKKaa 1 441    p p pp p pp pp a a aa p p a aa p p aa                         ))( 1 ())( 1 ( 11 1 1 1 1                      p pp p p p aa a p . (5) and      1 1 410 )( p j j jj Kaaaz   41221100 ...... Kaaaaaaa pp  
On The Zeros Of Polynomials... www.ijceronline.com ||April||2013|| Page 180   41221100 ..... Kaaaaaaa pp          4100 41221100 41221100 2 ..... ..... Kaaa Kaaaaaaa Kaaaaaaa p pp pp           10 1 0 2 1               p pp p aa p aa a p a  (6) Since, by hypothesis,   1 1 10 1 0 1 2 1                                   p pp p p p pp p p aa a p aa p aa a p a  , it follows from (5) and (6) that )()( zz   for 4 Kz  . Since )( z and )( z are analytic for 3 Kz  , it follows, by Rouche’s theorem, that )( z and )()( zz   i.e. F(z) have the same number of zeros in 3 Kz  . But the zeros of P(z) are also the zeros of F(z). Therefore, we conclude that P(z) has at least p zeros in Kz  , as the same is true of )( z . That proves Theorem 3. REFERENCES [1] N. K. Govil and Q. I. Rahman, On the Enestrom-Kakeya Theorem, Tohoku Math. J.20 (1968), 126- 136. [2] V.K. Jain, On the zeros of a polynomial, Proc.Indian Acad. Sci. Math. Sci.119 (1) ,2009, 37-43. [3] M. Marden, Geometry of Polynomials, Math. Surveys No. 3, Amer. Math. Soc. Providence, RI, 1966. [4] Roshan Lal, Susheel Kumar and Sunil Hans, On the zeros of polynomials and analytic functions, Annales Universitatis Mariae Curie –Sklodowska Lublin Polonia, Vol.LXV, No. 1, 2011, Sectio A, 97-108.

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Ae03401720180

  • 1. International Journal of Computational Engineering Research||Vol, 03||Issue, 4|| www.ijceronline.com ||April||2013|| Page 172 On The Zeros of Polynomials and Analytic Functions M. H. Gulzar Department of Mathematics University of Kashmir, Srinagar 190006 I. INTRODUCTION AND STATEMENT OF RESULTS Regarding the zeros of a polynomial , Jain [2] proved the following results: Theorem A: Let n n p p p p zazazazaazP    ............)( 1 110 be a polynomial of degree n such that pp aa 1 for some  np ,......,2,1 , , 1 1 n n pj jjp aaaMM     nn aMnp  ),1( , 1 1 1     p j jjp aamm .0),2( 1  mnp Then P(z) has at least p zeros in 1 )( 1     p aa M p z pp , provided 1 1 )( 1     p aa M p pp and                              11 11 0 p aa M p p aa M p ma pp p pp . Theorem B: Let n n p p p p zazazazaazP    ............)( 1 110 be a polynomial of degree n such that pp aa 1 for some  np ,......,2,1 , ,,.......,1,0, 2 arg nja j    for some real  and  and 011 ...... aaaa nn   . Then P(z) has at least p zeros in 1 1     p aa L p z pp , where     n pj jjpnnp aaaaaLL 1 1 sin)(cos)(  and Abstract In this paper we obtain some results on the zeros of polynomials and related analytic functions, which generalize and improve upon the earlier well-known results. Mathematics Subject Classification: 30C10, 30C15 Key-words and phrases: Polynomial, Analytic Function , Zero.
  • 2. On The Zeros Of Polynomials... www.ijceronline.com ||April||2013|| Page 173      1 1 101 sin)(cos)( p j jjpp aaaall  0),12( 1  lnp , provided 1 11 0 11                       p pp p pp p aa L p p aa L p la . In this paper, we prove the following results: Theorem 1: Let n n p p p p zazazazaazP    ............)( 1 110 be a polynomial of degree n such that pp aa 1 for some  1,......,2,1  np and ,0 0111 ............ aaaaaa ppnn    . Then P(z) has at least p zeros in 1 )( )2( 1 0 0      p aa M p Kz aKaaK a pp n n n  , where pnn aaaM  2 , provided   1 1 01 1 0 11                                 p pp p p pp p aa M p aa p aa M p a and K<1. Remark 1: Taking 0 in Theorem 1, we get the following result: Corollary 1: Let n n p p p p zazazazaazP    ............)( 1 110 be a polynomial of degree n such that pp aa 1 for some  1,......,2,1  np 0111 ............ aaaaaa ppnn   . Then P(z) has at least p zeros in 1 )( 2 1 1 1 10 0      p aa M p Kz aKaa a pp n n n  , where pnn aaaM 1 , provided   1 1 1 01 1 1 0 11                                   p pp p p pp p aa M p aa p aa M p a and 11 K . This result was earlier proved by Roshan Lal et al [4] . If the coefficients are positive in Theorem 1, we have the following result: Corollary 2: Let n n p p p p zazazazaazP    ............)( 1 110 be a polynomial of degree n such that pp aa 1 for some  1,......,2,1  np and ,0 .0............ 0111   aaaaaa ppnn  Then P(z) has at least p zeros in 1 )( )})1(2{ 1 2 2 202 0      p aa M p Kz aKaK a pp n n  , where pn aaM  )(22  , provided
  • 3. On The Zeros Of Polynomials... www.ijceronline.com ||April||2013|| Page 174   1 1 2 01 1 2 0 11                                   p pp p p pp p aa M p aa p aa M p a and 12 K . If the coefficients of the polynomial P(z) are complex, we prove the following result: Theorem 2: Let n n p p p p zazazazaazP    ............)( 1 110 be a polynomial of degree n such that pp aa 1 for some  1,......,2,1  np and ,0 0111 ............ aaaaaa ppnn    and for some real  and  , .,......,1,0, 2 arg nja j    Then P(z) has at least p zeros in             1 1 3 3 p aa M p Kz pp , where      1 1 13 sin)(cos)1sin)(cos( n pj jjpn aaaaM  , provided 1 1 3 1 3 0 11                                 p pp p pp p aa M p m p aa M p a Remark 2: Taking 0 in Theorem 2, it reduces to Theorem B. Remark 3: It is easy to see that 3 K <1. Next, we prove the following result on the zeros of analytic functions : Theorem 3: Let 0)( 0    j j j zazf be analytic in 4 Kz  and for some natural number p with pa a p p 1 2 1   , ............ 11210   ppp aaaaaa , for some 0 ,and   1 1 10 1 0 1 2 1                                   p pp p p p pp p p aa a p aa p aa a p a  . Then f(z) has at least p zeros in             p pp a aa p p Kz 1 4 1 . Remark 4: Taking 0 , Theorem 3 reduces to the following result: Corollary 3: Let 0)( 0    j j j zazf be analytic in 4 Kz  and for some natural number p with pa a p p 1 2 1   ,
  • 4. On The Zeros Of Polynomials... www.ijceronline.com ||April||2013|| Page 175 ............ 11210   ppp aaaaaa , for some 0 ,and   1 1 10 1 0 11                                   p pp p p p pp p p aa a p aa p aa a p a . Then f(z) has at least p zeros in             p pp a aa p p Kz 1 4 1 . Cor.3 was earlier proved by Roshan Lal et al [4]. II. LEMMA For the proofs of the above results , we need the following lemma due to Govil and Rahman [1]: Lemma : If 1 a and 2 a are complex numbers such that ,2,1, 2 arg  ja j   for some real numbers  and  , then  sin)(cos)( 212121 aaaaaa  . III. PROOFS OF THE THEOREMS 3.1 Proof of Theorem 1: Consider the polynomial )()1()( zPzzF  j n pj jj p pp p j j jj n n p p p p zaazaazaaa zazazazaaz           1 11 1 1 10 1 110 )()()( )............)(1( 1  n n za )()( zz   , where )( z      1 1 10 )( p j j jj zaaa and 1 1 11 )()()(      n n n pj jj p pp zaaazaaz . For Kz  (<1), we have, by using the hypothesis,              1 1 11 )( n n pj n j jj p pp KaKaaKaaz                   1 1 )1( 1 1 1 1 1 )( n pj pn jj pn nn pn n pp pp KaaKaaKaKKaa  pnnnn pp pp aaaaaKKaa     11 1 1 )(   pnnnn pp pp aaaaaKKaa     11 1 1 )(   pnn pp pp aaaKKaa    2)( 1 1
  • 5. On The Zeros Of Polynomials... www.ijceronline.com ||April||2013|| Page 176 p pnn pp pp aaa aap aa              2 )( )( 1 1  pnn p pnn pp aaa aaa aap                 2 2 )( 1 1 1 1 12                     p pp p pnn p aa aaa p  1 1 1                  p pp p p aa M p (1) Also for Kz  (<1),      1 1 10 )( p j j jj Kaaaz      1 1 10 )( p j jj aaKa )( 010 aaKa p                         12 )( 011 0 p aa aaa aap a p pnn pp   01 1 0 1 aa p aa M p a p pp               (2) Since, by hypothesis   1 1 01 1 0 11                                 p pp p p pp p aa M p aa p aa M p a , it follows from (1) and (2) that )()( zz   for Kz  . Hence, by Rouche’s theorem, )( z and )()( zz   i.e. F(z) have the same number of zeros in Kz  Since the zeros of P(z) are also the zeros of F(z) and since )( z has at least p zeros in Kz  ,it follows that P(z) has at least p zeros in Kz  .That prove the first part of Theorem 1. To prove the second part, we show that P(z) has no zero in n nn Kaaa a z   0 0 2  . Let )()1()( zPzzF  1 1 10 1 110 )( )............)(1(         n n j n j jj n n p p p p zazaaa zazazazaaz )(0 zga  , where
  • 6. On The Zeros Of Polynomials... www.ijceronline.com ||April||2013|| Page 177 1 1 1 1 )()(      n n n j j jj zazaazg . For Kz  (<1), we have, by using the hypothesis, 1 1 1 )(      n n j n j jj zazaazg 1 1 1      n n j n j jj KaKaa            n n n j jjnn KaaaaaK 1 1 11 )(  n nnnn KaaaaaK   011   n nnnn KaaaaaK   011   n nn KaaaK  0 2  . Since g(z) is analytic for Kz  , g(0)=0, we have, by Schwarz’s lemma,  zKaaaKzg n nn  0 2)(  for Kz  . Hence, for Kz  , )()( 0 zgazF   zKaaaKa zga n nn   00 0 2 )(  0 if )2( 0 0 n nn KaaaK a z    . This shows that F(z) and therefore P(z) has no zero in )2( 0 0 n nn KaaaK a z    . That proves Theorem 1 completely. Proof of Theorem 2: Consider the polynomial )()1()( zPzzF  j n pj jj p pp p j j jj n n p p p p zaazaazaaa zazazazaaz           1 11 1 1 10 1 110 )()()( )............)(1( 1  n n za )()( zz   , where )( z      1 1 10 )( p j j jj zaaa and
  • 7. On The Zeros Of Polynomials... www.ijceronline.com ||April||2013|| Page 178 1 1 11 )()()(      n n n pj jj p pp zaaazaaz .. For 3 Kz  (<1), we have, by using the hypothesis,              1 3 1 3131 )( n n pj n j jj p pp KaKaaKaaz                   1 1 )1( 31 1 313 1 331 n pj pn jj pn nn pn n pp pp KaaKaaKaKKaa                1 1 11 1 331 n pj jjnnn pp pp aaaaaKKaa                 1 1 11 1 331 n pj jjnnn pp pp aaaaaKKaa        cos}){( 1 1 331 nnn pp pp aaaKKaa   sin)(cos)(...... sin)(cos)(sin}){( 2121 111     nnnn ppppnn aaaa aaaaaa   cos)1sin)(cos( 1 331 pn pp pp aaKKaa         1 1 1 sin)( n pj jj aa  3 1 331 MKKaa pp pp    3 1 1 3 1 3 1 11 M p aa M p p aa M p aa p pp p pp pp                          1 1 3 1                      p pp p p aa M p m p aa M p a pp              1 1 3 0 mKa  30 (3) Also , for 3 Kz  , we have, by using the lemma and the hypothesis, jp j jj zaaaz      1 1 10 )(           1 1 130 1 1 310 p j jj p j j jj aaKa Kaaa
  • 8. On The Zeros Of Polynomials... www.ijceronline.com ||April||2013|| Page 179  .......sin)(cos)( 010130   aaaaKa      sin)(cos)( 2121   pppp aaaa      sin)(cos)( 10130 jjp aaaaKa mKa  30 (4) Thus, for 3 Kz  , we have from (3) and (4), )()( zz   . Since )( z and )( z are analytic for 3 Kz  , it follows by Rouche’s theorem that )( z and )()( zz   i.e. F(z) have the same number of zeros in 3 Kz  . But the zeros of P(z) are also the zeros of F(z). Therefore, we conclude that P(z) has at least p zeros in Kz  , as the same is true of )( z . That proves Theorem 2. Proof of Theorem 3: Consider the function )()1()( zfzzF           1 11 1 1 10 2 210 )()()( ......))(1( pj j jj p pp p j j jj zaazaazaaa zazaaz ),()( zz   where      1 1 10 )()( p j j jj zaaaz ,      1 11 )()()( pj j jj p pp zaazaaz . For 4 Kz  ( 4 K <1, by hypothesis for pa a p p 1 2 1   ), we have                 )1( 4 1 1 1 441 )( pj pj jj pp pp KaaKKaaz                1 1 1 441 pj jj pp pp aaKKaa p pp pp aKKaa 1 441    p p pp p pp pp a a aa p p a aa p p aa                         ))( 1 ())( 1 ( 11 1 1 1 1                      p pp p p p aa a p . (5) and      1 1 410 )( p j j jj Kaaaz   41221100 ...... Kaaaaaaa pp  
  • 9. On The Zeros Of Polynomials... www.ijceronline.com ||April||2013|| Page 180   41221100 ..... Kaaaaaaa pp          4100 41221100 41221100 2 ..... ..... Kaaa Kaaaaaaa Kaaaaaaa p pp pp           10 1 0 2 1               p pp p aa p aa a p a  (6) Since, by hypothesis,   1 1 10 1 0 1 2 1                                   p pp p p p pp p p aa a p aa p aa a p a  , it follows from (5) and (6) that )()( zz   for 4 Kz  . Since )( z and )( z are analytic for 3 Kz  , it follows, by Rouche’s theorem, that )( z and )()( zz   i.e. F(z) have the same number of zeros in 3 Kz  . But the zeros of P(z) are also the zeros of F(z). Therefore, we conclude that P(z) has at least p zeros in Kz  , as the same is true of )( z . That proves Theorem 3. REFERENCES [1] N. K. Govil and Q. I. Rahman, On the Enestrom-Kakeya Theorem, Tohoku Math. J.20 (1968), 126- 136. [2] V.K. Jain, On the zeros of a polynomial, Proc.Indian Acad. Sci. Math. Sci.119 (1) ,2009, 37-43. [3] M. Marden, Geometry of Polynomials, Math. Surveys No. 3, Amer. Math. Soc. Providence, RI, 1966. [4] Roshan Lal, Susheel Kumar and Sunil Hans, On the zeros of polynomials and analytic functions, Annales Universitatis Mariae Curie –Sklodowska Lublin Polonia, Vol.LXV, No. 1, 2011, Sectio A, 97-108.