On the maximum modulus of a polynomial and its polar derivative

Abstract

For a polynomial p(z) of degree n, having all zeros in |z| ≤ 1, Jain is shown that

$$\begin{array}{r}\underset{\left|z\right|=1}{\max }\left|D_{{\alpha }_t}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}p(z)\right|\ge \frac{n(n-1)\cdots (n-t+1)}{2^t}\times [\\ \left\{\left(\left|{\alpha }_1\right|-1\right)\cdots \left(\left|{\alpha }_t\right|-1\right)\right\}\underset{\left|z\right|=1}{\max }\left|p(z)\right|+\\ \left\{2^t\left(\left|{\alpha }_1\right|\cdots \left|{\alpha }_t\right|\right)-\left\{\left(\left|{\alpha }_1\right|-1\right)\cdots \left(\left|{\alpha }_t\right|-1\right)\right\}\right\}\underset{\left|z\right|=1}{\min }\left|p(z)\right|],\\ \left|{\alpha }_1\right|\ge 1,\left|{\alpha }_2\right|\ge 1,\cdots \left|{\alpha }_t\right|\ge 1,(t<n).\end{array}$$

In this paper, the above inequality is extended for the polynomials having all zeros in |z| ≤ k, where k ≤ 1. Our result generalizes certain well-known polynomial inequalities.

(2010) Mathematics Subject Classification. Primary 30A10; Secondary 30C10, 30D15.

1. Introduction and statement of results

Let p(z) be a polynomial of degree n, then according to the well-known Bernstein's inequality [1] on the derivative of a polynomial, we have

$$\underset{\left|z\right|=1}{max}\left|p^{\prime }\left(z\right)\right|\le n\underset{\left|z\right|=1}{max}\left|p\left(z\right)\right|.$$

(1.1)

This result is best possible and equality holding for a polynomial that has all zeros at the origin.

If we restrict to the class of polynomials which have all zeros in |z| ≤ 1, then it has been proved by Turan [2] that

$$\underset{\left|z\right|=1}{max}\left|p^{\prime }\left(z\right)\right|\ge \frac{n}{2}\underset{\left|z\right|=1}{max}\left|p\left(z\right)\right|.$$

(1.2)

The inequality (1.2) is sharp and equality holds for a polynomial that has all zeros on |z| = 1.

As an extension to (1.2), Malik [3] proved that if p(z) has all zeros in |z| ≤ k, where k ≤ 1, then

$$\underset{\left|z\right|=1}{max}\left|p^{\prime }\left(z\right)\right|\ge \frac{n}{1+k}\underset{\left|z\right|=1}{max}\left|p\left(z\right)\right|.$$

(1.3)

This result is best possible and equality holds for p(z) = (z - k)ⁿ.

Aziz and Dawood [4] obtained the following refinement of the inequality (1.2) and proved that if p(z) has all zeros in |z| ≤ 1, then

$$\underset{\left|z\right|=1}{max}\left|p^{\prime }\left(z\right)\right|\ge \frac{n}{2}\left\{\underset{\left|z\right|=1}{max}\left|p\left(z\right)\right|+\underset{\left|z\right|=1}{min}\left|p\left(z\right)\right|\right\}.$$

(1.4)

This result is best possible and equality attains for a polynomial that has all zeros on |z| = 1.

Let D _α p(z) denote the polar differentiation of the polynomial p(z) of degree n with respect to α ∈ ℂ. Then, D _α p(z) = np(z) + (α - z)p'(z). The polynomial D _α p(z) is of degree at most n - 1, and it generalizes the ordinary derivative in the sense that

$$\underset{\alpha \to \infty }{lim}\left[\frac{D_{\alpha }p\left(z\right)}{\alpha }\right]=p^{\prime }\left(z\right).$$

Shah [5] extended (1.2) to the polar derivative of p(z) and proved that if all zeros of the polynomial p(z) lie in |z| ≤ 1, then for every α with |α| ≥ 1, we have

$$\underset{\left|z\right|=1}{max}\left|D_{\alpha }p\left(z\right)\right|\ge \frac{n}{2}\left(\left|\alpha \right|-1\right)\underset{\left|z\right|=1}{max}\left|p\left(z\right)\right|.$$

(1.5)

This result is best possible and equality holds as p(z) = (z - 1)ⁿwith α ≥ 1.

Aziz and Rather [6] generalized (1.5) by extending (1.3) to the polar derivative of a polynomial. In fact, they proved that if all zeros of p(z) lie in |z| ≤ k, where k ≤ 1, then for every α with |α| ≥ k, we get

$$\underset{\left|z\right|=1}{max}\left|D_{\alpha }p\left(z\right)\right|\ge \frac{n}{1+k}\left(\left|\alpha \right|-k\right)\underset{\left|z\right|=1}{max}\left|p\left(z\right)\right|.$$

(1.6)

This result is best possible and equality holds for p(z) = (z - k)ⁿwith α ≥ k.

In the same paper, Aziz and Rather [6] sharpened the inequality (1.5) by proving that if all the zeros of p(z) lie in |z| ≤ 1, then for every α with |α| ≥ 1, we would obtain

$$\underset{\left|z\right|=1}{max}\left|D_{\alpha }p\left(z\right)\right|\ge \frac{n}{2}\left\{\left(\left|\alpha \right|-1\right)\underset{\left|z\right|=1}{max}\left|p\left(z\right)\right|+\left(\left|\alpha \right|-1\right)\underset{\left|z\right|=1}{min}\left|p\left(z\right)\right|\right\}.$$

(1.7)

This result is best possible and equality attains for p(z) = (z - 1)ⁿwith α ≥ 1.

As an extension to the inequality (1.7), Jain [7] proved that if p(z) has all zeros in |z| ≤ 1, then for all α₁,... α _t ∈ ℂ with |α₁| ≥ 1, |α₂| ≥ 1, ..., |α _t | ≥ 1, (1 ≤ t < n), we have

$$\begin{array}{r}\underset{\left|z\right|=1}{\max }\left|D_{{\alpha }_t}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}p(z)\right|\ge \frac{n(n-1)\cdots (n-t+1)}{2^t}[\\ \left\{\left(\left|{\alpha }_1\right|-1\right)\cdots \left(\left|{\alpha }_t\right|-1\right)\right\}\underset{\left|z\right|=1}{\max }\left|p(z)\right|+\\ \left\{2^t\left(\left|{\alpha }_1\right|\cdots \left|{\alpha }_t\right|\right)-\left\{\left(\left|{\alpha }_1\right|-1\right)\cdots \left(\left|{\alpha }_t\right|-1\right)\right\}\right\}\underset{\left|z\right|=1}{\min }\left|p(z)\right|],\end{array}$$

(1.8)

where

$$\begin{array}{c}{D}_{{\alpha }_j}{D}_{{\alpha }_{j-1}}\cdots D_{{\alpha }_1}p\left(z\right)=p_j\left(z\right)=\\ \left(n-j+1\right)p_{j-1}\left(z\right)+\left({\alpha }_j-z\right)p_{j-1^{\prime }}\left(z\right),j=1,2,\cdots ,t,\\ p_0\left(z\right)=p\left(z\right).\end{array}$$

This result is best possible and equality holds as p(z) = (z - 1)ⁿwith α₁ ≥ 1, α₂ ≥ 1,..., α _t ≥ 1.

The following result proposes an extension to (1.8). In a precise set up, we have

Theorem 1.1. Let p(z) be a polynomial of degree n having all zeros in |z| ≤ k, where k ≤ 1, then for all α₁, ... α _t ∈ ℂ with |α₁| ≥ k, |α₂| ≥ k,..., |α _t | ≥ k, (1 ≤ t < n),

$$\begin{array}{r}\underset{\left|z\right|=1}{\max }\left|D_{{\alpha }_t}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}p(z)\right|\ge \frac{n(n-1)\cdots (n-t+1)}{{(1+k)}^t}[\\ \left\{\left(\left|{\alpha }_1\right|-k\right)\cdots \left(\left|{\alpha }_t\right|-k\right)\right\}\underset{\left|z\right|=1}{\max }\left|p(z)\right|+\\ \left\{{\left(1+k\right)}^t\left(\left|{\alpha }_1\right|\cdots \left|{\alpha }_t\right|\right)-\left\{\left(\left|{\alpha }_1\right|-k\right)\cdots \left(\left|{\alpha }_t\right|-k\right)\right\}\right\}{k}^{-n}\underset{\left|z\right|=k}{\min }\left|p(z)\right|].\end{array}$$

(1.9)

This result is best possible and equality holds for p(z) = (z - k)ⁿwith α₁ ≥ k, α₂ ≥ k,..., α _t ≥ k.

If we take k = 1 in Theorem 1.1, then inequality (1.9) reduces to inequality (1.8).

If we take t = 1 in Theorem 1.1, the following refinement of inequality (1.6) can be obtained.

Corollary 1.2. Let p(z) be a polynomial of degree n, having all zeros in |z| ≤ k, where k ≤ 1, then for every α ∈ ℂ with |α| ≥ k,

$$\underset{\left|z\right|=1}{max}\left|D_{\alpha }p\left(z\right)\right|\ge \frac{n}{1+k}\left\{\left(\left|\alpha \right|-k\right)\underset{\left|z\right|=1}{max}\left|p\left(z\right)\right|+\left(\left|\alpha \right|+1\right)k^{-\left(n-1\right)}\underset{\left|z\right|=k}{min}\left|p\left(z\right)\right|\right\}.$$

(1.10)

This result is best possible and equality occurs if p(z) = (z - k)ⁿwith α ≥ k.

If we divide both sides of the above inequality in (1.10) by |α| and make |α| → ∞, we obtain a result proved by Govil [8].

2. Lemmas

For proof of the theorem, the following lemmas are needed. The first lemma is due to Laguerre [9].

Lemma 2.1. If all the zeros of an nth degree polynomial p(z) lie in a circular region C and w is any zero of D _α p(z), then at most one of the points w and α may lie outside C.

Lemma 2.2. If p(z) is a polynomial of degree n, having all zeros in the closed disk |z| ≤ k, k ≤ 1, then on |z| = 1,

$$\left|p^{\prime }\left(z\right)\right|\ge \frac{n}{1+k}\left|p\left(z\right)\right|.$$

(2.1)

This lemma is due to Govil [10].

Lemma 2.3. If p(z) is a polynomial of degree n, having no zeros in |z| < k, k ≥ 1, then on |z| = 1,

$$k\left|p^{\prime }\left(z\right)\right|\le \left|q^{\prime }\left(z\right)\right|,$$

(2.2)

where$q\left(z\right)=z^n\overline{p\left(1∕\bar{z}\right)}$.

The above lemma is due to Chan and Malik [11].

Lemma 2.4. If p(z) is a polynomial of degree n, having all zeros in the closed disk |z| ≤ k, k ≤ 1, then on |z| = 1,

$$\left|q^{\prime }\left(z\right)\right|\le k\left|p^{\prime }\left(z\right)\right|,$$

(2.3)

where$q\left(z\right)=z^n\overline{p\left(1∕\bar{z}\right)}$.

Proof. Since p(z) has all its zeros in |z| ≤ k, k ≤ 1, therefore q(z) has no zero in |z| < 1/k, 1/k ≥ 1. Now applying Lemma 2.3 to the polynomial q(z) and the result follows.

Lemma 2.5. If p(z) is a polynomial of degree n, having all zeros in the closed disk |z| ≤ k, k ≤ 1, then for every real or complex number α with |α| ≥ k and |z| = 1, we have

$$\left|D_{\alpha }p\left(z\right)\right|\ge \frac{n}{1+k}\left(\left|\alpha \right|-k\right)\left|p\left(z\right)\right|.$$

(2.4)

Proof. Let $q\left(z\right)=z^n\overline{p\left(1∕\bar{z}\right)}$, then |q'(z)| = |np(z) - zp'(z)| on |z| = 1. Thus, on |z| = 1, we get

$$\begin{array}{c}\left|D_{\alpha }p\left(z\right)\right|=\left|np\left(z\right)+\left(\alpha -z\right)p^{\prime }\left(z\right)\right|=\left|\alpha p^{\prime }\left(z\right)+np\left(z\right)-z{p}^{\prime }\left(z\right)\right|\ge \\ \left|\alpha p^{\prime }\left(z\right)-|np\left(z\right)-z{p}^{\prime }\left(z\right)\right|,\end{array}$$

that implies

$$\left|D_{\alpha }p\left(z\right)\right|\ge \left|\alpha \right|\left|p^{\prime }\left(z\right)\right|-\left|q^{\prime }\left(z\right)\right|.$$

(2.5)

By combining (2.3) and (2.5), we obtain

$$\left|D_{\alpha }p\left(z\right)\right|\ge \left(\left|\alpha \right|-k\right)\left|p^{\prime }\left(z\right)\right|.$$

that along Lemma 2.2, yields

$$\left|D_{\alpha }p\left(z\right)\right|\ge \frac{n}{1+k}\left(\left|\alpha \right|-k\right)\left|p\left(z\right)\right|.$$

Lemma 2.6. If$p\left(z\right)=a_0+a_{1}z+{\sum }_{i=2}^n{a}_i{z}^i$is a polynomial of degree n, having no zeros in |z| < k, k ≥ 1, then

$$\frac{k\left|a_1\right|}{\left|a_0\right|}\le n.$$

(2.6)

The above lemma is due to Gardner et al. [12].

Lemma 2.7. If$p\left(z\right)={\sum }_{i=0}^n{a}_i{z}^i$is a polynomial of degree n, having all zeros in |z| ≤ k, k ≤ 1, then

$$\frac{\left|a_{n-1}\right|}{\left|a_n\right|}\le nk.$$

(2.7)

Proof. Since p(z) has all zeros in |z| ≤ k, k ≤ 1, therefore

$$q\left(z\right)=z^n\overline{p\left(1∕\bar{z}\right)}=\overline{a_n}+\overline{a_{n-1}}z+\cdots +\overline{a_1}{z}^{n-1}+\overline{a_0}{z}^n,$$

is a polynomial of degree at most n, which does not vanish in |z| < 1/k, 1/k ≥ 1. By applying Lemma 2.6 for q(z), we get

$$\frac{\frac{1}{k}\left|a_{n-1}\right|}{\left|a_n\right|}\le \mathsf{\text{degree}}\left\{q\left(z\right)\right\}\le n,$$

which completes the proof.

Lemma 2.8. If p(z) is a polynomial of degree n having all zeros in |z| ≤ k, k ≤ 1, then for all α₁, ... α _t ∈ ℂ with |α₁| ≥ k, |α₂| ≥ k,..., |α _t | ≥ k, (1 ≤ t < n), and |z| = 1 we have

$$\begin{array}{c}\left|D_{{\alpha }_t}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}p\left(z\right)\right|\ge \frac{n\left(n-1\right)\cdots \left(n-t+1\right)}{{\left(1+k\right)}^t}\times \\ \left\{\left(\left|{\alpha }_1\right|-k\right)\cdots \left(\left|{\alpha }_t\right|-k\right)\right\}\left|p\left(z\right)\right|.\end{array}$$

(2.8)

Proof. If |α _j | = k for at least one j; 1 ≤ j ≤ t, then inequality (2.8) is trivial. Therefore, we assume that |α _j | > k for all j; 1 ≤ j ≤ t.

In the rest, we proceed by mathematical induction. The result is true for t = 1, by Lemma 2.5, that means if |α₁| > k then

$$\left|D_{{\alpha }_1}p\left(z\right)\right|\ge \frac{n}{1+k}\left(\left|{\alpha }_1\right|-k\right)\left|p\left(z\right)\right|.$$

(2.9)

Now for t = 2, since $D_{{\alpha }_1}p\left(z\right)=\left(n{a}_n{\alpha }_1+a_{n-1}\right)z^{n-1}+\cdots +\left(n{a}_0+{\alpha }_1{a}_1\right)$, and |α₁| > k, then $D_{{\alpha }_1}p\left(z\right)$ will be a polynomial of degree (n - 1). If it is not true, then the coefficient of z^n-1must be equal to zero, which implies

$$n{a}_n{\alpha }_1+a_{n-1}=0,$$

i.e,

$$\left|{\alpha }_1\right|=\frac{\left|a_{n-1}\right|}{n\left|a_n\right|}.$$

Applying Lemma 2.7, we get

$$\left|{\alpha }_1\right|=\frac{\left|a_{n-1}\right|}{n\left|a_n\right|}\le k.$$

But this result contradicts the fact that |α₁| > k. Hence, the polynomial $D_{{\alpha }_1}p\left(z\right)$ must be of degree (n - 1).

On the other hand, since all the zeros of p(z) lie in |z| ≤ k, therefore by applying Lemma 2.1, all the zeros of $D_{{\alpha }_1}p\left(z\right)$ lie in |z| ≤ k, then using Lemma 2.5 for the polynomial $D_{{\alpha }_1}p\left(z\right)$ of degree n - 1, and |α₂ | > k, it concludes that

$$\left|D_{{\alpha }_2}\left\{D_{{\alpha }_1}p\left(z\right)\right\}\right|\ge \frac{\left(n-1\right)}{1+k}\left(\left|{\alpha }_2\right|-k\right)\left|D_{{\alpha }_1}p\left(z\right)\right|.$$

Substituting the term $D_{{\alpha }_1}p\left(z\right)$ from (2.9) in the above inequality, we obtain

$$\left|D_{{\alpha }_2}{D}_{{\alpha }_1}p\left(z\right)\right|\ge \frac{n\left(n-1\right)}{{\left(1+k\right)}^2}\left(\left|{\alpha }_1\right|-k\right)\left(\left|{\alpha }_2\right|-k\right)\left|p\left(z\right)\right|.$$

This implies result is true for t = 2.

At this stage, we assume that the result is true for t = s < n; it means that for |z| = 1, we have

$$\begin{array}{c}\left|D_{{\alpha }_s}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}p\left(z\right)\right|\ge \frac{n\left(n-1\right)\cdots \left(n-s+1\right)}{{\left(1+k\right)}^s}\times \\ \left\{\left(\left|{\alpha }_1\right|-k\right)\cdots \left(\left|{\alpha }_s\right|-k\right)\right\}\left|p\left(z\right)\right|,\end{array}$$

(2.10)

and we will prove that the result is true for t = s + 1 < n.

According to the above procedure, using Lemmas 2.7 and 2.1, the polynomial $D_{{\alpha }_2}{D}_{{\alpha }_1}p\left(z\right)$ must be of degree (n - 2) for |α₁| > k, |α₂| > k, and has all zeros in |z| ≤ k. One can continue that $D_{{\alpha }_s}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}p\left(z\right)$ will be a polynomial of degree (n - s) for all α₁,... α _s ∈ ℂ with |α₁| ≥ k, |α₂| ≥ k,..., |α _s | ≥ k, (s < n), and has all zeros in |z| ≤ k. Therefore, for |α_s+1| > k, by applying Lemma 2.5 to $D_{{\alpha }_s}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}p\left(z\right)$, we get

$$\left|D_{{\alpha }_{s+1}}\left\{D_{{\alpha }_s}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}p\left(z\right)\right\}\right|\ge \frac{\left(n-s\right)}{1+k}\left(\left|{\alpha }_{s+1}\right|-k\right)\left|D_{{\alpha }_s}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}p\left(z\right)\right|.$$

(2.11)

By combining the terms (2.10) and (2.11), we obtain

$$\begin{array}{c}\left|D_{{\alpha }_{s+1}}{D}_{{\alpha }_s}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}p\left(z\right)\right|\ge \frac{n\left(n-1\right)\cdots \left(n-s\right)}{{\left(1+k\right)}^{s+1}}\times \\ \left\{\left(\left|{\alpha }_1\right|-k\right)\cdots \left(\left|{\alpha }_{s+1}\right|-k\right)\right\}\left|p\left(z\right)\right|.\end{array}$$

This implies that the result is true for t = s + 1. The proof is complete.

Lemma 2.9. If$p\left(z\right)={\sum }_{i=0}^n{a}_i{z}^i$ is a polynomial of degree n, p(z) ≠ 0 in |z| < k, then m < |p(z)| for |z| < k, and in particular m < |a₀|, where m = min_|z|=k|p(z)|.

The above lemma is due to Gardner et al. [13].

Lemma 2.10. If$p\left(z\right)={\sum }_{i=0}^n{a}_i{z}^i$is a polynomial of degree n having all zeros in |z| ≤ k, then

$$m\le k^n\left|a_n\right|,$$

(2.12)

where m = min_|z|=k|p(z)|.

Proof. If k = 0, then inequality (2.12) is trivial. Now we suppose that k > 0. Since the polynomial $p\left(z\right)={\sum }_{i=0}^n{a}_i{z}^i$ has all zeros in |z| ≤ k, the polynomial q(z) = zⁿp(1/z) = a _n + ⋯ + a₀zⁿhas no zero in $\left|z\right|<\frac{1}{k}$. Thus, by applying Lemma 2.9 for the polynomial q(z), we get

$$\underset{\left|z\right|=\frac{1}{k}}{min}\left|q\left(z\right)\right|<\left|a_n\right|.$$

(2.13)

Since ${min}_{\left|z\right|=\frac{1}{k}}\left|q\left(z\right)\right|=\frac{1}{k^n}{min}_{\left|z\right|=k}\left|p\left(z\right)\right|$, (2.13) implies that $\frac{m}{k^n}<\left|a_n\right|$.

3. Proof of the theorem

Proof of Theorem 1.1. Let m = min_|z|=k|p(z)|. If p(z) has a zero on |z| = k, then m = 0 and the result follows from Lemma 2.8. Henceforth, we suppose that all the zeros of p(z) lie in |z| < k, so that m > 0. Now m ≤ |p(z)| for |z| = k, therefore if λ is any real or complex number such that |λ| < 1, then $\left|\lambda m{\left(\frac{z}{k}\right)}^n\right|<\left|p\left(z\right)\right|$ for |z| = k. Since all zeros of p(z) lie in |z| < k, by Rouche's theorem we can deduce that all zeros of the polynomial $G\left(z\right)=p\left(z\right)-\lambda m{\left(\frac{z}{k}\right)}^n$ lie in |z| < k. Also it follows from Lemma 2.10, that $\left|\lambda \right|\frac{m}{k^n}<\left|a_n\right|$, hence the polynomial $G\left(z\right)=p\left(z\right)-\lambda \left(\frac{m}{k^n}\right)z^n$ is of degree n. Now we can apply Lemma 2.8 for the polynomial G(z) of degree n which has all zeros in |z| ≤ k. This implies that for all α₁,... α _t ∈ ℂ with |α₁| ≥ k, |α₂| ≥ k, ..., |α _t | ≥ k, (t < n), on |z| = 1,

$$\begin{array}{c}\left|D_{{\alpha }_t}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}G\left(z\right)\right|\ge \frac{n\left(n-1\right)\cdots \left(n-t+1\right)}{{\left(1+k\right)}^t}\times \\ \left\{\left(\left|{\alpha }_1\right|-k\right)\cdots \left(\left|{\alpha }_t\right|-k\right)\right\}\left|G\left(z\right)\right|.\end{array}$$

Equivalently

$$\begin{array}{c}\left|D_{{\alpha }_t}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}p\left(z\right)-\lambda \frac{m}{k^n}\left\{n\left(n-1\right)\cdots \left(n-t+1\right){\alpha }_1{\alpha }_2\cdots {\alpha }_1\right\}{z}^{n-t}\right|\ge \\ \frac{n\left(n-1\right)\cdots \left(n-t+1\right)}{{\left(1+k\right)}^t}\left\{\left(\left|{\alpha }_1\right|-k\right)\cdots \left(\left|{\alpha }_t\right|-k\right)\right\}\left|p\left(z\right)-\lambda m{\left(\frac{z}{k}\right)}^n\right|.\end{array}$$

(3.1)

But by Lemma 2.1, the polynomial $T\left(z\right)=D_{{\alpha }_t}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}G\left(z\right)$ has all zeros in |z| ≤ k. That is,

$$T\left(z\right)=D_{{\alpha }_t}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}G\left(z\right)\ne 0,for\left|z\right|>k.$$

Then, substituting G(z) in the above, we conclude that for every λ with |λ| < 1, and |z| > k,

$$\begin{array}{c}T\left(z\right)=D_{{\alpha }_t}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}p\left(z\right)-\\ \lambda \frac{m}{k^n}\left\{n\left(n-1\right)\cdots \left(n-t+1\right){\alpha }_1{\alpha }_2\cdots {\alpha }_t\right\}{z}^{n-t}\ne 0.\end{array}$$

(3.2)

Thus, for |z| > k,

$$\left|D_{{\alpha }_t}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}p\left(z\right)\right|\ge \frac{m}{k^n}\left\{n\left(n-1\right)\cdots \left(n-t+1\right)\left|{\alpha }_1\right|\left|{\alpha }_2\right|\cdots \left|{\alpha }_t\right|\right\}\left|z^{n-t}\right|.$$

(3.3)

If the inequality (3.3) is not true, then there is a point z = z₀ with |z₀| > k such that

$$\left|D_{{\alpha }_t}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}p\left(z_0\right)\right|<\frac{m}{k^n}\left\{n\left(n-1\right)\cdots \left(n-t+1\right)\left|{\alpha }_1\right|\left|{\alpha }_2\right|\cdots \left|{\alpha }_t\right|\right\}\left|z_0^{n-t}\right|.$$

Now take

$$\lambda =\frac{D_{{\alpha }_t}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}p\left(z_0\right)}{\frac{m}{k^n}\left\{n\left(n-1\right)\cdots \left(n-t+1\right){\alpha }_1{\alpha }_2\cdots {\alpha }_t\right\}{z}_0^{n-t}},$$

then |λ| < 1 and with this choice of λ, we have, T(z₀) = 0 for |z₀| > k, from (3.2). But it contradicts the fact that T(z) ≠ 0 for |z| > k. Hence, for |z| > k, we have

$$\left|D_{{\alpha }_t}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}p\left(z\right)\right|\ge \frac{m}{k^n}\left\{n\left(n-1\right)\cdots \left(n-t+1\right)\left|{\alpha }_1\right|\left|{\alpha }_2\right|\cdots \left|{\alpha }_t\right|\right\}\left|z^{n-t}\right|.$$

Taking a relevant choice of argument of λ, arg $\lambda =arg\left\{D_{{\alpha }_t}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}p\left(z\right)\right\}-arg\left\{{\alpha }_1{\alpha }_2\cdots {\alpha }_t{z}^{n-t}\right\}$, we have

$$\begin{array}{c}\left|D_{{\alpha }_t}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}p\left(z\right)\right.-\\ \left.\lambda \frac{m}{k^n}\left\{n\left(n-1\right)\cdots \left(n-t+1\right){\alpha }_1{\alpha }_2\cdots {\alpha }_t\right\}{z}^{n-t}\right|=\\ \left|D_{{\alpha }_t}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}p\left(z\right)\right.-\\ \left.\left|\lambda \right|\frac{m}{k^n}\left\{n\left(n-1\right)\cdots \left(n-t+1\right)\left|{\alpha }_1\right|\left|{\alpha }_2\right|\cdots \left|{\alpha }_t\right|\right\}{|z}^{n-t}\right|,\end{array}$$

where |z| = 1.

Therefore, we can rewrite (3.1) as

$$\begin{array}{c}\left|D_{{\alpha }_t}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}p\left(z\right)\right|-\\ \left.\left|\lambda \right|\frac{m}{k^n}\left\{n\left(n-1\right)\cdots \left(n-t+1\right)\left|{\alpha }_1\right|\left|{\alpha }_2\right|\cdots \left|{\alpha }_t\right|\right\}{z}^{n-t}\right|\ge \\ \frac{n\left(n-1\right)\cdots \left(n-t+1\right)}{{\left(1+k\right)}^t}\left\{\left(\left|{\alpha }_1\right|-k\right)\cdots \left(\left|{\alpha }_t\right|-k\right)\right\}\left(p\left(z\right)-\left|\lambda \right|\frac{m}{k^n}{\left|z\right|}^n\right),\end{array}$$

where |z| = 1.

In an equivalent way

$$\begin{array}{c}\left|D_{{\alpha }_t}\cdots D_{{\alpha }_2}{D}_{{\alpha }_1}p(z)\right|\ge \frac{n(n-1)\cdots (n-t+1)}{{(1+k)}^t}[\\ \left\{\left(\left|{\alpha }_1\right|-k\right)\cdots \left(\left|{\alpha }_t\right|-k\right)\left|p(z)\right|\right\}+\\ \left|\text{λ}\right|\left\{{\left(1+k\right)}^t\left(\left|{\alpha }_1\right|\left|{\alpha }_2\right|\cdots \left|{\alpha }_t\right|\right)-\left\{\left(\left|{\alpha }_1\right|-k\right)\cdots \left(\left|{\alpha }_t\right|-k\right)\right\}\right\}\frac{m}{k^n}].\end{array}$$

Making |λ| → 1, Theorem 1.1 follows.