INEQUALITIES FOR THE sth DERIVATIVE OF A POLYNOMIAL WITH PRESCRIBED ZEROS

. Let P(z) be a polynomial of degree n which does not vanish outside the closed disk | z | < k, where k ≤ 1 . According to a famous result known as Turans Theorem for k=1,we have In this paper we shall present several interesting generalizations and a reﬁnement of this result which include some results due to Malik,Govil and others.we extend Turans Theorem for the sth derivatives of a polynomial having t-fold zeros at origin and thereby obtain an another generalization of this beautiful result.


Introduction
Let P(z) be a polynomial of degree n,then according to a famous result known as Bernstein's inequality(for refrence,see [6,p-531]or [7]), The result is best possible and equality holds for the polynomial having all its zeros at origin.In the reverse direction it was proved by Turan[8] Both the inequalities (1.2) and (1.3) are sharp and equality holds for P (z) = αz n + β where |α| = |β|.
a j z j is a polynomial of degree n having all its zeros in the disk |z| ≤ k ≤ 1 with s-fold zeros at origin,then for |z| = 1, The result is sharp and extremal polynomial P (z) = z s (z + k) n−s , 0 < s ≤ n.
In this paper we shall first present the following refinement and a generalization of Theorem 1.1. Theorem 1.2. If P(z) is a polynomial of degree n ≥ 1 having all its zeros in |z| ≤ k, k ≤ 1 with t-fold zeros at the origin then for 1 ≤ s ≤ t + 1 ≤ n, The result is best possible and equality holds for the polynomial P (z) = (z + k) n , k ≤ 1.
Remark. For t=0 and s=1,this reduces to the result due to Malik.
For k=1,we get the following result.

Corollary. If P(z) is a polynomial of degree n having all its zeros in |z| ≤ 1, with t-fold zeros at the origin then for
Next we prove the following result which extends inequality(1.4) to the sth derivative.
The result is best possible with equality in (1.7) for the polynomial P (z) = (z + k) n .

Lemmas
For the proofs of these theorems,we need the following Lemmas.

Lemma 2.1. If P(z) is a polynomial of degree n having all its zeros in
The result is best possible with equality for the polynomial P (z) = me iα z n , m > 0.
Lemma 2.1 is due to Aziz and Dawood [1].
The result is best possible and equality in (2.2) holds for the polynomial P (z) = (z + k) n .
Proof. Since P(z) has all its zeros in |z| ≤ The result is sharp and equality holds for the polynomial P (z) = z t (z + k) n−t , 0 < t ≤ n.
Proof. If m = min |z|=k |P (z)|,then m ≤ |P (z)| for |z| = k,which gives m| z k | t ≤ |P (z)| for |z| = k.since all the zeros of P(z) lie in |z| ≤ k ≤ 1, with t-fold zeros at the origin,therefore for every complex number α such that |α| < 1, it follows (by Rouches Theorem for m > 0) that the polynomial G(z) = P (z) + αm k t z t has all its zeros in |z| ≤ k, k ≤ 1, with t-fold zeros at the origin.So that we can write Where H(z) is a polynomial of degree n-t having all its zeros in |z| ≤ k, k ≤ 1.
Now if |w| ≤ k ≤ 1,then it can be easily verified that Using this fact in (2.7),we see that which gives, for points e iθ , 0 ≤ θ < 2π, which are not the zeros of G(z).Since inequality (2.8) is trivally true for points e iθ , 0 ≤ θ < 2π, which are the zeros of P(z), it follows that Replacing G(z) by P (z) + αm k t z t in (2.9),then we get and for every α with |α| < 1. Choosing the argument of α such that it follows from (2.10), that Letting |α| → 1,we obtain Which is the desired result.

Proof of the theorems
Proof of Theorem 1.2. We prove this result with the help of mathematical induction.We use induction on s.For s=1,the result follows by Lemma 2.3.Assume that inequality (1.6) is true for s=r, that is we assume for 1 ≤ r ≤ t + 1, Where L r = 1 f or r = 1 = n(n − 1) · · · (n − r + 2) f or r ≥ 2.
(3.3) (3.3) shows that the result is true for s=r+1 also.We conclude with the help of mathematical induction that (1.6) holds for all 1 ≤ s < n.This completes the proof of Theorem 1.2.
Proof of Theorem 1.3.