Zeros of a Polynomial in a given Circle

International Journal of Engineering Inventions
e-ISSN: 2278-7461, p-ISSN: 2319-6491
Volume 3, Issue 2 (September 2013) PP: 38-46
www.ijeijournal.com Page | 38
Zeros of a Polynomial in a given Circle
M. H. Gulzar
Department of Mathematics, University of Kashmir, Srinagar 190006
Abstract: In this paper we discuss the problem of finding the number of zeros of a polynomial in a given circle
when the coefficients of the polynomial or their real or imaginary parts are restricted to certain conditions. Our
results in this direction generalize some well- known results in the theory of the distribution of zeros of
polynomials.
Mathematics Subject Classification: 30C10, 30C15
Keywords and phrases: Coefficient, Polynomial, Zero.
I. Introduction and Statement of Results
In the literature many results have been proved on the number of zeros of a polynomial in a given circle.
In this direction Q. G. Mohammad [5] has proved the following result:
Theorem A: Let 



0
)(
j
j
j zazP be a polynomial o f degree n such that
0...... 011   aaaa nn ,
Then the number of zeros of P(z) in
2
1
z does not exceed
0
log
2log
1
1
a
an
 .
K. K. Dewan [2] generalized Theorem A to polynomials with complex coefficients and proved the following
results:
Theorem B: Let 



0
)(
j
j
j zazP be a polynomial o f degree n such that jja )Re( , jja )Im( and
0...... 011    nn ,
Then the number of zeros of P(z) in
2
1
0
0
log
2log
1
1
a
n
j
jn 



.
Theorem C: Let 



0
)(
j
j
j zazP be a polynomial o f degree n with complex coefficients such that for some
real , ,
nja j ,......,2,1,0,
2
arg 


and
011 ...... aaaa nn   .
Then the number f zeros of P(z) in
2
1
0
1
0
sin2)1sin(cos
log
2log
1
a
aa
n
j
jn 


 
.

The above results were further generalized by researchers in various ways. M. H. Gulzar[4] proved the
following results:
Theorem D: Let 



0
)(
j
j
01111 ............    knknknnn ,
for some real numbers .0,1,10,0,  knnk 
If knkn   1 , then the number f zeros of P(z) in 10,  z , does not exceed
0
0
000 22)(1)1(2
log
1
log
1
a
n
j
jknknnn 
  

and if 1  knkn  , then the number f zeros of P(z) in 10,  z , does not exceed
0
0
000 22)(1)1(2
log
1
log
1
a
n
j
jknknnn 
  

Theorem E: Let 



0
)(
j
j
real , ,
nja j ,......,2,1,0,
2
arg 


and
01111 ............ aaaaaaa knknknnn    ,
for some 10,0,1,0,0    knnk .
If 1..1   eiaa knkn , then the number of zeros of P(z) in 10,  z , does not exceed
0
1
log
1
log
1
a
M

,
where
)1sincossin(cos)1sin)(cos(1    knn aaM




1
,1
00 sin22)1sin(cos
n
knjj
jaaa  ,
and
if 1..1   eiaa knkn , then the number of zeros of P(z) in 10,  z , does not exceed
0
2
log
1
log
1
a
M

,
where
)1sincossin(cos)1sin)(cos(2    knn aaM




1
,1
00 sin22)1sin(cos
n
knjj
jaaa  .

In the present paper , we find bounds for the number of zeros of P(z) of the
above results in a circle of any positive radius. More precisely, we prove the following resuls:
Theorem 1: Let 



0
)(
j
j
01111 ............    knknknnn ,
for some real numbers .0,1,10,0,  knnk  .
If knkn   1 , then the number f zeros of P(z) in )1,0(  cR
c
R
z , does not exceed
0
0
000
1
]22)(1)1(2[
log
log
1
a
R
c
n
j
jknknnn
n



 
for 1R
and
0
0
00000 ]2)(1)1(2[
log
log
1
a
Ra
c
n
j
jknknnn 
  
for 1R .
If 1  knkn  , then the number f zeros of P(z) in )1,0(  cR
c
R
z does not exceed
0
0
000
1
]22)(1)1(2[
log
log
1
a
R
c
n
j
jknknnn
n



 
for 1R
and
0
0
00000 ]2)(1)1(2[
log
log
1
a
Ra
c
n
j
jknknnn 
  
for 1R .
Remark 1: Taking R=1and 10,
1
 

c , Theorem 1 reduces to Theorem D.
If the coefficients ja are real i.e. jj  ,0 , then we get the following result from Theorem 1:
Corollary 1: Let 



0
)(
j
j
01111 ............ aaaaaaa knknknnn    ,
for some real numbers .0,1,10,0,  knank .
If knkn aa  1 , then the number f zeros of P(z) in )1,0(  cR
c
R
z does not exceed
0
000
1
2)(1)1(2[
log
log
1
a
aaaaaaaR
c
knknnn
n
 


for 1R

and
0
0000 )(1)1(2[
log
log
1
a
aaaaaaaRa
c
knknnn   
for 1R .
If 1  knkn aa , then the number f zeros of P(z) in )1,0(  cR
c
R
z does not exceed
0
000
1
2)(1)1(2[
log
log
1
a
aaaaaaaR
c
knknnn
n
 


for 1R
and
0
0000 )(1)1(2[
log
log
1
a
aaaaaaaRa
c
knknnn   
for 1R .
Applying Theorem 1 to the polynomial –iP(z) , we get the following result:
Theorem 2: : Let 



0
)(
j
j
01111 ............    knknknnn ,
for some real numbers .0,1,10,0,  knnk  .
If knkn    1 , then the number f zeros of P(z) in )1,0(  cR
c
R
z does not exceed
0
0
000
1
]22)(1)1(2[
log
log
1
a
R
c
n
j
jknknnn
n



 
for 1R
and
0
0
00000 ]2)(1)1(2[
log
log
1
a
Ra
c
n
j
jknknnn 
  
for 1R .
If 1  knkn  , then the number f zeros of P(z) in )1,0(  cR
c
R
z does not exceed
0
0
000
1
]22)(1)1(2[
log
log
1
a
R
c
n
j
jknknnn
n



 
for 1R
and
0
0
00000 ]2)(1)1(2[
log
log
1
a
Ra
c
n
j
jknknnn 
  
for 1R .

Theorem 3: Let 



0
)(
j
j
real , ,
nja j ,......,2,1,0,
2
arg 


and
01111 ............ aaaaaaa knknknnn    ,
for some 10,0,1,0,0    knnk .
If 1..1   eiaa knkn , then the number of zeros of P(z) in )1,0(  cR
c
R
z , does not exceed
0
3
log
log
1
a
M
c
,
where
)1sincossin(cos)1sin)(cos[(1
3  

 knn
n
aaRM
]sin22)1sin(cos
1
,1
00 



n
knjj
jaaa  , for 1R
and
)1sincossin(cos)1sin)(cos[(03    knn aaRaM
]sin22)1sin(cos
1
,1
00 



n
knjj
jaaa  , for 1R .
If 1..1   eiaa knkn , then the number of zeros of P(z) in )1,0(  cR
c
R
z does not exceed
0
4
log
log
1
a
M
c
,
Where
4  

 knn
n
aaRM
]sin22)1sin(cos
1
,1
00 



n
knjj
jaaa  for 1R
and
)1sincossin(cos)1sin)(cos[(04    knn aaRaM
]sin2)1sin(cos
1
,1
00 



n
knjj
jaaa  for 1R .
Remark 2: Taking R=1and 10,
1
 

c , Theorem 1 reduces to Theorem E.
For different values of the parameters R, c, k, , , we get many other interesting results.
2. Lemmas
For the proofs of the above results we need the following results:
Lemma 1: If f(z) is analytic in Rz  ,but not identically zero, f(0)  0 and
nkaf k ,......,2,1,0)(  , 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: If f(z) is analytic and )()( rMzf  in rz  , then the number of zeros of F(z) in 1,  c
c
r
z
does not exceed
)0(
)(
log
log
1
f
rM
c
.
Lemma 2 is a simple deduction from Lemma 1.
Lemma 3: Let 



0
)(
j
j
real , , ,0,
2
arg nja j 

 and
,0,1 njaa jj   then any t>0,
 sin)(cos)( 111   jjjjjj aataatata .s
Lemma 3 is due to Govil and Rahman [3].
3.Proofs of Theorems
Proof of Theorem 1: Consider the polynomial
)()1()( zPzzF 
kn
knkn
kn
knkn
n
nn
n
n
n
n
n
n
zaazaazaaza
azazazaz









)()(......)(
)......)(1(
1
1
11
1
01
1
1
001
1
21 )(......)( azaazaa kn
knkn  

1
11
1
)(......)()( 


 kn
knkn
n
nn
n
nn zzzi 
0
1
100
01
1
211
)()1(
)(......)()(


iziz
zzz
n
j
j
jj
kn
knkn
kn
knkn








If ,1 knkn    then
1
11
1
)(......)()()( 


 kn
knkn
n
nn
nn
nn zzzzizF 
.)()1()(
......)()1()(
0
1
10001
1
211


izizz
zzz
n
j
j
jj
kn
knkn
kn
kn
kn
knkn










For Rz  , we have by using the hypothesis
 

......)( 1
11 n
nn
nn
n
n
n RRRRzF  1
1

  kn
knkn R
+
kn
kn
kn
knkn RR 


   11
RRR kn
knkn 001
1
21 )1(......   

j
j
n
j
j R)( 1
0
00 

  
]22)(1)1(2[
0
000
1




n
j
jknknnn
n
R 
for 1R
and


 
n
j
jknknnnRa
1
00000 2)(1)1(2[ 
for 1R .
Hence by Lemma 2, the number of zeros of F(z) in )1,0(  cR
c
R
z , does not exceed
0
0
000
1
]22)(1)1(2[
log
log
1
a
R
c
n
j
jknknnn
n



 
for 1R
and
0
0
00000 ]2)(1)1(2[
log
log
1
a
Ra
c
n
j
jknknnn 
  
for 1R .
If ,1  knkn  then
1
11
1
)(......)()()( 


 kn
knkn
n
nn
nn
nn zzzzizF 
.)()(
......)()1()(
0
1
1001
1
21
1
1


iziz
zzz
n
j
j
jj
kn
knkn
kn
kn
kn
knkn










For Rz  , we have by using the hypothesis
 

......)( 1
11 n
nn
nn
n
n
n RRRRzF  1
1

  kn
knkn R
+



  kn
kn
kn
knkn RR  11
RRR kn
knkn 001
1
21 )1(......   

j
j
n
j
j R)( 1
0
00 

  
]22)(1)1(2[
0
000
1




n
j
jknknnn
n
R 
for 1R
and

 
n
j
jknknnnRa
1
00000 2)(1)1(2[ 
for 1R .
Hence by Lemma 2, the number of zeros of F(z) in )1,0(  cR
c
R
z does not exceed
0
0
000
1
]22)(1)1(2[
log
log
1
a
R
c
n
j
jknknnn
n



 
for 1R
and
0
0
00000 ]2)(1)1(2[
log
log
1
a
Ra
c
n
j
jknknnn 
  
for 1R .
That proves Theorem 1 completely.

Proof of Theorem 4: Consider the polynomial
)()1()( zPzzF 
kn
knkn
kn
knkn
n
nn
n
n
n
n
n
n
zaazaazaaza
azazazaz









)()(......)(
)......)(1(
1
1
11
1
01
1
1
001
1
21 )(......)( azaazaa kn
knkn  

If knkn aa  1 , i.e. 1 , then
1
11
1
)(......)()( 


 kn
knkn
n
nn
nn
n zaazaazzazF 
......)()1()( 1
211  





kn
knkn
kn
kn
kn
knkn zaazazaa 
0001 )1()( azazaa  
so that for Rz  , we have by using Lemma 3,
kn
knkn
kn
knkn
n
nn
nn
n RaaRaaRaaRRazF 




 1
1
11
1
......)( 
0001
1
21 )1(......1 aRaRaaRaaRa kn
knkn
kn
kn  


 
......sin)(cos)([ 11
1
 

 nnnnn
n
aaaaaR
])1(sin)(cos)(
......sin)(cos)(
sin)(cos)(
)1(sin)(cos)(
000101
2121
11
11
aaaaaa
aaaa
aaaa
aaaaa
knknknkn
knknknkn
knknknknkn











 cossin(cos)1sin)(cos[(1
 

knn
n
aaR




1
,1
00 sin22)1sin(cos)1sin
n
knjj
jaaa  ]
for 1R
and
 cossin(cos)1sin)(cos[(0  knn aaRa




1
,1
00 sin2)1sin(cos)1sin
n
knjj
jaaa  ]
for 1R
Hence , by Lemma 2, the number of zeros of F(z) and hence P(z) in )1,0(  cR
c
R
z does not exceed
0
3
log
log
1
a
M
c
,
where
3  

 knn
n
aaRM
]sin22)1sin(cos
1
,1
00 



n
knjj
jaaa  , for 1R
and
)1sincossin(cos)1sin)(cos[(03    knn aaRaM
]sin22)1sin(cos
1
,1
00 



n
knjj
jaaa  , for 1R .
If 1  knkn aa , i.e. 1 , then

1
11
1
)(......)()( 


 kn
knkn
n
nn
nn
n zaazaazzazF 
......)()1()( 1
21
1
1  





kn
knkn
kn
kn
kn
knkn zaazazaa 
001 )( azaa 
so that for Rz  , we have by hypothesis and Lemma 3,
kn
knkn
kn
knkn
n
nn
nn
n RaaRaaRaaRRazF 




 1
1
11
1
......)( 
......1 1
21
1
 



kn
knkn
kn
kn RaaRa
0001 )1( aRaRaa  
......sin)(cos)([ 11
1
 

 nnnnn
n
aaaaaR
])1(sin)(cos)(
......sin)(cos)(
sin)(cos)(
1sin)(cos)(
000101
2121
11
11
aaaaaa
aaaa
aaaa
aaaaa
knknknkn
knknknkn
knknknknkn











 cossin(cos)1sin)(cos[(1
 

knn
n
aaR
)1sin   ]sin22)1sin(cos
1
,1
00 



n
knjj
jaaa 
for 1R
and
 cossin(cos)1sin)(cos[(0  knn aaRa
)1sin   ]sin2)1sin(cos
1
,1
00 



n
knjj
jaaa 
for 1R .
Hence the number of zeros of P(z) in )1,0(  cR
c
R
z does not exceed
0
4
log
log
1
a
M
c
,
Where
4  

 knn
n
aaRM
]sin22)1sin(cos
1
,1
00 



n
knjj
jaaa  for 1R
and
)1sincossin(cos)1sin)(cos[(04    knn aaRaM
]sin2)1sin(cos
1
,1
00 



n
knjj
jaaa  for 1R .
That completes the proof of Theorem 3.
References
[1] L.V.Ahlfors, Complex Analysis, 3rd
edition, Mc-Grawhill
[2] K. K. Dewan, Extremal Properties and Coefficient estimates for Polynomials with restricted Zeros and on location of Zeros of
Polynomials, Ph.D. Thesis, IIT Delhi, 1980.
[3] N. K. Govil and Q. I. Rahman, On the Enestrom- Kakeya Theorem, Tohoku Math. J. 20(1968),126-136.
[4] M. H. Gulzar, On the Zeros of a Polynomial inside the Unit Disk, IOSR Journal of Engineering, Vol.3, Issue 1(Jan.2013) IIV3II,
50-59.
[5] Q. G. Mohammad, On the Zeros of Polynomials,Amer. Math. Monthly 72(1965), 631-633.

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