1 Introduction

Let \({\mathscr {P}}_n\) be the class of polynomials \(p(z):= {\sum _{j=0}^{n}}a_j z^j\) of degree at most n. For \(p \in {\mathscr {P}}_n\) and a positive real number k, we write:

$$\begin{aligned} {\mathcal {D}}_{k}:=\{z:|z|=k\},\,{\mathcal {D}}^+_k:=\{z:|z|>k\},\, {\mathcal {D}}^-_k:=\{z:|z|<k\},\, \\ \quad M(p,k):= \displaystyle {\max _{z\in {\mathcal {D}}_{k}}}|p(z)|\,\, \text {and} \,\, m(p,k):=\min _{z \in {\mathcal {D}}_{k}}|p(z)|. \end{aligned}$$

If \(p \in {\mathscr {P}}_n\) and \(p^{'}\) is the derivative of p,  then

$$\begin{aligned} M(p',1)\le n M(p,1). \end{aligned}$$
(1)

(1) is a famous sharp inequality due to Bernstein [2] (see also [6]). If we restrict ourselves to the class of polynomials \(p \in {\mathscr {P}}_n,\) such that \(p(z)\ne 0\) for \(z \in {\mathcal {D}}^{-}_1\), then inequality (1) can be sharpened. Infact, it was conjectured by Erdös and latter proved by Lax [3], that if p(z) does not vanish in \({\mathcal {D}}^{-}_1\), then

$$\begin{aligned} M(p^{'},1)\le \frac{n}{2}M(p,1). \end{aligned}$$
(2)

In case p(z) is a polynomial of degree n and does not vanish in \({\mathcal {D}}_{1}^{+},\) then it was shown by Turán [7] that

$$\begin{aligned} M(p^{'},1)\ge \frac{n}{2}M(p,1). \end{aligned}$$
(3)

For the polynomials \(p \in {\mathscr {P}}_n,\) with \(p(z)\ne 0\), \(z \in {\mathcal {D}}^-_k,\) \(k \ge 1\), Malik [4] proved

$$\begin{aligned} M(p^{'},1)\le \frac{n}{1+k}M(p,1), \end{aligned}$$
(4)

where for the \(n^{th}\) degree polynomial \(p(z)\ne 0,\) for \(z \in {\mathcal {D}}_{k}^{+},\) \(k \le 1,\) he obtained

$$\begin{aligned} M(p^{'},1) \ge \frac{n}{1+k} M(p,1). \end{aligned}$$
(5)

Aziz and Shah [1] generalized (5) and proved that if p(z),  the polynomial of degree n has all its zeros in \({\mathcal {D}}_{k}\cup {\mathcal {D}}_{k}^{-},\) \(k \le 1,\) with s-fold zeros at the origin, then

$$\begin{aligned} M(p^{'},1)\ge \frac{n+sk}{1+k}M(p,1). \end{aligned}$$
(6)

Nakprasit and Somsuwan [5] investigated \(M(p^{'},1)\) in terms of M(p, 1) for a polynomial \(p \in {\mathscr {P}}_n,\) having a zero of order s at some point \(z_0,\) where \(z_0 \in {\mathcal {D}}^{-}_1\) and proved:

Theorem A

If \(p(z):=(z-z_0)^s(a_0+ {\sum _{\nu =\mu }^{n-s}a_\nu z^\nu )}\), \(1 \le \mu \le n-s,\) \(0 \le s \le n-1,\) is a polynomial of degree n,  having a zero of order s at \(z_0,\) where \(z_0 \in {\mathcal {D}}_1^{-}\) and the remaining \(n-s\) zeros are outside \({\mathcal {D}}^-_k\), \(k \ge 1\), then

$$\begin{aligned} M(p^{'},1)\le \Bigg \{\frac{s}{(1-|z_0|)}+\frac{A}{(1-|z_0|)^s}\Bigg \}M(p,1)-\frac{A}{(k+|z_0|)^s}m(p,k), \end{aligned}$$
(7)

where

$$\begin{aligned} A=\frac{(1+|z_0|)^{s+1}(n-s)}{(1+k^\mu )(1-|z_0|)}. \end{aligned}$$

Observation

In Theorem A, if we put \(s=0,\) that is, if we assume, there is no zero inside \({\mathcal {D}}^-_k,\) then

$$\begin{aligned} M(p^{'},1)\le \frac{n(1+|z_0|)}{(1+k^\mu )(1-|z_0|)}M(p,1)-\frac{n(1+|z_0|)}{(1+k^\mu )(1-|z_0|)}m(p,k). \end{aligned}$$
(8)

The presence of \(z_0\) in the R.H.S of (8) as well as \(\Big (\frac{1+|z_0|}{1-|z_0|}\Big )\ge 1,\) shows that their attempt of obtaining the desired result is not only incomplete but incorrect.

In the light of Theorem A followed by the observation, we are in a position to prove the following results.

2 Main results

Theorem 1

If p(z) is a polynomial of degree n having no zeros in \({\mathcal {D}}^-_k\), \(k>1,\) except a zero of multiplicity s,  \(0\le s <n\) at \(z_0,\) where \(|z_{0}|\le 1- \dfrac{2s(k+1)}{n(k-1)+2s}\) then for \(n > \dfrac{2sk}{k-1},\)

$$\begin{aligned} M(p^{'},1)\le \frac{1}{2} \Bigg [n+ \frac{2s(k+|z_{0}|)}{(k+1)(1-|z_{0}|)}-\frac{n(k-1)}{(k+1)}\frac{|p(z)|^{2}}{(M(p,1))^{2}}\Bigg ]M(p,1). \end{aligned}$$

The result is best possible for \(z_{0}=0\) and equality holds for \(p(z):=z^s(z+k)^{n-s}\), \(0\le s <n,\) evaluated at \(z=1.\)

In particular if \(z_0=0\), then we have the following sharp result.

Corollary 1

If p(z) is a polynomial of degree n having all zeros outside \({\mathcal {D}}^-_k\), \(k> 1,\) except a zero of multiplicity s,  \(0 \le s < n\) at origin, then

$$\begin{aligned} M(p^{'},1)\le \frac{1}{2} \Bigg [ n+\frac{2sk}{k+1}- \frac{n(k-1)}{k+1}\frac{|p(z)|^{2}}{(M(p,1))^{2}}\Bigg ] M(p,1), \end{aligned}$$

where \(n > \dfrac{2sk}{k-1}.\)

Theorem 1 reduces to the following result, by taking \(s=0.\)

Corollary 2

If p(z) is a polynomial of degree n having no zeros in \({\mathcal {D}}^-_k\), \(k> 1,\) then

$$\begin{aligned} M(p^{'},1) \le \frac{1}{2} \Bigg [ n- \frac{n(k-1)}{k+1} \frac{|p(z)|^{2}}{(M(p,1))^{2}}\Bigg ]M(p,1). \end{aligned}$$

Equality sign holds for the polynomial \(p(z):=(z+k)^{n},\) evaluated at \(z=1.\)

Theorem 2

If p(z) is a polynomial of degree n having no zeros in \({\mathcal {D}}_{k}^{+}\), \(k \le 1,\) except a zero of multiplicity s,  \(0\le s< n\) at \(z_0,\) where \(z_0 \in {\mathcal {D}}^{-}_1\), then for \(z \in {\mathcal {D}}_{1}\)

$$\begin{aligned} |p^{'}(z)| \ge \frac{1}{1+k}\Bigg \{n+s\Bigg (\frac{k-|z_{0}|}{1+|z_{0}|}\Bigg )\Bigg \}|p(z)|. \end{aligned}$$

The result is best possible for \(z_{0}=0\) and equality holds for \(p(z):=z^s(z+k)^{n-s}\), \(0\le s <n,\) evaluated at \(z=1.\)

For \(k=1,\) we have the following result from Theorem 2.

Corollary 3

 If p(z) is a polynomial of degree n having no zeros in \({\mathcal {D}}_{1}^{+}\), except a zero of multiplicity s,  \(0\le s < n\) at \(z_0,\) where \(z_0 \in {\mathcal {D}}^{-}_1\), then for \(z \in {\mathcal {D}}_{1}\)

$$\begin{aligned} |p^{'}(z)| \ge \frac{1}{2}\Bigg \{n+s\Bigg (\frac{1-|z_{0}|}{1+|z_{0}|}\Bigg )\Bigg \}|p(z)|. \end{aligned}$$

Remark 1

In particular if \(z_{0}=0,\) then Theorem 2 reduces to inequality (6) due to Aziz and Shah [1].

3 Lemmas

For the proof of Theorem 1, we need the following Lemma due to Malik [4].

Lemma 1

If p(z) is a polynomial of degree at most n and \(p^{*}(z)=z^{n}\overline{p\Bigg (\displaystyle {\frac{1}{ {\overline{z}}}\Bigg )}},\) then for |z|=1

$$\begin{aligned} |(p^{*'}(z)|+|p^{'}(z)|\le n M(p,1). \end{aligned}$$
(9)

The result is best possible and equality is attained at \(p(z)= \alpha z^{n}\), \(\alpha\) being a complex number.

4 Proofs of theorems

Proof of Theorem 1

Since p(z) has all its zeros in \({\mathcal {D}}_{k} \cup {\mathcal {D}}_{k}^{+},\) except a zero of multiplicity s at \(z_{0},\) \(z_{0}\in {\mathcal {D}}_{1}^{-},\) \(0 \le s < n,\) therefore

$$\begin{aligned} p(z)=(z-z_{0})^{s}u(z), \end{aligned}$$

where u(z) is a polynomial of degree \(n-s\) having all zeros in \({\mathcal {D}}_{k} \cup {\mathcal {D}}_{k}^{+}.\) Therefore, if \(z_{1}, z_{2}, \ldots ,z_{n-s}\) be the zeros of u(z),  then \(|z_{j}| \ge k,\) \(k > 1,\) \(j=1,2, \ldots , n-s.\) Hence, we have

$$\begin{aligned} \frac{zp^{'}(z)}{p(z)}= \frac{sz}{z-z_{0}}+ \sum _{j=1}^{n-s} \frac{z}{z-z_{j}}. \end{aligned}$$

This, in particular, gives

$$\begin{aligned} {\mathrm {Re}} \Bigg (\frac{zp^{'}(z)}{p(z)}\Bigg )= {\mathrm {Re}}\Bigg (\frac{sz}{z-z_{0}}\Bigg )+{\mathrm {Re}}\Bigg (\sum _{j=1}^{n-s} \frac{z}{z-z_{j}}\Bigg ). \end{aligned}$$

For the points \(e^{i \theta },\) \(0 \le \theta < 2 \pi\) which are not the zeros of p(z),  we have

$$\begin{aligned} {\mathrm {Re}} \Bigg (\frac{e^{i \theta }p^{'}(e^{i \theta })}{p(e^{i \theta })}\Bigg )&= {\mathrm {Re}}\Bigg (\frac{se^{i \theta }}{e^{i \theta }-z_{0}}\Bigg )+{\mathrm {Re}}\Bigg (\sum _{j=1}^{n-s} \frac{e^{i \theta }}{e^{i \theta }-z_{j}}\Bigg )\\&= {\mathrm {Re}}\Bigg (\frac{se^{i \theta }}{e^{i \theta }-z_{0}}\Bigg )+{\mathrm {Re}}\Bigg (\sum _{j=1}^{n-s} \frac{1}{1-e^{-i \theta }z_{j}}\Bigg ). \end{aligned}$$

Using the fact that for \(|w| \ge k > 1,\)

$$\begin{aligned} {\mathrm {Re}} \Bigg ( \frac{1}{1-w}\Bigg ) \le \frac{1}{1+k} \end{aligned}$$

and

$$\begin{aligned} {\mathrm {Re}} \Bigg (\frac{se^{i \theta }}{e^{i\theta }-z_{0}}\Bigg )\le \Bigg |\frac{se^{i \theta }}{e^{i \theta } -z_{0}}\Bigg |\le \frac{s}{1-|z_{0}|} \end{aligned}$$

we get, for \(0 \le \theta <2 \pi ,\)

$$\begin{aligned} {\mathrm {Re}} \Bigg (\frac{e^{i \theta }p^{'}(e^{i \theta })}{p(e^{i \theta })}\Bigg ) \le \frac{s}{1-|z_{0}|}+\frac{n-s}{1+k}, \end{aligned}$$

Now, for \(p^{*}(z)=z^{n}\overline{p\Bigg (\dfrac{1}{{\overline{z}}}\Bigg )},\) it can be easily verified that

$$\begin{aligned} |(p^{*}(z))^{'}|=|np(z)-zp^{'}(z)|, \quad z \in {\mathcal {D}}_{1} \end{aligned}$$

This gives, for \(z \in {\mathcal {D}}_{1}\)

$$\begin{aligned} \Bigg |\frac{(p^{*}(z))^{'}}{p(z)}\Bigg |^{2}&= \Bigg | n- \frac{zp^{'}(z)}{p(z)}\Bigg |^{2}\\&=n^{2}+\Bigg |\frac{zp^{'}(z)}{p(z)}\Bigg |^{2}-2n \mathrm {Re \Bigg ( \frac{zp{'}(z)}{p(z)}\Bigg )}\\&\ge n^{2}+\Bigg |\frac{zp^{'}(z)}{p(z)}\Bigg |^{2}-2n \Bigg ( \frac{s}{1-|z_{0}|}+\frac{n-s}{1+k}\Bigg ). \end{aligned}$$

That is

$$\begin{aligned} |(p^{*}(z))^{'}|^{2} \ge |zp^{'}(z)|^{2}+\Bigg \{ n^{2} -2n \Bigg (\frac{s}{1-|z_{0}|}+ \frac{n-s}{1+k}\Bigg )\Bigg \}|p(z)|^{2}. \end{aligned}$$

Since

$$\begin{aligned} |z_{0}|\le 1-\dfrac{2s(k+1)}{n(k-1)+2s} \, \, \text {and} \, \, n > \dfrac{2sk}{k-1}, \end{aligned}$$

it can be easily verified that

$$\begin{aligned} n^{2}-2n \Bigg (\frac{s}{1-|z_{0}|}+\frac{n-s}{1+k}\Bigg ) \ge 0. \end{aligned}$$
(10)

From this we get, for \(z \in {\mathcal {D}}_{1}\)

$$\begin{aligned} |(p^{*}(z))^{'}| \ge \Bigg [ |p^{'}(z)|^{2}+\Bigg \{n^{2}-2n\Bigg ( \frac{s}{1-|z_{0}|}+\frac{n-s}{1+k}\Bigg ) \Bigg \}|p(z)|^{2}\Bigg ]^{\frac{1}{2}}. \end{aligned}$$
(11)

Inequality (11) together with Lemma 1, gives

$$\begin{aligned} |p^{'}(z)|+\Bigg [ |p^{'}(z)|^{2}+\Bigg \{n^{2}-2n\Bigg ( \frac{s}{1-|z_{0}|}+\frac{n-s}{1+k}\Bigg ) \Bigg \}|p(z)|^{2}\Bigg ]^{\frac{1}{2}} \le n M(p,1). \end{aligned}$$

This gives, for \(z \in {\mathcal {D}}_{1}\)

$$\begin{aligned} |p^{'}(z)|^{2}+\Bigg \{n^{2}-2n\Bigg (\frac{s}{1-|z_{0}|}+\frac{n-s}{1+k}\Bigg )\Bigg \}|p(z)|^{2}\le \Big (n M(p,1)-|p^{'}(z)|\Big )^{2}. \end{aligned}$$

On simplifying we get, for \(z \in {\mathcal {D}}_{1}\)

$$\begin{aligned} |p^{'}(z)|&\le \frac{1}{2} \Bigg [n- \Bigg \{\frac{n(k-1)}{(k+1)}- \frac{2s(k+|z_{0}|)}{(k+1)(1-|z_{0}|)}\Bigg \} \frac{|p(z)|^{2}}{(M(p,1))^{2}}\Bigg ]M(p,1)\\&\le \frac{1}{2}\Bigg [n+ \frac{2s(k+|z_{0}|)}{(k+1)(1-|z_{0}|)}-\frac{n(k-1)}{(k+1)}\frac{|p(z)|^{2}}{(M(p,1))^{2}}\Bigg ]M(p,1). \end{aligned}$$

From which the result follows.

Proof of Theorem 2

Since p(z) has all its zeros in \({\mathcal {D}}_{k} \cup {\mathcal {D}}_{k}^{-},\) except a zero of multiplicity s at \(z_{0},\) \(z_{0} \in {\mathcal {D}}_{1}^{-},\) \(0 \le s < n,\) therefore

$$\begin{aligned} p(z)=(z-z_{0})^{s}u(z), \end{aligned}$$

where u(z) is a polynomial of degree \(n-s\) having all its zeros in \({\mathcal {D}}_{k} \cup {\mathcal {D}}_{k}^{-}.\) Therefore, if \(z_{1}, z_{2}, \ldots ,z_{n-s}\) be the zeros of u(z),  then \(|z_{j}| \le k,\) \(k \le 1,\) \(j=1,2, \ldots , n-s.\) Hence, we have

$$\begin{aligned} \frac{zp^{'}(z)}{p(z)}= \frac{sz}{z-z_{0}}+ \sum _{j=1}^{n-s} \frac{z}{z-z_{j}}. \end{aligned}$$

This, in particular, gives

$$\begin{aligned} {\mathrm {Re}} \Bigg (\frac{zp^{'}(z)}{p(z)}\Bigg )= {\mathrm {Re}}\Bigg (\frac{sz}{z-z_{0}}\Bigg )+{\mathrm {Re}}\Bigg (\sum _{j=1}^{n-s} \frac{z}{z-z_{j}}\Bigg ). \end{aligned}$$

Therefore, for the points \(e^{i \theta },\) \(0\le \theta < 2 \pi\) which are not the zeros of p(z),  we have

$$\begin{aligned} {\mathrm {Re}} \Bigg (\frac{e^{i \theta }p^{'}(e^{i \theta })}{p(e^{i \theta })}\Bigg )&= {\mathrm {Re}}\Bigg (\frac{se^{i \theta }}{e^{i \theta }-z_{0}}\Bigg )+{\mathrm {Re}}\Bigg (\sum _{j=1}^{n-s} \frac{e^{i \theta }}{e^{i \theta }-z_{j}}\Bigg )\nonumber \\&= {\mathrm {Re}}\Bigg (\frac{se^{i \theta }}{e^{i \theta }-z_{0}}\Bigg )+{\mathrm {Re}}\Bigg (\sum _{j=1}^{n-s} \frac{1}{1-e^{-i \theta }z_{j}}\Bigg ). \end{aligned}$$
(12)

Using the facts that for \(|w|\le k \le 1\) we have

$$\begin{aligned} {\mathrm {Re}} \Bigg ( \frac{1}{1-w}\Bigg ) \ge \frac{1}{1+k} \end{aligned}$$

and for \(z_{0} \in {\mathcal {D}}_{1}^{-}\)

$$\begin{aligned} {\mathrm {Re}} \Bigg (\frac{se^{i \theta }}{e^{i \theta }-z_{0}}\Bigg ) \ge \frac{s}{1+|z_{0}|}. \end{aligned}$$

We get from (12)

$$\begin{aligned} |p^{'}(z)|&\ge \Bigg \{\frac{s}{1+|z_{0}|}+\frac{n-s}{1+k}\Bigg \}|p(z)|\\&= \frac{1}{1+k}\Bigg \{n+s\Bigg (\frac{k-|z_{0}|}{1+|z_{0}|}\Bigg )\Bigg \}|p(z)|. \end{aligned}$$

This proves the desired result.