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On the Number of Zeros of a Polynomial in a Given
Domain
M. H. Gulzar
Department of Mathematics University of Kashmir, Srinagar 190006
Email: [email protected]
Abstract: In this paper we obtain bounds for the number of zeros of a polynomial in a given domain when the coefficients of the polynomial or their real or imaginary parts are restricted to certain conditions. Our results generalize some known results in the theory of the distribution of zeros of polynomials.
Mathematics Subject Classification: 30C10, 30C15
Keywords and phrases: Coefficient, Polynomial, Zero.
1. Introduction and Statement of Results In the literature we find a large number of published research papers
(e.g.[2,4,5,7,8,9,10,11,12,13,16,17,19,20]) concerning the number of zeros of a polynomial in a circle.
S. K. Singh [18] proved the following result on the number of zeros of a polynomial:
Theorem A: Let
0
) (
j j jz a z
P be a polynomial o f degree n such that min0jn aj 1
andmax0jn aj an , then the number of zeros of P(z) in
k R
z does not exceed
k R a
n n n
log
} )
1 log{(
2
where
max{ 1 , 2 ,. 3 ,...}
n n
n n
n n
a a a a a a
R .
For the class of polynomials with real coefficients, Q. G. Mohammad [14] proved the following result:
Theorem B: Let
0
) (
j j jz a z
P be a polynomial o f degree n such that
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Then the number of zeros of P(z) in
2 1
z does not exceed
0 log 2 log
1 1
a an
.
Bidkham an d Dewan [3] generalized Theorem B in the following way:
Theorem C: Let
0
) (
j j jz a z
P be a polynomial o f degree n such that
an an1...ak1ak ak1...a1a0 0,
for some k,0k n.Then the number of zeros of P(z) in
2 1
z does not exceed
0 0
0 2
log 2 log
1
a
a a a a
an n k
.
Theorem D: Let
0
) (
j j jz a z
P be a polynomial o f degree n with complex coefficients such that
for some real , , ,0 ,
2
argaj jn and for some 0t 1,0k n,
n
n k
k k k
a t a
t a t a
t
a ... 1 1 ...
1
0 .
Then the number of zeros of P(z) in
2 1
z does not exceed
0 1
0 1
) 1 sin (cos
sin 2 cos 2
log 2 log
1
a t
a t a t a
t n
n n
j j j k
k
.
Ebadian et al [6] generalized the above results by proving the following
results:
Theorem E: Let
0
) (
j j jz a z
P be a polynomial o f degree n such that
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for some k, 0k n.Then the number of zeros of P(z) in
2
R z ,
R>0, does not exceed
0 0 0
1
) (
) (
log 2 log
1
a
a a R a a R a R
an n k k n k n
for R1
and
0 0 0
1
) (
) (
log 2 log
1
a
a a R a a R a R
an n k n k n
for R1 .
Theorem F: Let
0 ) (
j j jz
a z
P be a polynomial o f degree n with complex coefficients such that
for some real , , ,0 ,
2
argaj jn
and for some R0,0kn,
n
n k
k k k
a R a
R a R a
R
a0 1 ... 1 1 ... .
Then the number of zeros of P(z) in
2
R
z does not exceed
0
1
0 1
) 1 sin (cos
sin 2 cos 2
log 2 log
1
a R
a R a R R
a
R n
n n
j
j j k
k
.
In this paper we give generalizations of Theorems E and F. More precisely we prove the following results:
Theorem 1: Let
0
) (
j j jz a z
P be a polynomial o f degree n with Re(aj)j, Im(aj)j such
that for some k,,,0k1,0 1, 0n,
kn n1...1 1...0.
Then the number of zeros of P(z) in (R0,c1)
c R
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0
1
1
1
1 0
0 0 0 0
1
] 2
) (
[
] 2
) (
[
log log
1
a k
R
R a R a
c
n
j j n
n n n
n
j j n
n
for R1
and
0
1
1
1
1 0
0 0 0 0
1
] 2
) (
[
] 2
) (
[
log log
1
a k
R
R a R
a
c
n
j j n
n n n
n
j j n
n
for R1.
Remark 1: Taking k1, 1, c=2,Theorem 1 reduces to Theorem E if aj are real i.e. j 0,j.
For different values of the parameters k,,etc. we get many interesting results e.g. if we take
1
, we get the following result:
Corollary 1: Let
0
) (
j j jz a z
P be a polynomial o f degree n with Re(aj)j, Im(aj)j such
that for some k,0k 1,
kn n1...1 1...0,
n
0 . Then the number of zeros of P(z) in (R0,c1)
c R
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0 1 1 1 1 0 0 0 1 ] 2 ) ( [ ] 2 [ log log 1 a k R R a R a c n j j n n n n n j j n n for R1
and
0 1 1 1 1 0 0 0 1 ] 2 ) ( [ ] 2 [ log log 1 a k R R a R a c n j j n n n n n j j n n for R1.
Taking k1, n in Theorem 1, we get the following result:
Corollary 2: Let
0 ) ( j j jz a zP be a polynomial o f degree n with Re(aj)j, Im(aj)j such
that for some ,0 1,
n n1 ...0.
Then the number of zeros of P(z) in (R0,c1)
c R
z does not exceed
0 1 0 0 0 0 10 [ ( ) 2 ]
log log 1 a R a c n j j n n
n
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and
0
1 0
0 0 0
0 [ ( ) 2 ]
log log
1
a R
a
c
n
j j n
n
for R1.
Taking 1, 0in Theorem 1, we get the following result:
Corollary 3: Let
0 ) (
j j jz
a z
P be a polynomial o f degree n with Re(aj)j, Im(aj)j such
that for some k,0k1,
kn n1 ...a1 0,
then the number of zeros of P(z) in (R0,c1)
c R
z does not exceed
0
1 0
0 1
0 [2 ( ) 2 ]
log log
1
a k R
a
c
n
j j n
n n
n
for R1
and
0
1 0
0
0 [2 ( ) 2 ]
log log
1
a k
R a
c
n
j j n
n
n
for R1.
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Theorem 2: Let
0 ) ( j j jz a zP be a polynomial o f degree n with Re(aj)j, Im(aj)j such
that for some k,0k 1,0 1,
kn n1...1 1...0,
n
0 , then the number of zeros of P(z) in (R0,c1)
c R
z does not exceed
0 1 1 1 1 0 0 0 0 0 1 ] 2 ) ( [ ] 2 ) ( [ log log 1 a k R R a R a c n j j n n n n n j j n n for R1
and
0 1 1 1 1 0 0 0 0 0 1 ] 2 ) ( [ ] 2 ) ( [ log log 1 a k R R a R a c n j j n n n n n j j n n for R1.
Theorem 3: Let
0 ) ( j j jz a zP be a polynomial o f degree n with complex coefficients such that
for some real , , ,0 ,
2
argaj jn and for some R0,0n,0k 1,0 1,
n n a kR a R a R a R
a ... 1 ...
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Then the number of zeros of P(z) in (R0,c1)
c R
z does not exceed
n n n
n n
j
j j
a R R
a k a R R
a R a R c
1 1
1
1 1
0
2 ) 1 sin (cos
sin 2 cos 2
{ 1 log[ log
1
Ra0 (cossin1)2Ra0}].
Remark 2: For different values of the parameters k,,etc., we get many interesting results from Theorem 3 as in case of Theorem 1. For k1, 1, c=2,Theorem 3 reduces to Theorem F.
2.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 and
n k
a
f( k)0, 1,2,..., , then
n
j 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)in z r, then the number of zeros of f(z) in
c r z
(r>0,c>1)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 jz
a z
P be a polynomial o f degree n with complex coefficients such that
for some real , , ,0 ,
2
argaj jn and
, 0
,
1 j n
a
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taj aj1 (taj aj1)cos(taj aj1)sin.
Lemma 3 is due to Govil and Rahman [7].
3.Proofs of Theorems
Proof of Theorem 1: Consider the polynomial
F(z) =(1-z)P(z)
(1z)(anzn an1zn1 ...a1za0)
1 1 0 0
1
) (
... )
(a a z a a z a
z
a n
n n n
n
1
1
1 1
0 1
) (
] ) 1 ( ) [(
n
j
j j j n
n n
n n
nz a ka a k a z a a z
a
a a z a a a z
j
j j
j ) [( ) ( 1) ]
( 1 0 0
2
1
For z R, we have, by using the hypothesis
1
1 1 1
0 1
) (
] ) 1 ( ) [(
) (
n
j
j j j n
n n
n n
n R a k k R R
a z F
n
j
j j j j
j j
j R R R
1
1 0
0 1 1
1) [( ) (1 ) ] ( )
(
.
Which gives
( ) [ ( ) 2 ]
1
1 0
1
n
j j n
n n n
n n
n R a R k
a z F
] 2
) (
[
1
1 0
0 0
0
j j
R
for R1
and
( ) [ ( ) 2 ]
1
1 0
1
n
j j n
n n n
n n
n R a R k
a z F
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[ ( ) 2 ]
1
1 0
0 0
0
j j R
for R1.
Thus
1 [ { ( ) 2 }
) 0 (
)
( 1
1 0
1
0
n
j j n
n n n
n n
n R a R k
a a F
z F
[ ( ) 2 ]
1
1 0
0 0
0
j j
R
for R1
and
1 [ { ( ) 2 }
) 0 (
)
( 1
1 0
1
0
n
j j n
n n n
n n
n R a R k
a a F
z F
[ ( ) 2 ]
1
1 0
0 0
0
j j
R
for R1.
Hence , by Lemma 2,it follows that the number of zeros of F(z) in (R0,c1)
c R
z does not
exceed
0
1
1
1
1 0
0 0 0 0
1
] 2
) (
[
] 2
) (
[
log log
1
a k
R
R a R
a
c
n
j j n
n n n
n
j j n
n
for R1
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0 1 1 1 1 0 0 0 0 0 1 ] 2 ) ( [ ] 2 ) ( [ log log 1 a k R R a R a c n j j n n n n n j j n n for R1.
As the number of zeros of P(z) in (R0,c1)
c R
z does not exceed the
number of zeros of F(z) in (R0,c1)
c R
z , the theorem follows.
Proof of Theorem 3: Consider the polynomial
F(z) =(R-z)P(z)
(Rz)(anzn an1zn1 ...a1za0)
a z Ra Ra a z Ran an zn Ra a z
n n n n
n ( ) ( ) ... ( 1 0)
1 2 1 1
0
1
1 2 1 1 0 1 ) ( )] ( ) [( n n n n n n n n n
nz Ra kRa a kRa Ra z Ra a z
a z a a a Ra z a Ra z a Ra z a Ra )] ( ) [( ) ( ... ) ( ) ( ... 0 0 0 1 2 1 2 1 1 1
For z R, we have
1 2 1 1 1 0 1 1 )
(z an Rn Ra k Rn an kRan an Rn Ran an Rn
F , ) 1 ( ... ... 0 0 1 2 1 2 1 1 1 R a R a Ra R a Ra R a Ra R a Ra
which gives ,by using Lemma 3and the hypothesis
n n n n n n n n
n R Ra k R a a kRa a kRa R
a z
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R a R
a a R
a a R R
a a R a
a R
R a
a R a
a R
R a
R a a
R a
R a
R a
a R
an n n n n
0 0
1
0 1 2
1 2 1
2
1 1
1 1
1
1 1
2 1
2
) 1 ( ] sin ) (
cos ) [(
] sin ) (
cos ) [(
... ]
sin ) (
cos ) [(
] sin ) (
cos ) [(
... ]
sin ) (
cos ) [(
n
n n
n R R a
a k a
R 1 cos 1(cos sin 1) 2 1
2
1
1 0
0(cos sin 1) 2 2 sin
n
j
j j
a R R
a R a
R
n n n n
n
j
j j
a R R
a k a R R
a
R 1 1
1
1 1
2 ) 1 sin (cos
sin 2 cos
2
Ra0 (cos sin 1)2Ra0 .
The rest of the proof can be completed on the same lines as in the proof of Theorem 1.
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