In this paper, restricting the coefficients of a polynomial to certain conditions, we locate a region containing all of its zeros. Our results generalize many known results in addition to some interesting results which can be obtained by choosing certain values of the parameters. Mathematics Subject Classification: 30C10, 30C15
1. ISSN (e): 2250 – 3005 || Volume, 07 || Issue, 03|| March – 2017 ||
International Journal of Computational Engineering Research (IJCER)
www.ijceronline.com Open Access Journal Page 9
Location of Zeros of Polynomials
M.H. Gulzar1
, B.A.Zargar2
, A.W.Manzoor3
Department of Mathematics, University of Kashmir, Srinagar
I. INTRODUCTION
On the location of zeros of a polynomial with real coefficients Enestrom and Kakeya proved the following
theorem named after them as the Enestrom - Kakeya Theorem [9,10] :
Theorem A: all the zeros of a polynomial
n
j
j
j
zazP
0
)( satisfying 0....... 011
aaaa nn
lie
in 1z .
The above theorem has been generalized in different ways by various authors [2-11].
Recently Gulzar et al [7] proved the following result
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 10,10,, nn ,
,......
,......
1
1
nn
nn
and
001211
......
L ,
001211
......
M .
Then all the zeros of P(z) lie in
][
1
ML
a
z nn
n
.
II. MAIN RESULTS
In this paper we prove the following generalization of Theorem B:
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 10,10,, nn
and for some 1, 21
kk ,
,......
,......
2
2
212
1
2
11
2
111
1
1
kkkk
kkkk
n
n
n
n
n
n
n
n
and
001211
......
L ,
ABSTRACT
In this paper, restricting the coefficients of a polynomial to certain conditions, we locate a region
containing all of its zeros. Our results generalize many known results in addition to some interesting
results which can be obtained by choosing certain values of the parameters.
Mathematics Subject Classification: 30C10, 30C15.
Key Words and Phrases: Coefficients, Polynomial, Zeros.
2. Location of Zeros of Polynomials
www.ijceronline.com Open Access Journal Page 10
001211
......
M .
Then all the zeros of P(z) lie in
)()1()()1([
1)1()1(
2
1
1
21
jj
n
j
jjnn
nn
nn
kk
aa
kik
z
].)1()1( 21
MLkk nn
Further the number of zeros of P(z) in 1,1,
0
Rc
c
R
z
X
a
does not exceed
)1log(
log
1
log
log
1
00
0
a
X
ca
aX
c
and the number of zeros of P(z) in 1,1,
0
Rc
c
R
z
Y
a
does not exceed )1log(
log
1
log
log
1
00
0
a
Y
ca
aY
c
, where R is any positive number and
n
j
jjnn
nn
n
kMLRRaX
1
100
1
)()1()()[(
])()1(
1
2
n
j
jj
k
,
n
j
jjnn
n
n
kMLRRaY
1
100
1
)()1()()[(
])()1(
1
2
n
j
jj
k
.
Remark 1: For
121
kk
, Theorem 1 reduces to Theorem B.
Taking 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 10,10,, nn
and for some 11
k ,
,......
,......
1
11
2
111
1
1
nn
n
n
n
n
kkkk
001211
......
L ,
001211
......
M .
Then all the zeros of P(z) lie in
n
n
j
jjnn
n
n
kk
a
kz
)1()()1([
1
)1( 1
1
11
].
ML
3. Location of Zeros of Polynomials
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Also the number of zeros of P(z) in 1,1,
0
Rc
c
R
z
X
a
does not exceed
)1log(
log
1
log
log
1
00
0
a
X
ca
aX
c
and the number of zeros of P(z) in 1,1,
0
Rc
c
R
z
Y
a
does not exceed )1log(
log
1
log
log
1
00
0
a
Y
ca
aY
c
, where R is any positive number and
])()1()()[(
1
100
1
n
j
jjnn
nn
n
kMLRRaX
,
])()1()()[(
1
100
1
n
j
jjnn
n
n
kMLRRaY
Taking 0 in Theorem 1, we get the following result:
Corollary 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 10,10,, nn
and for some 1, 21
kk ,
,......
,......
0121
1
22
0111
1
11
kkk
kkk
n
n
n
n
n
n
n
n
Then all the zeros of P(z) lie in
n
j
jj
n
j
jjnn
nn
nn
kk
aa
kik
z
1
2
1
1
21
)()1()()1([
1)1()1(
].)1()1( 000021
nn
kk
Also the number of zeros of P(z) in 1,1,
0
Rc
c
R
z
X
a
does not exceed
)1log(
log
1
log
log
1
00
0
a
X
ca
aX
c
and the number of zeros of P(z) in 1,1,
0
Rc
c
R
z
Y
a
does not exceed )1log(
log
1
log
log
1
00
0
a
Y
ca
aY
c
, where R is any positive number and
])()1()()[(
1
10000
1
n
j
jjnn
nn
n
kMLRRaX ,
])()1(
1
2
n
j
jj
k , 1R
])()1()()[(
1
10000
1
n
j
jjnn
n
n
kMLRRaY
])()1(
1
2
n
j
jj
k , 1R
Similarly for other values of the parameters in Theorem 1, we get many other interesting results.
4. Location of Zeros of Polynomials
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III. LEMMAS
For the proof of Theorem 2, we make use of the following lemmas:
Lemma 1: Let f(z) (not identically zero) be analytic for 0)0(, fRz and ,0)( k
af
nk ,......,2,1 . 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: Let f(z) be analytic , 0)0( f and Mzf )( for Rz . Then the number of zeros of f(z) in
1, c
c
R
z is less than or equal to
)0(
log
log
1
f
M
c
.
Lemma 2 is a simple deduction from Lemma 1.
IV. PROOF OF THEOREM 1
Consider the polynomial
)()1()( zPzzF
zaazaazaaza
azazazaz
n
nn
n
n
n
n
n
n
)()(......)(
)......)(1(
1
1
11
1
01
1
1
001
)(...... azaa
......)()()1(
1
211111
1
n
nn
n
nn
n
n
n
n
zkzkzkza
0011
1
11
)(......)()(
zzzk
).......)(1(
1
1
2
2
1
11
zzzk
n
n
n
n
})(......
)()......)(1()(
......)()1(){(
001
1
1
1
1
12
1
12
1
212212
z
zzzkzk
zkzkzki
n
n
n
nn
n
n
n
nn
For 1z so that nj
z
j
,......,2,1,1
1
, we have, by using the hypothesis
......[)1()1()(
1
2111121
n
nn
n
nn
n
nnn
zkzkzkikzazF
)]......)(1(
............
)......)(1(
......
1
1
1
12
0011
1
12
1
21212
1
1
1
11
0011
1
11
zzk
zzzk
zkzkzzk
zzzk
n
n
n
nn
n
nn
n
n
2
321211
1121
[)1()1([
z
k
z
k
kkikzaz
nnnn
nnnnn
n
5. Location of Zeros of Polynomials
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)]......)(1(......
......)
......)(1(............
1
11
2
0
1
01
1
1
12212
121
1
1
1
0
1
011
1
11
n
n
nn
nn
nn
nnn
n
nnnn
zz
k
zz
zz
k
z
k
k
z
z
k
zzzz
k
3212111121
[)1()1([
nnnnnnnnn
n
kkkkikzaz
)]......)(1(......
......)
......)(1(............
112001
112212121
11001111
n
nnnn
n
k
kkk
kk
3212111121
{)1()1([
nnnnnnnnn
n
kkkkikzaz
.....)(1(............ 11001111
n
kk
)}]......)(1(......
......)
112001
112212121
n
nnnn
k
kkk
3212111121
{)1()1([
nnnnnnnnn
n
kkkkikzaz
)}]......)(1(......
)......)(1(......
11212212
1211111
nnn
nnn
kMkk
kkLk
n
j
jjnnnnn
n
kkikzaz
1
121
)()1({)1()1([
}])1()1()()1( 21
1
2
MLkkk nn
n
j
jj
0
if
n
j
jj
n
j
jjnnnnn
kkkikza
1
2
1
121
)()1()()1({)1()1(
}])1()1( 21
MLkk nn
i.e.if
n
j
jj
n
j
jjnn
nn
nn
kk
aa
kik
z
1
2
1
1
21
)()1()()1([
1)1()1(
].)1()1( 21
MLkk nn
This shows that those zeros of F(z) whose modulus is greater than 1 lie in
n
j
jj
n
j
jjnn
nn
nn
kk
aa
kik
z
1
2
1
1
21
)()1()()1([
1)1()1(
].)1()1( 21
MLkk nn
6. Location of Zeros of Polynomials
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Since the zeros of F(z) whose modulus is less than or equal to 1 already satisfy the above inequality and since
the zeros of P(z) are also the zeros of F(z) , it follows that all the zeros of P(z) lie in
n
j
jj
n
j
jjnn
nn
nn
kk
aa
kik
z
1
2
1
1
21
)()1()()1([
1)1()1(
].)1()1( 21
MLkk nn
Again
)()( 0
zGazF ,
where
......)()()1()(
1
211111
1
n
nn
n
nn
n
n
n
n
zkzkzkzazG
zzzk )(......)()( 011
1
11
).......)(1(
1
1
2
2
1
11
zzzk
n
n
n
n
}.)(......
)()......)(1()(
......)()1(){(
01
1
1
1
1
12
1
12
1
212212
z
zzzkzk
zkzkzki
n
n
n
nn
n
n
n
nn
For 0, RRz , we have
......)()()1()(
1
211111
1
n
nn
n
nn
n
n
n
n
RkRkRkRazG
RRRk 011
1
11
......)(
).......)(1(
1
1
2
2
1
11
RRRk
n
n
n
n
}......
)......)(1()(
......)()1()(
01
1
1
1
1
12
1
12
1
212212
R
RRRkRk
RkRkRk
n
n
n
nn
n
n
n
nn
LkkkkRRa nnnnn
nn
n
11211111
1
......)1[(
)]......)(1(......
)......)(1(
11212212
12111
nnn
nnn
kMkk
kk
n
j
jjnn
nn
n
kMLRRa
1
100
1
)()1()()[(
])()1(
1
2
n
j
jj
k
X
for 1R and
for 1R
n
j
jjnn
n
n
kMLRRazG
1
100
1
)()1()()[()(
])()1(
1
2
n
j
jj
k
Y
Since G(0)=0 and G(z) is analytic for Rz , it follows, by Schwarz Lemma , that for Rz ,
7. Location of Zeros of Polynomials
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zXzG )( for 1R and zYzG )( for 1R .
Hence, for Rz , 1R
)()( 0
zGazF
0
)(
0
0
zXa
zGa
if
X
a
z
0
.
Similarly, for Rz , 1R , 0)( zF if
Y
a
z
0
.
In other words, F(z) does not vanish in
X
a
z
0
for 1R and F(z) does not vanish in
Y
a
z
0
for 1R
in Rz . That means all the zeros of F(z) and hence all the zeros of P(z) lie in
X
a
z
0
for 1R and in
Y
a
z
0
for 1R in Rz .
Since for Rz , 0
)( aXzF for 1R and 0
)( aYzF for 1R and since
0)0( 0
aF , it follows by Lemma 2 that the number of zeros of P(z) in 1,1,
0
Rc
c
R
z
X
a
does
not exceed )1log(
log
1
log
log
1
00
0
a
X
ca
aX
c
and the number of zeros of P(z) in
1,1,
0
Rc
c
R
z
Y
a
does not exceed )1log(
log
1
log
log
1
00
0
a
Y
ca
aY
c
.
That proves Theorem 1 completely.
REFERENCES
[1]. L. V. Ahlfors, Complex Analysis, 3rd
edition, Mc-Grawhill.
[2]. A. Aziz and Q. G. Mohammad, Zero-free regions for polynomials and some generalizations of Enestrom-Kakeya Theorem, Canad.
Math. Bull.,27(1984),265- 272.
[3]. A. Aziz and B. A. Zargar, Some Extensions of Enestrom-Kakeya Theorem, Glasnik Mathematicki, 51 (1996), 239-244.
[4]. Y.Choo, On the zeros of a family of self-reciprocal polynomials, Int. J. Math. Analysis , 5 (2011),1761-1766.
[5]. N. K. Govil and Q. I. Rahman, On Enestrom-Kakeya Theorem, Tohoku J. Math. 20 (1968), 126-136.
[6]. M. H. Gulzar, Some Refinements of Enestrom-Kakeya Theorem, International Journal of Mathematical Archive , Vol 2(9), 2011,
1512-1519..
[7]. M. H. Gulzar,B.A. Zargar and A. W. Manzoor, Zeros of a Certain Class of Polynomials, International Journal of Advanced
Research in Engineering and Management, Vol.3 No. 2(2017).
[8]. A. Joyal, G. Labelle and Q. I. Rahman, On the location of the zeros of a polynomial, Canad. Math. Bull. ,10 (1967), 53-63.
[9]. M. Marden, Geometry of Polynomials, Math. Surveys No. 3, Amer. Math.Soc.(1966).
[10]. Q. I. Rahman and G. Schmeisser, Analytic Theory of Polynomials, Oxford University Press, New York (2002).
[11]. B. A. Zargar, On the zeros of a family of polynomials, Int. J. of Math. Sci. &Engg. Appls. , Vol. 8 No. 1(January 2014), 233-237.