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ISSN (e): 2250 – 3005 || Volume, 07 || Issue, 03|| March – 2017 ||
International Journal of Computational Engineering Research (IJCER)
www.ijceronline.com Open Access Journal Page 9
Location of Zeros of Polynomials
M.H. Gulzar1
, B.A.Zargar2
, A.W.Manzoor3
Department of Mathematics, University of Kashmir, Srinagar
I. INTRODUCTION
On the location of zeros of a polynomial with real coefficients Enestrom and Kakeya proved the following
theorem named after them as the Enestrom - Kakeya Theorem [9,10] :
Theorem A: all the zeros of a polynomial 


n
j
j
j
zazP
0
)( satisfying 0....... 011
 
aaaa nn
lie
in 1z .
The above theorem has been generalized in different ways by various authors [2-11].
Recently Gulzar et al [7] proved the following result
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 10,10,,  nn  ,
,......
,......
1
1








nn
nn
and
001211
......  
 
L ,
001211
......  
 
M .
Then all the zeros of P(z) lie in
][
1

  ML
a
z nn
n
.
II. MAIN RESULTS
In this paper we prove the following generalization of Theorem B:
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 10,10,,  nn 
and for some 1, 21
kk ,
,......
,......
2
2
212
1
2
11
2
111
1
1






kkkk
kkkk
n
n
n
n
n
n
n
n






and
001211
......  
 
L ,
ABSTRACT
In this paper, restricting the coefficients of a polynomial to certain conditions, we locate a region
containing all of its zeros. Our results generalize many known results in addition to some interesting
results which can be obtained by choosing certain values of the parameters.
Mathematics Subject Classification: 30C10, 30C15.
Key Words and Phrases: Coefficients, Polynomial, Zeros.
Location of Zeros of Polynomials
www.ijceronline.com Open Access Journal Page 10
001211
......  
 
M .
Then all the zeros of P(z) lie in
 



)()1()()1([
1)1()1(
2
1
1
21
jj
n
j
jjnn
nn
nn
kk
aa
kik
z 


].)1()1( 21 
  MLkk nn
Further the number of zeros of P(z) in 1,1,
0
 Rc
c
R
z
X
a
does not exceed
)1log(
log
1
log
log
1
00
0
a
X
ca
aX
c


and the number of zeros of P(z) in 1,1,
0
 Rc
c
R
z
Y
a
does not exceed )1log(
log
1
log
log
1
00
0
a
Y
ca
aY
c


, where R is any positive number and




n
j
jjnn
nn
n
kMLRRaX
1
100
1
)()1()()[(



])()1(
1
2 


n
j
jj
k

 ,




n
j
jjnn
n
n
kMLRRaY
1
100
1
)()1()()[(



])()1(
1
2 


n
j
jj
k

 .
Remark 1: For
121
 kk
, Theorem 1 reduces to Theorem B.
Taking 12
k in Theorem 1, we get 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 10,10,,  nn 
and for some 11
k ,
,......
,......
1
11
2
111
1
1










nn
n
n
n
n
kkkk
001211
......  
 
L ,
001211
......  
 
M .
Then all the zeros of P(z) lie in
n
n
j
jjnn
n
n
kk
a
kz 

)1()()1([
1
)1( 1
1
11
 

].
  ML
Location of Zeros of Polynomials
www.ijceronline.com Open Access Journal Page 11
Also the number of zeros of P(z) in 1,1,
0
 Rc
c
R
z
X
a
does not exceed
)1log(
log
1
log
log
1
00
0
a
X
ca
aX
c


and the number of zeros of P(z) in 1,1,
0
 Rc
c
R
z
Y
a
does not exceed )1log(
log
1
log
log
1
00
0
a
Y
ca
aY
c


, where R is any positive number and
])()1()()[(
1
100
1




n
j
jjnn
nn
n
kMLRRaX


 ,
])()1()()[(
1
100
1




n
j
jjnn
n
n
kMLRRaY



Taking 0  in Theorem 1, we get 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 10,10,,  nn 
and for some 1, 21
kk ,
,......
,......
0121
1
22
0111
1
11








kkk
kkk
n
n
n
n
n
n
n
n
Then all the zeros of P(z) lie in





n
j
jj
n
j
jjnn
nn
nn
kk
aa
kik
z
1
2
1
1
21
)()1()()1([
1)1()1(


].)1()1( 000021
  nn
kk
Also the number of zeros of P(z) in 1,1,
0
 Rc
c
R
z
X
a
does not exceed
)1log(
log
1
log
log
1
00
0
a
X
ca
aX
c


and the number of zeros of P(z) in 1,1,
0
 Rc
c
R
z
Y
a
does not exceed )1log(
log
1
log
log
1
00
0
a
Y
ca
aY
c


, where R is any positive number and
])()1()()[(
1
10000
1




n
j
jjnn
nn
n
kMLRRaX  ,
])()1(
1
2 


n
j
jj
k  , 1R
])()1()()[(
1
10000
1




n
j
jjnn
n
n
kMLRRaY 
])()1(
1
2 


n
j
jj
k  , 1R
Similarly for other values of the parameters in Theorem 1, we get many other interesting results.
Location of Zeros of Polynomials
www.ijceronline.com Open Access Journal Page 12
III. LEMMAS
For the proof of Theorem 2, we make use of the following lemmas:
Lemma 1: Let f(z) (not identically zero) be analytic for 0)0(,  fRz and ,0)( k
af
nk ,......,2,1 . Then



n
j j
i
a
R
fdf
1
2
0
log)0(log(Relog
2
1 



.
Lemma 1 is the famous Jensen’s Theorem (see page 208 of [1]).
Lemma 2: Let f(z) be analytic , 0)0( f and Mzf )( for Rz  . Then the number of zeros of f(z) in
1,  c
c
R
z is less than or equal to
)0(
log
log
1
f
M
c
.
Lemma 2 is a simple deduction from Lemma 1.
IV. 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(
1
211111
1



 n
nn
n
nn
n
n
n
n
zkzkzkza 
0011
1
11
)(......)()( 




 


zzzk
).......)(1(
1
1
2
2
1
11









 zzzk
n
n
n
n
})(......
)()......)(1()(
......)()1(){(
001
1
1
1
1
12
1
12
1
212212





















z
zzzkzk
zkzkzki
n
n
n
nn
n
n
n
nn
For 1z so that nj
z
j
,......,2,1,1
1
 , we have, by using the hypothesis
......[)1()1()(
1
2111121



n
nn
n
nn
n
nnn
zkzkzkikzazF 
)]......)(1(
............
)......)(1(
......
1
1
1
12
0011
1
12
1
21212
1
1
1
11
0011
1
11




































zzk
zzzk
zkzkzzk
zzzk
n
n
n
nn
n
nn
n
n
2
321211
1121
[)1()1([
z
k
z
k
kkikzaz
nnnn
nnnnn
n 







Location of Zeros of Polynomials
www.ijceronline.com Open Access Journal Page 13
)]......)(1(......
......)
......)(1(............
1
11
2
0
1
01
1
1
12212
121
1
1
1
0
1
011
1
11















































n
n
nn
nn
nn
nnn
n
nnnn
zz
k
zz
zz
k
z
k
k
z
z
k
zzzz
k
3212111121
[)1()1([ 
 nnnnnnnnn
n
kkkkikzaz 
)]......)(1(......
......)
......)(1(............
112001
112212121
11001111












n
nnnn
n
k
kkk
kk
3212111121
{)1()1([ 
 nnnnnnnnn
n
kkkkikzaz 
.....)(1(............ 11001111
  n
kk  
)}]......)(1(......
......)
112001
112212121








n
nnnn
k
kkk
3212111121
{)1()1([ 
 nnnnnnnnn
n
kkkkikzaz 
)}]......)(1(......
)......)(1(......
11212212
1211111








nnn
nnn
kMkk
kkLk



n
j
jjnnnnn
n
kkikzaz
1
121
)()1({)1()1([


}])1()1()()1( 21
1
2 

  

MLkkk nn
n
j
jj
0
if



n
j
jj
n
j
jjnnnnn
kkkikza
1
2
1
121
)()1()()1({)1()1(


}])1()1( 21 
  MLkk nn
i.e.if





n
j
jj
n
j
jjnn
nn
nn
kk
aa
kik
z
1
2
1
1
21
)()1()()1([
1)1()1(



].)1()1( 21 
  MLkk nn
This shows that those zeros of F(z) whose modulus is greater than 1 lie in





n
j
jj
n
j
jjnn
nn
nn
kk
aa
kik
z
1
2
1
1
21
)()1()()1([
1)1()1(



].)1()1( 21 
  MLkk nn
Location of Zeros of Polynomials
www.ijceronline.com Open Access Journal Page 14
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
j
jj
n
j
jjnn
nn
nn
kk
aa
kik
z
1
2
1
1
21
)()1()()1([
1)1()1(



].)1()1( 21 
  MLkk nn
Again
)()( 0
zGazF  ,
where
......)()()1()(
1
211111
1



 n
nn
n
nn
n
n
n
n
zkzkzkzazG 
zzzk )(......)()( 011
1
11





 


).......)(1(
1
1
2
2
1
11









 zzzk
n
n
n
n
}.)(......
)()......)(1()(
......)()1(){(
01
1
1
1
1
12
1
12
1
212212
z
zzzkzk
zkzkzki
n
n
n
nn
n
n
n
nn





















For 0,  RRz , we have
......)()()1()(
1
211111
1



 n
nn
n
nn
n
n
n
n
RkRkRkRazG 
RRRk 011
1
11
......)( 




 


).......)(1(
1
1
2
2
1
11









 RRRk
n
n
n
n
}......
)......)(1()(
......)()1()(
01
1
1
1
1
12
1
12
1
212212
R
RRRkRk
RkRkRk
n
n
n
nn
n
n
n
nn





















LkkkkRRa nnnnn
nn
n
 


 11211111
1
......)1[(
)]......)(1(......
)......)(1(
11212212
12111








nnn
nnn
kMkk
kk




n
j
jjnn
nn
n
kMLRRa
1
100
1
)()1()()[(



])()1(
1
2 


n
j
jj
k


X
for 1R and
for 1R




n
j
jjnn
n
n
kMLRRazG
1
100
1
)()1()()[()(



])()1(
1
2 


n
j
jj
k


Y
Since G(0)=0 and G(z) is analytic for Rz  , it follows, by Schwarz Lemma , that for Rz  ,
Location of Zeros of Polynomials
www.ijceronline.com Open Access Journal Page 15
zXzG )( for 1R and zYzG )( for 1R .
Hence, for Rz  , 1R
)()( 0
zGazF 
0
)(
0
0



zXa
zGa
if
X
a
z
0
 .
Similarly, for Rz  , 1R , 0)( zF if
Y
a
z
0
 .
In other words, F(z) does not vanish in
X
a
z
0
 for 1R and F(z) does not vanish in
Y
a
z
0
 for 1R
in Rz  . That means all the zeros of F(z) and hence all the zeros of P(z) lie in
X
a
z
0
 for 1R and in
Y
a
z
0
 for 1R in Rz  .
Since for Rz  , 0
)( aXzF  for 1R and 0
)( aYzF  for 1R and since
0)0( 0
 aF , it follows by Lemma 2 that the number of zeros of P(z) in 1,1,
0
 Rc
c
R
z
X
a
does
not exceed )1log(
log
1
log
log
1
00
0
a
X
ca
aX
c


and the number of zeros of P(z) in
1,1,
0
 Rc
c
R
z
Y
a
does not exceed )1log(
log
1
log
log
1
00
0
a
Y
ca
aY
c


.
That proves Theorem 1 completely.
REFERENCES
[1]. L. V. Ahlfors, Complex Analysis, 3rd
edition, Mc-Grawhill.
[2]. A. Aziz and Q. G. Mohammad, Zero-free regions for polynomials and some generalizations of Enestrom-Kakeya Theorem, Canad.
Math. Bull.,27(1984),265- 272.
[3]. A. Aziz and B. A. Zargar, Some Extensions of Enestrom-Kakeya Theorem, Glasnik Mathematicki, 51 (1996), 239-244.
[4]. Y.Choo, On the zeros of a family of self-reciprocal polynomials, Int. J. Math. Analysis , 5 (2011),1761-1766.
[5]. N. K. Govil and Q. I. Rahman, On Enestrom-Kakeya Theorem, Tohoku J. Math. 20 (1968), 126-136.
[6]. M. H. Gulzar, Some Refinements of Enestrom-Kakeya Theorem, International Journal of Mathematical Archive , Vol 2(9), 2011,
1512-1519..
[7]. M. H. Gulzar,B.A. Zargar and A. W. Manzoor, Zeros of a Certain Class of Polynomials, International Journal of Advanced
Research in Engineering and Management, Vol.3 No. 2(2017).
[8]. A. Joyal, G. Labelle and Q. I. Rahman, On the location of the zeros of a polynomial, Canad. Math. Bull. ,10 (1967), 53-63.
[9]. M. Marden, Geometry of Polynomials, Math. Surveys No. 3, Amer. Math.Soc.(1966).
[10]. Q. I. Rahman and G. Schmeisser, Analytic Theory of Polynomials, Oxford University Press, New York (2002).
[11]. B. A. Zargar, On the zeros of a family of polynomials, Int. J. of Math. Sci. &Engg. Appls. , Vol. 8 No. 1(January 2014), 233-237.

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Location of Zeros of Polynomials

  • 1. ISSN (e): 2250 – 3005 || Volume, 07 || Issue, 03|| March – 2017 || International Journal of Computational Engineering Research (IJCER) www.ijceronline.com Open Access Journal Page 9 Location of Zeros of Polynomials M.H. Gulzar1 , B.A.Zargar2 , A.W.Manzoor3 Department of Mathematics, University of Kashmir, Srinagar I. INTRODUCTION On the location of zeros of a polynomial with real coefficients Enestrom and Kakeya proved the following theorem named after them as the Enestrom - Kakeya Theorem [9,10] : Theorem A: all the zeros of a polynomial    n j j j zazP 0 )( satisfying 0....... 011   aaaa nn lie in 1z . The above theorem has been generalized in different ways by various authors [2-11]. Recently Gulzar et al [7] proved the following result 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 10,10,,  nn  , ,...... ,...... 1 1         nn nn and 001211 ......     L , 001211 ......     M . Then all the zeros of P(z) lie in ][ 1    ML a z nn n . II. MAIN RESULTS In this paper we prove the following generalization of Theorem B: 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 10,10,,  nn  and for some 1, 21 kk , ,...... ,...... 2 2 212 1 2 11 2 111 1 1       kkkk kkkk n n n n n n n n       and 001211 ......     L , ABSTRACT In this paper, restricting the coefficients of a polynomial to certain conditions, we locate a region containing all of its zeros. Our results generalize many known results in addition to some interesting results which can be obtained by choosing certain values of the parameters. Mathematics Subject Classification: 30C10, 30C15. Key Words and Phrases: Coefficients, Polynomial, Zeros.
  • 2. Location of Zeros of Polynomials www.ijceronline.com Open Access Journal Page 10 001211 ......     M . Then all the zeros of P(z) lie in      )()1()()1([ 1)1()1( 2 1 1 21 jj n j jjnn nn nn kk aa kik z    ].)1()1( 21    MLkk nn Further the number of zeros of P(z) in 1,1, 0  Rc c R z X a does not exceed )1log( log 1 log log 1 00 0 a X ca aX c   and the number of zeros of P(z) in 1,1, 0  Rc c R z Y a does not exceed )1log( log 1 log log 1 00 0 a Y ca aY c   , where R is any positive number and     n j jjnn nn n kMLRRaX 1 100 1 )()1()()[(    ])()1( 1 2    n j jj k   ,     n j jjnn n n kMLRRaY 1 100 1 )()1()()[(    ])()1( 1 2    n j jj k   . Remark 1: For 121  kk , Theorem 1 reduces to Theorem B. Taking 12 k in Theorem 1, we get 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 10,10,,  nn  and for some 11 k , ,...... ,...... 1 11 2 111 1 1           nn n n n n kkkk 001211 ......     L , 001211 ......     M . Then all the zeros of P(z) lie in n n j jjnn n n kk a kz   )1()()1([ 1 )1( 1 1 11    ].   ML
  • 3. Location of Zeros of Polynomials www.ijceronline.com Open Access Journal Page 11 Also the number of zeros of P(z) in 1,1, 0  Rc c R z X a does not exceed )1log( log 1 log log 1 00 0 a X ca aX c   and the number of zeros of P(z) in 1,1, 0  Rc c R z Y a does not exceed )1log( log 1 log log 1 00 0 a Y ca aY c   , where R is any positive number and ])()1()()[( 1 100 1     n j jjnn nn n kMLRRaX    , ])()1()()[( 1 100 1     n j jjnn n n kMLRRaY    Taking 0  in Theorem 1, we get 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 10,10,,  nn  and for some 1, 21 kk , ,...... ,...... 0121 1 22 0111 1 11         kkk kkk n n n n n n n n Then all the zeros of P(z) lie in      n j jj n j jjnn nn nn kk aa kik z 1 2 1 1 21 )()1()()1([ 1)1()1(   ].)1()1( 000021   nn kk Also the number of zeros of P(z) in 1,1, 0  Rc c R z X a does not exceed )1log( log 1 log log 1 00 0 a X ca aX c   and the number of zeros of P(z) in 1,1, 0  Rc c R z Y a does not exceed )1log( log 1 log log 1 00 0 a Y ca aY c   , where R is any positive number and ])()1()()[( 1 10000 1     n j jjnn nn n kMLRRaX  , ])()1( 1 2    n j jj k  , 1R ])()1()()[( 1 10000 1     n j jjnn n n kMLRRaY  ])()1( 1 2    n j jj k  , 1R Similarly for other values of the parameters in Theorem 1, we get many other interesting results.
  • 4. Location of Zeros of Polynomials www.ijceronline.com Open Access Journal Page 12 III. LEMMAS For the proof of Theorem 2, we make use of the following lemmas: Lemma 1: Let f(z) (not identically zero) be analytic for 0)0(,  fRz and ,0)( k af nk ,......,2,1 . Then    n j j i a R fdf 1 2 0 log)0(log(Relog 2 1     . Lemma 1 is the famous Jensen’s Theorem (see page 208 of [1]). Lemma 2: Let f(z) be analytic , 0)0( f and Mzf )( for Rz  . Then the number of zeros of f(z) in 1,  c c R z is less than or equal to )0( log log 1 f M c . Lemma 2 is a simple deduction from Lemma 1. IV. 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( 1 211111 1     n nn n nn n n n n zkzkzkza  0011 1 11 )(......)()(          zzzk ).......)(1( 1 1 2 2 1 11           zzzk n n n n })(...... )()......)(1()( ......)()1(){( 001 1 1 1 1 12 1 12 1 212212                      z zzzkzk zkzkzki n n n nn n n n nn For 1z so that nj z j ,......,2,1,1 1  , we have, by using the hypothesis ......[)1()1()( 1 2111121    n nn n nn n nnn zkzkzkikzazF  )]......)(1( ............ )......)(1( ...... 1 1 1 12 0011 1 12 1 21212 1 1 1 11 0011 1 11                                     zzk zzzk zkzkzzk zzzk n n n nn n nn n n 2 321211 1121 [)1()1([ z k z k kkikzaz nnnn nnnnn n        
  • 5. Location of Zeros of Polynomials www.ijceronline.com Open Access Journal Page 13 )]......)(1(...... ......) ......)(1(............ 1 11 2 0 1 01 1 1 12212 121 1 1 1 0 1 011 1 11                                                n n nn nn nn nnn n nnnn zz k zz zz k z k k z z k zzzz k 3212111121 [)1()1([   nnnnnnnnn n kkkkikzaz  )]......)(1(...... ......) ......)(1(............ 112001 112212121 11001111             n nnnn n k kkk kk 3212111121 {)1()1([   nnnnnnnnn n kkkkikzaz  .....)(1(............ 11001111   n kk   )}]......)(1(...... ......) 112001 112212121         n nnnn k kkk 3212111121 {)1()1([   nnnnnnnnn n kkkkikzaz  )}]......)(1(...... )......)(1(...... 11212212 1211111         nnn nnn kMkk kkLk    n j jjnnnnn n kkikzaz 1 121 )()1({)1()1([   }])1()1()()1( 21 1 2       MLkkk nn n j jj 0 if    n j jj n j jjnnnnn kkkikza 1 2 1 121 )()1()()1({)1()1(   }])1()1( 21    MLkk nn i.e.if      n j jj n j jjnn nn nn kk aa kik z 1 2 1 1 21 )()1()()1([ 1)1()1(    ].)1()1( 21    MLkk nn This shows that those zeros of F(z) whose modulus is greater than 1 lie in      n j jj n j jjnn nn nn kk aa kik z 1 2 1 1 21 )()1()()1([ 1)1()1(    ].)1()1( 21    MLkk nn
  • 6. Location of Zeros of Polynomials www.ijceronline.com Open Access Journal Page 14 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 j jj n j jjnn nn nn kk aa kik z 1 2 1 1 21 )()1()()1([ 1)1()1(    ].)1()1( 21    MLkk nn Again )()( 0 zGazF  , where ......)()()1()( 1 211111 1     n nn n nn n n n n zkzkzkzazG  zzzk )(......)()( 011 1 11          ).......)(1( 1 1 2 2 1 11           zzzk n n n n }.)(...... )()......)(1()( ......)()1(){( 01 1 1 1 1 12 1 12 1 212212 z zzzkzk zkzkzki n n n nn n n n nn                      For 0,  RRz , we have ......)()()1()( 1 211111 1     n nn n nn n n n n RkRkRkRazG  RRRk 011 1 11 ......)(          ).......)(1( 1 1 2 2 1 11           RRRk n n n n }...... )......)(1()( ......)()1()( 01 1 1 1 1 12 1 12 1 212212 R RRRkRk RkRkRk n n n nn n n n nn                      LkkkkRRa nnnnn nn n      11211111 1 ......)1[( )]......)(1(...... )......)(1( 11212212 12111         nnn nnn kMkk kk     n j jjnn nn n kMLRRa 1 100 1 )()1()()[(    ])()1( 1 2    n j jj k   X for 1R and for 1R     n j jjnn n n kMLRRazG 1 100 1 )()1()()[()(    ])()1( 1 2    n j jj k   Y Since G(0)=0 and G(z) is analytic for Rz  , it follows, by Schwarz Lemma , that for Rz  ,
  • 7. Location of Zeros of Polynomials www.ijceronline.com Open Access Journal Page 15 zXzG )( for 1R and zYzG )( for 1R . Hence, for Rz  , 1R )()( 0 zGazF  0 )( 0 0    zXa zGa if X a z 0  . Similarly, for Rz  , 1R , 0)( zF if Y a z 0  . In other words, F(z) does not vanish in X a z 0  for 1R and F(z) does not vanish in Y a z 0  for 1R in Rz  . That means all the zeros of F(z) and hence all the zeros of P(z) lie in X a z 0  for 1R and in Y a z 0  for 1R in Rz  . Since for Rz  , 0 )( aXzF  for 1R and 0 )( aYzF  for 1R and since 0)0( 0  aF , it follows by Lemma 2 that the number of zeros of P(z) in 1,1, 0  Rc c R z X a does not exceed )1log( log 1 log log 1 00 0 a X ca aX c   and the number of zeros of P(z) in 1,1, 0  Rc c R z Y a does not exceed )1log( log 1 log log 1 00 0 a Y ca aY c   . That proves Theorem 1 completely. REFERENCES [1]. L. V. Ahlfors, Complex Analysis, 3rd edition, Mc-Grawhill. [2]. A. Aziz and Q. G. Mohammad, Zero-free regions for polynomials and some generalizations of Enestrom-Kakeya Theorem, Canad. Math. Bull.,27(1984),265- 272. [3]. A. Aziz and B. A. Zargar, Some Extensions of Enestrom-Kakeya Theorem, Glasnik Mathematicki, 51 (1996), 239-244. [4]. Y.Choo, On the zeros of a family of self-reciprocal polynomials, Int. J. Math. Analysis , 5 (2011),1761-1766. [5]. N. K. Govil and Q. I. Rahman, On Enestrom-Kakeya Theorem, Tohoku J. Math. 20 (1968), 126-136. [6]. M. H. Gulzar, Some Refinements of Enestrom-Kakeya Theorem, International Journal of Mathematical Archive , Vol 2(9), 2011, 1512-1519.. [7]. M. H. Gulzar,B.A. Zargar and A. W. Manzoor, Zeros of a Certain Class of Polynomials, International Journal of Advanced Research in Engineering and Management, Vol.3 No. 2(2017). [8]. A. Joyal, G. Labelle and Q. I. Rahman, On the location of the zeros of a polynomial, Canad. Math. Bull. ,10 (1967), 53-63. [9]. M. Marden, Geometry of Polynomials, Math. Surveys No. 3, Amer. Math.Soc.(1966). [10]. Q. I. Rahman and G. Schmeisser, Analytic Theory of Polynomials, Oxford University Press, New York (2002). [11]. B. A. Zargar, On the zeros of a family of polynomials, Int. J. of Math. Sci. &Engg. Appls. , Vol. 8 No. 1(January 2014), 233-237.