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 A. Liman  et al. / BIBECHANA 10 (2014) 71-81 : BMHSS, p.71 (Online Publication: Dec., 2013) 

 

 

BIBECHANA 
A Multidisciplinary Journal of Science, Technology and Mathematics 

ISSN 2091-0762 (online) 
Journal homepage: http://nepjol.info/index.php/BIBECHANA 

 

On The Eneström-Kakeya Theorem  

A. Liman 
1  
, 
 
Tawheeda Rasool

1 
,   W.M Shah

2  
 
 

1Department of Mathematics, National Institute of Technology, India –190006  
2 
Department of Mathematics, Jammu and Kashmir Institute of Mathematical Sciences-190008 

e-mail: wmshah@rediffmail.com,  abliman22@yahoo.com    

Article history: Received 25 September, 2013; Accepted 23 November, 2013 

 

Abstract 

In this paper we present some interesting generalizations of Eneström-Kakeya type results concerning 

the location of zeros of a polynomial in the complex plane. We relax the hypothesis and put less 

restrictive conditions on the coefficients of the polynomial, and thereby generalize some classical 

results. 

  

Keywords: polynomials, zeros, Eneström-Kakeya theorem 
 

1. Introduction 

The following result known as Eneström-Kakeya Theorem (for Reference see [1-3]) is well-known 

in the theory of distribution of zeros of polynomials.  

Theorem A. If ∑
=

=
n

j

j

j zazP
0

)(  is a polynomial of degree n such that 

(1)                           0aa....aa
011nn
>≥≥≥≥ − , 

then all the zeros of P(z) lie in 1.z ≤  

Joyal, Labelle and Rahman [4] dropped the restriction on the hypothesis that all the 

coefficients be positive and proved the following: 

Theorem B. If ∑
=

=
n

j

j

j zazP
0

)( is a polynomial of degree n such that  

 1 1 0....n na a a a−≥ ≥ ≥ ≥ , 

then all the zeros of P(z) lie in  

 
0 0n

n

a a a
z

a

− +
≤ . 

More recently Aziz and Zargar [5] relaxed the hypothesis of Eneström Kakeya theorem and proved 

some interesting extensions of Theorem B. In fact they proved 

THEOREM C. If 
1

1 1 0( ) ...n n

n nP z a z a z a z a−
−= + + + + , is a polynomial of degree n such that for 

some 1k ≥ , 



 

 

A. Liman  et al. / BIBECHANA 10 (2014) 71-81 : BMHSS, p.72 (Online Publication: Dec., 2013) 

 

    1 2 1 0... ,n nka a a a a−≥ ≥ ≥ ≥ ≥  

then all the zeros of P(z) lie in  

   
0 0

1 .
n

n

ka a a
z k

a

− +
+ − ≤  

More recently, Shah and Liman [6] while assuming the coefficients of the polynomials to be complex 

numbers among other things proved the following generalization of Theorem C. 

Theorem D.  Let ( )P z =
0

n
j

j

j

a z
=
∑  be a polynomial with complex coefficients. If Re j ja α=  and 

Im j ja β= for 0,1, 2,...,j n= , 0na ≠  such that for some k 1≥  

                    1 2 1 0...n n nkα α α α α− −≥ ≥ ≥ ≥ ≥
, 
     

                    1 1 0.... 0n nβ β β β−≥ ≥ ≥ ≥ > , 

then all the zeros of ( )P z  lie in 

            
0 0

( 1)
n nn

n n

k
z k
a a

α α α βα − + +
+ − ≤  

In this paper, we prove some Enström-Kakeya type results which besides generalizing some earlier 

results proved in this direction, include Theorem B, Theorem C and Theorem D as special cases. We 

start by proving the following: 

      

2. Theorems and Proofs 

Theorem 1: Let ( )P z =
0

n
j

j

j

a z
=
∑  be a polynomial of degree n. If Re j ja α= , Im j ja β= for 

0,1, 2,...,j n= , 0na ≠  and if for some positive integer nλ ≤  and 1k ≥  

1 1

1 2 1 1 0... ...n n n

n n nk k k kλ λ λ
λ λα α α α α α α− + − − −

− − −≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥
 

then all the zeros of P(z) lie in 

 

0 0

0

( 1) ( | |) 2

1 .

n n

n j j n j

j j

n

k

z k
λ

α α α α α α β

α
= =

   
− + + − + − +  

   + − ≤
∑ ∑

 

 

If we take nλ =  we obtain the following : 

Corollary 1.1: Let ( )P z =
0

n
j

j

j

a z
=
∑  be a polynomial of degree n with 

 
Re j ja α= , Im j ja β= for 

0,1, 2,...,j n= , 0na ≠  and if for some 1k ≥  

1 2 1 0...n n nkα α α α α− −≥ ≥ ≥ ≥ ≥
 

then all the zeros of P(z)  lie in 
 

 



 

 

A. Liman  et al. / BIBECHANA 10 (2014) 71-81 : BMHSS, p.73 (Online Publication: Dec., 2013) 

 

0 0

0

2

1 .

n

n j

j

n

k

z k

α α α β

α
=

− + +

+ − ≤
∑

 
For  1nλ = −  and assuming all coefficients real and positive in Theorem 1, we obtain the following

 

Corollary 1.2: Let ( )P z =
0

n
j

j

j

a z
=
∑  be a polynomial of degree n and if for some  1k ≥  

2

1 2 1 0... 0n n nk a ka a a a− −≥ ≥ ≥ ≥ ≥ >
 

then all the zeros of P(z) lie in 

11 2( 1) n

n

a
z k k k

a

−+ − ≤ + −
 

or equivalently  

1 (2 1)z k k k+ − ≤ −
 Remark 1.1: Theorem C is a special case of Theorem 1, if we take all coefficients real and nλ = . 

Further for all coefficients real, nλ =  and 1k =  we obtain Theorem B.  

Theorem 2: Let ( )P z =
0

n
j

j

j

a z
=
∑  be a polynomial of degree n. If Re j ja α=  and Im j ja β= for 

0,1, 2,...,j n= , 0na ≠ and if for some positive integer nλ ≤  and 1k ≥
 

1 1

1 2 1 1 0... ...n n n

n n nk k k kλ λ λ
λ λβ β β β β β β− + − − −

− − −≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥
 

then all the zeros of P(z) lie in 

                     

0 0

0

( 1) ( | |) 2

1

n n

n j j n j

i j

n

k

z k
λ

β β β β β β α

β
= =

  
− + + − + − +  

  + − ≤
∑ ∑

 

Next as a generalization of Therorem D, we prove 

Theorem 3: Let ( )P z =
0

n
j

j

j

a z
=
∑  be a polynomial of degree n. If Re j ja α=  and Im j ja β= for 

0,1, 2,...,j n= , 0na ≠
 
and if for some positive integer nλ ≤  and 1k ≥  

1 1

1 2 1 1 0... ...n n n

n n nk k k kλ λ λ
λ λα α α α α α α− + − − −

− − −≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥
 

1 1 0... 0n nβ β β β−≥ ≥ ≥ ≥ >
 

then all the zeros of P(z) lie in 

 

0 0 ( 1) ( )

( 1) .

n

n j j n n

in

n n

k

z k
a a

λ

α α α α α α β
α =

  
− + + − + − +  

  + − ≤
∑

 
 

Remark 3.1: If we take nλ =  in above we obtain Theorem D. 
 

 

 

 



 

 

A. Liman  et al. / BIBECHANA 10 (2014) 71-81 : BMHSS, p.74 (Online Publication: Dec., 2013) 

 

Theorem 4: Let ( )P z =
0

n
j

j

j

a z
=
∑  be a polynomial of degree n. If for some positive integer 

nλ ≤ , 1k ≥  and 1µ ≥  

1 1

1 2 1 1 0... ...
n n n

n n nk k k k
λ λ λ

λ λα α α α α α α− + − − −
− − −≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥

 

1 1 0... 0n nµβ β β β−≥ ≥ ≥ ≥ >
 

then all the zeros of P(z) lie in 

( )0 0 1 ( )

1

n

n j j n n

in n

n n

k
k i

z
a a

λ

α α α α α α µβ
α µβ =

  
− + + − + − +  +   + − ≤

∑

 

 

If we take nλ =  in above we obtain the following: 

Corollary 4.1: Let ( )P z =
0

n
j

j

j

a z
=
∑  be a polynomial of degree n. If for some positive integer nλ ≤  

, 1k ≥  and 1µ ≥  

1 1 0...n nkα α α α−≥ ≥ ≥ ≥
 

                                          1 1 0... 0n nµβ β β β−≥ ≥ ≥ ≥ >
 

then all the zeros of ( )P z  lie in 
 

0 0
{ }

1
n nn n

n n

kk i
z

a a

α α α µβα µβ − + ++
+ − ≤

 

Theorem 5:  Let ( )P z =
0

n
j

j

j

a z
=
∑  be a polynomial of degree n. If for some positive integer nλ ≤  and 

1k ≥  
1 1

1 2 1 1 0... ...n n n

n n nk a k a k a k a a a aλ λ λ
λ λ

− + − − −
− − −≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥

 
and for some real β ,         

arg
2

ja
π

β α− ≤ ≤  , 0,1, 2,...j =  

then all the zeros of  lie in 
1 1

1 1

1 2sin
1 (cos sin ) {( 1) cos ( 1)sin } ( 1) | | .

n n

j i j

i i jn n

z k k k k a k a a
a aλ λ λ

α
α α α α

−

= + = = −

+ − ≤ + + − + + + − +∑ ∑ ∑

Proof of Theorem 1: Consider a polynomial 

       ( ) (1 ) ( )F z z P z= −
 

    
1

1 1 0 0( ) ... ( )n n

n n na z a a z a a z a+
−= − + − + + − +

 

   

{ }
{ }

1

1 1 0 0

1

1 1 0 0

( ) ... ( )

( ) ... ( )

n n

n n n

n n

n n n

z z z

i z z z

α α α α α α

β β β β β β

+
−

+
−

= − + − + + − +

+ − + − + + − +
  

 

 



 

 

A. Liman  et al. / BIBECHANA 10 (2014) 71-81 : BMHSS, p.75 (Online Publication: Dec., 2013) 

 

1

1 1

1 1 2 1

1 2

1 1 0 0 1 2

1

1 1 1 0 0

{ ( 1) ( ) ( ) ... ( )

( ) ... ( ) ( 1)(

... )} { ( ) ... ( ) }

n n n

n n n n n

n n

n n

n n

n n n

z z k k z k z k z

k z z k z z

z z i z z zλ

λ
λ λ

λ
λ λ

λ
λ λ

α α α α α α α

α α α α α α α

α α β β β β β β+

− +
− − − +

− −
− − −

+
+ −

= − + − + − + − + + −

+ − + + − + − − +

+ + + + − + − + + − +

 

1

1 1

1 1 2 1

1 2

1 1 0 0 1 2

1

1 0 1

1

{ ( 1) ( ) ( ) ... ( )

( ) ... ( ) ( 1)(

... )} ( ) }

n n n

n n n n n

n n

n n

n
n j

n j j

j

z z k k z k z k z

k z z k z z

z z i z i i zλ

λ
λ λ

λ
λ λ

λ
λ λ

α α α α α α α

α α α α α α α

α α β β β β+

− +
− − − +

− −
− − −

+
+ −

=

= − + − + − + − + + −

+ − + + − + − − +

+ + + − + + −∑
 

       

1 1 2 1 11

1 2 0
1 0 0 11 2

1

1 1 1
( 1) ( ) ( ) ... ( ) ( )

1 1 1
... ( ) ( 1) ... ( )

n

n n n n n n n

n
n n

n j jn n n n n j
j

z z k k k k k
z z z

k i z i i
z z z z z z z

λ λ λ λλ λ

λ
λ

α α α α α α α α α

α α α β
α α α β β β

− − − + −− − −

− −
−− − −

=

= − + − + − + − + + − + − +


 + + − + − − + + + − + + −  
  

∑
 

 
Let 1z >  so that

1
1

n j
z

− < , 0,1, 2,...j =
 

1 2 1 1

1 1

1 0 0 1 2 1

1 2 1

0

1

1

( ) 1 ...

... 1 ...

1
(| | | |)

n n n

n n n n n

n n

n n n n

n

n j jn n j
j

k k k
F z z z k k

z z z

k
zz z z z z

z
z z

λ λ λ λ
λ λ

λ λ
λ λ

α α α α α α
α α α

α α α α α α α

β
β β β

− − + −
− − − −

− − +

− − − −

− −
=

  − − −
≥ + − − − + + + + + 



 −
 + + + + − + + + +
 
 


+ + + + 


∑  

      

{ 1 1 2 1 1

1

1 0 0 0 1

1

1 ...

... 1 (| | | |)

n

n n n n n

n n

j n j j

j j

z z k k k k k

k

λ λ λ λ

λ

α α α α α α α α α

α α α α β β β β

− − − + −

−

−
= =

> + − −  − + − + + − + − +


+ + − + + − + + + + 


∑ ∑  

     

{ 1 1 2 1 1

1

1 0 0 0 1

1

1 ...

... 1 (| | | |)

n

n n n n n

n n

j n j j

j j

z z k k k k k

k

λ λ λ λ

λ

α α α α α α α α α

α α α α β β β β

− − − + −

−

−
= =

= + − −  − + − + + − + − +


+ + − + + − + + + + 


∑ ∑

     

0 0

0

1 ( 1) ( ) 2 | |
n n

n

n n j j n j

j j

z z k k
λ

α α α α α α α β
= =

        
= + − − − + + − + − +     

        
∑ ∑

 

   



 

 

A. Liman  et al. / BIBECHANA 10 (2014) 71-81 : BMHSS, p.76 (Online Publication: Dec., 2013) 

0,>
if  

0 0

0

( 1) ( | |) | | 2 | |

1

n n

n j j n j

j j

n

k

z k
λ

α α α α α α β

α
= =

 
− + + − + − + 

 + − >
∑ ∑

 

This shows that those zeros of ( )F z  for which 1z >
 
lie in 

       

0 0

0

( 1) ( | |) | | 2 | |

1 .

n n

n j j n j

j j

n

k

z k
λ

α α α α α α β

α
= =

 
− + + − + − + 

 + − ≤
∑ ∑

 

But those zeros of ( )F z  for which 1z ≤
 
already satisfy the above equation. We conclude that all 

the zeros of  ( )F z  and hence of ( )P z
 
lie in the disc 

             

0 0

0

( 1) ( | |) | | 2 | |

1

n n

n j j n j

i j

n

k

z k
λ

α α α α α α β

α
= =

 
− + + − + − + 

 + − ≤
∑ ∑

 

This completes the proof. 

 Proof of Theorem 2: The proof follows on the same lines of Theorem 1. 

Proof of Theorem 3:  Consider a polynomial
 

      
( ) (1 ) ( )F z z P z= −

 

    

1

1 1 0 0( ) ... ( )
n n

n n na z a a z a a z a
+

−= − + − + + − +

                      

               

1

1 1 0 0

1 1 0 0

{ ( ) ... ( ) }

{( ) ... ( ) }

n n

n n n

n

n n

a z z z

i z z

α α α α α

β β β β β

+
−

−

= − + − + + − +

+ − + + − +
            

              
1

1 1 1

1 1 2 1

1 2

1 1 0 0 1 2

1 1 1 0 0

{ ( 1) ( ) ( ) ... ( )

( ) ... ( ) ( 1)(

... )} {( ) ... ( ) }

n n n n

n n n n n n

n n

n n

n

n n

a z z k k z k z k z

k z z k z z

z z i z zλ

λ
λ λ

λ
λ λ

λ
λ λ

α α α α α α α

α α α α α α α

α α β β β β β+

+ − +
− − − +

− −
− − −

+ −

= − − − + − + − + + −

+ − + + − + − − +

+ + + + − + + − +  

               

1 1 2 1 1

1 2
1 1 0 01 2

1 1 0 01

1 1
( 1) ( ) ( ) ... ( )

1 1 1
( ) ... ( ) ( 1) ...

1 1
( ) ... ( )

n

n n n n n n n

n n

n n n n

n n n n

z a z k k k k
z z

k k
z z z z z z

i
z z

λ λ λ

λ
λ λ λ λ

α α α α α α α

α α α
α α α α α

β β β β β

− − − + − −

− −
− − − −

− −

= − − − + − + − + + −


 + − + + − + − − + + + + 
 

 + − + + − + 
       

                   



 

 

A. Liman  et al. / BIBECHANA 10 (2014) 71-81 : BMHSS, p.77 (Online Publication: Dec., 2013) 

1 1 2 1 1

1 1
( 1) ( ) ( ) ... ( )n

n n n n n n n
z a z k k k k

z z
λ λ λα α α α α α α− − − + − −

= − − − + − + − + + −


 

                       

1 2

1 1 0 01 2

1 1 1
( ) ... ( ) ( 1) ...n n

n n n n
k k

zz z z z z

λ
λ λ λ λ

α α α
α α α α α − −

− − − −

 
+ − + + − + − − + + + 

 
 

                

1 1 0 01

1 1
( ) ... ( )n n n n

i
z z

β β β β β− −

 + − + + − + 
 

 

       

 

Let 1z >  so that
1

1
n j
z

− < , 0,1, 2,...j =

 1 2 1

1

1 0 0 1 2 2

1 2 2

1 0 0

1 1

( ) ( 1) ...

( 1) ...

...

n n n

n n n n n

n n n

n n

n n n n

k k
F z z a z k k

z z

k
zz z z z

i
z z

λ λ
λ

α α α α
α α α

α α α α α α

β β β
β β

− − −
− −

− − −

−

− −

  − − 
≥ + − − − + + + + 

 

 −
 + + + − + + +
 
 

 −  + − + + +   
 

          

0 0( 1) ( 1) ( )
n

n n

n n j j n n

in

z a z k k
a λ

α
α α α α α α β

=

   
= + − − − + + − + − +   

   
∑

 

           
0,>
 

if 

 

0 0 ( 1) ( )

( 1)

n

n j j n n

in

n n

k

z k
a a

λ

α α α α α α β
α =

  
− + + − + − +  

  + − >
∑

 

This shows that the zeros of ( )F z , for which 1z >  lie in 

 
                   

0 0 ( 1) ( )

( 1) .

n

n j j n n

in

n n

k

z k
a a

λ

α α α α α α β
α =

  
− + + − + − +  

  + − ≤
∑

 

 

But those zeros of ( )F z  for which 1z ≤
 
already satisfy the above equation. We conclude that all 

the zeros of ( )F z  and hence of ( )P z  lie in the disc 

0 0 ( 1) ( )

( 1)

n

n j j n n

in

n n

k

z k
a a

λ

α α α α α α β
α =

  
− + + − + − +  

  + − ≤
∑

 

  



 

 

A. Liman  et al. / BIBECHANA 10 (2014) 71-81 : BMHSS, p.78 (Online Publication: Dec., 2013) 

 
Proof of Theorem 4: Consider a polynomial

 
 

( ) (1 ) ( )F z z P z= −
 

          

1

1 1 0(1 )( ... )n n

n nz a z a z a z a−
−= − + + + +

 

                     

1

1 1 0 0( ) ... ( )
n n

n n na z a a z a a z a
+

−= − + − + + − +

                

                     

1

1 1 0 0

1 1 0 0

{ ( ) ... ( ) }

{ ( ) ... ( ) }

n n

n n n

n

n n

a z z z

i z z

α α α α α

β β β β β

+
−

−

= − + − + + − +

+ + − + + − +
 

                

1

1 1 1

1 1 2 1

1 2

1 1 0 0 1 2 1

1 1 0 0

{ ( 1) ( ) ( ) ... ( )

( ) ... ( ) ( 1)( ...

)} {( ) ... ( ) }

n n n n

n n n n n n

n n

n n

n

n n n n

a z z k k z k z k z

k z z k z z z

z i z z

λ

λ
λ λ

λ
λ λ λ

λ
λ

α α α α α α α

α α α α α α α α

α µβ β µβ β β β β

+

+ − +
− − − +

− −
− − − +

−

= − − − + − + − + + −

+ − + + − + − − + + +

+ + + − − + + − +
 

            

1 1 2

1 2
1 1 0 01 2

1 1 0 01

1
( ) ( ) ( ) ...

1 1 1
( ) ( ) ( 1) ...

1 1
( ) ... ( )

n

n n n n n n n n n

n n

n n n n

n n n n

z a z k i i k k
z

k k
z z z z z z

i
z z

λ
λ λ λ λ

α α µβ β α α α α

α α α
α α α α α

µβ β β β β

− − −

− −
− − − −

− −

= − − + − + + − + − +


 + − + + − + − − + + + 
 

 + − + + − + 
   

1 1 2

1 2
1 1 0 01 2

1 1 0 01

1
( ) ( ) ( ) ...

1 1 1
( ) ... ( ) ( 1) ...

1 1
( ) ... ( )

n

n n n n n n n n

n n

n n n n

n n n n

z a z k i a k k
z

k k
z z z z z z

i
z z

λ
λ λ λ λ

α µβ α α α α

α α α
α α α α α

µβ β β β β

− − −

− −
− − − −

− −

 = − − − + + − + − + 


 + − + + − + − − + +  
  

 + − + + − +  
   

             

 

Let 1z >  so that
1

1
n j
z

− < , 0,1, 2,...j =  

    

{

}

1 1 2 1

1 0 0 1 2

1 1 0 0

( ) 1 ...

... ( 1)( ... )

...

n n n
n n n n n

n

n n

n n

k i
F z z a z k k k

a

k

λ λ

λ

α µβ
α α α α α α

α α α α α α

µβ β β β β

− − − −

− −

−

+
≥  + − − − + − + + − +

+ − + + − + + +

+ − + + − + 

 

             

0 0
1 ( 1) ( | |)

n
n n n

n n j j n n

in

k i
z a z k

a λ

α µβ
α α α α α α µβ

=

  +  
= + − − − + + − + − +    

   
∑

 

           

            
0,>
 

 



 

 

A. Liman  et al. / BIBECHANA 10 (2014) 71-81 : BMHSS, p.79 (Online Publication: Dec., 2013) 

         

0 0 ( 1) ( | |)

1

n

n j j n n

in n

n n

k
k i

z
a a

λ

α α α α α α µβ
α µβ =

  
− + + − + − +  +   + − >

∑

 

This shows that the zeros of ( )F z , for which 1z >  lie in 

 
         

0 0 ( 1) ( | |)

1 .

n

n j j n n

in n

n n

k
k i

z
a a

λ

α α α α α α µβ
α µβ =

  
− + + − + − +  +   + − ≤

∑

 

 

But those zeros of ( )F z  for which 1z ≤  already satisfy the above equation. We conclude that all the 

zeros of ( )F z and hence of ( )P z  lie in the disc. 

         

0 0 ( 1) ( | |)

1

n

n j j n n

in n

n n

k
k i

z
a a

λ

α α α α α α µβ
α µβ =

  
− + + − + − +  +   + − ≤

∑

 
   

 Proof of Theorem 5: Consider a polynomial
 

            
( ) (1 ) ( )F z z P z= −

 

          

1

1 1 0(1 )( ... )
n n

n nz a z a z a z a
−

−= − + + + +
 

         
1

1 1 0 0( ) ... ( )n n

n n na z a a z a a z a+
−= − + − + + − +

 

                     

1 1 2 1 1

1 2
1 1 0 01 2

1 1
( 1) ( ) ( ) ... ( )

1 1 1
( ) ( ) ( 1) ...

n

n n n n n n

n n

n n n n

z a z k ka a ka a ka a
z z

a a a
ka a a a a k

z z z z z z

λ λ λ

λ
λ λ λ λ

− − − + − −

− −
− − − −

= − + − + − + − + + −


 + − + − + − − + + + 
 

 Let 1z > so that
1

1
n j
z

− < , 0,1, 2,...j =    

    

{
}

1 1 2 1

1 1 0 0 1 2

( ) 1 { ...

... ( 1)(| | | | ... | |) .

n

n n n n n

n n

F z z a z k ka a ka a ka a

ka a a a a k a a a

λ λ

λ λ λ

− − − +

− − −

≥ + − − − + − + + −

+ − + + − + + − + + +
   

Using the fact (for reference see [2]) that for any two complex numbers 0b  and 1b  such that 

0 1 ,b b≥ and   arg
2

ja
π

β α− ≤ ≤  , 0,1, 2j = and β  real 

0 1 0 1 0 1(| | | |) cos (| | | |) sin .b b b b b bα α− ≤ − + +
 

 

 

 

 

 

 

 



 

 

A. Liman  et al. / BIBECHANA 10 (2014) 71-81 : BMHSS, p.80 (Online Publication: Dec., 2013) 

     

[ { 1 1

1 2 1 2

1 1 1 2 1

2 1 0 1 0

( ) | | | 1 | cos cos sin sin

cos cos sin sin ... cos

cos sin sin ... cos cos sin

sin ... cos cos sin sin

n

n n n n n

n n n n

F z z a z k k a a k a a

k a a k a a k a

a k a a a a a

a a a a a

λ

λ λ λ λ λ λ

λ

α α α α

α α α α α

α α α α α α

α α α α α

− −

− − − −

− − − − −

−

≥ + − − − + +

+ − + + + +

− + + + + − +

+ + + − + +
1

0... ( 1) | |
n

i

i

a k a
λ

−

=


+ + + − 


∑

        

[ { 1 1

1 2

1

1 0 0 0

| | | 1 | (cos sin ) ( 1) cos ( 1) sin

... ( 1) cos ( 1) sin 2 sin 2 sin ...

2 sin cos sin ( 1) | |

n

n n n n

n

i

i

z a z k k a k a k a

k a k a a a

a a a a k a

λ λ λ λ

λ

α α α α

α α α α

α α α

− −

− −

−

=

= + − − + + − + +

+ + − + + + + +


+ − + + + − 


∑            

    

[
1

1

1

1

1 0 0

1 1

| | | 1 | { (cos sin ) ( 1) cos | |

( 1)sin 2sin (sin cos )

n n
n

n n j i

i i

n

j j

i j

z a z k k a k a a

k a a a a

λ λ

λ λ

α α α

α α α α

−

−
= + =

−
= + = −

 
= + − − + + − + 

 


+ + + + − + 



∑ ∑

∑ ∑  

      

{
1

1 1

0

1

1 (cos sin ) (( 1)cos ( 1)sin )

2sin (1 sin cos ) ( 1) | |

n
n

n n j

i

n

j i

j i

z a z k k a k k a

a a k a

λ

λ λ

α α α α

α α α

= +

−

= − =

=  + − − + + − + +


+ + + − + − 



∑

∑ ∑
 

 

         

{
1

1 1

1

1 (cos sin ) (( 1) cos ( 1)sin )

2sin ( 1) | |

n
n

n n j

i

n

j i

j i

z a z k k a k k a

a k a

λ

λ λ

α α α α

α

= +

−

= − =


≥ + − − + + − + +




+ + − 



∑

∑ ∑
 

         
0,>
 

if  

    
1 1

1 1

1 2sin
1 (cos sin ) {( 1) cos ( 1)sin } ( 1) | |

n n

j j i

i j in n

z k k k k a a k a
a aλ λ λ

α
α α α α

−

= + = − =

+ − > + + − + + + + −∑ ∑ ∑

 

This shows that the zeros of ( )F z , for which 1z >  lie in  

 

 

 



 

 

A. Liman  et al. / BIBECHANA 10 (2014) 71-81 : BMHSS, p.81 (Online Publication: Dec., 2013) 

1 1

1 1

1 2sin
1 (cos sin ) {( 1) cos ( 1)sin } ( 1) | |.

n n

j j i

i j in n

z k k k k a a k a
a aλ λ λ

α
α α α α

−

= + = − =

+ − ≤ + + − + + + + −∑ ∑ ∑

But those zeros of ( )F z  for which 1z ≤
 
already satisfy the above equation the above  

equation. We conclude that all the zeros of ( )F z  and hence of ( )P z
 
lie in the disc 

 
1 1

1 1

1 2sin
1 (cos sin ) {( 1) cos ( 1)sin } ( 1) | |.

n n

j j i

i j in n

z k k k k a a k a
a aλ λ λ

α
α α α α

−

= + = − =

+ − ≤ + + − + + + + −∑ ∑ ∑

References 

1.   M. Marden, Geometry of Polynomials, Math. Surveys, No. 3; Amer. Math. Soc.     

             Providence R.I., (1966). 

2.   Q. I. Rahman and G. Schmeisser, Analytic Theory of Polynomials. Oxford Univ.  Press,     

      (2002). 
3.   W.M Shah and A. Liman, On Eneström Kakeya Theorem and related Analytic   Functions, Proc.   

      Indian Acad. Sci.( Math Sci.) vol.117,No. 3,(2007) 359-370. 

4. A. Joyal, G. Labelle and Q.I-Rahman, On the location of zeros of polynomials,  Canad.  Math. 
Bull.,10(1967) 53-63.  

5.  A.Aziz and B.A Zargar, Some extension of Eneström-Kakeya theorem,Glasnik Matematicki, 

31(1996) 239-244.