ABSTRACT: According to a Cauchy’s classical result all the zeros of a polynomial   n j j j P z a z 0 ( ) of degree n lie in z 1 A , where n j j n a a A  max0  1 . In this paper we find a ring-shaped region containing all or a specific number of zeros of P(z). Mathematics Subject Classification: 30C10, 30C15.

1. International Journal of Engineering Research and Development
e-ISSN: 2278-067X, p-ISSN: 2278-800X, www.ijerd.com
Volume 13, Issue 3 (March 2017), PP.01-08
1
A Ring-Shaped Region Containing All or A Specific Number of
The Zeros of A Polynomial
M. H. Gulzar1
, A. W. Manzoor2
Department Of Mathematics, University Of Kashmir, Srinagar
ABSTRACT: According to a Cauchy’s classical result all the zeros of a polynomial 

n
j
j
j zazP
0
)( of
degree n lie in Az 1 , where
n
j
nj
a
a
A 10max  . In this paper we find a ring-shaped region containing
all 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
j
j
j zazP
0
)( of degree n lie in the circle
Mz 1 , where
n
j
nj
a
a
M 10max  .
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
j
j
j zazP
0
)( of degree n lie in
na
M
z  if Man  ,
where 1
1
010
1
1
......max......max 




 n
n
z
n
nz
azaazaM .
Recently Gulzar [3] proved the following result :
Theorem C.All the zeros of the polynomial 

n
j
j
j zazP
0
)( of degree n lie in the ring-shaped region
21 rzr  , where
2
0
2
1
2
1
322
1
2
0
2
1
4
1
2
)(]4)([
M
aMRaMRaaMaR
r

 ,
)(]4)([
2
1
2
1
2
1
322
1
2
1
4
2
1
2
nnnnn aMRaMRaaMaR
M
r



,
zazaM n
nRz 1......max  
,
zazaM n
n
Rz 101 ......max 
 ,

2. A Ring-Shaped Region Containing All Or A Specific Number Of The Zeros Of A Polynomial
2
R being any positive number.
Theorem C.The number of zeros of the polynomial 

n
j
j
j zazP
0
)( of degree n in the ring-shaped region
Rc
c
R
zr  1,1 ,does not exceed
)1log(
log
1
0a
M
c
 ,
where M is as in Theorem 1.
II. MAIN RESULTS
In this paper we prove the following results:
Theorem 1. Let 

n
j
j
j zazP
0
)( be a polynomial of degree n and let 



1
01
,
n
j
j
n
j
j aLaL . Then all the
zeros of P(z) lie in 21 rzr  , where for 1R ,
0
32
0
2
1
2
1
323
0
2
0
2
1
4
1
2
)(]4)([
aLR
aLRRaLRaaLRaR
r n
nnn



,
)(]4)([
2
0
2
1
2
1
323
0
2
0
2
1
4
0
32
2
aLRRaLRaaLRaR
aLR
r
nnn
n




and for 1R
0
3
01
2
1
3
0
2
0
2
1
1
2
)(]4)([
aRL
aRLaRLaaRLa
r

 ,
)(]4)([
2
01
2
1
3
0
2
0
2
1
0
3
2
aLRaLRaaLRa
aLR
r


 ,
R being any positive number.
Theorem2. Let 

n
j
j
j zazP
0
)( be a polynomial of degree n and let 

n
j
jaL
1
. Then the number of zeros
of P(z) in c
c
R
zr  1,1 does not exceed
0
0
log
log
1
a
aLR
c
n

= )1log(
log
1
0a
LR
c
n
 for 1R
and
0
0
log
log
1
a
aRL
c

= )1log(
log
1
0a
RL
c
 for 1R .
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
j
j
j zazP
0
)( be a polynomial of degree n and let 



1
01
,
n
j
j
n
j
j aLaL . Then all
the zeros of P(z) lie in 21 rzr  , where,

3. A Ring-Shaped Region Containing All Or A Specific Number Of The Zeros Of A Polynomial
3
0
3
01
2
1
3
0
2
0
2
1
1
2
)(]4)([
aL
aLaLaaLa
r

 ,
)(]4)([
2
01
2
1
3
0
2
0
2
1
0
3
2
aLaLaaLa
aL
r


 .
Corollary 2. Let 

n
j
j
j zazP
0
)( be a polynomial of degree n and let 

n
j
jaL
1
. Then the number of zeros
of P(z) in c
c
R
zr  1,1 does not exceed
0
0
log
log
1
a
aL
c

= ).1log(
log
1
0a
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 1z ,f(0)=a, where 1a , bf  )0( and 1)( zf for 1z . Then, for
1z ,
)1()1(
)1()1(
)( 2
2
azbzaa
aazbza
zf


 .
The example
2
2
1
1
1)(
azz
a
b
zz
a
b
a
zf







shows that the estimate is sharp.
Lemma 1 is due to Govil, Rahman and Schmeisser [2].
Lemma 2.Let f(z) be analytic for Rz  ,f(0)=0, bf  )0( and Mzf )( for Rz  .
Then, for Rz  ,
zbM
bRzM
R
zM
zf



2
2
.)( .
Lemma 2 is a simple deduction from Lemma 1 .
Lemma 3.If f(z) is analytic for Rz  , 0)0( f and Mzf )( for Rz  , then the number of zeros of
f(z)in Rc
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)= zazaza n
n
n
n 1
1
1 ...... 
 .
Then G(z) is analytic for Rz  ,G(0)=0, 1)0( aG  and for Rz 
zazazazG n
n
n
n 1
1
1 ......)(  


4. A Ring-Shaped Region Containing All Or A Specific Number Of The Zeros Of A Polynomial
4
LR
aaaR
RaRaRa
zazaza
n
nn
n
n
n
n
n
n
n
n
n









]......[
......
......
11
1
1
1
1
1
1
for 1R
and for 1R ,
RLzG )( .
Hence, by Lemma 2, for Rz  ,
zaLR
RazLR
R
zLR
zG n
nn
1
2
1
2
.)(


 ,
for 1R
and for 1R ,
zaRL
RazRL
R
zRL
zG
1
2
1
2
.)(


 .
Therefore, for Rz  , 1R ,
0)()( azGzP 
0
.
)(
1
2
1
20
0





zaLR
RazLR
R
zLR
a
zGa
n
nn
if
0)()( 2
11
2
0  RazLRzLRzaLRRa nnn
i.e. if
0)( 2
00
2
1
222
 
LRazaLRRazLR nnn
which is true if
0
32
0
2
1
2
1
323
0
2
0
2
1
4
2
)(]4)([
aLR
aLRRaLRaaLRaR
z n
nnn



.
Similarly, for Rz  , 0)(,1  zPR if
0)()( 2
11
2
0  RazRLzRLzaRLRa
i.e. if
0)( 001
22
 RLazaRLazL
which is true if
0
3
01
2
1
3
0
2
0
2
1
2
)(]4)([
aRL
aRLaRLaaRLa
z

 .
This shows that P(z) does not vanish in

5. A Ring-Shaped Region Containing All Or A Specific Number Of The Zeros Of A Polynomial
5
0
32
0
2
1
2
1
323
0
2
0
2
1
4
2
)(]4)([
aLR
aLRRaLRaaLRaR
z n
nnn



for Rz  , 1R
and P(z) does not vanish in
0
3
01
2
1
3
0
2
0
2
1
2
)(]4)([
aRL
aRLaRLaaRLa
z


for Rz  , 1R .
In other words, all the zeros of P(z) lie in
0
32
0
2
1
2
1
323
0
2
0
2
1
4
2
)(]4)([
aLR
aLRRaLRaaLRaR
z n
nnn



i.e. 1rz  for Rz  , 1R
and in
0
3
01
2
1
3
0
2
0
2
1
2
)(]4)([
aRL
aRLaRLaaRLa
z


i.e. 1rz  for Rz  , 1R .
On the other hand, let
nn
nnn
azazaza
z
PzzQ  

1
1
10 ......)
1
()(
nazH  )( ,
where
zazazazH n
nn
1
1
10 ......)( 

 .
Then H(z) is analytic and LRzH n
)( for Rz  , H(0)=0, 1)0(  naH . Hence, by Lemma 2, for
Rz  ,
zaLR
RazLR
R
zLR
zH n
nn
1
2
1
2
.)(


 ,
for 1R
and for 1R ,
zaLR
RazLR
R
zLR
zH
1
2
1
2
.)(


 ..
Therefore, for Rz  , 1R ,
0)()( azHzQ 
0
.
)(
1
2
1
20
0





zaLR
RazLR
R
zLR
a
zHa
n
nn
if

6. A Ring-Shaped Region Containing All Or A Specific Number Of The Zeros Of A Polynomial
6
0)()( 2
11
2
0  RazLRzLRzaLRRa nnn
i.e. if
0)( 2
00
2
1
222
 
LRazaLRRazLR nnn
which is true if
0
32
0
2
1
2
1
323
0
2
0
2
1
4
2
)(]4)([
aLR
aLRRaLRaaLRaR
z n
nnn




.
Similarly, for Rz  , 0)(,1  zQR if
0)()( 2
11
2
0  RazLRzLRzaLRRa
i.e. if
0)( 001
22
 LRazaLRazL
which is true if
0
3
01
2
1
3
0
2
0
2
1
2
)(]4)([
aLR
aLRaLRaaLRa
z


 .
This shows that Q(z) does not vanish in
0
32
0
2
1
2
1
323
0
2
0
2
1
4
2
)(]4)([
aLR
aLRRaLRaaLRaR
z n
nnn




for Rz  , 1R
and Q(z) does not vanish in
0
3
01
2
1
3
0
2
0
2
1
2
)(]4)([
aLR
aLRaLRaaLRa
z



for Rz  , 1R .
In other words, all the zeros of )
1
()(
z
PzzQ n
 lie in
0
32
0
2
1
2
1
323
0
2
0
2
1
4
2
)(]4)([
aLR
aLRRaLRaaLRaR
z n
nnn




for Rz  , 1R
and in
0
3
01
2
1
3
0
2
0
2
1
2
)(]4)([
aRL
aRLaRLaaRLa
z


for Rz  , 1R .
Hence all the zeros of )
1
(
z
P lie in
0
32
0
2
1
2
1
323
0
2
0
2
1
4
2
)(]4)([
aLR
aLRRaLRaaLRaR
z n
nnn





7. A Ring-Shaped Region Containing All Or A Specific Number Of The Zeros Of A Polynomial
7
for Rz  , 1R
and in
0
3
01
2
1
3
0
2
0
2
1
2
)(]4)([
aLR
aLRaLRaaLRa
z



for Rz  , 1R .
Therefore , all the zeros of P(z) lie in
)(]4)([
2
0
2
1
2
1
323
0
2
0
2
1
4
0
32
aLRRaLRaaLRaR
aLR
z
nnn
n




i.e. 2rz  for Rz  , 1R and in
)(]4)([
2
01
2
1
3
0
2
0
2
1
0
3
aLRaLRaaLRa
aLR
z



i.e. 2rz  for Rz  , 1R .
Thus we have proved that all the zeros of P(z) lie in 21 rzr  . 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) in
c
c
R
z  1, does not exceed )1log(
log
1
0a
LR
c
n
 for 1R and )1log(
log
1
0a
RL
c
 for 1R .
Since P(z) is analytic for Rz  , P(0)= 0a and for Rz  ,
01
1
1 .......)( azazazazP n
n
n
n  

01
1
1 ....... azazaza
n
n
n
n 


0
01
1
1 .......
aLR
aRaRaRa
n
n
n
n
n

 

for 1R
and
0)( aRLzP 
for 1R ,
it follows , by Lemma 3, that the number of zeros of P(z) in c
c
R
zr  1,1 does not exceed
0
0
log
log
1
a
aLR
c
n

= )1log(
log
1
0a
LR
c
n
 for 1R
and
0
0
log
log
1
a
aRL
c

= )1log(
log
1
0a
RL
c
 for 1R and Theorem 2 follows.
REFERENCES
[1]. Ahlfors L. V. ,Complex Analysis, 3rd
edition, Mc-Grawhill.
[2]. Govil N. K., Rahman Q. I and Schmeisser G.,On the derivative of a polynomial,
Illinois Math. Jour. 23, 319-329(1979).
[3]. Gulzar M. H. ,Manzoor A. W. ,A Ring-shaped Region for the Zeros of a Polynomial,

8. A Ring-Shaped Region Containing All Or A Specific Number Of The Zeros Of A Polynomial
8
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).
[5]. Mohammad Q. G., On the Zeros of a Polynomial, Amer. Math. Monthly, 72, 631-633
(1966).