**INTRODUCTION**

If:

is a polynomial of degree at most n and P'(z) is its derivative, then:

and:

Inequality Eq. 1 is an immediate consequence of S. Bernstein’s inequality on the derivative of a trigonometric polynomial (Bernstein, 1930; Rahman and Schmeisser, 2002), where as inequality (Eq. 2) is a simple deduction from the maximum modulus principle (Riesz, 1916). In both inequalities Eq. 1 and 2 equality holds only when P(z) is a constant multiple of z^{n}.

If we restrict ourselves to a class of polynomials having no zero in |z|≤1, then the above inequality can be sharpened. In fact, Erdös conjectured and latter (Lax, 1944) proved that if P(z)≠0 in |z|≤1, then:

and:

Turan (1939) proved that, if P(z) has all its zeros in |z|≤1, then:

Concerning the minimum modulus of a polynomial P(z) and its derivative P’(z), Aziz and Dawood (1988) proved that, if P(z) has all its zeros in |z|≤1, then:

min_{|z| = 1} |P'(z)|≥n min_{|z| = 1} |P(z)| |
(6) |

Let, α be any complex number, the polynomial D_{α}P(z) = nP(z)+(α-z) P'(z) denote the polar derivative of the polynomial P(z) of degree at most n with respect to α. The polynomial D_{α}P(z) is of degree at most n-1 and it generalizes the ordinary derivative in the sense that:

Aziz (1988) extended inequalities Eq. 3 and 5 to the polar derivative of a polynomial and proved that if P(z) is a polynomial of degree n which does not vanish in |z|<1, then for every complex number α with |α|≥1:

Rahman and Schmeisser (2002) introduced a class B_{n} of operators B that map PεP_{n} into itself. That is, the operator B carries PεP_{n} into:

where, λ_{0}, λ_{1} and λ_{2} are real or complex numbers such that all the zeros of:

lie in the half plane:

Concerning this operator Shah and Liman (2008) proved.

**Theorem A:** If P(z)εP_{n} and P(z) ≠ 0 in |z|>1, then for |z|≥1:

|B[P(z)]|≥|B[z^{n}]|min_{|z|= 1}|P(z)| |
(9) |

**Theorem B:** If P(z)εP_{n} and P(z) ≠ 0 in |z|<1, then for |z|≥1:

Concerning the dependence of |P(Rz)-P(z)| on |P(z)| Aziz and Rather (1999) proved.

**Theorem C:** If P(z) is a polynomial of degree n, then for every real or complex number β with |β|≤1 and R≥1:

|P(Rz)-βP(z)|≤|R^{n}-β||z|^{n} max_{|z| = 1} |P(z)|, for |z|≥1 |
(11) |

**Theorem D:** If P(z) is a polynomial of degree n which does not vanish in |z|<1, then for every real or complex number β with |β|≤1 and R≥1:

Recently Bidkham and Mezerji (2011) have generalized some of the above inequalities by combining B and D_{α} operators and proved the following results.

**Theorem E:** If P(z) is a polynomial of degree at most n, having all its zeros in |z|≤1, then for every complex number α with |α|≥1:

**Theorem F:** If P(z) is a polynomial of degree at most n, having no zero in |z|<1, then for every α with |α|≥1:

In this study, we combine the different ideas and techniques used above and consider the operator B and D_{α} such that the operator B carries D_{α}P(z) into:

where, 0≤m≤n-1 and λ_{0}, λ_{1} and λ_{2} are real or complex numbers such that all zeros of:

lie in the half plane:

and obtain compact generalizations of some well-known polynomial inequalities. We first prove the following:

**Theorem 1:** If P(z) is a polynomial of degree n, then for every real or complex numbers α, β with |α|≥1, |β|≤1 and R>r≥1:

|B[D_{α}(P(Rz)-β P(rz))]|≤|α|n|R^{n}-r^{n}β||B[z^{n-1}]| max_{|z| = 1}|P(z)|, for |z|≥1 |
(16) |

The result is sharp and equality holds in inequality Eq. 16 for P(z) = az^{n}, a ≠ 0.

Substituting for B[D_{α}(P(Rz)-βP(rz)], we have for |z| = 1:

where, 0≤m≤n-1 and λ_{0}, λ_{1} and λ_{2 }are such that all the zeros of u(z) defined by inequality Eq. 15 lie in the half plane:

If we choose β = 0 and let R→1 in inequality Eq. 16 we get the following result.

**Corollary 1:** If P(z) is a polynomial of degree n, then for every real or complex number α with |α|≥1:

|B[D_{α}P(z)]|≤|α|n|B[z^{n-1}]| max_{|z| = 1}|P(z)|, for |z|≥1

The result is sharp and equality holds for the polynomial P(z) = az^{n}, a ≠ 0.

**Remark 1:** If we choose λ_{1} = 0 = λ_{2} with β = 0 and letting R→1 inequality Eq. 17 will reduce to:

|D_{α}P(z)|≤|α|n|z^{n-1}| max_{|z| = 1}|P(z)|, for |z|≥1 |
(18) |

Dividing both side of inequality Eq. 18 by |α| and letting |α|→∞, inequality Eq. 18 will reduce to inequality Eq. 1.

Choosing λ_{0} = 0 = λ_{2}, inequality Eq. 17 will give the following result.

**Corollary 2:** If P(z) is a polynomial of degree n, then for every real or complex numbers α, β with |α|≥1, |β|≤1 and R>r≥1:

Dividing both side of inequality Eq. 19 by |α| and letting |α|→∞, then m = n-1 and for β = 0 and R→1, inequality Eq. 19 will reduce to:

|P"(z)|≤n(n-1)|z^{n-2}| max_{|z| = 1}|P(z)|, for |z|≥1 |
(20) |

The result is best possible and equality holds in inequality Eq. 20 for P(z) = az^{n}.

We now prove the theorem which gives the extension of (Shah and Liman, 2008, lemma (2.3) to the polar derivative.

**Theorem 2:** If P(z) is a polynomial of degree n, then for every real or complex numbers α, β with |α| ≥1, |β| ≤1 and R>r≥1:

|B[D_{α}(P(Rz)-βP(rz))]|+|B[D_{α}(Q(Rz)-βQ(rz))]|≤n(|α||R^{n}-r^{n}β||B[z^{n}^{-1}]|+|1-β||λ_{0}|) max_{|z| = 1}|P(z)| | (21) |

for |z|≥1, where:

The result is best possible and the equality holds in inequality (Eq. 21) for P(z) = z^{n}+1.

Substituting for B[D_{α}(P(Rz)-βP(rz))] in inequality (Eq. 21), we have for |z|≥1:

where, 0≤m≤n-1 and λ_{0}, λ_{1} and λ_{2} are such that all the zeros of u(z) defined by inequality Eq. 15 lie in the half plane:

If, β = 0 was chosen and let R→1 in inequality (Eq. 21) we get the following extension of (Shah and Liman, 2008) (Lemma (2.3)) to polar derivatives.

**Corollary 3:** If P(z) is a polynomial of degree n, then for every real or complex numbers α with |α|≥1 and for |z|≥1:

|B[D_{α}P(z)]|+|B[D_{α}Q(z)]≤n(|α|B[z^{n-1}]|+|λ_{0}|) max_{|z| = 1}|P(z)|

Which implies:

Taking α = z in the above inequality, we get (Shah and Liman, 2008, Lemma (2.3)) that is:

|B[P(z)]|+|B[Q(z)]≤(|B[z^{n}]|+|λ_{0}|max_{|z| = 1}|P(z)|, for |z|≤1

If we take λ_{1} = 0 = λ_{2} with β = 0 and letting R→1 in inequality Eq. 22, we get the following result:

**Corollary 4:** If P(z) is a polynomial of degree n, then for every real or complex number α with |α|≥1:

|D_{α}P(z)|+|D_{α}Q(z)|≤n{|α||z^{n-1}|+1}max_{|z| = 1}|P(z)|, for |z|≥1 |
(23) |

Dividing both sides by |α| and letting |α|→4, inequality (Eq. 23) will reduce to:

|P’(z)|+|Q '(z)|≤ n|z^{n-1}|max_{|z|= 1}|P(z)|, for |z|≥1 |
(24) |

The result is best possible and equality holds in inequality (Eq. 24) for P(z) = z^{n}+1.

The above result is a special case of the result due to Govil and Rahman (1969), Inequality (3.2).

If we take λ_{0} = 0 = λ_{2} with β = 0 and letting R→1 in inequality (Eq. 22), we get the following result:

**Corollary 5:** If P(z) is a polynomial of degree n, then for every real or complex number α with |α|≥1:

m{|D_{α}P'(z)|+|D_{α}Q'(z)|}≤n|α|(n-1)^{2}|z^{n-2}|max_{|z| = 1}|P(z), for |z|≥1 |
(25) |

Dividing both sides by |α| and letting |α|→4, then m = n-1, inequality (Eq. 25) will reduce to:

|P''(z)|+|Q"(z)|≤n(n-1)|z^{n-2}|max_{|z| = 1}P(z)|, for |z|≥1 |
(26) |

The result is best possible and equality holds in inequality (Eq. 26) for P(z) = z^{n}+1.

Next, we prove a result for the class of polynomials not vanishing in a unit disc and obtain compact generalization of inequality Eq. 7. In fact we prove:

**Theorem 3:** If P(z) is a polynomial of degree n which does not vanish in |z|<1, then for every real or complex numbers α, β with |α|≥1, |β|≤1 and R>r≥1:

for |z|≥1.

The result is best possible and equality in inequality (Eq. 27) holds for P(z) = z^{n}+1.

Substituting for B[D_{α}(P(Rz)-βP(rz))] in inequality (Eq. 27), we have for |z|≥1:

where, 0≤m≤n-1 and λ_{0}, λ_{1} and λ_{2} are such that all the zeros of u(z) defined by inequality (Eq. 15) lie in the half plane Re z≤ m/4:

**Remark 2:** If we take β = 0 and let R→1, inequality (Eq. 27) will reduce to the following result due to Bidkham and Mezerji (2011).

If P(z) is a polynomial of degree at most n, having no zero in |z|±1, then for every α with |α|≥1:

**Remark 3:** If we take λ_{1} = 0 = λ_{2} with β = 0 and letting R→1, inequality (Eq. 28) reduces to inequality (7) that is:

On dividing both sides of above inequality by |α| and letting |α|→∞ we get inequality (3). Choosing λ_{1} = 0 = λ_{2} with β = 0 and letting R→1 in inequality (Eq. 28), we get the following result:

**Corollary 6:** If P(z) is a polynomial of degree n which does not vanish in |z|<1, then for every real or complex number α with |α|≥1:

Dividing both sides of inequality Eq. 29 by |α| and letting |α|→∞ then m = n-1 and we have:

The result is best possible and equality in inequality Eq. 30 holds for P(z) = z^{n}+1.

We now prove the following interesting result, which provides the compact generalization of inequality Eq. 13.

**Theorem 4:** If P(z) is a polynomial of degree n having all its zeros in |z|≤1, then for every real or complex numbers α, β with |α|≥1, |β|≤1 and R>r≥1:

The result is sharp and equality holds in inequality Eq. 31 for P(z) = az^{n}. Substituting for B[D_{α}(P(Rz)-βP(rz))], we have for |z|≥1:

where, 0≤m≤n-1 and λ_{0}, λ_{1} and λ_{2} are such that all the zeros of u(z) defined by inequality Eq. 15 lie in the half plane:

**Remark 4:** If we take β = 0 and let R→1, inequality Eq. 31 will reduce to inequality Eq. 13.

If we take λ_{0} = 0 = λ_{2} with β = 0 and letting R→1 in inequality Eq. 32, we will get the following result from which result of Aziz and Dawood (1988) follows as a special case.

**Corollary 7:** If P(z) is a polynomial of degree at most n having all its zeros in |z| = 1, then for every real or complex number α with |α|≥1:

The result is best possible and equality holds in inequality Eq. 33 for P(z) = az^{n}.

Dividing the inequality Eq. 33 both sides by |α| and letting |α|→∞ then m = n-1, we obtain the inequality Eq. 5 as a special case.

Choosing λ_{0} = 0 = λ_{2} with β = 0 and letting R→1 in inequality Eq. 32, we will get the following result.

**Corollary 8:** If P(z) is a polynomial of degree at most n having all its zeros in |z|≤1, then for every real or complex number α with |α|≥1:

Dividing both sides of the inequality Eq. 34 by |α| and letting |α|→∞, then m = n-1 we obtain:

The result is best possible and the equality holds in inequality Eq. 35 for P(z) = az^{n}.

As an improvement of inequality Eq. 31 and generalization of inequality Eq. 10, we prove the following result.

**Theorem 5:** If P(z) is a polynomial of degree at most n which does not vanish in |z|<1, then for every real or complex numbers α,β with |α|≥1, |β| ≤1 and R>r≥1:

The result is sharp and equality in inequality Eq. 36 holds for the polynomial having all the zeros on the unit disk.

Substituting for B[D_{α}(P(Rz)-βP(rz)))] in inequality Eq. 36, we have for |z|≥1:

where 0≤m≤n-1 and λ_{0}, λ_{1} and λ_{2} are such that all the zeros of u(z) defined by inequality Eq. 15 lie in the half plane:

**Remark 5:** If we take β = 0 and letting R→1, inequality Eq. 36 will reduce to inequality Eq. 14.

**Remark 6:** If we choose λ_{1} = 0 = λ_{2} with β = 0 and let R→1, inequality Eq. 37 will reduce to the following result due to Aziz and Shah (1998).

If P(z) is a polynomial of degree n which does not vanish in |z|<1, then for every complex number α with |α|≥1:

Dividing both sides by |α| and letting |α|→ ∞, in the above inequality, it follows that if P(z) ≠ 0 in |z|<1, then:

The above result is an interesting refinement of Erdö s-Lax Theorem inequality Eq. 3 and was proved by Aziz and Dawood (1988).

If we take λ_{0} = 0 = λ_{2} with β = 0 and let R→1 in inequality Eq. 37, we get the following result:

**Corollary 9:** If P(z) is a polynomial of degree at most n, having no zero in |z|≤1, then for every α with |α|≥1 and |z|≥1:

The result is best possible and equality holds in inequality Eq. 38 for P(z) = z^{n}+1.

Dividing both sides of the inequality Eq. 38 by |α| and letting |α| → ∞, then m = n-1 and we get:

**LEMMAS**

For the proof of above theorems we need the following lemmas. The first lemma follows from Laguerre (1989).

**Lemma 1:** If all the zeros of polynomial P(z) of degree n lie in |z|≤k, where k≤1, then for |α|≥k, the polar derivative D_{α} [P(z)] of P(z) at the point α also has all its zeros in |z|≤k.

The following lemma which we need is in fact implicit in (Rahman and Schmeisser, 2002, Lemma 14.5.7, p.540).

**Lemma 2:** If all the zeros of the polynomial P(z) of degree n lie in a circle |z|≤1, then all the zeros of the polynomial B[P(z)] also lie in |z|≤1.

As an application of lemmas 2 and 3 we have the following lemma.

**Lemma 3:** If all the zeros of polynomial P(z) of degree n lie in |z|≤1, then for |α|≥1, all the zeros of the polynomial B[D_{α}P(z)] also lie in |z|≤1.

**Proof:** From lemma 1 for k = 1 all the zeros of the polynomial D_{α}P(z) lie in |z|≤1 and so from lemma 2 the polynomial B[D_{α}P(z)] has all its zeros in |z|≤1.

The next lemma is due to Aziz and Rather (1998).

**Lemma 4:** If P(z) is a polynomial of degree at most n having all its zeros in |z|<k where k≤1, then |P(Rz)|>|P(z)|, for |z|≥1 and R>1.

**Lemma 5:** If P(z) is a polynomial of degree n which does not vanish in |z|<1, then for every real or complex numbers α, β with |α|≥1,|β|≤1 and R≥r≥1:

|B[D_{α}(P(Rz)-βP(rz))]|≤ |B[D_{α}(Q(Rz)-βQ(rz))]| |
(39) |

for |z| = 1, where:

**Proof:** For R = r = 1, the result reduces to Bidkham and Mezerji (2011) [Lemma 4, p.597]. Now we will prove the result for R>r≥1. Since all the zeros of P(z) lie in |z|≥1 and for every real or complex number λ with |λ|>1, the polynomial G(z) = P(z)-λQ(z) where:

has all its zeros in |z|≤1. Applying lemma 4 to the polynomial G(z) with k = 1, we get:

|G(rz)|<|G(Rz)|, for |z| = 1 and R>r≥1

Since all the zeros of G(z) lie in:

therefore for any real or complex number β with |β|≤1, the polynomial H(z) = G(Rz)-βG(rz), has all its zeros in |z|<1, for every λ with |λ|>1 and R>r≥1, by lemma 3 all the zeros of B[D_{α}H(z)] lie in |z|<1. This implies:

B[D_{α}(G(Rz)-βG(rz))] = B[D_{α}(P(Rz)-βP(rz))]-λB[D_{α}(Q(Rz)-βQ(rz))], for |z|≥1 and R>r≥1 |
(40) |

Inequality Eq. 40 implies:

B[D_{α}(G(Rz)-βP(rz))]|-βP(rz))]|≤|B[D_{α}(Q(Rz)-βQ(rz))] |
(41) |

for |z| = 1 and R > r≥1.

For if it is not true, then there is a point z = z_{o} with |z_{o}|≥1, such that:

B[D_{α}(P(Rz_{o})-βP(rz_{o}))]|≥|B[D_{α}(Q(Rz_{o})-βQ(rz_{o}) |
(42) |

for |z_{0}| ≥1 and R>r≥1.

Since all the zeros of Q(z) lie in |z|≤1, therefore it follows that all the zeros of Q(Rz)-βQ(rz), lie in |z|≤1 for every β with |β|≤1. Hence Q(Rz_{0})-βQ(rz_{0})≠0, for|z_{o}|≥1. Which implies:

B[D_{α}(Q(Rz_{o})-βQ(rz_{o}))]≠0, for |z_{o}|≥1 and R>r≥1

We take:

so that |λ|>1.

Which shows that B[D_{α}H(z)] has a zero in |z|≥1. Which is contradiction to the fact that all the zeros of B[D_{α}H(z)] lie in |z|<1. Thus:

B[D_{α}(P(Rz)-βP(rz))]|≤|B[D_{α}(Q(Rz)-βQ(rz))]|

for |z|≥1 and R≥r≥1.

**Proof of theorems**

**Proof of Theorem 1:** Let M = max_{|z| = 1}|P(z)| then |P(z)|≤M, for |z| = 1. Therefore, by Rouche’s Theorem we have all the zeros of the polynomial G(z) = P(z)+λz^{n}M, lie in |z|<1 for every λ with |λ|>1. Now from lemma 4, we have:

|G(rz)|<|G(Rz)|, for |z| = 1 and R>r≥1

Since all the zeros of G(Rz) lie in:

therefore if β is any real or complex number with |β|≤1, we have all the zeros of the polynomial:

G(Rz)-βG(rz) = (P(Rz)-βP(rz)) + λ(R^{n}-r^{n}β)z^{n}M

also lie in |z|<1 for every R>r≥1 and |λ|>1.

Therefore by lemma 3, all the zeros of B[D_{α}(G(Rz)-βG(rz))], lie in |z|<1 for every R>r≥1 and |λ|>1. Which implies:

B[D_{α}(G(Rz)-βG(rz))] = B[D_{α}(P(Rz)-βP(rz)] +λ_{α}n(R^{n}-r^{n}β)MB[z^{n}^{-1}] |
(43) |

for |z|<| and R>r≥1, inequality (43) implies

|B[Dα(P(Rz)-β(P(rz))]|≤|α|n|R^{n}-r^{n}β||B[z^{n}^{-1}]M| |
(44) |

for |z|≥1 and R>r≥1, if this is not true, then there is a point z = z_{o} with |z_{o}|≥1, such that:

| B[D_{α}(P(Rz_{o})-βP(rz_{o}))] |>| α | n |R^{n}-r^{n}β| |B[z_{o}^{n-1}]| M

We take:

So that | λ | >1, for this choice of λ we have B[D_{α}(G(Rz_{o})-βG(rz_{o}))] = 0, for | z_{o}| ≥1 |. Which is a contradiction to the fact that all the zeros of B[D_{α}(G(Rz)-βG(rz))] lie in | z | <1. Thus:

for | z | ≥1 and R>r≥1.

**Proof of theorem 2:** Let M = max_{|z| = 1} |P(z)|, then |P(z) |≤M for |z| = 1. Now for every real or complex number γ with | γ | >1, it follows from Rouche’s Theorem, the polynomial G(z) = P(z)+γM does not vanish in |z|<1. Now applying lemma 4 and 5 to the polynomial G(z) we have for every real or complex number β with |β| ≤1:

for |z|≥1 and R>r≥1, where:

Inequality Eq. 45 implies:

for |z|≥1 and R>r≥1.

Now choosing the argument of on the R.H.S of inequality (Eq. 46), such that:

for |z|≥1 and R>r≥1.

Therefore, we get from inequality Eq. 46:

for |z|≥1 and R>r≥1.

Therefore, inequality Eq. 48 implies:

for |z|≥1 and R>r≥1.

Letting|γ|→1, in inequality Eq. 49, we get:

for |z|≥1 and R>r≥1. Which proves the theorem.

**Proof of Theorem 3: **We have from lemma 5:

for |z|≥1 and R≥r≥1, where:

Also from Theorem 2, we have:

for |z|≥1 and R>r≥1, where:

Combining the above two inequalities, we get:

for |z|≥1 and R>r≥1, Which proves the theorem.

**Proof of Theorem 4: **If P(z) has a zero on |z| = 1, then the result is trivial. So, we suppose that P(z) has all its zeros in |z|<1. If m = min_{|z| = 1}|P(z)||, then m>0 and m≤|P(z)| for |z| = 1. Therefore, if γ is any complex number with |γ|<1, we have the polynomial G(z) = P(z)-γmz^{n} of degree n has all its zeros in |z|<1. Now from lemma 4, we have:

|G(rz)|<|G(Rz)| for |z| = 1 and R>r≥1

Since all the zeros of G(Rz) lie in:

therefore, for any real or complex number β with |β|≤1 and R>r≥1, it follows from Rouche’s Theorem, the polynomial H(z) = G(Rz)-βG(rz) has all its zeros in |z|<1. Therefore, from Lemma 3, all the zeros of B[D_{α}H(z)] lie in |z|<1. This implies:

for |z|≥1 and R>r≥1.

Inequality Eq. 50 implies for |z|≥1 and R>r≥1:

If inequality Eq. 51 is not true, then there is a point z = z_{0} with |z_{0}|≥1, such that:

We take:

so that |γ|<1.

For this choice of |γ|, we have B[D_{α}H(z)] = 0, for |z|≥1. Which is a contradiction to the fact that all the zeros of B[D_{α}H(z)] lie in |z|<1.

Thus, we have:

for |z|≥1 and R>r≥1.

Hence, the theorem follows.

**Proof of Theorem 5:** Since the polynomial P(z) does not vanish in |z|<1, therefore if m = min_{|z| = 1} |P(z)|, then m≤|P(z)|, for |z|≤1. Now for any real or complex number λ with |λ|≤1, the polynomial G(z) = P(z)+λmz^{n} does not vanish in |z|<1. For if this is not true, then there is a point z = z_{0}, with |z_{0}|<1, such that G(z_{0}) = P(z_{0})+λmz_{0}^{n} = 0. Which implies |P(z_{0}) | = |λm z_{0}^{n}|≤m, contradicting the fact that m≤|P(z)|for |z|≤1. Thus G(z) has no zero in |z|<1 for every λ with |λ|≤1. Applying lemma 5 to the polynomial G(z) we have for |β|≤1 and R>r≥1:

for |z|≥1 and R>r≥1. Where:

Inequality Eq. 50 implies:

for |z|≥1 and R>r≥1. Choosing λ in inequality Eq. 53 such that:

for |z|≥1 and R>r≥1. Inequality Eq. 54 implies:

for |z|≥1 and R>r≥1. Inequality (Eq. 55) implies:

for |z|≥1 and R>r≥1.

Letting |λ|→1, we have for |z|≥1 and R>r≥1:

for |z|≥1 and R>r≥1.

Applying Theorem 2, we get from inequality (Eq. 57):

for |z|≤1*. *Hence, the Theorem follows.