e-ISSN: 2278-067X, p-ISSN: 2278-800X, www.ijerd.com
Volume 13, Issue 3 (March 2017), PP.01-08
A Ring-Shaped Region Containing All or A Specific Number of
The Zeros of A Polynomial
M. H. Gulzar
1, A. W. Manzoor
2Department Of Mathematics, University Of Kashmir, Srinagar
ABSTRACT:
According to a Cauchy’s classical result all the zeros of a polynomial
n jj j
z
a
z
P
0
)
(
ofdegree n lie in
z
1
A
, wheren j n j
a
a
A
max
0 1 . In this paper we find a ring-shaped region containingall or a specific number of zeros of P(z).
Mathematics Subject Classification: 30C10, 30C15. Keywords and Phrases: Polynomial, Region, Zeros.
I.
INTRODUCTION
A classical result giving a region containing all the zeros of a polynomial is the following known as Cauchy’s Theorem [4]:
Theorem A. All the zeros of the polynomial
n jj j
z
a
z
P
0
)
(
of degree n lie in the circleM
z
1
, wheren j n j
a
a
M
max
0 1 .The above result has been generalized and improved in various ways. Mohammad [5] used Schwarz Lemma and proved the following result:
Theorem B. All the zeros of the polynomial
n jj j
z
a
z
P
0
)
(
of degree n lie inn
a
M
z
ifa
n
M
,where
max
1 1
...
0
max
1 0 1
...
n1 nz n
n
z
a
z
a
a
z
a
M
.Recently Gulzar [3] proved the following result :
Theorem C.All the zeros of the polynomial
n jj j
z
a
z
P
0
)
(
of degree n lie in the ring-shaped region2 1
z
r
r
, where
2
0 2
1 2 1 3 2 2 1 2 0 2
1 4
1
2
)
(
]
4
)
(
[
M
a
M
R
a
M
R
a
a
M
a
R
r
,
)
(
]
4
)
(
[
2
1 2 1 2
1 3 2 2
1 2 1 4
2 1 2
n n
n n
n
M
a
a
R
M
a
R
M
a
a
R
M
r
,
M
max
zRa
nz
n
...
a
1z
,R being any positive number.
Theorem C.The number of zeros of the polynomial
n jj j
z
a
z
P
0
)
(
of degree n in the ring-shaped regionR
c
c
R
z
r
1
,
1
,does not exceed
log(
1
)
log
1
0
a
M
c
,where M is as in Theorem 1.
II.
MAIN RESULTS
In this paper we prove the following results:
Theorem 1. Let
n jj j
z
a
z
P
0
)
(
be a polynomial of degree n and let
10 1
,
n
j j n
j
j
L
a
a
L
. Then all thezeros of P(z) lie in
r
1
z
r
2, where forR
1
,
0 3 2
0 2
1 2 1 3 2 3 0 2 0 2
1 4
1
2
)
(
]
4
)
(
[
a
L
R
a
L
R
R
a
L
R
a
a
L
R
a
R
r
nn n
n
,
)
(
]
4
)
(
[
2
0 2
1 2 1 3 2 3 0 2 0 2
1 4
0 3 2
2
a
L
R
R
a
L
R
a
a
L
R
a
R
a
L
R
r
n n
n
n
and for
R
1
0 3
0 1
2 1 3 0 2 0 2
1 1
2
)
(
]
4
)
(
[
a
RL
a
RL
a
RL
a
a
RL
a
r
,
)
(
]
4
)
(
[
2
0 1
2 1 3 0 2 0 2
1
0 3
2
a
L
R
a
L
R
a
a
L
R
a
a
L
R
r
,R being any positive number.
Theorem2. Let
n jj j
z
a
z
P
0
)
(
be a polynomial of degree n and let
n jj
a
L
1
. Then the number of zeros
of P(z) in
c
c
R
z
r
1
,
1
does not exceed0 0
log
log
1
a
a
L
R
c
n
=
log(
1
)
log
1
0
a
L
R
c
n
forR
1
and
0 0
log
log
1
a
a
RL
c
=
log(
1
)
log
1
0
a
RL
c
forR
1
.For different values of the parameters in Theorems 1 and 2, we get many interesting results. For example for R=1, we get the following results:
Corollary 1. Let
n jj j
z
a
z
P
0
)
(
be a polynomial of degree n and let
10 1
,
n
j j n
j
j
L
a
a
L
. Then all
0 3
0 1
2 1 3 0 2 0 2
1 1
2
)
(
]
4
)
(
[
a
L
a
L
a
L
a
a
L
a
r
,
)
(
]
4
)
(
[
2
0 1
2 1 3 0 2 0 2
1
0 3
2
a
L
a
L
a
a
L
a
a
L
r
.Corollary 2. Let
n jj j
z
a
z
P
0
)
(
be a polynomial of degree n and let
n jj
a
L
1
. Then the number of zeros
of P(z) in
c
c
R
z
r
1
,
1
does not exceed0 0
log
log
1
a
a
L
c
=
log(
1
).
log
1
0
a
L
c
III.
LEMMAS
For the proofs of the above theorems we make use of the following results:
Lemma1. Let f(z) be analytic in
z
1
,f(0)=a, wherea
1
,f
(
0
)
b
andf
(
z
)
1
forz
1
. Then, for1
z
,
)
1
(
)
1
(
)
1
(
)
1
(
)
(
22
a
z
b
z
a
a
a
a
z
b
z
a
z
f
.
The example
2 2
1
1
1
)
(
az
z
a
b
z
z
a
b
a
z
f
shows that the estimate is sharp.
Lemma 1 is due to Govil, Rahman and Schmeisser [2].
Lemma 2.Let f(z) be analytic for
z
R
,f(0)=0,f
(
0
)
b
andf
(
z
)
M
forz
R
.Then, for
z
R
,
z
b
M
b
R
z
M
R
z
M
z
f
2
2
.
)
(
.Lemma 2 is a simple deduction from Lemma 1 .
Lemma 3.If f(z) is analytic for
z
R
,f
(
0
)
0
andf
(
z
)
M
forz
R
, then the number of zeros off(z)in
c
R
c
R
z
,
1
is less than or equal to)
0
(
log
log
1
f
M
c
.Lemma 3 is a simple deduction from Jensen’s Theorem (see [1]).
IV.
PROOFS OF THEOREMS
Proof of Theorem 1. Let G(z)=
a
nz
n
a
n1z
n1
...
a
1z
.Then G(z) is analytic for
z
R
,G(0)=0,G
(
0
)
a
1and forz
R
z
a
z
a
z
a
z
L
R
a
a
a
R
R
a
R
a
R
a
z
a
z
a
z
a
n
n n n
n n n n
n n n n
]
...
[
...
...
1 1
1 1
1
1 1
1
for
R
1
and for
R
1
,
G
(
z
)
RL
.Hence, by Lemma 2, for
z
R
,
z
a
L
R
R
a
z
L
R
R
z
L
R
z
G
n n n
1 2 1 2
.
)
(
,for
R
1
and for
R
1
,
z
a
RL
R
a
z
RL
R
z
RL
z
G
1 2 1 2
.
)
(
.Therefore, for
z
R
,R
1
,
P
(
z
)
G
(
z
)
a
0
0
.
)
(
1 2 1 2
0 0
z
a
L
R
R
a
z
L
R
R
z
L
R
a
z
G
a
n n n
if
a
0R
2(
R
nL
a
1z
)
R
nL
z
(
R
nL
z
a
1R
2)
0
i.e. if
R
2nL
2z
2
a
1R
2(
R
nL
a
0)
z
a
0R
n2L
0
which is true if
0 3 2
0 2
1 2 1 3 2 3 0 2 0 2
1 4
2
)
(
]
4
)
(
[
a
L
R
a
L
R
R
a
L
R
a
a
L
R
a
R
z
nn n
n
.
Similarly, for
z
R
,R
1
,
P
(
z
)
0
if
a
0R
2(
RL
a
1z
)
RL
z
(
RL
z
a
1R
2)
0
i.e. if
L
2z
2
a
1(
RL
a
0)
z
a
0RL
0
which is true if
0 3
0 1
2 1 3 0 2 0 2
1
2
)
(
]
4
)
(
[
a
RL
a
RL
a
RL
a
a
RL
a
z
.
0 3 2
0 2
1 2 1 3 2 3 0 2 0 2
1 4
2
)
(
]
4
)
(
[
a
L
R
a
L
R
R
a
L
R
a
a
L
R
a
R
z
nn n
n
for
z
R
,R
1
and P(z) does not vanish in
0 3
0 1
2 1 3 0 2 0 2
1
2
)
(
]
4
)
(
[
a
RL
a
RL
a
RL
a
a
RL
a
z
for
z
R
,R
1
.In other words, all the zeros of P(z) lie in
0 3 2
0 2
1 2 1 3 2 3 0 2 0 2
1 4
2
)
(
]
4
)
(
[
a
L
R
a
L
R
R
a
L
R
a
a
L
R
a
R
z
nn n
n
i.e.
z
r
1 forz
R
,R
1
and in
0 3
0 1
2 1 3 0 2 0 2
1
2
)
(
]
4
)
(
[
a
RL
a
RL
a
RL
a
a
RL
a
z
i.e.
z
r
1 forz
R
,R
1
. On the other hand, letn
a
z
na
z
na
nz
a
nz
P
z
z
Q
1
1 1
0
...
)
1
(
)
(
H
(
z
)
a
n, where
H
(
z
)
a
0z
n
a
1z
n1
...
a
n1z
.Then H(z) is analytic and
H
(
z
)
R
nL
forz
R
, H(0)=0,H
(
0
)
a
n1. Hence, by Lemma 2, forR
z
,
z
a
L
R
R
a
z
L
R
R
z
L
R
z
H
nn n
1 2 1 2
.
)
(
,for
R
1
and for
R
1
,
z
a
L
R
R
a
z
L
R
R
z
L
R
z
H
1 2 1 2
.
)
(
..Therefore, for
z
R
,R
1
,
Q
(
z
)
H
(
z
)
a
0
0
.
)
(
1 2 1 2
0 0
z
a
L
R
R
a
z
L
R
R
z
L
R
a
z
H
a
a
0R
2(
R
nL
a
1z
)
R
nL
z
(
R
nL
z
a
1R
2)
0
i.e. if
R
2nL
2z
2
a
1R
2(
R
nL
a
0)
z
a
0R
n2L
0
which is true if
0 3 2 0 2 1 2 1 3 2 3 0 2 0 2 1 4
2
)
(
]
4
)
(
[
a
L
R
a
L
R
R
a
L
R
a
a
L
R
a
R
z
n n n n
.Similarly, for
z
R
,R
1
,
Q
(
z
)
0
if
a
0R
2(
R
L
a
1z
)
R
L
z
(
R
L
z
a
1R
2)
0
i.e. if
L
2z
2
a
1(
R
L
a
0)
z
a
0R
L
0
which is true if
0 3 0 1 2 1 3 0 2 0 2 1
2
)
(
]
4
)
(
[
a
L
R
a
L
R
a
L
R
a
a
L
R
a
z
.This shows that Q(z) does not vanish in
0 3 2 0 2 1 2 1 3 2 3 0 2 0 2 1 4
2
)
(
]
4
)
(
[
a
L
R
a
L
R
R
a
L
R
a
a
L
R
a
R
z
n n n n
for
z
R
,R
1
and Q(z) does not vanish in
0 3 0 1 2 1 3 0 2 0 2 1
2
)
(
]
4
)
(
[
a
L
R
a
L
R
a
L
R
a
a
L
R
a
z
for
z
R
,R
1
.In other words, all the zeros of
(
)
(
1
)
z
P
z
z
Q
nlie in 0 3 2 0 2 1 2 1 3 2 3 0 2 0 2 1 4
2
)
(
]
4
)
(
[
a
L
R
a
L
R
R
a
L
R
a
a
L
R
a
R
z
n n n n
for
z
R
,R
1
and in 0 3 0 1 2 1 3 0 2 0 2 1
2
)
(
]
4
)
(
[
a
RL
a
RL
a
RL
a
a
RL
a
z
for
z
R
,R
1
.Hence all the zeros of
(
1
)
z
P
lie infor
z
R
,R
1
and in
0 3
0 1
2 1 3 0 2 0 2
1
2
)
(
]
4
)
(
[
a
L
R
a
L
R
a
L
R
a
a
L
R
a
z
for
z
R
,R
1
.Therefore , all the zeros of P(z) lie in
)
(
]
4
)
(
[
2
0 2
1 2 1 3 2 3 0 2 0 2
1 4
0 3 2
a
L
R
R
a
L
R
a
a
L
R
a
R
a
L
R
z
n n
n
n
i.e.
z
r
2 forz
R
,R
1
and in
)
(
]
4
)
(
[
2
0 1
2 1 3 0 2 0 2
1
0 3
a
L
R
a
L
R
a
a
L
R
a
a
L
R
z
i.e.
z
r
2 forz
R
,R
1
.Thus we have proved that all the zeros of P(z) lie in
r
1
z
r
2. That completes the proof of Theorem 1. Proof of Theorem 2. To prove Theorem 2, we need to show only that the number of zeros of P(z) inc
c
R
z
,
1
does not exceedlog(
1
)
log
1
0
a
L
R
c
n
forR
1
andlog(
1
)
log
1
0
a
RL
c
forR
1
.Since P(z) is analytic for
z
R
, P(0)=a
0 and forz
R
,
P
(
z
)
a
z
a
n 1z
n 1...
a
1z
a
0 nn
a
nz
n
a
n1z
n1
...
a
1z
a
0
0
0 1
1
1
...
a
L
R
a
R
a
R
a
R
a
n
n n n n
for
R
1
and
P
(
z
)
RL
a
0 forR
1
,it follows , by Lemma 3, that the number of zeros of P(z) in
c
c
R
z
r
1
,
1
does not exceed0 0
log
log
1
a
a
L
R
c
n
=
log(
1
)
log
1
0
a
L
R
c
n
forR
1
and
0 0
log
log
1
a
a
RL
c
=
log(
1
)
log
1
0
a
RL
c
forR
1
and Theorem 2 follows.REFERENCES
[1]. Ahlfors L. V. ,Complex Analysis, 3rd edition, Mc-Grawhill.
Mathematical Journal of Interdisciplinary Sciences, Vol.4, No.2(March 2016), 177-182. [4]. Marden M., Geometry of Polynomials, Mathematical Surveys Number 3, Amer.
Math.. Soc. Providence, RI, (1966).