## Abstract

If $ P(z)=a_nz^n+\sum_{j=\mu}^{n}a_{n-j}z^{n-j}$, $1\leq\mu\leq n$, is a polynomial of degree $n$ having all its zeros in $|z|\le k$, $k\leq 1$, then it was recently claimed by K. K. Dewan, Naresh Singh, Abdullah Mir [\textit{Extensions of some polynomial inequalities to the polar derivative}, {J. Math. Anal. Appl.} \textbf{352} (2009), 807--815] that for every real or complex number $\alpha$, with $|\alpha|\geq k^\mu$, \begin{align*} \underset{\left|z\right|=1}{\max}\left|D_\alpha P(z)\right|&\geq \frac{n\left(|\alpha|-k^\mu\right)}{1+k^\mu}\underset{\left|z\right|=1}{\max}\left|P(z)\right|+\frac{n\left(|\alpha|+1\right)}{k^{n-\mu}\left(1+k^\mu\right)}m\\ &\quad +n\left(\dfrac{k^\mu-A_\mu}{1+k^\mu}\right)\underset{\left|z\right|=1}{\max}|P(z)|+\dfrac{n(A_\mu-k^\mu)}{k^n(1+k^\mu)}m \end{align*} where $m=\min_{|z|=k}|P(z)|$, $D_\alpha P(z)$ is a polar derivative of $P(z)$ with respect to the point $\alpha\in\mathbb{C}$ and $A_\mu$ is given by (1.11). The proof of this result is not correct. In this paper, we present certain more refined results which not only provides a correct proof of above inequality as a special case but also yields a refinement of above and other related result.

## Citation

Suhail Gulzar. Nisar A. Rather. "Refinements of some inequalities concerning the polar derivative of a polynomial." Funct. Approx. Comment. Math. 51 (2) 269 - 283, December 2014. https://doi.org/10.7169/facm/2014.51.2.3

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