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1. International Journal of Engineering Research and Development
e-ISSN: 2278-067X, p-ISSN: 2278-800X, www.ijerd.com
Volume 10, Issue 6 (June 2014), PP.60-69
Zeros of Polynomials in Ring-shaped Regions
M. H. Gulzar
Department of Mathematics University of Kashmir, Srinagar 190006
Abstract:- In this paper we subject the coefficients of a polynomial and their real and imaginary parts to certain
conditions and give bounds for the number of zeros in a ring-shaped region. Our results generalize many
previously known results and imply a number of new results as well.
Mathematics Subject Classification: 30 C 10, 30 C 15
Keywords and Phrases:- Coefficient, Polynomial, Zero
I. INTRODUCTION AND STATEMENT OF RESULTS
A large number of research papers have been published so far on the location in the complex plane of
some or all of the zeros of a polynomial in terms of the coefficients of the polynomial or their real and
imaginary parts. The famous Enestrom-Kakeya Theorem [5] states that if the coefficients of the polynomial
( ) satisfy n n  a  a   a  a 0 1 1 0 ...... , then all the zeros of P(z) lie in the closed disk
( ) be a polynomial such that n n  a  a   a  a 0 1 1 0 ...... . Then the
( ) be a polynomial such that

arg a j n j    
z  does not exceed
n  j
a a
( ) b e a polynomial of degree n with j j Re(a )  ,
60
j
n

P z 
a z j j
0
z  1.
By putting a restriction on the coefficients of a polynomial similar to that of the Enestrom-Kakeya Theorem,
Mohammad [6] proved the following result:
Theorem A: Let
j
n

P z 
a z j j
0
number of zeros of P(z) in
1
z  does not exceed
2
an  .
0
log
1
log 2
1
a
For polynomials with complex coefficients, Dewan [1] proved the following results:
Theorem B: Let
j
n

P z 
a z j j
0
, 0,1,2,....., ,
2
 
for some real  and  and
n n  a  a   a  a 0 1 1 0 ...... .
Then the number of zeros of P(z) in
1
2
0
1
0
(cos sin 1) 2sin
log
1
log 2
a
n
j


     
.
Theorem C: Let
j
n

P z 
a z j j
0
a j n j j Im( )   ,  0,1,2,......, such that
n n      0 1 1 0 ...... .

2. Zeros of Polynomials in Ring-shaped Regions
z  does not exceed
( ) be a polynomial of degree n such that for some k 1,o  1 and for
n n n j M a k
1 0 0 0 0  2( 1)  (  )    2   .
( ) be a polynomial of degree n such that for some
    and for some k 1,o  1 and some integer
( ) be a polynomial of degree n with j j Re(a )  ,
n j        .
( ) be a polynomial of degree n such that for some 1, 1,0 1 1 2 k  k    ,
61
Then the number of zeros of P(z) in
1
2
j n 
0 log
0
1
log 2
1
a
n
j


 
.
Recently Gulzar [3,4] proved the following results:

Theorem D: Let 
n
j
j
j P z a z
0
some integer ,0   n 1,
1 1 1 1 0   ......    ......                 k n n .
Then P(z) has no zero in
0
M
1
a
z  , where

         
n
j
1

Theorem E: Let 
n
j
j
j P z a z
0

real , ; arg a    , j 
0,1,2,......,
n j 2
,0   n 1,
1 1 1 1 0 a a ...... a k a a ...... a a n n                .
Then P(z) has no zeros in
0
M
2
a
z  , where
2 0 0 M a (cos sin 1) 2 a (k k sin 1) a (cos sin 1) a n                


 

1
1,
2sin
n
j j
j a

 .

Theorem F: Let 
n
j
j
j P z a z
0
a j n j j Im( )   ,  0,1,......, such that for some 1, 1,0 1 1 2 k  k    ,
1 2 1 2 1 0     ......  n n n k k
Then P(z) has no zero in
0
M
3
a
z  , where
( ) ( ) ( ) 3 1 2 1 1 2 1 1 1         n n n n n n n n M a k  k  k  k    



     
1
1
0 0 0 0 ( ) 2
n
j

Theorem G: Let 
n
j
j
j P z a z
0
1 2 1 2 1 0 k a k a a ...... a a n n n       .

3. Zeros of Polynomials in Ring-shaped Regions
j  a   a  a .
( ) be a polynomial of degree n such that for some k 1,o  1 and for
z does not exceed
                  
1 [ 2( 1) ( ) 2 ]
n n j
                  
1 [ 2( 1) ( ) 2 ]
n n j
( ) be a polynomial of degree n such that for some k 1,o  1 and for
R
0   R  c 
                  
1 [ 2( 1) ( ) 2 ]
n n j
                  
1 [ 2( 1) ( ) 2 ]
n n j
n n n j M a k
1 0 0 0 0  2( 1)  (  )    2   .
( ) be a polynomial of degree n such that for some
62
Then P(z) has no zero in
0
M
4
a
z  , where
(cos sin 1) (sin cos 1) (1 cos ) 4 1 2 1 1              n n n M k a k a a



    
2
1
0 0 (cos sin 1) 2sin
n
j
In this paper, we prove the following results:

Theorem 1: Let 
n
j
j
j P z a z
0
some integer ,0   n 1,
1 1 1 1 0   ......    ......                 k n n .
R
Then the number of zeros of P(z) in  (R  0, c 1)
c
0
1
0 0 0 0 0
log
1
log
a
a R a R k
c
n
j
n n
n 
for R 1 and
0
1
0 0 0 0 0
log
1
log
a
a R a R k
c
n
j
n
n 
for R 1.
Combining Theorem 1 and Theorem D, we get a bound for the number of zeros of P(z) in a ring-shaped
region as follows:

Corollary 1: Let 
n
j
j
j P z a z
0
some integer ,0   n 1,
1 1 1 1 0   ......    ......                 k n n .
a
Then the number of zeros of P(z) in ( 0, 1)
1
c
z
M
does not exceed
0
1
0 0 0 0 0
log
1
log
a
a R a R k
c
n
j
n n
n 
for R 1 and
0
1
0 0 0 0 0
log
1
log
a
a R a R k
c
n
j
n
n 
for R 1, where

         
n
j
1

Theorem 2: Let 
n
j
j
j P z a z
0

4. Zeros of Polynomials in Ring-shaped Regions
    and for some k 1,o  1 and some integer
z does not exceed
a R a R a a k k
     
{ (cos  sin  ) 2 ( sin 
1)
a R a R a a k k
     
{ (cos  sin  ) 2 ( sin 
1)
( ) be a polynomial of degree n such that for some
    and for some k 1,o  1 and some integer
R
0   R  c 
a R a R a a k k
     
{ (cos  sin  ) 2 ( sin 
1)
a R a R a a k k
     
{ (cos  sin  ) 2 ( sin 
1)
( ) be a polynomial of degree n with j j Re(a )  ,
z does not exceed
63

real , ; arg a    , j 
0,1,2,......,
n j 2
,0   n 1,
1 1 1 1 0 a a ...... a k a a ...... a a n n                .
R
Then the number of zeros of P(z) in  (R  0, c 1)
c










0
1
    



 

(cos sin 1) 2sin }
1
log
1
log
1
1,
0 0
0
n
j j
j
n
n n
n
a a a
c a

   
for R 1
and










0
1
    



 

(cos sin 1) 2sin }
1
log
1
log
1
1,
0 0
0
n
j j
j
n
n n
n
a a a
c a

   
for R 1.
Combining Theorem 2 and Theorem E, we get a bound for the number of zeros of P(z) in a ring-shaped
region as follows:

Corollary 2: Let 
n
j
j
j P z a z
0

real , ; arg a    , j 
0,1,2,......,
n j 2
,0   n 1,
1 1 1 1 0 a a ...... a k a a ...... a a n n                .
a
Then the number of zeros of P(z) in ( 0, 1)
2
c
z
M
does not exceed










0
1
    



 

(cos sin 1) 2sin }
1
log
1
log
1
1,
0 0
0
n
j j
j
n
n n
n
a a a
c a

   
for R 1
and










0
1
    



 

(cos sin 1) 2sin }
1
log
1
log
1
1,
0 0
0
n
j j
j
n
n n
n
a a a
c a

   
for R 1,
where
2 0 0 M a (cos sin 1) 2 a (k k sin 1) a (cos sin 1) a n                


 

1
1,
2sin
n
j j
j a

 .

Theorem 3: Let 
n
j
j
j P z a z
0
a j n j j Im( )   ,  0,1,......, such that for some 1, 1,0 1 1 2 k  k    ,
1 2 1 2 1 0     ......  n n n k k
R
Then the number of zeros of P(z) in  (R  0, c 1)
c

5. Zeros of Polynomials in Ring-shaped Regions
      
{(   ) (   )  (  
)
n n n n n n n
1 2  1 1 2  1  1 
1
n j
      
{(   ) (   )  (  
)
n n n n n n n
1 2  1 1 2  1  1 
1
n j
( ) be a polynomial of degree n with j j Re(a )  ,
R
0   R  c 
      
{(   ) (   )  (  
)
n n n n n n n
1 2  1 1 2  1  1 
1
n j
      
{(   ) (   )  (  
)
n n n n n n n
1 2  1 1 2  1  1 
1
n j
n j        .
( ) be a polynomial of degree n such that for some 1, 1,0 1 1 2 k  k    ,
z does not exceed
a R  a  R { k a (cos   sin   1)  k a (sin   cos 
 1)

a
n n n
0 1 2 1
c a      
n j
64










1
     




( ) 2 }
1
log
1
log
1
1
0 0 0 0
0
n
j
n n
n a R R k k k k
c a       
for R 1 and










1
     




( ) 2 }
1
log
1
log
1
1
0 0 0 0
0
n
j
n
n a R R k k k k
c a       
for R 1 .
Combining Theorem 3 and Theorem F, we get a bound for the number of zeros of P(z) in a ring-shaped
region as follows:

Corollary 3: Let 
n
j
j
j P z a z
0
a j n j j Im( )   ,  0,1,......, such that for some 1, 1,0 1 1 2 k  k    ,
1 2 1 2 1 0     ......  n n n k k
a
Then the number of zeros of P(z) in ( 0, 1)
3
c
z
M
does not exceed










1
     




( ) 2 }
1
log
1
log
1
1
0 0 0 0
0
n
j
n n
n a R R k k k k
c a       
for R 1 and










1
     




( ) 2 }
1
log
1
log
1
1
0 0 0 0
0
n
j
n
n a R R k k k k
c a       
for R 1 , where
( ) ( ) ( ) 3 1 2 1 1 2 1 1 1         n n n n n n n n M a k  k  k  k    



     
1
1
0 0 0 0 ( ) 2
n
j

Theorem 4: Let 
n
j
j
j P z a z
0
1 2 1 2 1 0 k a k a a ...... a a n n n       .
R
Then the number of zeros of P(z) in  (R  0, c 1)
c










1
      






(1 cos sin ) (cos sin 1) 2sin }
1
log
1
log
2
1
1 0
0
n
j
n n
n
a a a
for R 1 and

6. Zeros of Polynomials in Ring-shaped Regions
a R  a  R { k a (cos   sin   1)  k a (sin   cos 
 1)

a
n n n
0 1 2 1
c a      
n j
( ) be a polynomial of degree n such that for some 1, 1,0 1 1 2 k  k    ,
R
0   R  c 
a R  a  R { k a (cos   sin   1)  k a (sin   cos 
 1)

a
n n n
0 1 2 1
c a      
n j
a R  a  R { k a (cos   sin   1)  k a (sin   cos 
 1)

a
n n n
0 1 2 1
c a      
n j

arg z    j  j
R
z  does not exceed
1 (1 )( ...... ...... 

65










1
      






(1 cos sin ) (cos sin 1) 2sin }
1
log
1
log
2
1
1 0
0
n
j
n
n
a a a
for R 1.
Combining Theorem 4 and Theorem G, we get a bound for the number of zeros of P(z) in a ring-shaped
region as follows:

Corollary 4: Let 
n
j
j
j P z a z
0
1 2 1 2 1 0 k a k a a ...... a a n n n       .
a
Then the number of zeros of P(z) in ( 0, 1)
4
c
z
M
does not exceed










1
      






(1 cos sin ) (cos sin 1) 2sin }
1
log
1
log
2
1
1 0
0
n
j
n n
n
a a a
for R 1 and










1
      






(1 cos sin ) (cos sin 1) 2sin }
1
log
1
log
2
1
1 0
0
n
j
n
n
a a a
for R 1, where
(cos sin 1) (sin cos 1) (1 cos ) 4 1 2 1 1              n n n M k a k a a
0 0  a (cos sin 1)  a .
For different values of the parameters in the above results, we get many interesting results including
generalizations of some well-known results.
II. LEMMAS
For the proofs of the above results, we need the following results:
Lemma 1: Let z z C 1 2 , with 1 2 z  z and , 1,2
  for some real numbers
2
 and  . Then
( )cos ( )sin 1 2 1 2 1 2 z  z  z  z  z  z .
The above lemma is due to Govil and Rahman [2].
Lemma 2: Let F(z) be analytic in z  R , F(z)  M for z  R and F(0)  0 . Then
for c>1, the number of zeros of F(z) in the disk
c
M
0
log
1
log
a
c
.
For the proof of this lemma see [7].
III. PROOFS OF THEOREMS
Proof of Theorem 1: Consider the polynomial
F(z)=(1-z)P(z)
n
n
n
n
n
n
n
  z a z  a z 
1
  a z 

1
 a z 
 a z 
1
  a z 
1
 a z n 


1


1

1

7. Zeros of Polynomials in Ring-shaped Regions
  a z a a z a a z a a z a a z n
n n
           
1 2  (a  a )z ...... (a  a )z  a 
 
 n
n
1        
1

     k k  z k 
       

             1 

2  
1 
F z  a R     R     R n
    k R 
 k   R n n
         
k R k R R R
   (  1)    ......
 

R R a
j j     
n n
 a R  1  a  R       k  k  n n  n  n
 
        
  ( 1)    ......
        

1 1 2 1 0
           
1 
n n j
n
 F z a R a R k k n n n n
1               
        
  ( 1)    ......
        

1 1 2 1 0
           
1 
n n j
                  
1 [ 2( 1) ( ) 2 ]
n n j
                  
1 [ 2( 1) ( ) 2 ]
n n j
66

 

n
n n
n
1
( ) ( ) ...... ( ) ( ) n 1
1
1
1 1 2
1




  
1 0 0
1

 
( ) ( ) ...... 1
n n
  
1 1 2
n n
n
a z   z   z
n  ......  {(  )  (  )} 
 {(  )   1  

1
1

(k )}z ( )z ...... {( ) ( )}z 1 0 0 0
0
1 i ( )z a j
 
  j j 1
n
j
  
For z  R , we have, by using the hypothesis,
1
( ) ....... ( 1) 1
1
1
1 1 2


  


n
n n
n
n
0
0 1
1
1 0
1
1 1 2
(1 ) ( ).
j
n
j


  
   
 



 
   [    .......   ( 1) 0 1 1 2 1
n
n

(1 ) ( )]
0 1
1

   
j
j j
k k
   
[ 2( 1) ( ) 2 ]
1
0 0 0 0 0
n
j
n n
n a R a R  k         
for R 1
and
( ) [    .......   ( 1) n
0 1 1 2  1
  n

(1 ) ( )]
0 1
1

   
j
j j
k k
   
[ 2( 1) ( ) 2 ]
1
0 0 0 0 0
n
j
n
n a R a R  k         
for R 1.
R
Hence, by Lemma 2, it follows that the number of zeros of F(z) in  , c 1
c
z
does not exceed
0
1
0 0 0 0 0
log
1
log
a
a R a R k
c
n
j
n n
n 
for R 1 and
0
1
0 0 0 0 0
log
1
log
a
a R a R k
c
n
j
n
n 

8. Zeros of Polynomials in Ring-shaped Regions
for R 1.
Since the zeros of P(z) are also the zeros of F(z), the proof of Theorem 1is complete.
Proof of Theorem 2: Consider the polynomial
F(z)=(1-z)P(z)
1 (1 )( ...... ...... 

  a z a a z a a z a a z a a z n
n n
           
1 2  (a  a )z ...... (a  a )z  a 
 
  F z a R a a R a a R a ka R k a R n
n n
( )   1      .......     (  1) 

         
ka a R k a R a a R a a R
         n n n n n n
a a a k a a k a n n
      
( ) sin ...... ( ) cos ( ) sin
  
   
1 2 1 1
k k k k
       
( 1) ( ) cos ( ) sin ( 1)
       
   1   1

a a a a a a
      
( ) cos ( ) sin ...... ( ) cos
   
   
   
1 2 1 2 1 0
n n
1             a R a R a a k k n
n
1             F z a R a R a a k k n
a R a R a a k k
     
{ (cos  sin  ) 2 ( sin 
1)
a R a R a a k k
     
{ (cos  sin  ) 2 ( sin 
1)
F z   z P z   z a z  a z n

  a z  a n n

1 1 a z (a a )z (a a )z ...... (a a )z a n
67
n
n
n
n
n
n
n
  z a z  a z 
1
  a z 

1
 a z 
 a z 
1
  a z 
1
 a z n 


1


1

1

 

n
n n
n
1
( ) ( ) ...... ( ) ( ) n 1
1
1
1 1 2
1




  
1 0 0
1

 
For z  R , we have, by using the hypothesis and Lemma 1,
1 1
1
1
1 1 2


  


n
n n
n
n
a R
0
1 0
1
1 1 2
(1 )
( 1) ......

 
 



 
 
  
[( ) cos ( ) sin ( ) cos 0 1 1 1 2
1
   
n n
n a R a R a a a a a a
a a a
(  ) sin  (1 
) ]
1 0 0
   
   
 
[ (cos sin ) 2 ( sin 1) 0
n

1
, 1 
0 0  a (cos  sin 1)  a 2sin ]
 

n
j j
j a


for R 1
and
( ) [ (cos sin ) 2 ( sin 1) n
0

1
, 1 
0 0  a (cos  sin 1)  a 2sin ]
 

n
j j
j a


for R 1 .
R
Hence, by Lemma 2, it follows that the number of zeros of F(z) in  , c 1
c
z
does not exceed










0
1
    



 

(cos sin 1) 2sin }
1
log
1
log
1
1,
0 0
0
n
j j
j
n
n n
n
a a a
c a

   
for R 1
and










0
1
    



 

(cos sin 1) 2sin }
1
log
1
log
1
1,
0 0
0
n
j j
j
n
n n
n
a a a
c a

   
for R 1.
Since the zeros of P(z) are also the zeros of F(z), Theorem 2 follows.
Proof of Theorem 3: Consider the polynomial
1
n
( ) (1 ) ( ) (1 )( ...... ) 1 0
1 0 0
1
n n
n
1 1 2
n n
n
  
    
    n   

9. Zeros of Polynomials in Ring-shaped Regions
         n
  n
1 {( ) ( ) ( )} ( ) 
n n n n n n
        
...... {(   ) (   )} {(  
) ......
1 0 0 0 1
 n
n
n
n
1           
1 0 0 1 R (1 ) R R a j
 
          
 j j  1              
           
j j
1 0 0 
         n n n n n n n

0 0 0 0 
n j       
         n n n n n n n

0 0 0 0 
n j       
      
{(   ) (   )  (  
)
n n n n n n n
1 2  1 1 2  1  1 
1
n j
      
{(   ) (   )  (  
)
n n n n n n n
1 2  1 1 2  1  1 
1
n j
F z   z P z   z a z  a z n

  a z  a n n

1 1 a z (a a )z (a a )z ...... (a a )z a n
         n
  n
1 {( ) ( ) ( )} ( ) 
n n n n n n
68
1
n n
1 2  1 1 2  1  1  1 
2
n
n a z k k  k  k   z   z
z a
( ) }
1 0 0
z i z n
n n
   

For z  R , we have, by using the hypothesis,
( ) ( 1) ( 1) ...... 1
n n
n
n
1 2  1 1 2  1  1 
2
n n
n
n F z a R k  k  R k  R k  R   R
0
1
n
j
[ ( 1) ( 1) ..... 1 2 1 1 2 1 1 2

n n n n n n
n n
n a R R k  k  k  k   
(1 ) ( )] 1

1
n
j
[( ) ( ) ( ) 1 2 1 1 2 1 1 1
1
   
n n
n a R R k  k  k  k    
1
( ) 2 ]

1
     
n
j
for R 1
and
1
( ) [( ) ( ) ( ) 1 2  1 1 2  1  1 
1
n
n F z a R R k  k  k  k    
1
( ) 2 ]

1
     
n
j
for R 1 .
R
Hence, by Lemma 2, it follows that the number of zeros of F(z) in  , c 1
c
z
does not exceed










1
     




( ) 2 }
1
log
1
log
1
1
0 0 0 0
0
n
j
n n
n a R R k k k k
c a       
for R 1 and










1
     




( ) 2 }
1
log
1
log
1
1
0 0 0 0
0
n
j
n
n a R R k k k k
c a       
for R 1 .
Since the zeros of P(z) are also the zeros of F(z), Theorem 3 follows.
Proof of Theorem 4: Consider the polynomial
1
n
( ) (1 ) ( ) (1 )( ...... ) 1 0
1 0 0
1
n n
n
1 1 2
n n
n
  
    
    n   
1
n n
1 2  1 1 2  1  1  1 
2
n
n a z k a k a k a a k a a z a a z
1 0 0 0 0  ...... {(a a )  (a  a )}z  a
For z  R , we have, by using the hypothesis and Lemma 1,

10. Zeros of Polynomials in Ring-shaped Regions
 n
( )  1    (  1)  (  1)   
1
 ...... 1 2  1 1 2  1  1 
2
  1       
   
k a a a a an n n n n
          
( 1) ( ) cos ( ) sin
a a a a a
 
2 1 1 2 1 2
( ) cos ( ) sin (1 )
     
    
 1 [ (cos  sin  1) (sin  cos 
1) 0 1 2 1
1 0 
n j a    a    a
( )  1 [ (cos  sin  1) (sin  cos 
1) 0 1 2 1
1 0 
n j a    a    a
a R  a  R { k a (cos   sin   1)  k a (sin   cos 
 1)

a
n n n
0 1 2 1
c a      
n j
a R  a  R { k a (cos   sin   1)  k a (sin   cos 
 1)

a
n n n
0 1 2 1
c a      
n j
69
n n
n
n
n
n
n
n n
n
n F z a R k a k a R k a R k a R a a R
1 0 0 0  a a R  (1 ) a R  a
n n n n n
n n
n a R a R [(k a k a ) cos (k a k a ) sin (k 1) a 0 1 2 1 1 2 1 1
1 0 1 0 0
n n n
n n
n  a R  a  R k a    k a    a 
2
(1 cos sin ) (cos sin 1) 2sin ]
1


       
n
j
for R 1
and
n n n
n
n F z  a R  a  R k a    k a    a 
2
(1 cos sin ) (cos sin 1) 2sin ]
1


       
n
j
for R 1 .
R
Hence, by Lemma 2, it follows that the number of zeros of F(z) in  , c 1
c
z
does not exceed










1
      






(1 cos sin ) (cos sin 1) 2sin }
1
log
1
log
2
1
1 0
0
n
j
n n
n
a a a
for R 1 and










1
      






(1 cos sin ) (cos sin 1) 2sin }
1
log
1
log
2
1
1 0
0
n
j
n
n
a a a
for R 1.
Since the zeros of P(z) are also the zeros of F(z), Theorem 4 follows.
REFERENCES
[1]. K.K. Dewan, Extremal Properties and Coefficient Estimates for Polynomials with Restricted
Coefficients and on Location of Zeros of Polynomials,Ph. Thesis, IIT Delhi, 1980.
[2]. N. K. Govil and Q. I. Rahman, On the Enestrom-Kakeya Theorem, Tohoku Mathematics Journal,
20(1968), 136.
[3]. M. H. Gulzar, The Number of Zeros of a Polynomial in a Disk,International Journal of Innovative
Science, Engineering & Technology, Vol.1, Issue 2,April 2014, 37-46.
[4]. M. H. Gulzar, Zeros of Polynomials under Certain Coefficient Conditions, Int. Journal of Advance
Foundation and Research in Science and Technology (to appear) .
[5] M. Marden, , Geometry of Polynomials,Math. Surveys No. 3, Amer. Math Society(R.I. Providence)
(1966).
[6] Q. G. Mohammad, On the Zeros of Polynomials, American Mathematical Monthly, 72 (6) (1965), 631-
633.
[7] E. C. Titchmarsh, The Theory of Functions, 2nd Edition, Oxford University Press,London, 1939.