some inequalities for polynomials vanishing inside a circle

Mathematical Journal of Interdisciplinary Sciences

Vol. 3, No. 1, September 2014

pp. 1–14

DOI: 10.15415/mjis.2014.31001

Some Inequalities for Polynomials Vanishing Inside a Circle

ABDULLAH MIR* AND BILAL AHMAD DAR†

Department of Mathematics, University of Kashmir, Srinagar-190006, India

*Email: [email protected]; †Email: [email protected]

Received: March 22, 2013| Revised: February 13, 2014| Accepted: February 13, 2014

Published online: September 20, 2014 The Author(s) 2014. This article is published with open access at www.chitkara.edu.in/publications

Abstract: If P z a zjn

jj( )= =Σ 0 is a polynomial of degree n having all its zeros

in | | ,z k k≤ ≥1, then it is known that

Max P zn

kMax P zz n z| | | || | | ( ) |= =

′ ≥+1 11

( )

In this paper we obtain a generalization as well as an improvement of the above inequality. Besides this some other results are also obtained.

Keywords and Phrases: Inequalities in the complex domain, Polynomials, Extremal prob lems, Zeros.

Mathematics Subject Classification (2010): 30A10, 30C10, 30C15.

1. Introduction and Statement of Results

If P(z) is a polynomial of degree n, then concerning the estimate of | ( ) |′P z , it was shown by Turan [8] that if P(z) has all its zeros in |z| ≤ 1, then

Max P zn

Max P zz z| | | || | | ( ) |= =′ ≥1 12( ) (1)

In (1) equality holds if all the zeros of P(z) lie in |z| = 1.More generally if the polynomial P(z) has all its zeros in |z| ≤ k ≤ 1, it was

proved by Malik [7] that the inequality (1) can be replaced by

Max P zn

kMax P zz z| | | || | | ( ) |= =

′ ≥+1 11

( ) (2)

The case when P(z) has all its zeros in | | ,z k k≤ ≥1 , was setteled by Govil [5], who proved

2

Mir, A Dar, BA

Theorem A. If P z a zjn

jj( )= =Σ 0 is a polynomial of degree n havihg all its

zeros in |z| ≤ k,k ≥ 1, then

Max P zn

kMax P zz n z| | | || | | ( ) |= =

′ ≥+1 11

( ) (3)

The result is best possible and equality holds for the polynomial P(z) = zn + kn.Aziz [1] improved upon the bound in (3) by taking into account the location

of all the zeros of the polynomial P(z) instead of concerning only the zero of largest modulus. More precisely, he proved the following result.

Theorem B. If all the zeros of the polynomial P z z zjn

j( ) ( )= −=Π 1 of degree n lie in |z| ≤ k, k ≥ 1, then

Max P zk

k

k zMax P zz n

jj

n

z| | | || || |

| (==

=′ ≥

+ +

∑11

1

2

1( ) )) | (4)

The result is best possible and equality holds in (4) for P(z) = zn + kn.Govil [3] also obtained the following improvement of Theorem B.

Theorem C. Let P z a z a z z ajn

j

j

n j nj

n( ) ( ),= = − ≠= =∏∑ 0 0

1, be a polynomial

of degree n z k j nj j≥ ≤ ≤ ≤2 1,| | , , and let k max k k kn= ≥{ , , , }1 2 1… . Then

Max p zk

k

k kP zz n

jj

n

z| | | || ( ) | max | ( ) |==

=′ ≥

+ +

∑11

1

2

1

+++ +

−−

−−

+ −

−−

=∑2

1

1 1 1

2

11

12

1

1

| |

| |

a

k k k

k

n

k

n

a

nn

j

n n

j

n

kk2

(5)

if n > 2 and

Max p zk

k

k kP zz n

jj

n

z| | | || ( ) | max | ( ) |==

=′ ≥

+ +

∑11

1

2

1

++−+ +

+ −

=

=∑( )

| |

| | .

k

ka

k k

ka if n

n

njj

n1

1

1

11

2

11

1

(6)

In (5) and (6) equality holds for P(z) = zn + kn.

3

Some Inequalities for Polynomials

Vanishing Inside a Circle

In this paper we prove the following result which gives an improvement of Theorem C, and thus as well of inequality (4).

Theorem D. Let P z a z a z z ajn

j

j

n j nj

n( ) ( ),= = − ≠= =∏∑ 0 0

1 be a polynomial of

degree n ≥ 2 and P z k j nj j( ) ,| | ,0 0 1≠ ≤ ≤ ≤ , and let k max k k kn= ≥{ , , , }1 2 1… . Then

Max P zk

k

k kMax P zz n

jj

n

z| | | || ( ) | | ( ) |==

=′ ≥

+ +

∑11

1

2

1

++−+ +

+′

==∑k

k k

k

k kP z

k P

n

n njj

n

z k

1

1

0

1( )min | ( ) |

| (

| |

)) | | ( ) |

( )

+ ′

+ +

−−

−−

=

−

∑k Q

k k

k

k k

k

n

kn

n njj

n n n1

1

20

1

1 1

nn

kP if n

−

+ −

′ >

2

11

0 22

| ( ) | .

(7)

and

Max P zk

k

k kMax P zz n

jj

n

z| | | || ( ) | | ( ) |==

=′ ≥

+ +

∑11

1

2

1

++−+ +

+′

==∑k

k k

k

k kP z

k P

n

n njj

n

z k

1

1

0

1( )min | ( ) |

| (

| |

)) | | ( ) |

( )

( )+ ′

+ +

−

+ −

−

=∑k Q

k k

k

k k

k

n

k

n

n njj

n n1

1

0

1

1

11

′ =| ( ) | .P if n0 2 (8)

where Q z z Pz

n( )=

1.

The result is best possible and equality holds in (7) and (8) for P(z) = zn + kn.

Since k

k kj+≥

1

2 for 1 ≤ j ≤ n, then above theorem gives in particular.

Corollary 1. If P z a z z an jn

j n( ) ( ),= − ≠=Π 1 0 , is a polynomial of degree n ≥ 2 having all its zeros in |z| ≤ k, k ≥ 1, P(0) ≠ 0, then

4

Mir, A Dar, BA

Max P zn

kMax P z

n k

k kz n z

n

n n z| | | | | || ( ) | | ( ) |( )

( )min= =

′ ≥+

+−+1 11

1

2 1 ==

− −

+′ + ′

+−−

−−

k

n

n n

n n

P z

n k P k Q

k k

k

n

k

n

| ( ) |

( | ( ) | | ( ) |)

( )

0 0

2 1

1 1

2

1 2

+ −

′ >1

10 2

2kP if n| ( ) | , (9)

and

Max P zn

kMax P z

n k

k kz n z

n

n n z| | | | | || ( ) | | ( ) |( )

( )min= =

′ ≥+

+−+1 11

1

2 1 ==

−

+− ′ + ′

+

+ −

k

n n

n n

P z

k k P k Q

k k

k

| ( ) |

( ) ( | ( ) | | ( ) |)

( )

1 0 0

2 1

11

1

′ =| ( ) | ,P if n0 2

(10)

where Q z z Pz

n( )=

1.

Equality holds in (9) and (10) for P(z) = zn + kn.

It is easy to verify that if k > 1 and n > 2, then k

n

k

n

n n−−

−−

>

−1 1

20

2

,

hence for polynomials of degree > 1, (9) and (10) together provide a refinement of Theorem A.

Now, on applying above theorem to Q z z Pz

n( )=

1, we obtain the

following result which gives a generalization as well as an improvement upon some well known inequalities.

Corollary 2. If P z a z z an jn

j n( ) ( ),= − ≠=Π 1 0 , is a polynomial of degree n ≥ 2 and |z

j| ≥ k

j and let k = min{k

1, k

2, ...,k

n} ≤ 1, then if | ( ) |′P z and | ( ) |′Q z attain

the maximum at the same point on |z| = 1, we have

Max P zk

n kk k

k kMaxz n

n j

jj

n

z| | | || ( ) |==

=′ ≤

+−

−

+∑

1

1

1

111

1

1

1

| ( ) |

min| |

P z

k

k

k

k kj

n

n

j

j

n

z−−

+ +

==∑ kk P z| ( ) |

5



−

′ + ′

+ +

−−

−

=∑

( | ( ) | | ( ) |)k P k Q

k

k

k kj

k

n

kn

n

j

j

n n0 0

1

11

1

2 (( )

( ) | ( ) | ,

1

2

2

1 0 22

−

−

−

− − ′ >

k

n

n

k Q if n

(11)

and

Max P zk

n kk k

k kMaxz n

n j

jj

n

z| | | || ( ) |==

=′ ≤

+−

−

+

∑11

1

1 11

1

1

1

| ( ) |

min| |

P z

k

k

k

k kj

n

n

j

j

n

z−−+

+

=∑ ==

−

=

−′ + ′

+ +

∑

k

n

n

j

j

n

P z

k P k Q

k

k

k kj

| ( ) |

( | ( ) | | ( ) |)0 0

1

1

1−

− − ′ =

( )

( ) | ( ) | ,

1

1 0 2

k

n

k Q if n

n

(12)

where Q z z Pz

n( )=

1.

The result is best possible for k k j nj = ≤ ≤,1 and equality holds in (11) and (12) for P(z) = zn + kn.

Proof of Corollary 2. Since the zeros of P(z) satisfy | | ,z k j nj j≥ ≤ ≤1 , such

that k min k k kn= { , , , }1 2 … . It follows that the zeros of Q z z Pz

n( )=

1 satisfy

1 11

z kj n

j j

≤ ≤ ≤, , such that 1 1 1 1

1 2k k k kn

=

max , , ,… or equivalently k =

min{k1, k

2,...,k

n}. Applying inequality (7) to the polynomial Q(z) and for the

case n > 2, we have

Max Q z

k

k

k k

z n

j

j

n

| | | ( ) |==

′ ≥

+

+

∑11

2

11

1

1 1

+−

+

=Max Q z

k

k

z

n

n

| | | ( ) |1

11

11 +

= =∑1

1

1 111

k

k

k k

Q

nj

j

n

zk

min | (| |

zz) |

6

Mir, A Dar, BA

+

′ +′

+

+

−

10

10

11 1

1

1 1

1kQ

k P

k k

k

k

n

n n

| ( ) || ( ) |

kk

k

n

k

j

j

n

n

=∑

×

−

−

1

11

1 −

−

+ −

−n

n

k

2

1

21

1

1

′

=+ +

=∑

2

1

0

2

1

| ( ) |Q

k

k

k

k k

n

njj

n

+′ + ′

+

=

−

Max P z

k k P k Q

k

k

z

n n

n

| | | ( ) |

( | ( ) | | ( ) |

( )

1

10 0

1jj

jj

n

n

n

n

n

k k

k

nk

k

n k

+

×−

−−−

=

−

−

∑1

2

2

1 1

2( ) + − ′( ) | ( ) | .1 02k Q (13)

Let Max P z P ezi

| | | ( ) | ( ) |=′ = ′

1α , where 0 ≤ α ≤ 2π. Since | ( ) |′P z and

| ( ) |′Q z attains their maximum at the same point on |z| = 1, it follows that Max Q z Q ez

i| | | ( ) | ( ) |=

′ = ′1

α .

We have by Lemma 1(stated in section 2) that

| ( ) | | ( ) | | ( ) || |′ + ′ ≤ =P e Q e nMax P zi i

zα α

1 (14)

Using (13) in (14), we get

nMax P z P ek

k

k

k kMz

in

n

j

jj

n

| | | ( ) | | ( ) |==

≥ ′ ++ +

∑11

2

1α aax P z

k

K

k

k kP z

z

n

n

j

jj

n

z k

| |

| |

| ( ) |

min | (

=

==+

−+

+∑

1

1

1

1)) |

( | ( ) | | ( ) |)

( )

(+

′ + ′

+ +−

−−−

=

−

∑k P k Q

k

k

k k

k

n

k kn

n

j

jj

n n n0 0

1

1 11

1

2 2 ))

( ) | ( ) |

n

k Q

−

+ − ′

2

1 02

implying

7



| ( ) | | ( ) || |′ ≤ −

+ +

−

==∑P e n

k

k

k

k kMax P zi

n

n

j

jj

n

zα 2

1 11

11

1

0

1

−+

+

−′ +

==∑k

K

k

k kP z

k P k

n

n

j

jj

n

z kmin | ( ) |

( | ( ) |

| |

nn

n

j

jj

n n nQ

k

k

k k

k

n

k k

n

−

=

−′

+ +−

−−−

∑

1

1

2 20

1

1 1

2

| ( ) |)

( )

( )

− − ′( ) | ( ) |1 02k Q

which proves the desired inequality for n > 2.The result for n = 2 follows on the same lines as that for n > 2 but instead

of using (7), we use (8).As it is easy to see, our Corollary 2 provides a generalization as well as an

improvement of the following result due to A. Aziz and N.Ahmad [2].

Theorem D. Let P z z zjn

j( ) ( )= −=Π 1 be a polynomial of degree n which does

not vanish in |z| < k, where k ≤ 1, and let Q z z Pz

n( )=

1. If | ( ) |′P z and

| ( ) |′Q z become maximum at the same point on |z| = 1, then

Max P zk

n kz k

z kMaxz n

n j

jj

n

| | | ( ) || |

| |==

′ ≤+

−−

+

∑11

1

1 || | | ( ) | .z P z=1 (15)

The result is best possible and equality holds for the polynomial P(z) = zn + kn.

2. Lemmas

For the proof of the theorem, we need the following lemmas.

Lemma 1. If P(z) is a polynomial of degree n, then on |z| = 1

| ( ) | | ( ) | | ( ) || |′ + ′ ≤ =P z Q z nMax P zz 1 (16)

where Q z z Pz

n( )=

1.

This is a special case of a result due to Govil and Rahman [6, Lemma10]. Lemma 2. If P(z) is a polynomial of degree n, then for R > 1,

Max P z R Max P z R R Pz Rn

zn n

| | | || ( ) | | ( ) | ( ) | ( ) | ,= =−≤ − − >1

2 0 1for n (17)

and

Max P z RMax P z R Pz R z| | | || ( ) | | ( ) | ( ) | ( ) | .= =≤ − − =1 1 0 1for n (18)

8

Mir, A Dar, BA

The above result is due to Frappier, Rahman and Ruscheweyh [3, Theorem2]. Lemma 3. If P(z) is a polynomial of degree n ≥ 2 and P( )0 0≠ , then for |z| = 1 and R ≥ 1,

| ( ) ( ) | | ( ) ( ) |

( ) | ( ) |

(| ( ) | |

| |

P Rz P z Q Rz Q z

R Max P z

P

nz

− + −

≤ −

− ′ + ′

=1

0

1

QQR

n

R

nfor n

n n

( ) |) ,01 1

22

2−−

−−

>−

(19)

and

| ( ) ( ) | | ( ) ( ) |

( ) | ( ) |

(| ( ) | |

| |

P Rz P z Q Rz Q z

R Max P z

P

nz

− + −

≤ −

− ′ + ′

=1

0

1

qqR

nfor n

n

( ) |)( )

,01

2−

=

(20)

where Q z z Pz

n( )=

1.

Proof of Lemma 3. If P(z) is a polynomial of degree n and p( )0 0≠ , then

Q z z Pz

n( )=

1 is also a polynomial of degree n. Applying inequality (17) of

Lemma 2 to the polynomials ′P z( ) and ′Q z( ) , which are of degree (n — 1), we obtain for all t≥1and 0 2≤ <θ π ,

| ( ) | | ( ) | ( ) | ( ) |′ ≤ ′ − − ′ >− − −P te t P e t t P for ni n i n nθ θ1 1 3 0 2 (21)

and

| ( ) | | ( ) | ( ) | ( ) | .′ ≤ ′ − − ′ >− − −Q te t Q e t t Q for ni n i n nθ θ1 1 3 0 2 (22)

Adding inequalities (21) and (22), we get for n > 2,

| ( ) | | ( ) | | ( ) | ( )

( ) |

′ + ′ ≤ ′ + ′( )− −

−

− −

P te Q te t P e Q e

t t

i i n i i

n n

θ θ θ θ1

1 3 ′′ + ′( )P Q( ) | | ( ) | .0 0

Using Lemma 1 in above inequality, we get

| ( ) | | ( ) | | ( ) |

( ) | (

| |′ + ′ ≤

− − ′

−=

− −

P te Q te nt Max P z

t t P

i i nz

n n

θ θ 11

1 3 0)) | | ( ) |+ ′( )Q 0 (23)

for 0 2≤ <θ π and t≥1.

9



Also for each θ θ π,0 2≤ < and R > 1, we have

P P e e P te dti i i iR

(Re ) ( ) ( )θ θ θ θ− = ′∫1

and

Q Q e e Q te dti i i iR

(Re ) ( ) ( )θ θ θ θ− = ′∫1

Hence for n > 2

| (Re ) ( ) | | (Re ) ( ) | (| ( ) | | ( ) |)P P e Q Q e P te Q te di i i i i iθ θ θ θ θ θ− + − ≤ ′ + ′ ttR

.1∫

On combining inequality (23) with the above inequality, we get for n > 2

| (Re ) ( ) | | (Re ) ( ) |

[ | ( ) | (| |

P P e Q Q e

nt Max P z t

i i i i

nz

n

θ θ θ θ− + −

≤ −−=

11

−− −

=

− ′ + ′

= − − ′

∫ 1 3

1

1

0 0

1

t P Q dt

R Max P z P

nR

nz

)(| ( ) | | ( ) |]

( ) | ( ) | (| (| | 00 01 1

2

2

) | | ( ) |)+ ′ −−

−−

−

QR

n

R

n

n n

for 0 2≤ <θ π and R > 1, which is equivalent to the desired result.The proof of (20) follows on the same lines as the proof of (19), but instead

of using (17), it uses (18).

Lemma 4. If P(z) is a polynomial of degree n≥ 2 having all its zeros in | | ,z k k≤ ≥1, andP( )0 0≠ , then

Max P zk

kMax P z

k P k q

z k

n

n z

n

| | | || ( ) | | ( ) |

( | ( ) | | ( ) |

= =

−

≥+

+′ + ′

2

1

0 0

1

1 ))

( )

,

1

1 1

22

2

+

×−−

−−

>−

k

k

n

k

nfor n

n

n n

(24)

and

Max P z

k

kMax P z

k P k q

z k

n

n z

n

| | | || ( ) | | ( ) |

( | ( ) | | ( ) |

= =

−

≥+

+′ + ′

2

1

0 0

1

1 ))

( )

( ),

1

12

+−

=k

k

nfor n

n

n

(25)

10

Mir, A Dar, BA where Q z z P

zn( )=

1.

Proof of Lemma 4. Since P(z) has all its zeros in | | ,z k k≤ ≥1, we write

p z a z r en j

i

j

nj( ) ,= −( )

=∏ θ

1

where an≠ 0 and r k j nj ≤ =, , , ,1 2… .

Then, clearly for points eiθ θ π,0 2≤ < , other than zeros of P(z), we have

P k e

P e

k e r e

e r e

k r k r

i

i

ij

i

ij

ij

n

j

j

j

( )

( )

22

1

4 2 22

θ

θ

θ θ

θ θ=

−( )−( )

=+ −

=∏

jj j

j j jj

n

n

j

n

r r

k k

cos( )

cos( )

.

θ θ

θ θ

−

+ − −

≥ =

=

=

∏ 1 221

1∏∏

This implies,

| ( ) | | ( ) |P k e k P ei n i2 θ θ≥

for points eiθ θ π,0 2≤ < , other than the zeros of P(z). Since this inequality is trivial for points eiθ , which are the zeros of P(z), it follows that

| ( ) | | ( ) | | | .P k z k P z for zn2 1≥ = (26)

Let G(z) = P(kz) and H z z Gz

z Pk

zn n( )==

1. Applying inequality (19) to

the polynomial G(z), we get for | |z =1 and k≥1,

| ( ) | | ( ) | ( ) | ( ) |

(| ( ) | | ( ) |)

| |G kz H kz k Max G z

G Hk

nz

n

+ ≤ +

− ′ + ′ −

=1

0 01

1

nn

k

nfor n

n

−−−

>−2 1

22.

Equivalently, for |z| = 1

| ( ) | | ( ) | ( ) | ( ) |

( | ( ) | | (

| |P k z k P z k Max P z

k P k Q

n nz k

n

2

1

1

0 0

+ ≤ +

− ′ + ′

=

− )) |) .k

n

k

nfor n

n n−−

−−

>−1 1

22

2

11



This gives with the help of (26) for |z| = 1

2 1

0 011

k P z k Max P z

k P k Qk

n

n nz k

nn

| ( ) | ( ) | ( ) |

( | ( ) | | ( ) |)

| |≤ +

− ′ + ′ −

=

− −−−−

−k

n

n 2 1

2,

From which inequality (24) follows for n > 2.The proof of the lemma for n = 2 follows on the same lines as that for n

> 2, but using inequality (20) instead of inequality (19).Next we prove the following result which gives an improvement of the

above lemma by involving the term min | ( ) || |z k P z= as follows.

Lemma 5. If P(z) is a polynomial of degree n≥ 2 having all its zeros in |z| < k, k > 1, and P( )0 0≠ , then

Max P zk

kMax P z

k

kMz k

n

n z

n

n| | | || ( ) | | ( ) |= =≥+

+−+

2

1

1

11 iin P z

k P k Q

k

k

n

k

n

z k

n

n

n n

| | | ( ) |

( | ( ) | | ( ) |)

( )

=

−

−

+′ + ′

+

×−−

−−

0 0

1

1 1

1

2

222

>for n ,

(27)

and

Max P zk

kMax P z

k

kMz k

n

n z

n

n| | | || ( ) | | ( ) |= =≥+

+−+

2

1

1

11 iin P z

k P k Q

k

k

nfor n

z k

n

n

n

| | | ( ) |

( | ( ) | | ( ) |)

( )

( ),

=

−

+′ + ′

+−

=0 0

1

12

1

(28)

where Q z z Pz

n( )=

1 .

Proof of Lemma 5. Since P(z) has all of its zeros in | | ,z k k≤ ≥1, then by Rouche’s theorem, the polynomial F(z) = P(z) + λm with | |λ ≤1 , where m= =Min P zz k| | | ( ) | , also has all its zeros in | | ,z k k≤ ≥1 . So, on applying Lemma 4 for n > 2 to the polynomial F(z), we have

Max F zk

kMax F z

k F k T

z k

n

n z

n

| | | || ( ) | | ( ) |

( | ( ) | | ( ) |

= =

−

≥+

+′ + ′

2

1

0 0

1

1 )),

1

1 1

2

2

+( )−−

−−

−

k

k

n

k

nn

n n

12

Mir, A Dar, BA Where T z

zz P

zm Q zn( ) ( )=

=

+

= +z Fn 1 1

λ λλmzn

This implies for n > 2

Max P z mk

kMax P z m

k P k

z k

n

n z

n

| | | || ( ) | | | | ( ) |

( | ( ) |

= =

−

+ ≥+

+

+′ +

λ λ2

1

0

1

11

2

0

1

1 1

2

| ( ) |)

( )

,

′

+

×−−

−−

−

Q

k

k

n

k

n

n

n n

(29)

If we choose z0 so that Max P z P zz| | | ( ) | | ( ) |= =1 0 and also choose argument of

λ suitably, such that

Max P z m P z mz| | | ( ) | | ( ) | | | ,= + = +1 0λ λ

the inequality (29) becomes

Max P z mk

kP z m

k P k Q

k

k

z k

n

n

n

n

n

=

−

( ) + ≥+

( ) +{ }

+′( ) + ′( )( )

+( )

λ λ2

10 0

1

0

1−−−

−−

−1 1

2

2

n

k

n

n

,

Which on simplification gives,

Max P z

k

kMax P

k

km

k

z k

n

n z

n

n| | | || ( ) | | ( ) | | |( )

( )

( |

= =≥+

+−+

+′

2

1

1

11 z λ

PP k Q

k

k

n

k

n

n

n

n n( ) | | ( ) |)

( ).

0 0

1

1 1

2

1 2+ ′

+−−

−−

− −

Finally letting λ→1, we get the desired result for n > 2.The proof of inequality (28) follows on the same lines as the proof of

(27), but instead of inequality(24) of Lemma 4, we use the inequality (25) of the same lemma.

3. Proof of the Theorem

The polynomial G z P kz a kz zn jj

n( ) ( ) ( )= = −

=∏ 1 has all its zeros in |z| ≤ 1

and we have

13



′=

−

=∑G z

G zz

z

kjj

n( )

( )

1

1

so that for 0 ≤ θ < 2π

′

≥′

=

−

≥G e

G e

e G e

G e

e

ez

k

i

i

i i

i

i

i j

( )

( )Re

( )

( )Re

θ

θ

θ θ

θ

θ

θ

kk

k z jj

n

j

n

+==∑∑ | |11

This implies

Max G zk

k zMax G zz

jz

j

n

| | | || ( ) || |

| ( ) |,= ==

′ ≥+∑1 1

1

equivalently

kMax P kzk

k zMax P kzz

jz

j

n

| | | || ( ) || |

| ( ) | .= ==

′ ≥+∑1 1

1

(30)

Since P'(z) is a polynomial of degree n — 1, it follows by (17) that

k k Max P z k k Pk

k zMaxn

zn n

jz k

−=

− −=

′ − −( ) ′ ≥+

11

1 3 0| | | || ( ) | ( | ( ) |)| |

| PP zj

n

( ) | .=∑

1

Now using inequality (27) of Lemma 5 in above inequality, we get

knMaxz

P z kn kn P

k

k zj

kn

knM

| || ( ) | ( | ( ) |

| |

=′ − − − ′

≥+ +

12 0

2

1aax

zP z

j

n

kn

kn z kP z

| || ( ) |

min| |

| ( ) |

==∑

+−

+

=

+

11

1

1

(( | ( ) | | ( ) |)

( )

k P kn Q

kn

kn

n

kn

n

′ + − ′

+

×−−

− −−

0 1 0

1

1 2 1

2

implying

14

Mir, A Dar, BA

Max P zk

k z kMax P z

k

zj

n zj

n

n

| | | || ( ) || |

| ( ) |

(

= ==

′ ≥+ +

+

∑1 11

2

1−−+

+′ + ′

+

=

−

1

10 0

1

1

)

( )min | ( ) |

( | ( ) | | ( ) |)

( )

| |k kP z

k P k Q

k k

k

n n z k

n

n n

nn n

n

k

n

kP

−−

−−

+ −

′

−1 1

2

11

0

2

2| ( ) |

from which the theorem follows for n > 2.The proof of the theorem for n = 2 follows on the same lines as that of

n > 2, but instead of using inequalities (17) and(27) of Lemma 2 and Lemma 5 respectively, we use inequalities (18) and (28) of the respective lemmas.

References

[1] Aziz, A. Inequalities for the derivative of a polynomial, Proc. Amer. Math. Soc., 89 (1983), 259-266. http://dx.doi.org/10.1090/S0002-9939-1983-0712634-5

[2] Aziz, A. and Rather, N. A. Inequalities for the derivative of a polynomial, Proc. Indian Acad. Sci.(Math. Sci.), 107, 2 (1997), 189-196.

[3] Frappier, C. Rahman Q. I. and Ruscheweyh, St. New inequalities for polynomials, Trans. Amer. Math. Soc., 288 (1985), 69-99.

http://dx.doi.org/10.1090/S0002-9947-1985-0773048-1

[4] Govil, N. K. Inequalities for the derivative of a polynomial, J. Approx. Theory, 63 (1990), 65-71. http://dx.doi.org/10.1016/0021-9045(90)90114-6

[5] Govil, N. K. On the derivative of a polynomial, Proc. Amer. Math. Soc., 41 (1973), 543-546. http://dx.doi.org/10.1090/S0002-9939-1973-0325932-8

[6] Govil N. K. and Rahman, Q. I. Functions of exponential type not vanishing in a half-plane and related polynomials, Trans. Amer. Math. Soc., 137 (1969), 501-517. http://dx.doi.org/10.1090/S0002-9947-1969-0236385-6

[7] Malik, M. A. On the derivative of a polynomial, J. London Math. Soc., 2 (1969), 57-60. http://dx.doi.org/10.1112/jlms/s2-1.1.57

[8] Turan, P. Uber die ableitung von polynomen, Composition Math., 7 (1939). 89-95.

some inequalities for polynomials vanishing inside a circle

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