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1. International Journal of Engineering Inventions
e-ISSN: 2278-7461, p-ISSN: 2319-6491
Volume 2, Issue 11 (July 2013) PP: 54-59
www.ijeijournal.com Page | 54
On the Zeros of Complex Polynomials
M. H. Gulzar
Department of Mathematics, University of Kashmir, Srinagar
Abstract: In the framework of the Enestrom-Kakeya Theorem various results have been proved on the
location of zeros of complex polynomials. In this paper we give some new results on the zeros of complex
polynomials by restricting the real and imaginary parts of their coefficients to certain conitions.
Mathematics Subject Classification: 30C10, 30C15
Key-words and Phrases: Coefficients, Polynomials, Zeros
I. Introduction and Statement of Results
The following result known as the Enestrom-Kakeya Theorem [10] is well-known in the theory of
distribution of zeros of polynomials:
Theorem A: Let 

n
j
j
j zazP
0
)( be a polynomial of degree n such that its coefficients satisfy
0... 011   aaaa nn .
Then all the zeros of P(z) lie in the closed unit disk 1z .
In the literature [1-9, 12] there exist several generalisations and extensions of this result .
Recently Y. Choo [3] proved the following results:
Theorem B: Let 

n
j
j
j zazP
0
)( be a polynomial of degree n with
njaa jjjj ,....,2,1,0,)Im(,)Re(   , such that for some  and  , and for some 1k , 2k ,
011111 ......    nnk
011112 ......    nnk .
Then P(z) has all its zeros in 21 RzR  where
1
0
1
M
a
R  and
na
M
R 2
2 
with
)(2)1()1( 211    nnn kkaM )()( 001   nnk
and
)(2)1()1( 212    nn kkM 00021 )()( akk nn   .
M. H. Gulzar [8] made an improvement on the above result by proving the following result:
Theorem C: Let 

n
j
j
j zazP
0
)( be a polynomial of degree n with
njaa jjjj ,....,2,1,0,)Im(,)Re(   , such that for some  and  , and for some 1k , 2k , 21, ,
011111 ......    nnk
021112 ......    nnk .
Then P(z) has all its zeros in 43 RzR  where
3
0
3
M
a
R  and
na
M
R 4
4 

2. On the Zeros of Complex Polynomials
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with
)()(2)1()1( 21213 nnnnn kkkkaM   
02020101 )1()1(  
and
)()(2)1()1( 21214 nnnn kkkkM   
002020101 )1()1( a  .
The aim of this paper is to find a ring-shaped region between two concentric circles with centre not necessarily
on the origin. More precisely, we prove the following results:
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  and  , and for some 1k , 2k , 21, ,
011111 ......    nnk
021112 ......    nnk .
Then P(z) has all its zeros in 615 RzR   where
n
n
a
k 

)1( 1
1

 ,
n
n
a
k
M
a
R
)1( 1
5
0
5

 and
na
M
R 6
6 
with
)()(2)1()1( 21215 nnnnn kkkkaM   
02020101 )1()1(  
and
nnn kkkM   )1()()(2 2216 
002020101 )1()1( a  .
Theorem 2: Let 

n
j
j
j zazP
0
)( be a polynomial of degree n with
njaa jjjj ,....,2,1,0,)Im(,)Re(   , such that for some  and  , and for some 1k , 2k , 21, ,
011111 ......    nnk
021112 ......    nnk .
Then P(z) has all its zeros in 827 RzR   where
n
n
a
ki 

)1( 2
2

 ,
n
n
a
k
M
a
R
)1( 2
7
0
7

 and
na
M
R 8
8 
with
)()(2)1()1( 21217 nnnnn kkkkaM   
02020101 )1()1(  
and
nnn kkkM   )1()()(2 1218 
002020101 )1()1( a  .
Remark 1: The bounds for the zeros of P(z) in both Theorem 1 and Theorem 2 are easily seen to be sharper
than those of Theorems B and C. For different values of the parameters 2121 ,,, kk , we get many other
interesting results. For example for

3. On the Zeros of Complex Polynomials
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11 k , 12  , in Theorem 2 ,we get a result due to B. L. Raina et al [11].
For 1,1 21   , Theorem 1 reduces to Theorem B.
For 11 k , 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  and  , and for some 21, ,
01111 ......    nn
02111 ......    nn .
Then P(z) has all its zeros in 109 RzR  where
9
0
9
M
a
R  and
na
M
R 10
10 
with
)()(29 nnnaM    02020101 )1()1(  
and
)()(210 nnM    002020101 )1()1( a  .
II. Proofs of Theorems
Proof of Theorem 1: Consider the polynomial
)()1()( zPzzF 
......)()(
)......)(1(
1
211
1
01
1
1







n
nn
n
nn
n
n
n
n
n
n
zaazaaza
azazazaz
001
2
12 )()( azaazaa 
......)()1()( 1
21111
1
 

 n
nn
n
n
n
nn
n
n zzkzkza 
})1()(......)(
)1(){()1()(
02021
1
21
212001011
zzz
zkzkiazz
n
nn
n
n
n
nn







......){(})1([{ 21
111 

 

z
kkzaz nn
nnnn
n 

}]
)1(
......
)1(
){(}
)1(
02
1
02121
2
12
0
1
01
1
011
nn
nn
n
nnnnn
zzz
z
k
ki
z
a
zz



















For 1z , we have nj
z
jn
,......,1,0,1
1

and, therefore ,
....{)1([)(
21
111 




z
kkzazzF
nn
nnnn
n 


4. On the Zeros of Complex Polynomials
www.ijeijournal.com Page | 57
}]
)1(
......
)1()1(
02
1
02121
2
12
0
1
01
1
011
nn
nn
n
nnnnn
zzz
z
k
k
z
a
zz



















01121111 ......{)1([    nnnnnn
n
kkzaz
21212001 )1()1(   nnnnn kka 
}])1( 02021  
)(......)(){()1([ 112111     nnnnnn
n
kkzaz
}])1()(
......)()(......)()1(
)()1()(......)(
02021
11122
210010111










nnn
nn
k
ka
nnnnn
n
kkkkzaz   )1()()(2{)1([ 2211 
}])1()1( 002010201 a 
0
if
nnnnn kkkkza   )1()()(2{)1( 2211 
})1()1( 002010201 a 
or
nnn
nn
n
kkk
aa
k
z 

 )1()()(2[{
1)1(
221
1



This shows that those zeros of F(z) and hence P(z) whose modulus is greater than 1 lie in
nnn
nn
n
kkk
aa
k
z 

 )1()()(2[{
1)1(
221
1



}])1()1( 02010201   .
Since the zeros of P(z) of modulus less than or equal to 1 already satisfy the above inequality, it follows that all
the zeros of P(z) lie in
61 Rz  .
To prove the other half of the theorem, we have
......)()1()()( 1
21111
1
 

 n
nn
n
n
n
nn
n
n zzkzkzazF 
002021
1
21
21201011
})1()(......)(
)1(){()1()(
azzz
zkzkizz
n
nn
n
n
n
nn







0)( azQ  ,
where
......)()1()()( 1
21111
1
 

 n
nn
n
n
n
nn
n
n zzkzkzazQ 

5. On the Zeros of Complex Polynomials
www.ijeijournal.com Page | 58
})1()(......)(
)1(){()1()(
02021
1
21
21201011
zzz
zkzkizz
n
nn
n
n
n
nn







For 1z ,
)-()-(......)()k-()1()( 1112n11-n1     nnnn kzazQ
0202111
1221201011
)1()(......)-()-(......
)()()1()1()(......


 


 nnnnn kk
nnnnn kkkkza   )1()()(2)1( 2211 
02010201 )1()1(   .
Rkza nn  )1( 1
51 )1( MRka nn   ,
where
nnn kkkR   )1()()(2 221 
02020101 )1()1(  
Since Q(z) is analytic and Q(0)=0, it follows by Rouche’s theorem that
zMzQ 5)(  for 1z .
Therefore , for 1z ,
)()( 0 zQazF 
)(0 zQa 
zMa 50 
0
if
5
0
M
a
z  .
This shows that F(z) does not vanish in
5
0
M
a
z  .It is easy to see that 05 aM  . In other words, it follows
that F(z) and hence P(z) has all its zeros in z
M
a

5
0
.
Since
1111   zzz ,
we have
11
5
0
  zz
M
a
.
Therefore
11
5
0
  z
M
a
i.e.

6. On the Zeros of Complex Polynomials
www.ijeijournal.com Page | 59
1
1
5
0 )1(




 z
a
k
M
a
n
n
.
Hence, it follows that P(z) has all its zeros in
1
1
5
0 )1(




 z
a
k
M
a
n
n
.
That completes the proof of Theorem 1.
Proof of Theorem 2: Similar to that of Theorem 2.
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