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ISSN (e): 2250 – 3005 || Vol, 04 || Issue, 5 || May – 2014 ||
International Journal of Computational Engineering Research (IJCER)
www.ijceronline.com Open Access Journal Page 9
Extensions of Enestrom-Kakeya Theorem
M. H. Gulzar
Department of Mathematics
University of Kashmir, Srinagar 190006
I. INTRODUCTION AND STATEMENT OF RESULTS
A famous result giving a bound for all the zeros of a polynomial with real positive monotonically
decreasing coefficients is the following result known as Enestrom-Kakeya theorem [8]:
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 the closed disk 1z .
If the coefficients are monotonic but not positive, Joyal, Labelle and Rahman [6] gave the following
generalization of Theorem A:
Theorem B: Let 


n
j
j
j
zazP
0
)( be a polynomial of degree n such that
011
...... aaaa nn
 
.
Then all the zeros of P(z) lie in the closed disk
n
n
a
aaa
z
00

 .
Aziz and Zargar [1] generalized Theorem B by proving the following result:
Theorem C: Let 


n
j
j
j
zazP
0
)( be a polynomial of degree n such that for some ,1k
011
...... aaaka nn
 
.
Then all the zeros of P(z) lie in the closed disk
n
n
a
aaka
kz
00
1

 .
Gulz ar [4,5] generalized Theorem C to polynomials with complex coefficients and proved the following results:
Theorem D: Let 


n
j
j
j
zazP
0
)( be a polynomial of degree n with jj
a )Re( ,
nja jj
,......,1,0,)Im(   such that for some 10,1  k ,
011
......   nn
k .
Then all the zeros of P(z) lie in the closed disk
ABSTRACT:
In this paper we give an extension of the famous Enestrom-Kakeya Theorem, which
generalizes many generalizations of the said theorem as well.
Mathematics Subject Classification: 30 C 10, 30 C 15
Keywords and Phrases: Coefficient, Polynomial, Zero
Extensions of Enestrom-Kakeya Theorem
www.ijceronline.com Open Access Journal Page 10
n
n
j
jn
n
n
a
k
a
kz




0
000
2)(2
)1(


.
Theorem E: Let 


n
j
j
j
zazP
0
)( be a polynomial of degree n with jj
a )Re( ,
nja jj
,......,1,0,)Im(   such that for some 10,1  k ,
011
......   nn
k .
Then all the zeros of P(z) lie in the closed disk
n
n
j
jn
n
n
a
k
a
kz




0
000
2)(2
)1(


.
Theorem F: : Let 


n
j
j
j
zazP
0
)( be a polynomial of degree n such that for some
real ,,......,2,1,0,
2
arg;, nja j


 and for some 10,1  k ,
011
...... aaaak nn
  .
Then all the zeros of P(z) lie in the closed disk
n
n
j
jnn
a
aaaak
z





1
1
00
sin2)1sin(cos2)sincos1( 
.
Some questions which have been raised by some researchers in connection with
the Enestrom-Kakeya Theorem are[2]:
What happens , if ( i) instead of the leading coefficient n
a , there is some j
a with
11 
 jjj
aaa such that for some 1k , 01111
............    jjjnn
akaaaa ,
j=1,2,…..,n and (ii) for some 011121
......;1,1
2
aaakakkk nn
 
.
In this direction, Liman and Shah [7, Cor.1] have proved the following result:
Theorem G: Let 


n
j
j
j
zazP
0
)( be a polynomial of degree n such that for some ,1k ,
01111
............ aaakaaaa nn
   .
Then P(z) has all its zeros in
n
nj
n
j
jn
a
aaakaaa
z










)()1(00

.
Unfortunately, the conclusion of the theorem is not correct and their claim that it follows from Theorem 1 in
[7] is false. The correct form of the result is as follows:
Theorem H: Let 


n
j
j
j
zazP
0
)( be a polynomial of degree n with jj
a )Re( ,
nja jj
,......,1,0,)Im(   such that for some ,1k ,
01111
............ aaakaaaa nn
   .
Then P(z) has all its zeros in
Extensions of Enestrom-Kakeya Theorem
www.ijceronline.com Open Access Journal Page 11
n
n
a
akaaa
z

)1(200

 .
In this paper , we are going to prove the following more general result:
Theorem 1: Let 


n
j
j
j
zazP
0
)( be a polynomial of degree n such that for some 10,1  k ,
01111
............ aaakaaaa nn

 
.
Then P(z) has all its zeros in
n
n
a
aaaaka
z
000
2)()1(2 


.
Remark 1: For 1 ,Theorem 1 reduces to Theorem H.
Taking in particular 1
1




a
a
k in Theorem 1, we get the following
Corollary 1: Let 


n
j
j
j
zazP
0
)( be a polynomial of degree n with jj
a )Re( ,
nja jj
,......,1,0,)Im(   such that
01111
............ aaaaaaa nn

 
.
Then P(z) has all its zeros in
n
n
a
aaaa
a
aa
a
z
000
1
2)()(2 







.
For 1 , Cor. 1 reduces to the following
Corollary 2: Let 


n
j
j
j
zazP
0
)( be a polynomial of degree n with jj
a )Re( ,
nja jj
,......,1,0,)Im(   such that
01111
............ aaaaaaa jnn
   .
Then P(z) has all its zeros in
n
n
a
aaa
a
aa
a
z
))( 00
1








.
Theorem 1 is a special case of the following more general result:
Theorem 2: Let 


n
j
j
j
zazP
0
)( be a polynomial of degree n with jj
a )Re( ,
nja jj
,......,1,0,)Im(   such that for some ,1k
01111
............  
 
knn
.
Then P(z) has all its zeros in
n
n
j
jn
a
k
z




0
000
22)()1(2  
.
Remark 2: If njj
,......,2,1,0,0  i.e. j
a is real, then Theorem 2 reduces to
Theorem 1.
Extensions of Enestrom-Kakeya Theorem
www.ijceronline.com Open Access Journal Page 12
Applying Theorem 2 to the polynomial –iP(z), we get the following result:
Theorem 3: Let 


n
j
j
j
zazP
0
)( be a polynomial of degree n with jj
a )Re( ,
nja jj
,......,1,0,)Im(   such that for some 10,1  k ,
01111
............    jjjnn
k .
Then P(z) has all its zeros in
n
n
j
jn
a
ka
z




0
000
22)()1(2  
.
For polynomials with complex coefficients, we have the following form of Theorem 1:
Theorem 4: Let 


n
j
j
j
zazP
0
)( be a polynomial of degree n such that for some 10,1  k ,
01111
............ aaaakaaa nn

  .
Then P(z) has all its zeros in
)1sin(2)1sin(cos)sin(cos[
1
  
kkaaka
a
z n
n
]2)1sin(cos 00
aa  
Remark 3: For k=1, Theorem 4 reduces to Theorem F with k=1.
Next, we prove the following result:
Theorem 5: Let 


n
j
j
j
zazP
0
)( be a polynomial of degree n with jj
a )Re( ,
nja jj
,......,1,0,)Im(   such that for some 10,1,1 21
 kk ,
012121
......    nnn
kk .
Then P(z) has all its zeros in
n
n
j
jnnnnnnn
a
kkkk
z





0
011121121
22)()()( 
.
Remark 4: If njj
,......,2,1,0,0  i.e. j
a is real, we get the following result:
Corollary 3: Let 


n
j
j
j
zazP
0
)( be a polynomial of degree n such that for some
10,1,1 21
 kk ,
012121
...... aaaakak nnn
 
.
Then P(z) has all its zeros in
n
nnnnnnn
a
aaaaakakakak
z
011121121
2)()()( 


.
Applying Theorem 2 to the polynomial –iP(z), we get the following result from Theorem 4:
Theorem 6: Let 


n
j
j
j
zazP
0
)( be a polynomial of degree n with jj
a )Re( ,
nja jj
,......,1,0,)Im(   such that for some 10,1,1 21
 kk ,
Extensions of Enestrom-Kakeya Theorem
www.ijceronline.com Open Access Journal Page 13
012121
......    nnn
kk .
Then P(z) has all its zeros in
n
n
j
jnnnnnnn
a
kkkk
z





0
011121121
22)()()( 
.
Theorem 7: Let 


n
j
j
j
zazP
0
)( be a polynomial of degree n such that for some
real ,,......,2,1,0,
2
arg;, nja j


 and for some 10,1,1 21
 kk ,
012121
...... aaaakak nnn
 
.
Then P(z) has all its zeros in

















2
1
0
01121
sin2)1sin(cos
2)cos1()sincos1()sincos1(
1
n
j
j
nnnn
n
aa
aaaakak
a
z


.
Remark 4: For 1, 21
 kkk , Theorem 6 reduces to Theorem F.
Taking 1 in Theorem 7, we get the following
Corollary 4: Let 


n
j
j
j
zazP
0
)( be a polynomial of degree n such that for some
real ,,......,2,1,0,
2
arg;, nja j


 and for some ,1,1 21
 kk ,
012121
...... aaaakak nnn
  .
Then P(z) has all its zeros in
nnn
n
aakak
a
z  
)sincos1()sincos1([
1
121

]sin22)sincos1()cos1(
1
1
001 




n
j
jn
aaaa  .
II. LEMMA
For the proof of Theorem 6, we need the following lemma:
Lemma: Let 1
a and 2
a be any two complex numbers such that 21
aa  and for some real numbers  and
 , 2,1,
2
arg  ja j

 ,then
 sin)(cos)( 212121
aaaaaa  .
The above lemma is due to Govil and Rahman [3] .
3. Proofs of Theorems
Proof of Theorem 2: Consider the polynomial
)......)(1()()1()( 01
1
1
azazazazzPzzF
n
n
n
n



1
1
1
211
1
)(......)()(








zaazaazaaza
n
nn
n
nn
n
n
0011
)(......)( azaazaa  


Extensions of Enestrom-Kakeya Theorem
www.ijceronline.com Open Access Journal Page 14
......)()(
1
211
1



 n
nn
n
nn
n
n
zzza 
})......(){()}(){(
......)}(){()}(){(
001100001
1
1
1












zziz
zkkzkk
n
nn
For ,1z we have, nj
z
j
,.......,2,1,1
1
 so that, by using the hypothesis,
1
1
1
111
1
......[)(








 zkzzzazF
n
nn
n
nn
n
n
0
1
100
011
1
)1(
......)1()1(















n
j
j
jj
zz
zazkzkzk
11111
1
......
1
{[[ 
 
 nnnnnn
n
z
k
z
zaz
}]
1111
)1(
1
......
1
)1(
11
)1(
0
1
1010
10111`
n
n
j
jnjjnn
nnnn
zzzz
z
a
z
k
z
k
z
k

 







 )1(......[[ 1211
 
kkzaz nnnnn
n
])(
)1(......)1(
0
1
1
00011

 






n
j
jj
kk
0
]22)()1(2[[
0
000

 

n
j
jnn
n
kzaz  
if
n
n
j
jn
a
k
z




0
000
22)()1(2  
.
This shows that the zeros of F(z) having modulus greater than 1 lie in
n
n
j
jn
a
k
z




0
000
22)()1(2  
.
But the zeros of F(z) whose modulus is less than or equal to 1 already satisfy the above inequality. Hence, it
follows that all the zeros of F(z) lie in
n
n
j
jn
a
k
z




0
000
22)()1(2  
.
Extensions of Enestrom-Kakeya Theorem
www.ijceronline.com Open Access Journal Page 15
Since the zeros of P(z) are also the zeros of F(z), the result follows.
Proof of Theorem 4: : Consider the polynomial
)......)(1()()1()( 01
1
1
azazazazzPzzF
n
n
n
n



1
1
1
211
1
)}(){(......)()(








zakakaazaazaaza
n
nn
n
nn
n
n
......)()}(){(
1
211







zaazakaaka
00001
)}(){( azaaaa   .
For ,1z we have, nj
z
j
,.......,2,1,1
1
 , so that, by using the hypothesis and the Lemma,
1
1
1
111
1
......[)(








zkaazaazaazazF
n
nn
n
nn
n
n
00
011
1
)1(
......)1()1(
aza
zaazakzakazak

 









11211
1
......
1
{[ 
  nnnnnn
n
z
kaa
z
aaaazaz
}]
11
)1(
1
......
1
)1(
11
)1(
010
10111`
nn
nnnn
z
a
z
a
z
aa
z
ak
z
aka
z
ak







kaaaaaazaz nnnnn
n
  1211
......{[
}])1(
......)1()1(
00
011
aa
aaakakaak

 


 cos)(sin)(cos){([ 2111 
 nnnnnnn
n
aaaaaazaz
}])1(sin)(
cos)......(sin)(cos)(
)1(sin)(cos)()1(
sin)(cos)(......sin)(
0001
012121
11
1121
aaaa
aaaaaa
akaakaakak
akaakaaa nn














)1sin(2)1sin(cos)sin(cos{[   
kkaakazaz nn
n
}]2)1sin(cos 00
aa  
0
if
)1sin(2)1sin(cos)sin(cos[
1
  
kkaaka
a
z n
n
]2)1sin(cos 00
aa  
Extensions of Enestrom-Kakeya Theorem
www.ijceronline.com Open Access Journal Page 16
This shows that the zeros of F(z) having modulus greater than 1 lie in
)1sin(2)1sin(cos)sin(cos[
1
  
kkaaka
a
z n
n
]2)1sin(cos 00
aa  
But the zeros of F(z) whose modulus is less than or equal to 1 already satisfy the above inequality. Hence, it
follows that all the zeros of F(z) lie in
)1sin(2)1sin(cos)sin(cos[
1
  
kkaaka
a
z n
n
]2)1sin(cos 00
aa  
Since the zeros of P(z) are also the zeros of F(z), the result follows.
Proof of Theorem 5: Consider the polynomial
)......)(1()()1()( 01
1
1
azazazazzPzzF
n
n
n
n



001
1
211
1
)(......)()( azaazaazaaza
n
nn
n
nn
n
n




1
211121121
1
)()}()(){(




n
nn
n
nnnnnn
n
n
zzkkkkza 
})(
......){()}(){(......
001
100001



 
z
ziz
n
nn
For ,1z we have, nj
z
j
,.......,2,1,1
1
 , so that, by using the hypothesis,
1
211121121
1
[)(




n
nn
n
nn
n
nn
n
nn
n
n
zzkzkzkkzazF 
])1(...... 0
1
10001
  


n
j
j
jj
zzza
......
1
[[ 211121121
z
kkkkzaz nnnnnnnnn
n

 
}]
1111
)1(
1
0
1
1010101 n
n
j
jnjjnnn
zzzzz
  


......)1()1({[ 21121121
  nnnnnnn
n
kkkkzaz 
}])()1( 0
1
10001
  


n
j
jj
 ......)1()1({[ 1121121
  nnnnnn
n
kkkkzaz 
}]22)(
0
000 


n
j
j

)()(){([ 11121121 
 nnnnnnnn
n
kkkkzaz 
}]22
0
0 


n
j
j

0
if
Extensions of Enestrom-Kakeya Theorem
www.ijceronline.com Open Access Journal Page 17
)()()[(
1
11121121 
 nnnnnnn
n
kkkk
a
z  ]22
0
0 


n
j
j

This shows that the zeros of F(z) having modulus greater than 1 lie in
)()()[(
1
11121121 
 nnnnnnn
n
kkkk
a
z  ]22
0
0 


n
j
j
 .
But the zeros of F(z) whose modulus is less than or equal to 1 already satisfy the above inequality. Hence, it
follows that all the zeros of F(z) lie in
)()()[(
1
11121121 
 nnnnnnn
n
kkkk
a
z  ]22
0
0 


n
j
j

Since the zeros of P(z) are also the zeros of F(z), Theorem 4 follows.
Proof of Theorem 6. Consider the polynomial
)......)(1()()1()( 01
1
1
azazazazzPzzF
n
n
n
n



001
1
211
1
)(......)()( azaazaazaaza
n
nn
n
nn
n
n




1
211121121
1
)()}()(){(




n
nn
n
nnnnnn
n
n
zaazaakaakakakza
00001
)}(){(...... azaaaa  
For ,1z we have, nj
z
j
,.......,2,1,1
1
 , so that, by using the hypothesis and the Lemma,
1
211121121
1
[)(




n
nn
n
nn
n
nn
n
nn
n
n
zaazaakzakzakakzazF 
])1(...... 0001
azazaa  
......
1
[[ 211121121
 
z
aaaakaakakakzaz nnnnnnnnn
n
]
11
)1(
1
010101 nnn
z
a
z
a
z
aa  

......{[ 211121121
  nnnnnnnnn
n
aaaakaakakakzaz
}])1( 0001
aaaa  
nnnnnn
n
akakakakakzaz )1(sin)(cos){([ 1121121
 

}])1(sin)(cos)(
......sin)(cos)()1(
00101
212112
aaaaa
aaaaak nnnnn



 
nnnn
n
aakakzaz  
)sincos1()sincos1({[ 121

}]sin22)sincos1()cos1(
1
1
001 




n
j
jn
aaaa 
>0
if
Extensions of Enestrom-Kakeya Theorem
www.ijceronline.com Open Access Journal Page 18
nnn
n
aakak
a
z  
)sincos1()sincos1([
1
121

]sin22)sincos1()cos1(
1
1
001 




n
j
jn
aaaa  .
This shows that the zeros of F(z) having modulus greater than 1 lie in
nnn
n
aakak
a
z  
)sincos1()sincos1([
1
121

]sin22)sincos1()cos1(
1
1
001 




n
j
jn
aaaa  .
But the zeros of F(z) whose modulus is less than or equal to 1 already satisfy the above inequality. Hence, it
follows that all the zeros of F(z) lie in
nnn
n
aakak
a
z  
)sincos1()sincos1([
1
121

]sin22)sincos1()cos1(
1
1
001 




n
j
jn
aaaa  .
Since the zeros of P(z) are also the zeros of F(z), the result follows.
REFERENCES
[1] A.Aziz and B.A.Zargar, Some Extensions of Enestrom-Kakeya Theorem, Glasnik Math. 31(1996), 239-244.
[2] N. K. Govvil, International Congress of ASIAM, Jammu University, India March 31-April 3, 2007.
[3] N. K. Govil and Q.I. Rahman, On the Enestrom-Kakeya Theorem, Tohoku Math.J.20 (1968), 126-136.
[4] M. H. Gulzar, Some Refinements of Enestrom-Kakeya Theorem, Int. Journal of Mathematical Archive -2(9, 2011),1512-1529.
[5] M. H. Gulzar, Ph.D thesis, Department of Mathematics, University of Kashmir Srinagar, 2012.
[6] A. Joyal, G. Labelle and Q. I. Rahman, On the Location of Zeros of Polynomials, Canad. Math. Bull., 10(1967), 53-66.
[7] A. Liman and W. M. Shah, Extensions of Enestrom-Kakeya Theorem, Int. Journal of Modern Mathematical Sciences, 2013, 8(2),
82-89.
[8] M. Marden, Geometry of Polynomials, Math. Surveys, No.3; Amer. Math. Soc. Providence R.I. 1966.

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C04502009018

  • 1. ISSN (e): 2250 – 3005 || Vol, 04 || Issue, 5 || May – 2014 || International Journal of Computational Engineering Research (IJCER) www.ijceronline.com Open Access Journal Page 9 Extensions of Enestrom-Kakeya Theorem M. H. Gulzar Department of Mathematics University of Kashmir, Srinagar 190006 I. INTRODUCTION AND STATEMENT OF RESULTS A famous result giving a bound for all the zeros of a polynomial with real positive monotonically decreasing coefficients is the following result known as Enestrom-Kakeya theorem [8]: 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 the closed disk 1z . If the coefficients are monotonic but not positive, Joyal, Labelle and Rahman [6] gave the following generalization of Theorem A: Theorem B: Let    n j j j zazP 0 )( be a polynomial of degree n such that 011 ...... aaaa nn   . Then all the zeros of P(z) lie in the closed disk n n a aaa z 00   . Aziz and Zargar [1] generalized Theorem B by proving the following result: Theorem C: Let    n j j j zazP 0 )( be a polynomial of degree n such that for some ,1k 011 ...... aaaka nn   . Then all the zeros of P(z) lie in the closed disk n n a aaka kz 00 1   . Gulz ar [4,5] generalized Theorem C to polynomials with complex coefficients and proved the following results: Theorem D: Let    n j j j zazP 0 )( be a polynomial of degree n with jj a )Re( , nja jj ,......,1,0,)Im(   such that for some 10,1  k , 011 ......   nn k . Then all the zeros of P(z) lie in the closed disk ABSTRACT: In this paper we give an extension of the famous Enestrom-Kakeya Theorem, which generalizes many generalizations of the said theorem as well. Mathematics Subject Classification: 30 C 10, 30 C 15 Keywords and Phrases: Coefficient, Polynomial, Zero
  • 2. Extensions of Enestrom-Kakeya Theorem www.ijceronline.com Open Access Journal Page 10 n n j jn n n a k a kz     0 000 2)(2 )1(   . Theorem E: Let    n j j j zazP 0 )( be a polynomial of degree n with jj a )Re( , nja jj ,......,1,0,)Im(   such that for some 10,1  k , 011 ......   nn k . Then all the zeros of P(z) lie in the closed disk n n j jn n n a k a kz     0 000 2)(2 )1(   . Theorem F: : Let    n j j j zazP 0 )( be a polynomial of degree n such that for some real ,,......,2,1,0, 2 arg;, nja j    and for some 10,1  k , 011 ...... aaaak nn   . Then all the zeros of P(z) lie in the closed disk n n j jnn a aaaak z      1 1 00 sin2)1sin(cos2)sincos1(  . Some questions which have been raised by some researchers in connection with the Enestrom-Kakeya Theorem are[2]: What happens , if ( i) instead of the leading coefficient n a , there is some j a with 11   jjj aaa such that for some 1k , 01111 ............    jjjnn akaaaa , j=1,2,…..,n and (ii) for some 011121 ......;1,1 2 aaakakkk nn   . In this direction, Liman and Shah [7, Cor.1] have proved the following result: Theorem G: Let    n j j j zazP 0 )( be a polynomial of degree n such that for some ,1k , 01111 ............ aaakaaaa nn    . Then P(z) has all its zeros in n nj n j jn a aaakaaa z           )()1(00  . Unfortunately, the conclusion of the theorem is not correct and their claim that it follows from Theorem 1 in [7] is false. The correct form of the result is as follows: Theorem H: Let    n j j j zazP 0 )( be a polynomial of degree n with jj a )Re( , nja jj ,......,1,0,)Im(   such that for some ,1k , 01111 ............ aaakaaaa nn    . Then P(z) has all its zeros in
  • 3. Extensions of Enestrom-Kakeya Theorem www.ijceronline.com Open Access Journal Page 11 n n a akaaa z  )1(200   . In this paper , we are going to prove the following more general result: Theorem 1: Let    n j j j zazP 0 )( be a polynomial of degree n such that for some 10,1  k , 01111 ............ aaakaaaa nn    . Then P(z) has all its zeros in n n a aaaaka z 000 2)()1(2    . Remark 1: For 1 ,Theorem 1 reduces to Theorem H. Taking in particular 1 1     a a k in Theorem 1, we get the following Corollary 1: Let    n j j j zazP 0 )( be a polynomial of degree n with jj a )Re( , nja jj ,......,1,0,)Im(   such that 01111 ............ aaaaaaa nn    . Then P(z) has all its zeros in n n a aaaa a aa a z 000 1 2)()(2         . For 1 , Cor. 1 reduces to the following Corollary 2: Let    n j j j zazP 0 )( be a polynomial of degree n with jj a )Re( , nja jj ,......,1,0,)Im(   such that 01111 ............ aaaaaaa jnn    . Then P(z) has all its zeros in n n a aaa a aa a z ))( 00 1         . Theorem 1 is a special case of the following more general result: Theorem 2: Let    n j j j zazP 0 )( be a polynomial of degree n with jj a )Re( , nja jj ,......,1,0,)Im(   such that for some ,1k 01111 ............     knn . Then P(z) has all its zeros in n n j jn a k z     0 000 22)()1(2   . Remark 2: If njj ,......,2,1,0,0  i.e. j a is real, then Theorem 2 reduces to Theorem 1.
  • 4. Extensions of Enestrom-Kakeya Theorem www.ijceronline.com Open Access Journal Page 12 Applying Theorem 2 to the polynomial –iP(z), we get the following result: Theorem 3: Let    n j j j zazP 0 )( be a polynomial of degree n with jj a )Re( , nja jj ,......,1,0,)Im(   such that for some 10,1  k , 01111 ............    jjjnn k . Then P(z) has all its zeros in n n j jn a ka z     0 000 22)()1(2   . For polynomials with complex coefficients, we have the following form of Theorem 1: Theorem 4: Let    n j j j zazP 0 )( be a polynomial of degree n such that for some 10,1  k , 01111 ............ aaaakaaa nn    . Then P(z) has all its zeros in )1sin(2)1sin(cos)sin(cos[ 1    kkaaka a z n n ]2)1sin(cos 00 aa   Remark 3: For k=1, Theorem 4 reduces to Theorem F with k=1. Next, we prove the following result: Theorem 5: Let    n j j j zazP 0 )( be a polynomial of degree n with jj a )Re( , nja jj ,......,1,0,)Im(   such that for some 10,1,1 21  kk , 012121 ......    nnn kk . Then P(z) has all its zeros in n n j jnnnnnnn a kkkk z      0 011121121 22)()()(  . Remark 4: If njj ,......,2,1,0,0  i.e. j a is real, we get the following result: Corollary 3: Let    n j j j zazP 0 )( be a polynomial of degree n such that for some 10,1,1 21  kk , 012121 ...... aaaakak nnn   . Then P(z) has all its zeros in n nnnnnnn a aaaaakakakak z 011121121 2)()()(    . Applying Theorem 2 to the polynomial –iP(z), we get the following result from Theorem 4: Theorem 6: Let    n j j j zazP 0 )( be a polynomial of degree n with jj a )Re( , nja jj ,......,1,0,)Im(   such that for some 10,1,1 21  kk ,
  • 5. Extensions of Enestrom-Kakeya Theorem www.ijceronline.com Open Access Journal Page 13 012121 ......    nnn kk . Then P(z) has all its zeros in n n j jnnnnnnn a kkkk z      0 011121121 22)()()(  . Theorem 7: Let    n j j j zazP 0 )( be a polynomial of degree n such that for some real ,,......,2,1,0, 2 arg;, nja j    and for some 10,1,1 21  kk , 012121 ...... aaaakak nnn   . Then P(z) has all its zeros in                  2 1 0 01121 sin2)1sin(cos 2)cos1()sincos1()sincos1( 1 n j j nnnn n aa aaaakak a z   . Remark 4: For 1, 21  kkk , Theorem 6 reduces to Theorem F. Taking 1 in Theorem 7, we get the following Corollary 4: Let    n j j j zazP 0 )( be a polynomial of degree n such that for some real ,,......,2,1,0, 2 arg;, nja j    and for some ,1,1 21  kk , 012121 ...... aaaakak nnn   . Then P(z) has all its zeros in nnn n aakak a z   )sincos1()sincos1([ 1 121  ]sin22)sincos1()cos1( 1 1 001      n j jn aaaa  . II. LEMMA For the proof of Theorem 6, we need the following lemma: Lemma: Let 1 a and 2 a be any two complex numbers such that 21 aa  and for some real numbers  and  , 2,1, 2 arg  ja j   ,then  sin)(cos)( 212121 aaaaaa  . The above lemma is due to Govil and Rahman [3] . 3. Proofs of Theorems Proof of Theorem 2: Consider the polynomial )......)(1()()1()( 01 1 1 azazazazzPzzF n n n n    1 1 1 211 1 )(......)()(         zaazaazaaza n nn n nn n n 0011 )(......)( azaazaa    
  • 6. Extensions of Enestrom-Kakeya Theorem www.ijceronline.com Open Access Journal Page 14 ......)()( 1 211 1     n nn n nn n n zzza  })......(){()}(){( ......)}(){()}(){( 001100001 1 1 1             zziz zkkzkk n nn For ,1z we have, nj z j ,.......,2,1,1 1  so that, by using the hypothesis, 1 1 1 111 1 ......[)(          zkzzzazF n nn n nn n n 0 1 100 011 1 )1( ......)1()1(                n j j jj zz zazkzkzk 11111 1 ...... 1 {[[     nnnnnn n z k z zaz }] 1111 )1( 1 ...... 1 )1( 11 )1( 0 1 1010 10111` n n j jnjjnn nnnn zzzz z a z k z k z k            )1(......[[ 1211   kkzaz nnnnn n ])( )1(......)1( 0 1 1 00011          n j jj kk 0 ]22)()1(2[[ 0 000     n j jnn n kzaz   if n n j jn a k z     0 000 22)()1(2   . This shows that the zeros of F(z) having modulus greater than 1 lie in n n j jn a k z     0 000 22)()1(2   . But the zeros of F(z) whose modulus is less than or equal to 1 already satisfy the above inequality. Hence, it follows that all the zeros of F(z) lie in n n j jn a k z     0 000 22)()1(2   .
  • 7. Extensions of Enestrom-Kakeya Theorem www.ijceronline.com Open Access Journal Page 15 Since the zeros of P(z) are also the zeros of F(z), the result follows. Proof of Theorem 4: : Consider the polynomial )......)(1()()1()( 01 1 1 azazazazzPzzF n n n n    1 1 1 211 1 )}(){(......)()(         zakakaazaazaaza n nn n nn n n ......)()}(){( 1 211        zaazakaaka 00001 )}(){( azaaaa   . For ,1z we have, nj z j ,.......,2,1,1 1  , so that, by using the hypothesis and the Lemma, 1 1 1 111 1 ......[)(         zkaazaazaazazF n nn n nn n n 00 011 1 )1( ......)1()1( aza zaazakzakazak             11211 1 ...... 1 {[    nnnnnn n z kaa z aaaazaz }] 11 )1( 1 ...... 1 )1( 11 )1( 010 10111` nn nnnn z a z a z aa z ak z aka z ak        kaaaaaazaz nnnnn n   1211 ......{[ }])1( ......)1()1( 00 011 aa aaakakaak       cos)(sin)(cos){([ 2111   nnnnnnn n aaaaaazaz }])1(sin)( cos)......(sin)(cos)( )1(sin)(cos)()1( sin)(cos)(......sin)( 0001 012121 11 1121 aaaa aaaaaa akaakaakak akaakaaa nn               )1sin(2)1sin(cos)sin(cos{[    kkaakazaz nn n }]2)1sin(cos 00 aa   0 if )1sin(2)1sin(cos)sin(cos[ 1    kkaaka a z n n ]2)1sin(cos 00 aa  
  • 8. Extensions of Enestrom-Kakeya Theorem www.ijceronline.com Open Access Journal Page 16 This shows that the zeros of F(z) having modulus greater than 1 lie in )1sin(2)1sin(cos)sin(cos[ 1    kkaaka a z n n ]2)1sin(cos 00 aa   But the zeros of F(z) whose modulus is less than or equal to 1 already satisfy the above inequality. Hence, it follows that all the zeros of F(z) lie in )1sin(2)1sin(cos)sin(cos[ 1    kkaaka a z n n ]2)1sin(cos 00 aa   Since the zeros of P(z) are also the zeros of F(z), the result follows. Proof of Theorem 5: Consider the polynomial )......)(1()()1()( 01 1 1 azazazazzPzzF n n n n    001 1 211 1 )(......)()( azaazaazaaza n nn n nn n n     1 211121121 1 )()}()(){(     n nn n nnnnnn n n zzkkkkza  })( ......){()}(){(...... 001 100001      z ziz n nn For ,1z we have, nj z j ,.......,2,1,1 1  , so that, by using the hypothesis, 1 211121121 1 [)(     n nn n nn n nn n nn n n zzkzkzkkzazF  ])1(...... 0 1 10001      n j j jj zzza ...... 1 [[ 211121121 z kkkkzaz nnnnnnnnn n    }] 1111 )1( 1 0 1 1010101 n n j jnjjnnn zzzzz      ......)1()1({[ 21121121   nnnnnnn n kkkkzaz  }])()1( 0 1 10001      n j jj  ......)1()1({[ 1121121   nnnnnn n kkkkzaz  }]22)( 0 000    n j j  )()(){([ 11121121   nnnnnnnn n kkkkzaz  }]22 0 0    n j j  0 if
  • 9. Extensions of Enestrom-Kakeya Theorem www.ijceronline.com Open Access Journal Page 17 )()()[( 1 11121121   nnnnnnn n kkkk a z  ]22 0 0    n j j  This shows that the zeros of F(z) having modulus greater than 1 lie in )()()[( 1 11121121   nnnnnnn n kkkk a z  ]22 0 0    n j j  . But the zeros of F(z) whose modulus is less than or equal to 1 already satisfy the above inequality. Hence, it follows that all the zeros of F(z) lie in )()()[( 1 11121121   nnnnnnn n kkkk a z  ]22 0 0    n j j  Since the zeros of P(z) are also the zeros of F(z), Theorem 4 follows. Proof of Theorem 6. Consider the polynomial )......)(1()()1()( 01 1 1 azazazazzPzzF n n n n    001 1 211 1 )(......)()( azaazaazaaza n nn n nn n n     1 211121121 1 )()}()(){(     n nn n nnnnnn n n zaazaakaakakakza 00001 )}(){(...... azaaaa   For ,1z we have, nj z j ,.......,2,1,1 1  , so that, by using the hypothesis and the Lemma, 1 211121121 1 [)(     n nn n nn n nn n nn n n zaazaakzakzakakzazF  ])1(...... 0001 azazaa   ...... 1 [[ 211121121   z aaaakaakakakzaz nnnnnnnnn n ] 11 )1( 1 010101 nnn z a z a z aa    ......{[ 211121121   nnnnnnnnn n aaaakaakakakzaz }])1( 0001 aaaa   nnnnnn n akakakakakzaz )1(sin)(cos){([ 1121121    }])1(sin)(cos)( ......sin)(cos)()1( 00101 212112 aaaaa aaaaak nnnnn      nnnn n aakakzaz   )sincos1()sincos1({[ 121  }]sin22)sincos1()cos1( 1 1 001      n j jn aaaa  >0 if
  • 10. Extensions of Enestrom-Kakeya Theorem www.ijceronline.com Open Access Journal Page 18 nnn n aakak a z   )sincos1()sincos1([ 1 121  ]sin22)sincos1()cos1( 1 1 001      n j jn aaaa  . This shows that the zeros of F(z) having modulus greater than 1 lie in nnn n aakak a z   )sincos1()sincos1([ 1 121  ]sin22)sincos1()cos1( 1 1 001      n j jn aaaa  . But the zeros of F(z) whose modulus is less than or equal to 1 already satisfy the above inequality. Hence, it follows that all the zeros of F(z) lie in nnn n aakak a z   )sincos1()sincos1([ 1 121  ]sin22)sincos1()cos1( 1 1 001      n j jn aaaa  . Since the zeros of P(z) are also the zeros of F(z), the result follows. REFERENCES [1] A.Aziz and B.A.Zargar, Some Extensions of Enestrom-Kakeya Theorem, Glasnik Math. 31(1996), 239-244. [2] N. K. Govvil, International Congress of ASIAM, Jammu University, India March 31-April 3, 2007. [3] N. K. Govil and Q.I. Rahman, On the Enestrom-Kakeya Theorem, Tohoku Math.J.20 (1968), 126-136. [4] M. H. Gulzar, Some Refinements of Enestrom-Kakeya Theorem, Int. Journal of Mathematical Archive -2(9, 2011),1512-1529. [5] M. H. Gulzar, Ph.D thesis, Department of Mathematics, University of Kashmir Srinagar, 2012. [6] A. Joyal, G. Labelle and Q. I. Rahman, On the Location of Zeros of Polynomials, Canad. Math. Bull., 10(1967), 53-66. [7] A. Liman and W. M. Shah, Extensions of Enestrom-Kakeya Theorem, Int. Journal of Modern Mathematical Sciences, 2013, 8(2), 82-89. [8] M. Marden, Geometry of Polynomials, Math. Surveys, No.3; Amer. Math. Soc. Providence R.I. 1966.