e
-ISSN: 2250-3021,
p
-ISSN: 2278-8719
Vol. 3, Issue 1 (Jan. 2013), ||V3|| PP 50-59
On the Zeros of a Polynomial inside the Unit Disk
M. H. Gulzar
Department of Mathematics University of Kashmir, Srinagar 190006
Abstract:In this paper we find an upper bound for the number of zeros of a polynomial inside the unit disk, when the coefficients of the polynomial or their real and imaginary parts are restricted to certain conditions.
Mathematics Subject Classification:30C10, 30C15
Key-words and phrases: Polynomial, Unit disk, Zero.
I. INTRODUCTION AND STATEMENT OF RESULTS
Regarding the number of zeros of a polynomial inside the unit disk, the following results were recently proved by M. H. Gulzar [2] :
Theorem A: Let P(z)=
nj j j
z
a
0be a polynomial of degree n such that for some
0
,
a
n
a
n1
...
a
1
a
0.Then the number of zeros of P(z) in
,
0
1
1
0
z
M
a
does not exceed
0
0 0
2
log
1
log
1
a
a
a
a
a
n
n
,
where
M
1
2
a
n
a
n
a
0.Theorem B: Let P(z)=
nj j j
z
a
0be a polynomial of degree n with complex coefficients .If
,
Re
a
j
jIm
a
j
j,
j
0
,
1
,...,
n
,
and for some
0
,
n
n1
...
1
0,
then the number of zeros of P(z) in
,
0
1
,
2
0
z
M
a
does not exceed
, 2
2 log 1 log
1
0
0 0 0
a
n
j j n
n
where
nj j n
n
M
1 0
0
2
2
2
.Theorem C: Let P(z)=
nj j j
z
a
0be a polynomial of degree n with complex coefficients .If
,
Re
a
j
jIm
a
j
j,
j
0
,
1
,...,
n
,
and for some
0
,
n
n1
...
1
0,
then the number of zeros of P(z) in
,
0
1
,
3
0
z
M
a
, 2
2 log 1 log
1
0
0 0 0
a
n
j j n
n
where
n
j j n
n M
1 0
0
3 2 2 .
TheoremD. Let P(z)=
n
j j jz a
0
be a polynomial of degree n with complex coefficients such that
argaj 2,j0,1,...n,
for some real
and
0 1 1 ... a a
a
an n
,for some
0
.Then the number of zeros of P(z) in ,0 1,
4
0
z M
a does not exceed
, sin
2 ) 1 sin (cos
) 1 sin )(cos
( log 1 log
1
0
1
1 0
a
a a
a
n
j j
n
where
1
1 0
4 ( )(cos sin 1) (cos sin ) 2sin
n
j j
n a a
a
M
In this paper, we try to give generalizations of the above results. In fact, we prove the following :
Theorem 1: Let
n
j j jz a z
P
0
)
( be a polynomial of degree n with
Re(
a
j)
j andIm(
a
j)
j, j=0,1,2,……,n. If for some real numbers
,
0
,1
k
n
,
nk
0
,
nk1
nk,
0
1
,
n
n1
...
nk1
nk
nk1
...
1
0,then the number of zeros of P(z) in
,
0
1
5
0
z
M
a
, does not exceed
0
0 0
0
0 2 2
( 1
) 1 ( 2
log 1 log
1
a
n
j j k
n k
n n
n
,
where
n
j j k
n k
n n
n M
1 0
0 0 0
5 2 ( 1) 1 ( ) 2 ,
and if
nk
nk1,
then the number of zeros of P(z) in,
0
1
60
z
M
a
, does not exceed
0
0 0
0
0 ) 2 2
( 1
) 1 ( 2
log 1 log
1
a
n
j j k
n k
n n
n
where
n
j j k
n k
n n
n
M
1 0
0 0 0
6 2 (1 ) 1 ( ) 2
Remark 1: Taking
1
,
1
,Theorem 1 reduces to Theorem B.Taking
1
in Theorem 1, we get the following result:Corollary 1: Let
n
j j jz
a z
P
0 )
( be a polynomial of degree n with
Re(
a
j)
j andIm(
a
j)
j, j=0,1,2,……,n. If for some real numbers
0
,0
1
,
then the number of zeros of P(z) in ,0 1
7
0
z M
a
, does not exceed
0
0 0
0
0 ) 2 2
( 2
log 1 log
1
a
n
j j n
n
,
where
n
j j n
n
M
1 0
0 0 0
7 2 ( ) 2
Remark 2: If
a
j are real i.e.
j
0
for all j , Theorem 1 gives the following result which reduces to Theorem A by taking
1
,
1
:Theorem 2: Let
n j
j jz
a z
P
0 )
( be a polynomial of degree n .If for some real numbers
,
0
,,
0
,
1
k
n
a
nk
a
nk1
a
nk,
0
1
,
a
n
a
n1
...
a
nk1
a
nk
a
nk1
...
a
1
a
0,then the number of zeros of P(z) in
,
0
1
8
0
z
M
a
, does not exceed
0
0 0
0 ) 2
( 1
) 1 ( 2
log 1 log
1
a
a a
a a
a a
an n n k n k
,
where
M
8
2
a
n
a
n
(
1
)
a
nk
1
a
nk
(
a
0
a
0)
a
0 ,and if
a
nk
a
nk1,
then the number of zeros of P(z) in,
0
1
9
0
z
M
a
, does not exceed
0
0 0
0 ) 2
( 1
) 1 ( 2
log 1 log
1
a
a a
a a
a a
an n n k n k
,
where
M
9
2
a
n
a
n
(
1
)
a
nk
1
a
nk
(
a
0
a
0)
a
0 .Applying Theorem 1 to the polynomial –iP(z), we get the following result, which reduces to Theorem C by taking
1
:Theorem 3: Let
nj j j
z
a
z
P
0
)
(
be a polynomial of degree n withRe(
a
j)
j andIm(
a
j)
j, j=0,1,2,……,n. If for some real numbers
,
0
,1
k
n
,
nk
0
,
nk1
nk,
0
1
,
n
n1
...
nk1
nk
nk1
...
1
0,then the number of zeros of P(z) in
,
0
1
10
0
z
M
a
, does not exceed
0
0 0 0
0 ) 2 2
( 1
) 1 ( 2
log 1 log
1
a
n
j j k
n k
n n
n
where
n
j j k
n k
n n
n
M
1 0
0 0 0
10 2 ( 1) 1 ( ) 2 ,
and if
nk
nk1,
then the number of zeros of P(z) in ,0 1 110
z M
, 2
2 ) (
1 )
1 ( 2
log 1 log
1
0
0 0
0 0
a
n
j j k
n k
n n
n
where
. 2
) (
1 )
1 ( 2
1 0
0 0 0
11
n
j j k
n k
n n
n
M Theore
m 4: Let
nj j j
z
a
z
P
0
)
(
be a polynomial of degree n .If for some real numbers
0
,
0
,,
0
,
1
k
n
a
nk
0
1
,
a
n
a
n1
...
a
nk1
a
nk
a
nk1
...
a
1
a
0 ,and for some real
,a
j,
j
0
,
1
,...,
n
2
arg
anda
nk1
a
nk , i.e.
1
,then thenumber of zeros of P(z) in
,
0
1
12
0
z
M
a
, does not exceed
, ]
sin 2 2 ) 1 sin (c os
) 1 sin
c os sin
(c os )
1 sin (c os [(
log 1 log
1
0
1
, 1 0
0
a
a a
a
a a
n
k n j j
j k
n n
where
M
12
(
a
n(cos
sin
1
)
a
nk(cos
sin
cos
sin
1
)
1, 1 0
0
(cos
sin
1
)
2
sin
n
k n j j
j
a
a
a
and if
a
nk
a
nk1 , i.e.
1
, then the number of zeros of P(z) in,
0
1
13
0
z
M
a
, does not
exceed
, ]
sin 2 2
) 1 sin (c os
) 1 sin c os
sin (c os
) 1 sin (c os
[(
log 1 log
1
0
1
, 1 0
0
a
a a
a
a a
n
k n j j
j k
n n
where
)
1
sin
cos
sin
(cos
)
1
sin
(cos
(
13
a
n
a
nk
M
1
, 1 0
0 (cos sin 1) 2sin
n
k n j j
j a a
a
.
Remark 4: Taking
1
,
1
, Theorem 4 reduces to Theorem D.II. LEMMAS
For the proofs of the above results , we need the following results:
Lemma 1: . Let P(z)=
nj j j
z
a
0be a polynomial of degree n with complex coefficients such that
for
n
j
a
j,
0
,
1
,...
,
2
arg
some real
,then for some t>0,
c os
sin .1 1
1
a ta a ta aj
taj j j j j
Lemma 2.If p(z) is regular ,p(0)
0 andp
(
z
)
M
inz
1
,
then the number of zeros of p(z) in,
1
0
,
z
does not exceed) 0 ( log 1 log 1 p M (see[4],p171).
III. PROOFS OF THEOREMS
Proof of Theorem 1: Consider the polynomial
F
(
z
)
(
1
z
)
P
(
z
)
k n k n k n k n k n k n n n n n n n n n n
z
a
a
z
a
a
z
a
a
z
a
a
z
a
z
a
z
a
z
)
(
)
(
...
)
(
)
...
)(
1
(
1 1 1 1 1 0 1 1 1
(
a
nk1
a
nk2)
z
nk1
...
(
a
1
a
0)
z
a
00 1 1 0 0 1 1 2 1 1 1 1 1 1
)
(
)
(
...
)
(
)
(
)
(
...
)
(
)
(
i
z
i
z
z
z
z
z
z
i
n j j j j k n k n k n k n k n k n k n k n k n n n n n n n
If
nk1
nk,
then1 1 1
1
)
(
...
)
(
)
(
)
(
nk
nk nkn n n n n n
n
i
z
z
z
z
z
F
.
)
(
)
1
(
)
(
...
)
(
)
1
(
)
(
0 1 1 0 0 0 1 1 2 1 1
i
z
i
z
z
z
z
z
n j j j j k n k n k n k n k n k n k n k n
For
z
1
,
...
)
(
z
n n n 1F
nk1
nk+
nk
nk1
1
nk
n j j k n k n 0 0 0 0 1 21
...
(
1
)
2
n j j k n k n n n 0 0 00
)
2
2
(
1
)
1
(
2
Hence by Lemma 2, the number of zeros of F(z) in
z
,
0
1
, does not exceed. 2 2 ) ( 1 ) 1 ( 2 log 1 log 1 0 0 0 0 0 a n j j k n k n n n
On the other hand, let
Q(z)=
a
nz
n1
(
a
n
a
n1)
z
n
...
(
a
nk1
a
nk)
z
nk1
(
a
nk
a
nk1)
z
nk
(
a
n ka
n k)
z
n k...
(
a
1a
0)
z
12
1
For
z
1
,k n k n k n k n k n n n n
z
Q
(
)
1
...
1
1
1
n j j k n k 1 0 0 0 1 21 ... (1 ) 2
5 1
0 0 0
0 ) 2
( 1
) 1 (
2 M
n
j j k
n n
n
.
Since Q(0)=0, we have, by Rouche’s theorem,
z
M
z
Q
(
)
5 , forz
1
. Thus
a
0
Q
(
z
)
a
0
M
5z
0
if
5 0
M a z .
This shows that F(z) has no zero in
5 0
M a
z . Consequently it follows that the number of zeros of F(z) and
hence P(z) in
,
0
1
5
0
z
M
a
, does not exceed
. 2
2 ) (
1 )
1 ( 2
log 1 log
1
0
0 0
0 0
a
n
j j k
n k
n n
n
If
nk
nk1,
then
(
)
(
)
1
(
1)
...
(
1
)
nk1 k n kn n
n n n
n n
n
i
z
z
z
z
z
F
.
)
(
)
1
(
)
(
...
)
(
)
1
(
)
(
0 1
1
0 0 0
1
1 2 1
1
i
z
i
z
z
z
z
z
n
j
j j j
k n k n k n
k n k n k
n k n k n
For
z
1
,
...
)
(
z
n n n 1F
nk1
nk+
nk
nk1
1
nk
nj j k
n k n
0 0
0 0 1
2
1
...
(
)
2
2
nj j k
n k
n n
n
0 0
0
0
)
2
2
(
1
)
1
(
2
Hence by Lemma 2, the number of zeros of F(z) in
z
,
0
1
, does not exceed
0
0 0
0
0 ) 2 2
( 1
) 1 ( 2
log 1 log
1
a
n
j j k
n k
n n
n
.
On the other hand, let
Q(z)= n k n k n k
k n k n k n n
n n n
n
z
a
a
z
a
a
z
a
a
z
a
(
)
...
(
)
(
1)
1 1
1 1
(
a
n ka
n k)
z
n k...
(
a
1a
0)
z
12
1
)
(
)
(
z
a
0Q
z
n
j
j j j
k n k n k n k
n k n k
n k n k n
k n k n k
n n
n n n
n n n
z
i
z
z
z
z
a
z
z
z
z
z
i
1
1 0
0 1
1 2 1
1 1
1 1
1 1
)
(
)
1
(
)
(
...
)
(
)
1
(
)
(
)
(
...
)
(
)
(
For
z
1
,k n k
n k n k n k
n n
n n
z
Q
(
)
1
...
1
1
1
nj j k
n k
1 0
0 0 0 1
2
1
...
(
)
2
6 1
0 0 0
0 ) 2
( 1
) 1 (
2 M
n
j j k
n n
n
.
Since Q(0)=0, we have, by Rouche’s theorem,
z
M
z
Q
(
)
6 , forz
1
. Thus
a
0
Q
(
z
)
a
0
M
6z
0
if
6 0
M
a
z
.This shows that F(z) has no zero in
6 0
M
a
z
. Consequently it follows that the number of zeros of F(z) andhence P(z) in
,
0
1
6
0
z
M
a
, does not exceed
0
0 0
0
0 ) 2 2
( 1
) 1 ( 2
log 1 log
1
a
n
j j k
n k
n n
n
.
That proves Theorem 1.
Proof of Theorem 4: Consider the polynomial
F
(
z
)
(
1
z
)
P
(
z
)
k n k n k n k
n k n k n n
n n n
n
n n n n
z
a
a
z
a
a
z
a
a
z
a
a
z
a
z
a
z
a
z
)
(
)
(
...
)
(
)
...
)(
1
(
1 1
1 1
1
0 1 1
1
1 0 0
1 2
1
)
...
(
)
(
a
a
z
n ka
a
z
a
k n k
n
If
a
nk1
a
nk , i.e.
1
, then1 1 1
1
)
(
...
)
(
)
(
nk
nk nk nn n n
n
n
z
z
a
a
z
a
a
z
a
z
F
(
a
nk
a
nk1)
z
nk
(
1
)
a
nkz
nk
(
a
nk1
a
nk2)
z
nk1
...
(
a
1
a
0)
z
(
1
)
a
0z
a
0so that for
z
1
, we have by using Lemma 1,)
(
)
(
z
a
0Q
z
1 1
1
...
)
(
z
a
n
a
n
a
n
a
nk
a
nk
a
nk
a
nkF
1
a
nk
a
nk1
a
nk2
...
a
1
a
0
1
a
0
a
0
a
n
(
a
n
a
n1)
cos
(
a
n
a
n1)
sin
...
0 0 0
1 0
1
2 1
2 1
1 1
1 1
)
1
(
sin
)
(
cos
)
(
...
sin
)
(
cos
)
(
sin
)
(
cos
)
(
)
1
(
sin
)
(
cos
)
(
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
k n k
n k
n k
n
k n k n k
n k n
k n k
n k
n k
n k
n
(
a
n)(cos
sin
1
)
a
nk(cos
sin
cos
sin
1
)
1, 1 0
0
(cos
sin
1
)
2
2
sin
n
k n j j
j
a
a
a
Hence , by Lemma 2, the number of zeros of F(z) in
z
,
0
1
, does not exceed0
1
, 1 0
0 (c os sin 1) 2 2sin ]
) 1 sin
c os sin
(c os )
1 sin (c os
[(
log 1 log
1
a
a a
a
a a
n
k n j j
j k
n n
On the
other hand, let
Q(z)= n k n k n k
k n k n k n n
n n n
n
z
a
a
z
a
a
z
a
a
z
a
(
)
...
(
)
(
1)
1 1
1 1
(
a
n ka
n k)
z
n k...
(
a
1a
0)
z
12
1
For
z
1
,
)
(
z
Q
(
a
n)(cos
sin
1
)
a
nk(cos
sin
cos
sin
1
)
1, 1 0
0
(cos
sin
1
)
2
sin
n
k n j j
j
a
a
a
M
12.Since Q(0)=0 , we have , by Rouche’s Theorem,
Q
(
z
)
M
11z
,
forz
1
.Thus, for
z
1
,)
(
)
(
z
a
0Q
z
F
a
0
Q
(
z
)
a
0
M
12z
0
if.
12 0
M
a
z
This shows that F(z) has all its zeros z with
z
1
in.
12 0
M
a
z
Thus, the number of zeros of F(z) and hence P(z) in
,
0
1
12
0
z
M
a
0 1
, 1 0
0 (c os sin 1) 2sin ]
) 1 sin
c os sin
(c os )
1 sin (c os
[(
log 1 log
1
a
a a
a
a a
n
k n j j
j k
n n
If
a
nk
a
nk1 , i.e.
1
, then
(
)
1
(
1)
...
(
1
)
nk1 k n kn n
n n n
n
n
z
z
a
a
z
a
a
z
a
z
F
(
a
nk
a
nk1)
z
nk
(
1
)
a
nkz
nk1
(
a
nk1
a
nk2)
z
nk1
...
(
a
1
a
0)
z
(
1
)
a
0z
a
0so that for
z
1
, we have by Lemma 1,1 1
1
...
)
(
z
a
n
a
n
a
n
a
nk
a
nk
a
nk
a
nkF
1
a
nk
a
nk1
a
nk2
...
a
1
a
0
1
a
0
a
0
a
n
(
a
n
a
n1)
cos
(
a
n
a
n1)
sin
...
0 0 0
1 0
1
2 1
2 1
1 1
1 1
) 1 ( sin ) (
c os ) (
... sin
) (
c os ) (
sin ) (
c os ) (
1 sin ) (
c os ) (
a a a
a a
a
a a
a a
a a
a a
a a
a a
a
k n k
n k
n k
n
k n k n k
n k n
k n k
n k
n k
n k
n
(
a
n)(cos
sin
1
)
a
nk(cos
sin
cos
sin
1
)
1, 1 0
0
(cos
sin
1
)
2
2
sin
n
k n j j
j
a
a
a
Hence , by Lemma 2, the number of zeros of F(z) in
z
,
0
1
, does not exceedlog
1
log
1
01
, 1 0
0(c os sin 1) 2 2sin ]
) 1 sin c os
sin (c os )
1 sin (c os [(
a
a a
a
a a
n
k n j j
j k
n n
On the other hand, let
Q(z)= n k n k n k
k n k n k n n
n n n
n
z
a
a
z
a
a
z
a
a
z
a
(
)
...
(
)
1(
1)
1 1
1
(
a
n ka
n k)
z
n k...
(
a
1a
0)
z
12
1
For
z
1
, by using Lemma 1,
)
(
z
Q
(
a
n)(cos
sin
1
)
a
nk(cos
sin
cos
sin
1
)
1
, 1 0
0 (cos sin 1) 2sin
n
k n j j
j a a
a
M
13.Since Q(0)=0 , we have , by Rouche’s Theorem, ( ) ,
13 z
M z
Q for
z
1
.Thus, for
z
1
,) ( )
(z a0 Q z
F
a
0
Q
(
z
)
0
if
.
13 0
M
a
z
This shows that F(z) has all its zeros z with
z
1
in.
13 0
M
a
z
Thus, the number of zeros of F(z) and hence P(z) in
,
0
1
13
0
z
M
a
, does not exceed
log
1
log
1
01
, 1 0
0(cos sin 1) 2sin ]
) 1 sin cos
sin (cos )
1 sin (cos [(
a
a a
a
a a
n k n j j
j k
n n
.
That proves Theorem 4.
REFERENCES
[1] N.K.Govil and Q.I.Rahman,On the Enestrom-Kakeya Theorem, Tohoku Math.J.20(1968) , 126-136. [2] M. H. Gulzar, On the number of Zeros of a polynomial in a prescribed Region, International journal of
Mathematical Archive-2(9), 2011, 35-46.
