C031012017

Research Inventy: International Journal Of Engineering And Science Vol.3, Issue 10 (Octoberr 2013), PP 12-17 Issn(e): 2278-4721, Issn(p):2319-6483, Www.Researchinventy.Com

Number of Zeros of a Polynomial in a Given Circle M. H. Gulzar Department of Mathematics University of Kashmir, Srinagar 190006

ABSTRACT: In this paper we consider 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 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 Regarding the number of zeros of a polynomial in a given circle, Q. G. Mohammad [6] proved the following result: Theorem A: Let P ( z ) 



a z j 0

j

j

be a polynomial o f degree n such that

an  an1  ......  a1  a0  0 .

1 Then the number of zeros of P(z) in z  does not exceed 2 a 1 1 log n . log 2 a0 K. K. Dewan [2] generalized Theorem A to polynomials with complex coefficients and proved the following results: Theorem B: 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

 n   n 1  ......   1   0  0 .

1 Then the number of zeros of P(z) in z  does not exceed 2 n

1 Theorem C: Let P ( z ) 

j 0

a0

.



a z j 0

real

1 log log 2

n    j

j

j

be a polynomial o f degree n with complex coefficients such that for some

,  , arg a j     

 2

, j  0,1,2,......,n

and

an  an1  ...... a1  a0 .

1 Then the number f zeros of P(z) in z  does not exceed 2

12

Number of Zeros of a Polynomial in a Given Circle… n 1

1 log log 2

a n (cos  sin   1)  2 sin   a j j 0

.

a0

C. M. Upadhye [3] found bounds for the number of zeros of P(z) in Theorems B and C with less restrictive conditions on the coefficients , which were further improved by M. H. Gulzar [5] by proving the following resuls: Theorem D: Let P ( z ) 



a z j 0

real

j

j

be a polynomial o f degree n with complex coefficients such that for some

,  , arg a j     

 2

, j  0,1,2,......,n

and

k an  an1  ...... a1   a0 ,

z   ,0    1, does not exceed

for k  1,0    1 .Then the number f zeros of P(z) in

n 1

k a n (cos   sin   1)  2 sin   a j  2 a 0   a 0 (cos   sin   1)

1 log

1

j 1

log



Theorem E: Let P ( z ) 

.

a0 

a z j 0

j

j

be a polynomial o f degree n such that Re( a j )   j , Im( a j ) 

 j and

k n   n 1  ......   1   0 , for k  1,0    1 ,then the number f zeros of P(z) in

z   ,0    1, does not exceed n

k (  n   n )  2  0   (  0   0 )  2  j

1 log

1

j 0

log

a0



.

In this paper we find bounds for the number of zeros of a polynomial in a circle of any positive radius of which the above results are easy consequences. More precisely we prove the following results: Theorem 1: 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

k n   n 1  ......   1   0 , for k  1,0    1 .Then the number f zeros of P(z) in

z

R ( R  0, c  1) does not exceed c

n 1 1 log [ R n1{k (  n   n )   (  0   0 )  2  0  2  j ] log c a0 j 0 for R  1

and

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Number of Zeros of a Polynomial in a Given Circle… n 1 1 log [ a 0  R{k (  n   n )   (  0   0 )   0   0  2  j ] log c a0 j 1 for R  1 .

Remark 1: Taking

c

1



and R=1, Theorem 1 reduces to Theorem E .

If the coefficients a j are real i.e.

Corollary 1: Let P ( z ) 

 j  0, j , we get the following result from Theorem 1:



a z j 0

j

j

be a polynomial o f degree n such that

kan  a n 1  ......  a1  a 0 , for k  1,0    1 .Then the number f zeros of P(z) in

z

R ( R  0, c  1) , does not exceed c

1 1 log [ R n 1 {k ( a n  a n )   ( a 0  a 0 )  2 a 0 ] log c a0 for

R 1

and

1 1 log [ a 0  R{k ( a n  a n )   ( a 0  a 0 )  a 0 ] log c a0

R  1.

for

Applying Theorem 1 to the polynomial –iP(z), we get the following result: 

j Theorem 2: Let P ( z )   a j z be a polynomial o f degree n such that Re( a j )   j , Im( a j )   j and j 0

k n   n 1  ......   1   0 , for k  1,0    1 .Then the number f zeros of P(z) in

z

R ( R  0, c  1) , does not exceed c

n 1 1 n 1 log [ R {k (  n   n )   (  0   0 )  2  0  2  j ] log c a0 j 0 for R  1

and n 1 1 log [ a 0  R{k (  n   n )   (  0   0 )   0   0  2  j ] log c a0 j 1 for R  1 .

Theorem 3: Let P ( z ) 



a z j 0

real

j

j

be a polynomial o f degree n with complex coefficients such that for some

,  , arg a j     

 2

, j  0,1,2,......,n

and

k an  an1  ...... a1   a0 , for k  1,0    1 ,then the number f zeros of P(z) in

z

14

R ( R  0, c  1) does not exceed c

Number of Zeros of a Polynomial in a Given Circle… n 1

1 log log c

R n 1 [k a n (cos  sin   1)  2 sin   a j   a0 (cos  sin   1)  2 a0 ] j 1

.

a0 for

R 1

and n 1

1 log log c

a 0  R[k a n (cos  sin   1)  2 sin   a j   a 0 (cos  sin   1)  a 0 ] j 1

.

a0 for

c

Remark 2: Taking

1



R  1.

,R=1, Theorem 3 reduces to Theorem D .

II. LEMMAS For the proofs of the above results we need the following results:

z  R ,but not identically zero, f(0)  0 and

Lemma 1: If f(z) is analytic in

f (a k )  0, k  1,2,......, n , then

1 2



2

0

n

R

j 1

aj

log f (Re d  log f (0)   log i

.

Lemma 1 is the famous Jensen’s theorem (see page 208 of [1]).

r f ( z)  M (r ) in z  r , then the number of zeros of f(z) in z  , c  1 c

Lemma 2: If f(z) is analytic and does not exceed

M (r ) 1 log . log c f ( 0) Lemma 2 is a simple deduction from Lemma 1.

Lemma 3: Let P ( z ) 



a z j 0

real

j

j

 ,  , arg a j     

 2

be a polynomial o f degree n with complex coefficients such that for some

,0  j  n, and

a j  a j 1 ,0  j  n, then for any t>0, ta j  a j 1  (t a j  a j 1 ) cos  (t a j  a j 1 ) sin  .s Lemma 3 is due to Govil and Rahman [4].

III. 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 n 1

 a n z n 1  a 0  [( k n   n 1 )  (k  1) n ]z n   ( j   j 1 ) z j j 2

15

Number of Zeros of a Polynomial in a Given Circle… n

 [( 1   0 )  (  1) 0 ]z  i  (  j   j 1 ) z j . j 1

For

z  R , we have by using the hypothesis n 1

F ( z )  a n R n 1  a 0  R n [( k  1)  n  (k n   n 1 )]    j   j 1 R j j 2

n

 R[  1   0  (1   )  0 ]    j   j 1 R j . j 1

Therefore n 1

F ( z )  R n 1 [  n   n  (k  1)  n  (k n   n 1 )   ( j   j 1 ) j 1

n

  0   0  ( 1   0 )  (1   )  0   (  j   j 1 )] j 1

n

 R n 1 [k (  n   n )   (  0   0 )  2  0  2  j ] j 1

for

R 1

and n

F (z)  a 0  R[k (  n   n )   (  0   0 )   0   0  2  j ] j 1

for

R 1 .

Hence , by Lemma 2, the number of zeros of F(z) and therefore P(z) in

z

R does not exceed c

n 1 1 log [ R n1{k (  n   n )   (  0   0 )  2  0  2  j ] log c a0 j 0 for R  1

and n 1 1 log [ a 0  R{k (  n   n )   (  0   0 )   0   0  2  j ] log c a0 j 1 for R  1 .

That proves Theorem 1.

Proof of Theorem 2: 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 n 1

 a n z n 1  a 0  [( kan  a n 1 )  (k  1)a n ]z n   (a j  a j 1 ) z j j 2

 [( a1  a 0 )  (  1)a 0 ]z .

For

z  R , we have by using the hypothesis and Lemma 3 n 1

F ( z )  a n R n 1  a 0  R n [( k  1) a n  kan  a n 1 ]   a j  a j 1 R j j 2

 [ a1  a0  (1   ) a0 ]R 16

Number of Zeros of a Polynomial in a Given Circle…

 an R n1  a0  R n [(k  1) an  R n [(k an  an1 ) cos n 1

 (k a n  a n 1 ) sin  ]   ( a j  a j 1 ) cos R j j 2

n 1

  ( a j  a j 1 ) sin R j  [( a1   a 0 ) cos   ( a1   a 0 ) sin  j 2

 (1   ) a0 ]R which implies n 1

F ( z )  R n 1 [k a n (cos   sin   1)  2 sin   a j   a 0 (cos   sin   1)  2 a 0 ] j 1

for

R 1

and n 1

F ( z )  a 0  R[k a n (cos   sin   1)  2 sin   a j   a 0 (cos   sin   1)  a 0 ] j 1

for

R  1.

Hence , by Lemma 2, it follows that the number of zeros of F(z) and therefore P(z) in

z

R does not exceed c

n 1

1 log log c

R n 1 [k a n (cos  sin   1)  2 sin   a j   a 0 (cos  sin   1)  2 a 0 ] j 1

.

a0 for

R 1

and n 1

1 log log c

a 0  R[k a n (cos  sin   1)  2 sin   a j   a 0 (cos  sin   1)  a 0 ] j 1

a0 for

R  1.

That proves Theorem 3.

REFERENCES [1].

[2].

[3].

[4].

[5].

[6].

L.V.Ahlfors, Complex Analysis, 3rd edition, Mc-Grawhill 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.

K. K. Dewan, Theory of Polynomials and Applications, Deep & Deep Publications Pvt. Ltd. New Delhi, 2007, Chapter 17.

N. K. Govil and Q. I. Rahman, On the Enestrom- Kakeya Theorem, Tohoku Math. J. 20(1968),126136.

M. H. Gulzar, On the Number of Zeros of a Polynomial in a Prescribed Region, Int. Research Journal of Pure Algebra-2(2), February. 2012,35-46.

Q. G. Mohammad, On the Zeros of Polynomials,Amer. Math. Monthly 72(1965), 631-633.

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