- Open Access
Integral mean estimates for polynomials whose zeros are within a circle
© Singh and Shah; licensee Springer. 2011
- Received: 26 December 2010
- Accepted: 19 August 2011
- Published: 19 August 2011
Let P(z) be a polynomial of degree n having all its zeros in |z| ≤ K ≤ 1, then for each δ > 0, p > 1, q > 1 with , Aziz and Ahmad (Glas Mat Ser III 31:229-237, 1996) proved that
In this paper, we extend the above inequality to the class of polynomials , 1 ≤ μ ≤ n, having all its zeros in |z| ≤ K ≤ 1, and obtain a generalization as well as refinement of the above result.
Mathematics Subject Classification (2000)
30A10, 30C10, 30C15
- Derivative of a polynomial
- Integral mean estimates
- Inequalities in complex domain
Malik  also obtained a generalization of (1) in the sense that the right-hand side of (1) is replaced by a factor involving the integral mean of |P(z)| on |z| = 1. In fact, he proved the following:
The result is sharp, and equality in (3) holds for P(z) = (z + 1) n . If we let δ → ∞ in (3), we get (1).
As a generalization of Theorem A, Aziz and Shah  proved the following:
Aziz and Ahmad  generalized (3) in the sense that Max|z|=1|P′(z)| on |z| = 1 on the right-hand side of (3) is replaced by a factor involving the integral mean of |P′(z)| on |z| = 1 and proved the following:
If we let p → ∞ (so that q → 1) in (5), we get (3).
In this paper, we consider a class of polynomials , 1 ≤ μ ≤ n, having all the zeros in |z| ≤ K ≤ 1, and thereby obtain a more general result by proving the following:
where m = Min|z|=K|P (z)|.
If we take λ = 0 in Theorem 1, we get the following:
For μ = 1 in Theorem 1, we have the following:
where m = Min|z|=K|P (z)|.
It shows that for λ = 0, Corollary 2 provides a refinement of the result of Aziz and Ahmad .
The next result immediately follows from Theorem 1, if we let p → ∞ (so that q → 1)
From inequality (11), we conclude the following:
Further, if we take K = t = μ = 1 in the Corollary 4, we get a result of Aziz and Dawood .
For the proof of this theorem, we need the following lemmas.
The first lemma is due to Qazi .
Proof of Lemma 2
This proves Lemma 2.
Remark 1: Lemma 3 of Govil and Mc Tume  is a special case of this lemma when μ = 1.
Proof of Theorem 1
Let m = Min|z|=K|P (z)|, so that m ≤ |P (z)| for |z| = K. Therefore, for every complex number λ with |λ| < 1, we have |mλ| < |P(z)| on |z| = K. Since P(z) has all its zeros in |z| ≤ K ≤ 1, by Rouche's theorem, it follows that all the zeros of the polynomial G(z) = P(z) + λm lie in |z| ≤ K ≤ 1.
which proves the desired result.
The authors are grateful to the referee for useful comments.
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