e-ISSN: 2278-067X, p-ISSN: 2278-800X, www.ijerd.com
Volume 8, Issue 11 (October 2013), PP. 54-61
On the Zeros of Complex Polynomials in a Given Circle
M. H. Gulzar
Department of Mathematics University of Kashmir, Srinagar 190006
Abstract:- The problemof finding the number of zeros of a polynomial in a given circle is of great interest in the theory of the distribution of zeros of polynomials. In this paper we consider the said problem under certain conditions on the coefficients of the polynomial or their real and imaginary parts and prove certain results that generalize some well-known results on the subject.
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. Recently M. H. Gulzar [3] proved the following results:
Theorem A: : Let Let
0
)
(
j j j
z
a
z
P
be a polynomial o f degree n such thatRe(
a
j)
j,Im(
a
j)
jand for some positive integers
,
and for some real numbersk
1,
k
2,
1,
2,
.
...
...
...
...
0 2 1 1
1 2
0 1 1 1
1 1
n n
n n
k
k
Then the number of zeros of P(z) in
z
,
0
1
does not exceed0 *
log
1
log
1
a
M
, where
M
*
a
n
(
k
1
1
)
n
(
k
2
1
)
n
2
(
)
(
k
1
n
k
2
n)
(
1
1
)
0
1
0
(
2
1
)
0
2
0
a
0 .Theorem B: Let Let
0
)
(
j j j
z
a
z
P
be a polynomial o f degree n such thatRe(
a
j)
j,Im(
a
j)
jand for some positive integers
,
,
,
and for some real numbers1
0
,
1
0
,
1
0
,
1
0
;
,
,
,
;
,
,
,
2 3 4 1 2 3 4 1 2 3 41
k
k
k
k
k
k
k
k
,0
1
1
,
0
2
1
,
1
0
,
1
0
,
1
0
,
1
0
1
2
3
4
,
.
.
...
...
...
...
...
...
...
...
1 4 3 1
2 1 2 1
] 2 [ 2 2
0 3 2 2
2 2 ]
2 [ 2 3
1 2 3 1
2 1 2 1
] 2 [ 2 2
0 1 2 2
2 2 ]
2 [ 2 1
n n n n
k
k
k
k
M
a
r
1
0 andn
a
M
r
2
,with
1 4
3 1 2
1 1
1
(
1
)
(
1
)
(
1
)
(
1
)
a
na
na
k
nk
nk
nk
nM
1 4
0 3 1 2 0 1 0 3 0 1 1 4 1 2
1 4 1 2 3
1 1 2 2 1 2 2
)
1
(
)
1
(
)
1
(
)
1
(
)
(
)
(
)
(
)
(
)
(
2
k
nk
nk
nk
nand
1 4
3 1 2
1 0 1
1
(
1
)
(
1
)
(
1
)
(
1
)
a
na
a
k
nk
nk
nk
nM
.
)
1
(
)
1
(
)
1
(
)
1
(
)
(
)
(
)
(
)
(
)
(
2
1 4
0 3 1 2 0 1 0 3 0 1 1 4 1 2
1 4 1 2 3
1 1 2 2 1 2 2
k
nk
nk
nk
nIf n is odd, then P(z) has all its zeros in the disk
R
1
z
R
2, where
M
a
R
01 and
n
a
M
R
2
,and
M
andM
are respectively the same asM
andM
, except thatk
1,
k
2,
k
3,
k
4 are respectively replaced byk
2,
k
1,
k
4,
k
3.
In this paper we shall find a bound for the number of zeros of the polynomials of Theorems A and B in a circle of any positive radius. In fact, we prove the following results:
Theorem 1: Let Let
0
)
(
j j j
z
a
z
P
be a polynomial o f degree n such thatRe(
a
j)
j,Im(
a
j)
jand for some positive integers
,
and for some real numbers,
1
0
,
1
0
,
1
0
,
1
0
;
,
,
,
2 1 2 1 2 1 21
k
k
k
k
.
...
...
...
...
0 1 1 1
1 1
2
0 1 1 1
1 1
1
n n
n n
k
k
Then the number of zeros of P(z) in
(
R
0
,
c
1
)
c
R
z
does not exceed0 1
log
log
1
a
M
c
, where]
)
(
)
(
[
1 20 1
1
n n n
n n
n n n
n
R
a
R
k
k
a
M
R
[
1(
0
0)
0]
R
[
2(
0
0)
0]
forR
1
and]
)
(
)
(
[
1 20 1
1
n n n
n n
n n n
n
R
a
R
k
k
a
M
R
[
1(
0
0)
2(
0
0)
0
0]
forR
1
.Remark 1: Taking R=1and
1
,
0
1
c
, Theorem 1 reduces to Theorem A.For different values of
k
1,
k
2,
1,
2, Theorem 1 gives many other interesting results.Theorem 2: Let Let
0
)
(
j j j
z
a
z
P
be a polynomial o f degree n such thatRe(
a
j)
j,Im(
a
j)
jand for some positive integers
,
,
,
and for some real numbers1
0
,
1
0
,
1
0
,
1
0
;
,
,
,
;
,
,
,
k
k
k
k
k
k
k
1
0
,
1
0
3
4
,
.
.
...
...
...
...
...
...
...
...
1 4 3 1
2 1 2 1
] 2 [ 2 2
0 3 2 2
2 2 ]
2 [ 2 3
1 2 3 1
2 1 2 1
] 2 [ 2 2
0 1 2 2
2 2 ]
2 [ 2 1
n n n n
k
k
k
k
Then, for even n, the number of zeros of P(z) in in
(
R
0
,
c
1
)
c
R
z
does not exceed0 2
log
log
1
a
M
c
,where
]
)
(
)
(
[
1 20 1
2
n n n
n n
n n n
n
R
a
R
k
k
a
M
R
[
1(
0
0)
0]
R
[
2(
0
0)
0]
forR
1
and]
)
(
)
(
[
1 20 1
2
n n n
n n
n n n
n
R
a
R
k
k
a
M
R
[
1(
0
0)
2(
0
0)
0
0]
forR
1
.If n is odd, the number of zeros of P(z) in
(
R
0
,
c
1
)
c
R
z
does not exceed0 2
log
log
1
a
M
c
, where
2
M
is same asM
2except thatk
1,
k
2,
k
3,
k
4 are respectively replaced byk
2,
k
1,
k
4,
k
3. For different values ofk
1,
k
2,
1,
2, Theorem 2 gives many interesting results.Theorem 3: Let Let
0
)
(
j j j
z
a
z
P
be a polynomial o f degree n such that for some positive integers
,
,
and for some real numbers
k
1,
k
2,
1,
2,
0
k
1
1
,
0
k
2
1
,
0
1
1
,
0
2
1
,
.
...
...
...
...
1 2 3 1
2 1 2 1
] 2 [ 2 2
0 1 2 2
2 2 ]
2 [ 2 1
a
a
a
a
a
k
a
a
a
a
a
k
n n
Then for even n the number of zeros of P(z) in
(
R
0
,
c
1
)
c
R
z
does not exceed0 3
log
log
1
a
M
c
,where for
R
1
,
M
3
a
nR
n2
a
n1R
n1
a
0
R
n[
a
1
(
1
k
1)
a
n
(
1
k
2)
a
n1
)]
2
(
sin
)
2
2
(
cos
2 2 0
1 1 2 1 2 1
0 1 1 2 1 2 1
2 2
n
j j n
n
n n
a
a
a
a
k
a
k
a
a
a
k
a
k
a
a
and for
R
1
,
)].
2
(
sin
)
2
2
(
cos
2 2 0
1 1 2 1 2 1
0 1 1 2 1 2 1
2 2
n
j j n
n
n n
a
a
a
a
k
a
k
a
a
a
k
a
k
a
a
If n is odd, then the number of zeros of P(z) in
(
R
0
,
c
1
)
c
R
z
does not exceed0 4
log
log
1
a
M
c
,where
M
4 is same asM
3 except thatk
1,
k
2are respectively replaced byk
2,
k
1. For different values ofk
1,
k
2, Theorem 3 gives many interesting results.II.
LEMMAS
For the proofs of the above results we need the following results:
Lemma 1: If f(z) is analytic in
z
R
,but not identically zero, f(0)
0 andf
(
a
k)
0
,
k
1
,
2
,...,
n
, then
nj j
i
a
R
f
d
f
1 2
0
log
(Re
log
(
0
)
log
2
1
Lemma 1 is the famous Jensen’s theorem (see page 208 of [1]).
Lemma 2: If f(z) is analytic and
f
(
z
)
M
(
r
)
inz
r
, then the number of zeros of f(z) in
,
c
1
c
r
z
does not exceed
)
0
(
)
(
log
log
1
f
r
M
c
.Lemma 2 is a simple deduction from Lemma 1.
Lemma 3: Let
0
)
(
j j j
z
a
z
P
be a polynomial o f degree n with complex coefficients such that for somereal
,
,,
0
,
2
arg
a
j
j
n
and,
0
,
1
j
n
a
a
j
j
then for any t>0,
ta
j
a
j1
(
t
a
j
a
j1)
cos
(
t
a
j
a
j1)
sin
. Lemma 3 is due to Govil and Rahman [2].III.
PROOFS OF THEOREMS
Proof of Theorem 1: Consider the polynomial
F
(
z
)
(
1
z
)
P
(
z
)
1 2 1 1
1 1
0 1
0 0
1 1
1
0 1 1
1
)
(
)
(
)
1
(
)
(
...
)
(
)
...
)(
1
(
n n n n n n n
n n
n
n n n n
n
n n n n
z
z
k
z
k
a
z
a
a
z
a
a
z
a
a
z
a
a
z
a
z
a
z
a
z
}
)
1
(
)
(
...
)
(
)
(
...
)
(
)
(
)
1
(
{
)
1
(
)
(
...
)
(
)
(
...
0 2
0 2 1 1
1 1
1 2 1 1
2 2
0 1
0 1 1 1
1 1
z
z
z
z
z
z
k
z
k
i
z
z
z
z
n n n n n n n
n
For
z
R
, we have by using the hypothesis...
)
(
)
(
)
1
(
)
(
1
0
1
1
1
n2
n1 n1
n n n
n n n
n
R
a
k
R
k
R
R
a
z
R
R
R
R
R
R
k
R
k
R
R
R
R
n n n
n n n
n n
0 2 0
2 1 1
1 1
1 1 2 2
1 2
0 1 0
1 1 1
1 1
)
1
(
)
(
...
)
(
)
(
...
)
(
)
(
)
1
(
)
1
(
)
(
...
)
(
)
(
For
R
1
,...
)
1
[(
)
(
1
0
1 n
n1
1 n
n2
n1
nn
n
R
a
R
k
k
a
z
F
]
)
1
(
...
[
]
)
1
(
...
[
]
...
)
1
(
0 2 0
2 1
2 1
1 0
1 2
0 1 1
2 1
1 1
1 2
1 2 2
1 2
1 1
2
R
R
k
k
n n n n n
1
0
[
n
n
1(
n
n)
2(
n
n)
]
nn
n
R
a
R
k
k
a
R
[
1(
0
0)
0]
R
[
2(
0
0)
0]
ForR
1
,]
)
(
)
(
[
)
(
z
a
nR
n1
a
0
R
n
n
n
k
1
n
n
k
2
n
n
F
R
[
1(
0
0)
2(
0
0)
0
0]
.Hence by Lemma 2, the number of zeros of F(z) and therefore P(z) in
(
R
0
,
c
1
)
c
R
z
does not exceed0 1
log
log
1
a
M
c
, where]
)
(
)
(
[
1 20 1
1
n n n
n n
n n n
n
R
a
R
k
k
a
M
R
[
1(
0
0)
0]
R
[
2(
0
0)
0]
forR
1
and]
)
(
)
(
[
1 20 1
1
n n n
n n
n n n
n
R
a
R
k
k
a
M
R
[
1(
0
0)
2(
0
0)
0
0]
forR
1
. That proves Theorem 1.Proof of Theorem 2: Let n be even.Consider the polynomial
F
(
z
)
(
1
z
2)
P
(
z
)
(
1
z
2)(
a
nz
n
a
n1z
n1
...
a
1z
a
0)
az
2a
1z
1(
a
a
2)
z
(
a
n 1a
n 3)
z
n 1...
(
a
3a
1)
z
3 nn n n
n
n
(
a
2
a
0)
z
2
a
1z
a
0
2
1 1
(
1
1
)
(
1 n
n2)
n
(
2
1
)
n1n n n
n n
n
z
a
z
k
z
k
z
k
z
a
.
)
(
)
(
)
(
)
(
...
)
(
)
(
)
(
)
1
(
)
(
)
1
(
{
)
(
)
(
)
(
)
(
...
)
(
)
(
)
(
2 0 0 3 2 0 3 2 3 1 1 4 3 1 4 3
3 5 3 2
4 2
1 3 1 4
1 1 4
2 3
3 0
1
2 0 0 1 2 0 1 2 3 1 1 2 3 1 2 3
3 5 3 2
4 2 1
3 1 2
z
z
z
z
z
z
z
k
z
k
z
k
z
k
i
a
z
a
z
z
z
z
z
z
z
k
n n n n
n n n
n n
n n n
n n n
n
n n n n
n n
n n n
For
z
R
, we have by using the hypothesis1 1 2 1
2 1
1 1 2
)
1
(
)
(
)
1
(
)
(
n nn n n
n n n
n n
n
R
a
R
k
R
k
R
k
R
a
z
3 1 2 3 1 2 3 1 2 3 2 1 2 1 2 1 2 1 2 2 2 2 2 2 2 2 2 2 3 3 5 2 2 4 1 1 2 3
)
1
(
)
(
...
)
(
)
(
...
)
(
)
(
...
)
(
)
(
)
(
R
R
R
R
R
R
R
R
R
k
n n n n n n n nn
2 2 2 2 2 2 2 2 2 3 2 3 0 1 2 0 1 2 0 1 2)
(
)
(
...
)
(
)
1
(
)
1
(
)
(
R
R
R
k
R
k
a
R
a
R
R
n n n n n
2 0 3 2 0 3 2 3 1 4 3 1 4 3 1 2 3 2 1 2 1 2 1 2 1 2)
1
(
)
(
)
1
(
)
(
)
(
)
(
...
R
R
R
R
R
R
For
R
1
,1 2 1 2 1 0 1 1 2
)
1
(
)
1
[(
)
(
n
n
n
nn n
n n
n
R
a
R
a
R
k
k
k
a
z
F
]
)
1
(
)
(
)
1
(
...
...
)
1
(
)
1
(
)
(
)
1
(
...
...
...
1 0 3 0 3 2 1 4 1 4 3 3 2 1 2 1 2 1 2 2 2 2 2 2 2 3 2 3 0 0 1 0 1 2 1 2 1 2 3 3 2 1 2 1 2 1 2 2 2 2 2 2 2 3 5 2 4 1 2 3a
k
k
a
k
n n n n n n n n n
2
1 1
0
[
n
n
n1
n1 nn n n
n
R
a
R
a
R
a
].
)
1
(
)
1
(
)
1
(
)
1
(
)
(
)
(
)
(
)
(
)
(
2
1 0 4 1 3 0 1 1 2 0 3 0 1 1 4 1 2 1 4 1 2 3 1 1 2 2 1 2 2a
k
k
k
k
n n n n
For
R
1
,1 1 0 1 1 2
[
)
(
z
a
nR
n
a
nR
n
a
R
n
n
n
nF
].
)
1
(
)
1
(
)
1
(
)
1
(
)
(
)
(
)
(
)
(
)
(
2
1 0 4 1 3 0 1 1 2 0 3 0 1 1 4 1 2 1 4 1 2 3 1 1 2 2 1 2 2a
k
k
k
k
n n n n
Hence , by Lemma 2, the number of zeros of F(z) and therefore P(z) in
(
R
0
,
c
1
)
c
R
z
does notexceed 0 2
log
log
1
a
M
c
, where forR
1
,1 1 0 1 1 2
2
[
n n n n n nn n
n
R
a
R
a
R
a
M
2
(
2
21
2
21)
(
k
1
n
k
3
n)
],
)
1
(
)
1
(
)
1
(
)
1
(
)
(
)
(
)
(
1 0 4 1 3 0 1 1 2 0 3 0 1 1 4 1 2 1 4 1 2a
k
k
n n
and for
R
1
,1 1 0 1 1 2
2
[
n n n n nn n
n
R
a
R
a
R
a
].
)
1
(
)
1
(
)
1
(
)
1
(
)
(
)
(
)
(
)
(
)
(
2
1 0 4 1 3 0 1 1 2 0 3 0 1 1 4 1 2 1 4 1 2 3 1 1 2 2 1 2 2a
k
k
k
k
n n n n
The proof for odd n is similar and is omitted. That proves Theorem 2.
Proof of Theorem 3: Suppose that n is even and the coefficient conditions hold i.e.
.
...
...
...
...
1 2 3 1 2 1 2 1 2 0 1 2 2 2 2 1a
a
a
a
a
k
a
a
a
a
a
k
n n
Consider the polynomial
)
...
)(
1
(
)
(
)
1
(
)
(
1 01 1 2 2
a
z
a
z
a
z
a
z
z
P
z
z
F
n nn
n
az
n2
a
n1z
n1
(
a
n
a
n2)
z
n
(
a
n1
a
n3)
z
n1
...
(
a
3
a
1)
z
3
(
a
2
a
0)
z
2
a
1z
a
0
a
nz
n2
a
n1z
n1
(
k
1
1
)
a
nz
n
(
k
1a
n
a
n2)
z
n
(
k
2
1
)
a
nz
n10 1 2 0 0 1 2 0 1 2 3 1 1 2 3 1 2 3 1 2 3 2 1 2 1 2 1 2 1 2 2 2 2 2 2 2 2 2 2 3 5 3 2 4 2 1 3 1 2
)
(
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a
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n n n n n n n n n
For
z
R
, we have by using the hypothesis and Lemma 31 1 2 1 2 1 1 1 2
)
1
(
)
1
(
)
(
n nn n n n n n n n
n
R
a
R
k
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k
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k
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a
z
F
0 1 2 0 1 2 0 1 2 3 1 2 3 1 2 3 1 2 3 2 1 2 1 2 1 2 1 2 2 2 2 2 2 2 2 2 2 3 3 5 2 2 4 1 1 2 3)
1
(
)
1
(
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a
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k
a
n n n n n n n n n
2
1 1
0
[(
1
1)
n
(
1
2)
n1 nn n n
n
R
a
R
a
R
k
a
k
a
a
]
sin
)
(
cos
)
(
)
1
(
)
1
(
sin
)
(
cos
)
(
...
sin
)
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cos
)
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sin
)
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cos
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...
sin
)
(
cos
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sin
)
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cos
)
(
...
sin
)
(
cos
)
(
sin
)
(
cos
)
(
1 0 1 2 0 1 2 0 1 1 2 1 2 3 1 2 3 3 2 1 2 3 2 1 2 1 2 1 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 4 2 4 1 2 1 2a
a
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a
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k
a
a
k
a
n n n n n n n n
a
nR
n2
a
n1R
n1
a
0
R
n[
a
1
(
1
k
1)
a
n
(
1
k
2)
a
n1
