In this paper we locate the regions which contain all or some of the zeros of a polynomial when the coefficients of the polynomial are restricted to certain conditions. Mathematics Subject Classification: 30C10, 30C15.

1. ISSN (e): 2250 – 3005 || Volume, 07 || Issue, 02|| February – 2017 ||
International Journal of Computational Engineering Research (IJCER)
www.ijceronline.com Open Access Journal Page 18
Location of Regions Containing all or Some Zeros of a
Polynomial
M.H. Gulzar
Department of Mathematics, University of Kashmir, Srinagar
I. INTRODUCTION
The following result known as the Enestrom-Kakeya Theorem [3] is of fundamental importance on locating a
region containing all the zeros of a polynomial with monotonically increasing positive coefficients:
Theorem A: Let 


n
j
j
j
zazP
0
)( be a polynomial of degree n such that
0...... 011
 
aaaa nn
.
Then all the zeros of P(z) lie in 1z .
In the literature several generalizations and extensions of this result are available [].
Recently Gulzar et al [1,2] proved the following results:
Theorem B: Let 


n
j
j
j
zazP
0
)( be a polynomial of degree n with
njaa jjjj
,......,2,1,0,)Im(,)Re(   such that for some 10,  n and for some
1,1  ok ,

  
......1nn
k
and
001211
......  
 
L .
Then all the zeros of P(z) lie in
n
n
j
jn
n
n
a
Lk
a
k
z






0
2)1(
)1(



.
Theorem C: Let 


n
j
j
j
zazP
0
)( be a polynomial of degree n with
njaa jjjj
,......,2,1,0,)Im(,)Re(   such that for some 10,  n and for some
1,1  ok ,

   11
......nn
k
and
001211
......  
 
L .
Then the number f zeros of P(z) in 10,  z does not exceed
ABSTRACT
In this paper we locate the regions which contain all or some of the zeros of a polynomial when the
coefficients of the polynomial are restricted to certain conditions.
Mathematics Subject Classification: 30C10, 30C15.
Keywords and Phrases: Coefficients, Polynomial, Regions, Zeros.

2. Location of Regions Containing all or Some Zeros of a Polynomial
www.ijceronline.com Open Access Journal Page 19
0
0
2)1()1(
log
1
log
1
a
Lkka
n
j
jnnn 

 


.
II. MAIN RESULTS
In this paper we prove the following generalizations of Theorems B and C:
Theorem 1: Let 


n
j
j
j
zazP
0
)( be a polynomial of degree n with
njaa jjjj
,......,2,1,0,)Im(,)Re(   such that for some 10,  n and for some
1,1, 21
 okk ,

  
......121 nn
kk
and
001211
......  
 
L .
Then all the zeros of P(z) lie in
n
n
j
jnn
nn
n
a
Lkk
a
k
a
k
z










0
121
121
2)()1(
)1()1(



.
Theorem 2: Let 


n
j
j
j
zazP
0
)( be a polynomial of degree n with
njaa jjjj
,......,2,1,0,)Im(,)Re(   such that for some 10,  n and for some
1,1, 21
 okk ,

   1121
......nn
kk
and
001211
......  
 
L .
Then the number of zeros of P(z) in 10,  z does not exceed
.2)()1(2)1([
1
log
1
log
1
0
1211
0




n
j
jnnnn
Lkkka
a



Remark 1: Taking 12
k , Theorem 1 reduces to Theorem B and Theorem 2 reduces to Theorem C.
For different values of the parameters in Theorems 1 and 2 , we get many interesting results. For example taking
1 , Theorem 1 gives the following result:
Corollary 1: Let 


n
j
j
j
zazP
0
)( be a polynomial of degree n with
njaa jjjj
,......,2,1,0,)Im(,)Re(   such that for some 10,  n and for some
,1, 21
kk ,

  
......121 nn
kk
and
001211
......  
 
L .
Then all the zeros of P(z) lie in

3. Location of Regions Containing all or Some Zeros of a Polynomial
www.ijceronline.com Open Access Journal Page 20
n
n
j
jnnn
n
n
a
Lkk
a
k
z







0
1121
1
2))(1(
)1(



.
Taking 1 , Theorem 2 gives the following result:
Corollary 2: Let 


n
j
j
j
zazP
0
)( be a polynomial of degree n with
njaa jjjj
,......,2,1,0,)Im(,)Re(   such that for some 10,  n and for some
,1, 21
kk ,

   1121
......nn
kk
and
001211
......  
 
L .
Then the number f zeros of P(z) in 10,  z does not exceed
].2)1(2)1([
1
log
1
log
1
0
1211
0




n
j
jnnnn
Lkkka
a



III. LEMMA
For the proof of Theorem 2, we need the following result:
Lemma: Let f (z) be analytic for 0)0(,1  fz and Mzf )( for 1z . Then the number of zeros of
f(z) in 10,  z does not exceed
)0(
log
1
log
1
f
M

.
(for reference see [4] ).
3. Proofs of Theorems
Proof of Theorem 1: Consider the polynomial
)()1()( zPzzF 




zaazaazaaza
azazazaz
n
nn
n
n
n
n
n
n
)()(......)(
)......)(1(
1
1
11
1
01
1
1








001
)(...... azaa 
1
212121211
1
)()1()()1(




n
nn
n
n
n
nn
n
n
n
n
zkzkzkkzkza 
})(
......){()(......)(
)1()(......)()1(
001
10011
11
1
2
32
1
12



















z
zizz
zzzzk
n
nn
n
nn
n
n
For 1z so that nj
z
j
,......,2,1,1
1
 , we have, by using the hypothesis
1
212
1
12121121
.1[)1()1()(





n
nn
n
n
n
nn
n
nnn
zkzkzkkzkkzazF 
]......{......
1......)(
0011001
1
11
1
2
32
















zzz
zzzz
n
nn
n
nn

4. Location of Regions Containing all or Some Zeros of a Polynomial
www.ijceronline.com Open Access Journal Page 21
z
k
z
k
kkkkzaz
nnn
nnnnn
n 21212
121121
)1(
{)1()1([


























 nnn
nn
zzzz
1
11
1
2
32
)1(
......
............
21
1
0
1
01







zzz
nn
nnnn



}]
0
1
01
nn
zz



 
12121121
)1({)1()1([ 
 nnnnnn
n
kkkkkzaz 

 )1(..... 132212
  nnnn
k
}]
............
001
2110011

 

  nnnn
12121121
)1({)1()1([ 
 nnnnnn
n
kkkkkzaz 

 )1(..... 132212
  nnnn
k
}]
............
001
2110011

 

  nnnn
Lkkkkzaz nnnnn
n
  
 121121
)1({)1()1([
}]2)(
0



n
j
j
 
0
if




n
j
jnnnnn
Lkkkkza
0
121121
2)()1()1()1(  
i.e. if
n
n
j
jnn
nn
n
a
Lkk
a
k
a
k
z










0
121
121
2)()1(
)1()1(



.
This shows that those zeros of F(z) whose modulus is greater than 1 lie in
n
n
j
jnn
nn
n
a
Lkk
a
k
a
k
z










0
121
121
2)()1(
)1()1(



.
Since the zeros of F(z) whose modulus is less than or equal to 1 already satisfy the above inequality and since
the zeros of P(z) are also the zeros of F(z) , it follows that all the zeros of P(z) lie in
n
n
j
jnn
nn
n
a
Lkk
a
k
a
k
z










0
121
121
2)()1(
)1()1(



.

5. Location of Regions Containing all or Some Zeros of a Polynomial
www.ijceronline.com Open Access Journal Page 22
That proves Theorem 1.
Proof of Theorem 2: Consider the polynomial
)()1()( zPzzF 




zaazaazaaza
azazazaz
n
nn
n
n
n
n
n
n
)()(......)(
)......)(1(
1
1
11
1
01
1
1








001
)(...... azaa 
1
212121211
1
)()1()()1(




n
nn
n
n
n
nn
n
n
n
n
zkzkzkkzkza 
})(
......){()(......)(
)1()(......)()1(
001
10011
11
1
2
32
1
12



















z
zizz
zzzzk
n
nn
n
nn
n
n
For 1z , we have, by using the hypothesis
12212121211
)1()1()1()( 
 nnnnnnnn
kkkkkkazF 
001211
0011132
.......
......)1(......

 




nnnn
nn
12212121211
)1()1()1( 
 nnnnnnnn
kkkkkka 
001211
0011132
.......
......)1(......

 




nnn
nnn
)())1(2)1( 1211 
  
Lkkka nnnn



n
j
j
0
2 
Since F(z) is analytic for 1z , 0)0( 0
 aF , it follows by the Lemma that the number of zeros
of F(z) in 10,  z des not exceed
.2)()1(2)1([
1
log
1
log
1
0
1211
0




n
j
jnnnn
Lkkka
a



Since the zeros of P(z) are also the zeros of F(z) , it follows that the number of zeros
of P(z) in 10,  z des not exceed
.2)())1(2)1([
1
log
1
log
1
0
1211
0




n
j
jnnnn
Lkkka
a



That completes the proof of Theorem 2.
REFERENCES
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