G0323846

International Journal of Engineering Inventions e-ISSN: 2278-7461, p-ISSN: 2319-6491 Volume 3, Issue 2 (September 2013) PP: 38-46

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

P( z )   a j z j be a polynomial o f degree n such that j 0

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 j z j be a polynomial o f degree n such that Re( a j )   j , Im(a j )   j and j 0

 n   n1  ......  1   0  0 , Then the number of zeros of P(z) in

z

1 does not exceed 2 n

1

1 log log 2

n    j j 0

.

a0



Theorem C: Let

P( z )   a j z j be a polynomial o f degree n with complex coefficients such that for some j 0

real

,  , arg a j     

 2

, j  0,1,2,......, n

and

a n  a n1  ......  a1  a0 .

Then the number f zeros of P(z) in

z

1 does not exceed 2 n 1

1 log log 2

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

a0 www.ijeijournal.com

.

Page | 38

Zeros of a Polynomial in a given Circle The above results were further generalized by researchers in various ways. M. H. Gulzar[4] proved the following results: 

Theorem D: Let

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

   n   n1  ...... nk 1   nk   nk 1 ......  1   0 , for some real numbers  ,   0,0    1,1  k  n,  n k  0.

If  n  k 1   n  k , then the number f zeros of P(z) in z   ,0    1 , does not exceed n

2    n   n  (  1) n  k    1  n  k   (  0   0 )  2  0  2  j

1 log

1

j 0

log

a0

 and if  n  k   n  k 1 , then

the number f zeros of P(z) in

z   ,0    1, does not exceed n

2    n   n  (1   ) n  k  1    n  k   (  0   0 )  2  0  2  j

1 log

1

j 0

log

a0

 

Theorem E: Let

P( z )   a j z j be a polynomial o f degree n with complex coefficients such that for some j 0

real

,  , arg a j     

 2

, j  0,1,2,......, n

and

  a n  a n1  ......  a nk 1   a nk  a nk 1  ...... a1   a0 , for some   0,   0,1  k  n,  n  k  0,0    1 .

If a n  k 1  a n  k i.e.  1 , then the number of zeros of P(z) in z   ,0    1 , does not exceed 1 log

1

log



M1 , a0

where

M 1  (   a n )(cos   sin   1)  a nk (cos   sin    cos    sin     1)   a 0 (cos   sin   1)  2 a 0  2 sin 

n 1

a

j 1, j  n  k

j

,

and if

a n k  a n k 1 i.e.  1 , then the number of zeros of P(z) in z   ,0    1 , does not exceed

1 log

1



log

M2 , a0

where

M 2  (   a n )(cos   sin   1)  a nk (cos   sin    cos    sin     1)

  a 0 (cos   sin   1)  2 a 0  2 sin 

n 1

a

j 1, j  n  k

www.ijeijournal.com

j

.

Page | 39

Zeros of a Polynomial in a given Circle 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

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

   n   n1  ...... nk 1   nk   nk 1 ......  1   0 , for some real numbers  ,   0,0    1,1  k  n,  n k  0. .

If

 nk 1   nk , then

the number f zeros of P(z) in

R ( R  0, c  1) , does not exceed c

z

n

1 log log c

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

a0 for

R 1

and n

1 log log c

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

a0 for

If

 nk   nk 1 , then

the number f zeros of P(z) in

z

R  1.

R ( R  0, c  1) does not exceed c n

1 log log c

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

a0 for

R 1

and n

1 log log c

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

a0 for

R  1.

1

If the coefficients

,0    1 , Theorem 1 reduces to Theorem D.  a j are real i.e.  j  0, j , then we get the following result from Theorem 1:

Corollary 1: Let

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

Remark 1: Taking R=1and

c



j 0

  a n  a n1  ......a nk 1  a nk  a nk 1 ......  a1  a0 , for some real numbers  ,   0,0    1,1  k  n, a n k  0. .

If

R ( R  0, c  1) does not exceed c    1 a nk   ( a0  a0 )  2 a0

a n  k 1  a n  k , then the number f zeros of P(z) in z 

R 1 log log c

n 1

[2   a n  a n  (  1)a n  k

a0 for www.ijeijournal.com

R 1 Page | 40

Zeros of a Polynomial in a given Circle and

a 0  R[2   a n  a n  (  1)a n  k    1 a n  k   ( a 0  a 0 )  a 0 1 log log c a0 for R  1 .

R If a n  k  a n  k 1 , then the number f zeros of P(z) in z  ( R  0, c  1) does not exceed c n 1 R [2   a n  a n  (1   )a n  k  1   a n  k   ( a 0  a 0 )  2 a 0 1 log log c a0 for R  1 and

a 0  R[2   a n  a n  (1   )a n  k  1   a n  k   ( a 0  a 0 )  a 0 1 log log c a0 for R  1 .

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

Theorem 2: : Let

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

   n   n1  ...... nk 1   nk   nk 1 ......  1   0 , for some real numbers  ,   0,0    1,1  k  n,  n k  0. .

If

 n k 1   n k , then

the number f zeros of P(z) in

z

R ( R  0, c  1) does not exceed c n

1 log log c

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

a0 for

R 1

and n

1 log log c If

 nk

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

a0

for R  1 .

R   n  k 1 , then the number f zeros of P(z) in z  ( R  0, c  1) does not exceed c n

1 log log c

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

a0 for

R 1

and n

1 log log c

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

a0 for

www.ijeijournal.com

R  1.

Page | 41

Zeros of a Polynomial in a given Circle 

Theorem 3: Let

P( z )   a j z j be a polynomial o f degree n with complex coefficients such that for some j 0

real

,  , arg a j     

 2

, j  0,1,2,......, n

and

for some If

  a n  a n1  ......  a nk 1   a nk  a nk 1  ...... a1   a0   0,   0,1  k  n,  nk  0,0    1 .

a n k 1  a n k i.e.  1 , then 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 where

M 3  R n1 [(   a n )(cos   sin   1)  a nk (cos   sin    cos    sin     1)

  a 0 (cos   sin   1)  2 a 0  2 sin 

n 1

a

j 1, j  n  k

j

] , for R  1

and

M 3  a0  R[(   a n )(cos   sin   1)  a nk (cos   sin    cos    sin     1)

  a 0 (cos   sin   1)  2 a 0  2 sin 

If

n 1

a

j 1, j  n  k

j

] , for R  1 .

a n k  a n k 1 i.e.  1 , then the number of zeros of P(z) in z 

R ( R  0, c  1) does not exceed c

M 1 log 4 , log c a0 Where

M 4  R n1 [(   a n )(cos   sin   1)  a nk (cos   sin    cos    sin     1) n 1

a

  a 0 (cos   sin   1)  2 a 0  2 sin 

j 1, j  n  k

j

] for R  1

and

M 4  a0  R[(   a n )(cos   sin   1)  a nk (cos   sin    cos    sin     1)   a 0 (cos   sin   1)  a 0  2 sin  Remark 2: Taking R=1and

c

1



n 1

a

j 1, j  n  k

j

]

for

R  1.

,0    1 , 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

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

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

1 2



2

0

n

R

j 1

aj

log f (Re i d  log f (0)   log www.ijeijournal.com

.

Page | 42

Zeros of a Polynomial in a given Circle Lemma 1 is the famous Jensen’s theorem (see page 208 of [1]).

Lemma 2: If f(z) is analytic and

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

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 j z j be a polynomial o f degree n with complex coefficients such that for some j 0

real

 ,  , arg a j     

 2

,0  j  n, and

a j  a j 1 ,0  j  n, then 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 [3].

3.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  a0 )

 a n z n1  (a n  a n1 ) z n  ......  (a nk 1  a nk ) z nk 1  (a nk  a nk 1 ) z nk  (ank 1  ank 2 ) z nk 1  ......  (a1  a0 ) z  a0  ( n  i n ) z n1  ( n   n1 ) z n  ......  ( nk 1   nk ) z nk 1

 ( n k   n k 1 ) z n k  ( n k 1   n k 2 ) z n k 1  ......  (1   0 ) z n

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

If

 nk 1   nk , then F ( z)  ( n  i n ) z n1  z n  (    n   n1 ) z n  ......  ( nk 1   nk ) z nk 1  ( n  k   n  k 1 ) z n  k  (  1) n  k z n  k  ( n  k 1   n  k  2 ) z n  k 1  ...... n

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

For

z  R , we have by using the hypothesis

F ( z )   n R n1   n R n1   R n     n   n1 R n  ......   n k 1   n k R n k 1 +

 nk   nk 1 R nk    1  nk R nk

  nk 1   nk 2 R nk 1  ......   1   0 R  (1   )  0 R n

  0   0   (  j   j 1 ) R j j 0

n

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

for

R 1

and www.ijeijournal.com

Page | 43

Zeros of a Polynomial in a given Circle

n

 a 0  R[2    n   n  (  1) n  k    1  n  k   (  0   0 )   0   0  2  j j 1

for R  1 .

R Hence by Lemma 2, the number of zeros of F(z) in z  ( R  0, c  1) , does not exceed c n

1 log log c

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

a0 for

R 1

and n

1 log log c

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

a0 for

If

R  1.

 nk   nk 1 , then F ( z)  ( n  i n ) z n1  z n  (    n   n1 ) z n  ......  ( nk 1   nk ) z nk 1  ( n  k   n  k 1 ) z n  k  (1   ) n  k z n  k 1  ( n  k 1   n  k  2 ) z n  k 1  ...... n

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

For

z  R , we have by using the hypothesis

F ( z )   n R n1   n R n1   R n     n   n1 R n  ......   n k 1   n k R nk 1 +  nk

  nk 1 R nk  1    nk R nk 

  nk 1   nk 2 R nk 1  ......   1   0 R  (1   )  0 R n

  0   0   (  j   j 1 ) R j j 0

n

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

for

R 1

and n

 a 0  R[2    n   n  (1   ) n  k  1    n  k   (  0   0 )   0   0  2  j j 1

for

R  1.

Hence by Lemma 2, the number of zeros of F(z) in z  R ( R  0, c  1) does not exceed c n

1 log log c

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

a0 for

R 1

and n

1 log log c

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

a0

for

R  1.

That proves Theorem 1 completely. www.ijeijournal.com

Page | 44

Zeros of a Polynomial in a given Circle Proof of Theorem 4: Consider the polynomial

F ( z)  (1  z) P( z)  (1  z )(a n z n  a n1 z n1  ......  a1 z  a0 )

 a n z n1  (a n  a n1 ) z n  ......  (a nk 1  a nk ) z nk 1  (a nk  a nk 1 ) z nk  (ank 1  ank 2 ) z nk 1  ......  (a1  a0 ) z  a0 If

ank 1  ank , i.e.   1 , then F ( z)  an z n1  z n  (   an  an1 ) z n  ......  (ank 1  ank ) z nk 1

 (ank  ank 1 ) z nk  (  1)ank z nk  (ank 1  ank 2 ) z nk 1  ......  (a1  a0 ) z  (  1)a0 z  a0 so that for

z  R , we have by using Lemma 3,

F ( z )  a n R n1  R n    a n  a n1 R n  ......  a nk 1  a nk R nk 1  a nk  a nk 1 R nk    1 a nk R nk  a nk 1  a nk 2 R nk 1  ......  a1  a0 R  (1   ) a0 R  a0  R n1 [ a n    (   a n  a n1 ) cos   (   a n  a n1 ) sin   ......  ( a n  k 1  a n  k ) cos   ( a n  k 1  a n  k ) sin   (  1) a n  k  ( a n  k  a n  k 1 ) cos   ( a n  k  a n  k 1 ) sin   ( a n  k 1  a n  k  2 ) cos   ( a n  k 1  a n  k  2 ) sin   ......  ( a1   a 0 ) cos   ( a1   a 0 ) sin   (1   ) a 0  a 0 ]

 R n1 [(   a n )(cos   sin   1)  a nk (cos   sin    cos    sin     1)   a 0 (cos   sin   1)  2 a 0  2 sin  for

n 1

a

j 1, j  n  k

j

]

R 1

and

 a0  R[(   a n )(cos   sin   1)  a nk (cos   sin    cos    sin     1)   a 0 (cos   sin   1)  a 0  2 sin 

for

n 1

a

j 1, j  n  k

j

]

R 1

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

z

R ( R  0, c  1) does not exceed c

M 1 log 3 , log c a0 where

M 3  R n1 [(   a n )(cos   sin   1)  a nk (cos   sin    cos    sin     1)

  a 0 (cos   sin   1)  2 a 0  2 sin 

n 1

a

j 1, j  n  k

j

] , for R  1

and

M 3  a0  R[(   a n )(cos   sin   1)  a nk (cos   sin    cos    sin     1)

  a 0 (cos   sin   1)  2 a 0  2 sin  If

n 1

a

j 1, j  n  k

j

] , for R  1 .

ank  ank 1 , i.e.   1 , then www.ijeijournal.com

Page | 45

Zeros of a Polynomial in a given Circle

F ( z)  an z n1  z n  (   an  an1 ) z n  ......  (ank 1  ank ) z nk 1  (ank  ank 1 ) z nk  (1   )ank z nk 1  (ank 1  ank 2 ) z nk 1  ......

 (a1  a0 ) z  a0 so that for

z  R , we have by hypothesis and Lemma 3,

F ( z )  a n R n1  R n    a n  a n1 R n  ......  a nk 1  a nk R nk 1  a nk  a nk 1 R nk  1   a nk R nk 1  a nk 1  a nk 2 R nk 1  ......  a1  a0 R  (1   ) a0 R  a0 R

n 1

[ a n    (   a n  a n1 ) cos   (   a n  a n1 ) sin   ......

 ( a n  k 1   a n  k ) cos   ( a n  k 1   a n  k ) sin   1   a n  k  ( a n  k  a n  k 1 ) cos   ( a n  k  a n  k 1 ) sin   ( a n  k 1  a n  k  2 ) cos   ( a n  k 1  a n  k  2 ) sin   ......  ( a1   a 0 ) cos   ( a1   a 0 ) sin   (1   ) a 0  a 0 ]

 R n1 [(   a n )(cos   sin   1)  a nk (cos   sin    cos    sin   1   )   a 0 (cos   sin   1)  2 a 0  2 sin  for

n 1

a

j 1, j  n  k

j

]

R 1

and

 a0  R[(   a n )(cos   sin   1)  a nk (cos   sin    cos    sin   1   )   a 0 (cos   sin   1)  a 0  2 sin 

n 1

a

j 1, j  n  k

j

]

for R  1 .

R Hence the number of zeros of P(z) in z  ( R  0, c  1) does not exceed c M4 1 log log c a0

,

Where

M 4  R n 1 [(   a n )(cos   sin   1)  a n k (cos   sin    cos    sin     1)

  a 0 (cos   sin   1)  2 a 0  2 sin 

n 1

a

j 1, j  n  k

j

] for R  1

and

M 4  a0  R[(   a n )(cos   sin   1)  a nk (cos   sin    cos    sin     1)   a 0 (cos   sin   1)  a 0  2 sin 

n 1

a

j 1, j  n  k

j

]

for

R  1.

That completes the proof of Theorem 3.

References [1] [2] [3] [4] [5]

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.

N. K. Govil and Q. I. Rahman, On the Enestrom- Kakeya Theorem, Tohoku Math. J. 20(1968),126-136.

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.

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

www.ijeijournal.com

Page | 46