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1. 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

2. 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
13

3. 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

4. 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

5. 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

6. 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.
17