Inequalities Concerning The Maximum Modulus of Polynomials

B.A.Zargar,A.W.Manzoor and Shaista Bashir

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

Abstract.Let P(z)be a polynomial of degree n having all its zeros in|z|≤k,k≤1,then for every real or complex number β,with|β|≤1 and R≥1,it was shown by A.Zireh et al.[7]that for|z|=1,In this paper,we shall present a refinement of the above inequality.Besides,we shall also generalize some well-known results.

Key Words:Growth of polynomials,minimum modulus of polynomials,inequalities.

1 Introduction and statement of results

If P(z)is a polynomial of degree n then concerning the estimate of|P(z)|on the disk|z|=R,R>0,we have the following inequalities

and

Inequality(1.1)is a simple consequence of Maximum Modulus Principle[5]where as inequality(1.2)is due to Zarantonillo and Verga[6].Both the inequalities are sharp and equality holds for P(z)=λzn,|λ|=1.

For polynomials having no zero in|z|<1,an inequality analogus to(1.1)due to Ankeny and Rivilin[1]is the following:

The inequality is sharp and equality holds for the polynomial P(z)=αzn+β,where|α|=|β|.

As a refinement of inequality(1.3)Aziz and Dawood[3]have found that

If P(z)is a polynomial of degree n which does not vanish in|z|<1,then for R≥1,

The result is sharp and equality holds for the polynomial P(z)=αzn+β with|α|=|β|.

Recently,A.Zireh et al.[7]have generalised inequality(1.4)and some results due to Dewan and Hans[4].In fact they have considered the zeros of largest moduli and proved the following results.

Theorem 1.1.Let P(z)be a polynomial of degree n,having all its zeros in|z|≤k,k≤1,then for every real or complex number β with|β|≤1,R≥1 and|z|=1.

The result is best possible and equality holds for the polynomial

Theorem 1.2.If P(z)is a polynomial of degree n having no zeros in|z|

The ine quality(1.6)is sharp ande quality hold sfor the polynomial P(z)=αzn+βkn,with|α|=|β|.

In this paper,we consider the moduli of all the zeros of a polynomial and present some interesting results which provide refinements of Theorems A and B.we shall also generalize some well known results.

First,we shall prove the following refinement of Theorem 1.1.

Theorem1.3.Let P(z)beapolynomialofdegree n and z1,z2,···,znareitszerossuchthat|zj|≤k,k≤1,then for every real or complex number β with|β|≤1,R≥1 and|z|=1,we have

The result is best possible and equality holds for

If we take β=0 in Theorem 1.3,we get the following elegant result:

Corollary 1.1.If P(z)is a polynomial of degree n,having all its zeros in|z|≤k,k≤1,then for R≥1,

The result is best possible and equality holds for

Remark 1.1.To show Theorem 1.3 is a refinement of Theorem 1.1 we shall transfer the terms with βon the same side of inequality(1.7)and get

since

that is,k(R?1)≥|zj|(R?1)or,|zj|≤k,which is trivially true and hence Theorem 1.3 is a refinement of Theorem 1.1.

Next,we shall present the following refinement of Theorem 1.2.

Theorem 1.4.If P(z)is a polynomial of degree n having no zeros in|z|

The inequality(1.8)is sharp and equality holds for P(z)=αzn+βkn,with|α|=|β|.

If we take β=0 in Theorem 1.4,we get the following generalization of inequality(1.4).

Corollary 1.2.If P(z)is a polynomial of degree n having no zeros in|z|

The inequality(1.9)is sharp and equality holds for P(z)=αzn+βkn,|α|=|β|.

2 Lemmas

For the proof of these Theorems we need the following Lemmas.Lemma 2.1 is of independent interest.

Lemma 2.1.If P(z)is a polynomial of degree n,such that P(z)does not vanish in|z|

The result is best possible and equality holds for P(z)=(z+k)n.

Proof of Lemma 2.1.Since all the zeros of P(z)lie in|z|≥k,k≥1,we write

where Rj≥k,j=1,2,···,n.Therefore,for 0≤θ<2π,r≤1,we have

Which implies

or,

Thus,we complete the proof.

Applying Lemma 2.1 to the polynomialwe get

Lemma 2.2.If P(z)is a polynomial of degree n having all its zeros in the disk|z|

the result is best possible and equality holds for P(z)=(z+k)n.This result is a refinement of a result due to Aziz[2].

Lemma 2.3.Let F(z)be a polynomial of degree n having all its zeros in|z|≤k,k≤1 and P(z)be a polynomial of degree not exceeding that of F(z).If|P(z)|≤|F(z)|for|z|=k,k≤1,and if z1,z2,···,znare the zeros of P(z)then for every real or complex number β with|β|≤1 and|z|=1,R≥1,

Proof of Lemma 2.2.Let F(z)be a polynomial of degree n having all its zeros in|z|

therefore for any β with|β|<1,we have

that is for|z|=1,

Hence for an appropriate choice of the argument α,we get

Thus it follows from inequality(2.4)fo|z|=1 that

If the inequality(2.5)is not true,then there is a point z=z0with|z0|=1 such that for R≥1,

we take

then|α|<1 and with this choice of α,we have from(2.4),A(z0)=0 for|z0|=1.But this contradicts the fact that A(z)6=0 for|z|=1.Hence for every real or complex β with|β|=1,(2.5)follows.This completes the proof of Lemma 2.3. ?If we take

in Lemma 2.3 we get,the following:

Lemma 2.4.Let P(z)be a polynomial of degree n,and if z1,z2,···,znare the zeros of P(z)then for|β|≤1,|z|=1 and R≥1,we have

Lemma 2.5.Let P(z)be a polynomial of degree n,and if z1,z2,···,znare its zeros,then for every β with|β|≤1,R≥1 and|z|=1,we have

where

Also

Therefore inequality(2.8)implies,

which gives,

where

Choosing the argument of α suitably we get for|z|=1 and|β|≤1,

Combining(2.9)and(2.10),it follows that

This gives,

Which implies,

Letting|α|→1,the result follows.Which completes the proof of Lemma 2.5.

If we take β=0 in Lemma 2.5,we get the following generalization of inequality(1.3).

Corollary 2.1.Let P(z)be a polynomial of degree n,then for every R≥1 and|z|=1,we have

where

Taking β=0 in Lemma 2.4,we get the following generalization of inequality(1.1).

Corollary 2.2.Let P(z)be a polynomial of degree n,then for any R≥1 and k≤1,

3 Proofs of Theorems

Proof of Theorem 1.3.If P(z)has a zero on|z|=k,then inequality(1.7)is trivial.So we assume that P(z)has all its zeros in|z|

Therefore,for|β|<1,the polynomial

that is,

will have no zeros on|z|=1.As|α|<1,therefore for|β|<1

that is

For|β|=1 the inequality follows.Hence the proof of Theorem 1.3 is completed. ?Proof of Theorem 1.4.Lettherefore for every α with|α|<1,|P(z)|>|αm|.

So by Rouche’s Theorem the polynomial G(z)=P(z)?αm has no zero in|z|≤k.

Equivalently,the polynomial

has all its zeros in|z|≤k and|G(z)|=|H(z)|for|z|=k.Applying Lemma 2.3,it follows that for|β|≤1 and|z|=1,R≥1,

or,it gives by

that is

which implies

Since for|z|=1,|P(z)|=|Q(z)|,which gives

Applying Theorem 1.3 to the polynomial Q(z),we have

where|z|=1 and|β|≤1.

Choosing argument of α suitably in(3.3),we get for|z|=1 and|β|≤1,

Inequality(3.2)can be rewritten as

that is

Letting|α|→1,we get for|z|=1 and R≥1

Also by Lemma 2.5,we have for|z|=1 and R≥1

Combining inequalities(3.5)and(3.6),we get the required result.Hence the result follows.