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International Journal of Engineering Inventions
e-ISSN: 2278-7461, p-ISSN: 2319-6491
Volume 3, Issue 2 (September 2013) PP...
Zeros of a Polynomial in a given Circle
www.ijeijournal.com Page | 39
The above results were further generalized by resear...
Zeros of a Polynomial in a given Circle
www.ijeijournal.com Page | 40
In the present paper , we find bounds for the number...
Zeros of a Polynomial in a given Circle
www.ijeijournal.com Page | 41
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Zeros of a Polynomial in a given Circle
www.ijeijournal.com Page | 42
Theorem 3: Let 
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Zeros of a Polynomial in a given Circle
www.ijeijournal.com Page | 43
Lemma 1 is the famous Jensen’s theorem (see page 208...
Zeros of a Polynomial in a given Circle
www.ijeijournal.com Page | 44
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Zeros of a Polynomial in a given Circle
www.ijeijournal.com Page | 45
Proof of Theorem 4: Consider the polynomial
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Zeros of a Polynomial in a given Circle
www.ijeijournal.com Page | 46
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Zeros of a Polynomial in a given Circle

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Zeros of a Polynomial in a given Circle

  1. 1. International Journal of Engineering Inventions e-ISSN: 2278-7461, p-ISSN: 2319-6491 Volume 3, Issue 2 (September 2013) PP: 38-46 www.ijeijournal.com Page | 38 Zeros of a Polynomial in a given Circle M. H. Gulzar Department of Mathematics, University of Kashmir, Srinagar 190006 Abstract: In this paper we discuss the problem of finding the number of zeros of a polynomial in a given circle when the coefficients of the polynomial or their real or imaginary parts are restricted to certain conditions. Our results in this direction generalize some well- known results in the theory of the distribution of zeros of polynomials. Mathematics Subject Classification: 30C10, 30C15 Keywords and phrases: Coefficient, Polynomial, Zero. I. Introduction and Statement of Results In the literature many results have been proved on the number of zeros of a polynomial in a given circle. In this direction Q. G. Mohammad [5] has proved the following result: Theorem A: Let     0 )( j j j zazP be a polynomial o f degree n such that 0...... 011   aaaa nn , Then the number of zeros of P(z) in 2 1 z does not exceed 0 log 2log 1 1 a an  . K. K. Dewan [2] generalized Theorem A to polynomials with complex coefficients and proved the following results: Theorem B: Let     0 )( j j j zazP be a polynomial o f degree n such that jja )Re( , jja )Im( and 0...... 011    nn , Then the number of zeros of P(z) in 2 1 z does not exceed 0 0 log 2log 1 1 a n j jn     . Theorem C: Let     0 )( j j j zazP be a polynomial o f degree n with complex coefficients such that for some real , , nja j ,......,2,1,0, 2 arg    and 011 ...... aaaa nn   . Then the number f zeros of P(z) in 2 1 z does not exceed 0 1 0 sin2)1sin(cos log 2log 1 a aa n j jn      .
  2. 2. Zeros of a Polynomial in a given Circle www.ijeijournal.com Page | 39 The above results were further generalized by researchers in various ways. M. H. Gulzar[4] proved the following results: Theorem D: Let     0 )( j j j zazP be a polynomial o f degree n such that jja )Re( , jja )Im( and 01111 ............    knknknnn , for some real numbers .0,1,10,0,  knnk  If knkn   1 , then the number f zeros of P(z) in 10,  z , does not exceed 0 0 000 22)(1)1(2 log 1 log 1 a n j jknknnn      and if 1  knkn  , then the number f zeros of P(z) in 10,  z , does not exceed 0 0 000 22)(1)1(2 log 1 log 1 a n j jknknnn      Theorem E: Let     0 )( j j j zazP be a polynomial o f degree n with complex coefficients such that for some real , , nja j ,......,2,1,0, 2 arg    and 01111 ............ aaaaaaa knknknnn    , for some 10,0,1,0,0    knnk . If 1..1   eiaa knkn , then the number of zeros of P(z) in 10,  z , does not exceed 0 1 log 1 log 1 a M  , where )1sincossin(cos)1sin)(cos(1    knn aaM     1 ,1 00 sin22)1sin(cos n knjj jaaa  , and if 1..1   eiaa knkn , then the number of zeros of P(z) in 10,  z , does not exceed 0 2 log 1 log 1 a M  , where )1sincossin(cos)1sin)(cos(2    knn aaM     1 ,1 00 sin22)1sin(cos n knjj jaaa  .
  3. 3. Zeros of a Polynomial in a given Circle www.ijeijournal.com Page | 40 In the present paper , we find bounds for the number of zeros of P(z) of the above results in a circle of any positive radius. More precisely, we prove the following resuls: Theorem 1: Let     0 )( j j j zazP be a polynomial o f degree n such that jja )Re( , jja )Im( and 01111 ............    knknknnn , for some real numbers .0,1,10,0,  knnk  . If knkn   1 , then the number f zeros of P(z) in )1,0(  cR c R z , does not exceed 0 0 000 1 ]22)(1)1(2[ log log 1 a R c n j jknknnn n      for 1R and 0 0 00000 ]2)(1)1(2[ log log 1 a Ra c n j jknknnn     for 1R . If 1  knkn  , then the number f zeros of P(z) in )1,0(  cR c R z does not exceed 0 0 000 1 ]22)(1)1(2[ log log 1 a R c n j jknknnn n      for 1R and 0 0 00000 ]2)(1)1(2[ log log 1 a Ra c n j jknknnn     for 1R . Remark 1: Taking R=1and 10, 1    c , Theorem 1 reduces to Theorem D. If the coefficients ja are real i.e. jj  ,0 , then we get the following result from Theorem 1: Corollary 1: Let     0 )( j j j zazP be a polynomial o f degree n such that jja )Re( , jja )Im( and 01111 ............ aaaaaaa knknknnn    , for some real numbers .0,1,10,0,  knank . If knkn aa  1 , then the number f zeros of P(z) in )1,0(  cR c R z does not exceed 0 000 1 2)(1)1(2[ log log 1 a aaaaaaaR c knknnn n     for 1R
  4. 4. Zeros of a Polynomial in a given Circle www.ijeijournal.com Page | 41 and 0 0000 )(1)1(2[ log log 1 a aaaaaaaRa c knknnn    for 1R . If 1  knkn aa , then the number f zeros of P(z) in )1,0(  cR c R z does not exceed 0 000 1 2)(1)1(2[ log log 1 a aaaaaaaR c knknnn n     for 1R and 0 0000 )(1)1(2[ log log 1 a aaaaaaaRa c knknnn    for 1R . Applying Theorem 1 to the polynomial –iP(z) , we get the following result: Theorem 2: : Let     0 )( j j j zazP be a polynomial o f degree n such that jja )Re( , jja )Im( and 01111 ............    knknknnn , for some real numbers .0,1,10,0,  knnk  . If knkn    1 , then the number f zeros of P(z) in )1,0(  cR c R z does not exceed 0 0 000 1 ]22)(1)1(2[ log log 1 a R c n j jknknnn n      for 1R and 0 0 00000 ]2)(1)1(2[ log log 1 a Ra c n j jknknnn     for 1R . If 1  knkn  , then the number f zeros of P(z) in )1,0(  cR c R z does not exceed 0 0 000 1 ]22)(1)1(2[ log log 1 a R c n j jknknnn n      for 1R and 0 0 00000 ]2)(1)1(2[ log log 1 a Ra c n j jknknnn     for 1R .
  5. 5. Zeros of a Polynomial in a given Circle www.ijeijournal.com Page | 42 Theorem 3: Let     0 )( j j j zazP be a polynomial o f degree n with complex coefficients such that for some real , , nja j ,......,2,1,0, 2 arg    and 01111 ............ aaaaaaa knknknnn    , for some 10,0,1,0,0    knnk . If 1..1   eiaa knkn , then the number of zeros of P(z) in )1,0(  cR c R z , does not exceed 0 3 log log 1 a M c , where )1sincossin(cos)1sin)(cos[(1 3     knn n aaRM ]sin22)1sin(cos 1 ,1 00     n knjj jaaa  , for 1R and )1sincossin(cos)1sin)(cos[(03    knn aaRaM ]sin22)1sin(cos 1 ,1 00     n knjj jaaa  , for 1R . If 1..1   eiaa knkn , then the number of zeros of P(z) in )1,0(  cR c R z does not exceed 0 4 log log 1 a M c , Where )1sincossin(cos)1sin)(cos[(1 4     knn n aaRM ]sin22)1sin(cos 1 ,1 00     n knjj jaaa  for 1R and )1sincossin(cos)1sin)(cos[(04    knn aaRaM ]sin2)1sin(cos 1 ,1 00     n knjj jaaa  for 1R . Remark 2: Taking R=1and 10, 1    c , Theorem 1 reduces to Theorem E. For different values of the parameters R, c, k, , , we get many other interesting results. 2. Lemmas For the proofs of the above results we need the following results: Lemma 1: If f(z) is analytic in Rz  ,but not identically zero, f(0)  0 and nkaf k ,......,2,1,0)(  , then    n j j i a R fdf 1 2 0 log)0(log(Relog 2 1     .
  6. 6. Zeros of a Polynomial in a given Circle www.ijeijournal.com Page | 43 Lemma 1 is the famous Jensen’s theorem (see page 208 of [1]). Lemma 2: If f(z) is analytic and )()( rMzf  in rz  , then the number of zeros of F(z) in 1,  c c r z does not exceed )0( )( log log 1 f rM c . Lemma 2 is a simple deduction from Lemma 1. Lemma 3: Let     0 )( j j j zazP be a polynomial o f degree n with complex coefficients such that for some real , , ,0, 2 arg nja j    and ,0,1 njaa jj   then any t>0,  sin)(cos)( 111   jjjjjj aataatata .s Lemma 3 is due to Govil and Rahman [3]. 3.Proofs of Theorems Proof of Theorem 1: Consider the polynomial )()1()( zPzzF  kn knkn kn knkn n nn n n n n n n zaazaazaaza azazazaz          )()(......)( )......)(1( 1 1 11 1 01 1 1 001 1 21 )(......)( azaazaa kn knkn    1 11 1 )(......)()(     kn knkn n nn n nn zzzi  0 1 100 01 1 211 )()1( )(......)()(   iziz zzz n j j jj kn knkn kn knkn         If ,1 knkn    then 1 11 1 )(......)()()(     kn knkn n nn nn nn zzzzizF  .)()1()( ......)()1()( 0 1 10001 1 211   izizz zzz n j j jj kn knkn kn kn kn knkn           For Rz  , we have by using the hypothesis    ......)( 1 11 n nn nn n n n RRRRzF  1 1    kn knkn R + kn kn kn knkn RR       11 RRR kn knkn 001 1 21 )1(......     j j n j j R)( 1 0 00      ]22)(1)1(2[ 0 000 1     n j jknknnn n R  for 1R and
  7. 7. Zeros of a Polynomial in a given Circle www.ijeijournal.com Page | 44    n j jknknnnRa 1 00000 2)(1)1(2[  for 1R . Hence by Lemma 2, the number of zeros of F(z) in )1,0(  cR c R z , does not exceed 0 0 000 1 ]22)(1)1(2[ log log 1 a R c n j jknknnn n      for 1R and 0 0 00000 ]2)(1)1(2[ log log 1 a Ra c n j jknknnn     for 1R . If ,1  knkn  then 1 11 1 )(......)()()(     kn knkn n nn nn nn zzzzizF  .)()( ......)()1()( 0 1 1001 1 21 1 1   iziz zzz n j j jj kn knkn kn kn kn knkn           For Rz  , we have by using the hypothesis    ......)( 1 11 n nn nn n n n RRRRzF  1 1    kn knkn R +      kn kn kn knkn RR  11 RRR kn knkn 001 1 21 )1(......     j j n j j R)( 1 0 00      ]22)(1)1(2[ 0 000 1     n j jknknnn n R  for 1R and    n j jknknnnRa 1 00000 2)(1)1(2[  for 1R . Hence by Lemma 2, the number of zeros of F(z) in )1,0(  cR c R z does not exceed 0 0 000 1 ]22)(1)1(2[ log log 1 a R c n j jknknnn n      for 1R and 0 0 00000 ]2)(1)1(2[ log log 1 a Ra c n j jknknnn     for 1R . That proves Theorem 1 completely.
  8. 8. Zeros of a Polynomial in a given Circle www.ijeijournal.com Page | 45 Proof of Theorem 4: Consider the polynomial )()1()( zPzzF  kn knkn kn knkn n nn n n n n n n zaazaazaaza azazazaz          )()(......)( )......)(1( 1 1 11 1 01 1 1 001 1 21 )(......)( azaazaa kn knkn    If knkn aa  1 , i.e. 1 , then 1 11 1 )(......)()(     kn knkn n nn nn n zaazaazzazF  ......)()1()( 1 211        kn knkn kn kn kn knkn zaazazaa  0001 )1()( azazaa   so that for Rz  , we have by using Lemma 3, kn knkn kn knkn n nn nn n RaaRaaRaaRRazF       1 1 11 1 ......)(  0001 1 21 )1(......1 aRaRaaRaaRa kn knkn kn kn       ......sin)(cos)([ 11 1     nnnnn n aaaaaR ])1(sin)(cos)( ......sin)(cos)( sin)(cos)( )1(sin)(cos)( 000101 2121 11 11 aaaaaa aaaa aaaa aaaaa knknknkn knknknkn knknknknkn             cossin(cos)1sin)(cos[(1    knn n aaR     1 ,1 00 sin22)1sin(cos)1sin n knjj jaaa  ] for 1R and  cossin(cos)1sin)(cos[(0  knn aaRa     1 ,1 00 sin2)1sin(cos)1sin n knjj jaaa  ] for 1R Hence , by Lemma 2, the number of zeros of F(z) and hence P(z) in )1,0(  cR c R z does not exceed 0 3 log log 1 a M c , where )1sincossin(cos)1sin)(cos[(1 3     knn n aaRM ]sin22)1sin(cos 1 ,1 00     n knjj jaaa  , for 1R and )1sincossin(cos)1sin)(cos[(03    knn aaRaM ]sin22)1sin(cos 1 ,1 00     n knjj jaaa  , for 1R . If 1  knkn aa , i.e. 1 , then
  9. 9. Zeros of a Polynomial in a given Circle www.ijeijournal.com Page | 46 1 11 1 )(......)()(     kn knkn n nn nn n zaazaazzazF  ......)()1()( 1 21 1 1        kn knkn kn kn kn knkn zaazazaa  001 )( azaa  so that for Rz  , we have by hypothesis and Lemma 3, kn knkn kn knkn n nn nn n RaaRaaRaaRRazF       1 1 11 1 ......)(  ......1 1 21 1      kn knkn kn kn RaaRa 0001 )1( aRaRaa   ......sin)(cos)([ 11 1     nnnnn n aaaaaR ])1(sin)(cos)( ......sin)(cos)( sin)(cos)( 1sin)(cos)( 000101 2121 11 11 aaaaaa aaaa aaaa aaaaa knknknkn knknknkn knknknknkn             cossin(cos)1sin)(cos[(1    knn n aaR )1sin   ]sin22)1sin(cos 1 ,1 00     n knjj jaaa  for 1R and  cossin(cos)1sin)(cos[(0  knn aaRa )1sin   ]sin2)1sin(cos 1 ,1 00     n knjj jaaa  for 1R . Hence the number of zeros of P(z) in )1,0(  cR c R z does not exceed 0 4 log log 1 a M c , Where )1sincossin(cos)1sin)(cos[(1 4     knn n aaRM ]sin22)1sin(cos 1 ,1 00     n knjj jaaa  for 1R and )1sincossin(cos)1sin)(cos[(04    knn aaRaM ]sin2)1sin(cos 1 ,1 00     n knjj jaaa  for 1R . That completes the proof of Theorem 3. References [1] L.V.Ahlfors, Complex Analysis, 3rd edition, Mc-Grawhill [2] 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. [3] N. K. Govil and Q. I. Rahman, On the Enestrom- Kakeya Theorem, Tohoku Math. J. 20(1968),126-136. [4] M. H. Gulzar, On the Zeros of a Polynomial inside the Unit Disk, IOSR Journal of Engineering, Vol.3, Issue 1(Jan.2013) IIV3II, 50-59. [5] Q. G. Mohammad, On the Zeros of Polynomials,Amer. Math. Monthly 72(1965), 631-633.

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