Theory and Modern Applications

Aspects of univalent holomorphic functions involving multiplier transformation and Ruscheweyh derivative

Abstract

Making use multiplier transformation and Ruscheweyh derivative,we introduce a new class of analytic functions $\mathcal{RI}\left(\gamma ,\lambda ,l,\alpha ,\beta \right)$ defined on the open unit disc, and investigate its various characteristics. Further we obtain distortion bounds, extreme points and radii of close-to-convexity, starlikeness and convexity and neighborhood property for functions belonging to the class $\mathcal{RI}\left(\gamma ,\lambda ,l,\alpha ,\beta \right)$.

MSC:30C45, 30A20, 34A40.

1 Introduction

Let denote the class of functions of the form $f\left(z\right)=z+{\sum }_{j=2}^{\mathrm{\infty }}{a}_{j}{z}^{j}$, which are analytic and univalent in the open unit disc $U=\left\{z:z\in \mathbb{C}:|z|<1\right\}$. is a subclass of consisting of the functions of the form $f\left(z\right)=z-{\sum }_{j=2}^{\mathrm{\infty }}|{a}_{j}|{z}^{j}$. For functions $f,g\in \mathcal{A}$ given by $f\left(z\right)=z+{\sum }_{j=2}^{\mathrm{\infty }}{a}_{j}{z}^{j}$, $g\left(z\right)=z+{\sum }_{j=2}^{\mathrm{\infty }}{b}_{j}{z}^{j}$, we define the Hadamard product (or convolution) of f and g by $\left(f\ast g\right)\left(z\right)=z+{\sum }_{j=2}^{\mathrm{\infty }}{a}_{j}{b}_{j}{z}^{j}$, $z\in U$.

Definition 1.1 (Ruscheweyh [1])

For $f\in \mathcal{A}$, $n\in \mathbb{N}$, the operator ${R}^{n}$ is defined by ${R}^{n}:\mathcal{A}\to \mathcal{A}$,

$\begin{array}{c}{R}^{0}f\left(z\right)=f\left(z\right),\hfill \\ {R}^{1}f\left(z\right)=z{f}^{\prime }\left(z\right),\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\dots ,\hfill \\ \left(n+1\right){R}^{n+1}f\left(z\right)=z{\left({R}^{n}f\left(z\right)\right)}^{\prime }+n{R}^{n}f\left(z\right),\phantom{\rule{1em}{0ex}}z\in U.\hfill \end{array}$

Remark 1.1 If $f\in \mathcal{A}$, $f\left(z\right)=z+{\sum }_{j=2}^{\mathrm{\infty }}{a}_{j}{z}^{j}$, then ${R}^{n}f\left(z\right)=z+{\sum }_{j=2}^{\mathrm{\infty }}\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}{a}_{j}{z}^{j}$, $z\in U$.

If $f\in \mathcal{T}$, $f\left(z\right)=z-{\sum }_{j=2}^{\mathrm{\infty }}{a}_{j}{z}^{j}$, then ${R}^{n}f\left(z\right)=z-{\sum }_{j=t+1}^{\mathrm{\infty }}\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}{a}_{j}{z}^{j}$, $z\in U$.

Definition 1.2 [2, 3]

For $f\in \mathcal{A}$, $n\in \mathbb{N}\cup \left\{0\right\}$, $\lambda ,l\ge 0$, the operator $I\left(n,\lambda ,l\right)f\left(z\right)$ is defined by the following infinite series:

$I\left(n,\lambda ,l\right)f\left(z\right):=z+\sum _{j=2}^{\mathrm{\infty }}{\left(\frac{\lambda \left(j-1\right)+l+1}{l+1}\right)}^{n}{a}_{j}{z}^{j}.$

Remark 1.2 [4]

It follows from the above definition that

$\begin{array}{c}I\left(0,\lambda ,l\right)f\left(z\right)=f\left(z\right),\hfill \\ \left(l+1\right)I\left(n+1,\lambda ,l\right)f\left(z\right)=\left(l+1-\lambda \right)I\left(n,\lambda ,l\right)f\left(z\right)+\lambda z{\left(I\left(n,\lambda ,l\right)f\left(z\right)\right)}^{\prime },\phantom{\rule{1em}{0ex}}z\in U.\hfill \end{array}$

Remark 1.3 The operator $I\left(n,\lambda ,0\right)={D}_{\lambda }^{n}$ is the generalized Sălăgean operator introduced by Al-Oboudi [5], and $I\left(n,1,0\right)={S}^{n}$ is the Sălăgean differential operator [6].

Definition 1.3 [7, 8]

Let $\gamma ,\lambda ,l\ge 0$, $n\in \mathbb{N}$. Denote by $R{I}_{n,\lambda ,l}^{\gamma }$ the operator given by $R{I}_{n,\lambda ,l}^{\gamma }:\mathcal{A}\to \mathcal{A}$,

$R{I}_{n,\lambda ,l}^{\gamma }f\left(z\right)=\left(1-\gamma \right){R}^{n}f\left(z\right)+\gamma I\left(n,\lambda ,l\right)f\left(z\right),\phantom{\rule{1em}{0ex}}z\in U.$

Remark 1.4 If $f\in \mathcal{A}$, $f\left(z\right)=z+{\sum }_{j=2}^{\mathrm{\infty }}{a}_{j}{z}^{j}$, then

$R{I}_{n,\lambda ,l}^{\gamma }f\left(z\right)=z+\sum _{j=2}^{\mathrm{\infty }}\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}{a}_{j}{z}^{j},\phantom{\rule{1em}{0ex}}z\in U.$

If $f\in \mathcal{T}$, $f\left(z\right)=z-{\sum }_{j=2}^{\mathrm{\infty }}{a}_{j}{z}^{j}$, then

$R{I}_{n,\lambda ,l}^{\gamma }f\left(z\right)=z-\sum _{j=2}^{\mathrm{\infty }}\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}{a}_{j}{z}^{j},\phantom{\rule{1em}{0ex}}z\in U.$

Remark 1.5 The operator $R{I}_{n,\lambda ,0}^{\gamma }f\left(z\right)=R{D}_{\lambda ,\gamma }^{n}f\left(z\right)$ which was introduced in [9] and the operator $R{I}_{n,1,0}^{\gamma }f\left(z\right)={L}_{\gamma }^{n}f\left(z\right)$ which was introduced in [10].

Following the work of Najafzadeh and Pezeshki [11] we can define the class $\mathcal{RI}\left(\gamma ,\lambda ,l,\alpha ,\beta \right)$ as follows.

Definition 1.4 For $\gamma ,\lambda ,l\ge 0$, $0\le \alpha <1$ and $0<\beta \le 1$, let $\mathcal{RI}\left(\gamma ,\lambda ,l,\alpha ,\beta \right)$ be the subclass of consisting of functions that satisfying the inequality

$|\frac{R{I}_{n,\lambda ,l}^{\mu ,\gamma }f\left(z\right)-1}{2\nu \left(R{I}_{n,\lambda ,l}^{\mu ,\gamma }f\left(z\right)-\alpha \right)-\left(R{I}_{n,\lambda ,l}^{\mu ,\gamma }f\left(z\right)-1\right)}|<\beta ,$
(1.1)

where

$R{I}_{n,\lambda ,l}^{\mu ,\gamma }f\left(z\right)=\left(1-\mu \right)\frac{R{I}_{n,\lambda ,l}^{\gamma }f\left(z\right)}{z}+\mu {\left(R{I}_{n,\lambda ,l}^{\gamma }f\left(z\right)\right)}^{\prime },$
(1.2)

$0<\nu \le 1$.

Remark 1.6 If $f\in \mathcal{T}$, $f\left(z\right)=z-{\sum }_{j=2}^{\mathrm{\infty }}{a}_{j}{z}^{j}$, then

$\begin{array}{rcl}R{I}_{n,\lambda ,l}^{\mu ,\gamma }f\left(z\right)& =& 1-\sum _{j=t+1}^{\mathrm{\infty }}\left[1+\mu \left(j-1\right)\right]\\ ×\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}{a}_{j}{z}^{j-1},\phantom{\rule{1em}{0ex}}z\in U.\end{array}$

Remark 1.7 The class $\mathcal{RI}\left(\gamma ,\lambda ,0,\alpha ,\beta \right)=\mathcal{RD}\left(\gamma ,\lambda ,\alpha ,\beta \right)$ defined and studied in [12] and $\mathcal{RI}\left(\gamma ,1,0,\alpha ,\beta \right)=\mathcal{L}\left(\gamma ,\alpha ,\beta \right)$ defined and studied in [13].

2 Coefficient bounds

In this section we obtain coefficient bounds and extreme points for functions in $\mathcal{RI}\left(\gamma ,\lambda ,l,\alpha ,\beta \right)$.

Theorem 2.1 Let the function $f\in \mathcal{T}$. Then $f\in \mathcal{RI}\left(\gamma ,\lambda ,l,\alpha ,\beta \right)$ if and only if

$\begin{array}{c}\sum _{j=t+1}^{\mathrm{\infty }}\left(1+\mu \left(j-1\right)\right)\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}{a}_{j}\hfill \\ \phantom{\rule{1em}{0ex}}<2\beta \nu \left(1-\alpha \right).\hfill \end{array}$
(2.1)

The result is sharp for the function $F\left(z\right)$ defined by

$F\left(z\right)=z-\frac{2\beta \nu \left(1-\alpha \right)}{\left(1+\mu \left(j-1\right)\right)\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}}{z}^{j},\phantom{\rule{1em}{0ex}}j\ge t+1.$

Proof Suppose f satisfies (2.1). Then for $|z|<1$, we have

$\begin{array}{c}|R{I}_{n,\lambda ,l}^{\mu ,\gamma }f\left(z\right)-1|-\beta |2\nu \left(R{I}_{n,\lambda ,l}^{\mu ,\gamma }f\left(z\right)-\alpha \right)-\left(R{I}_{n,\lambda ,l}^{\mu ,\gamma }f\left(z\right)-1\right)|\hfill \\ \phantom{\rule{1em}{0ex}}=|-\sum _{j=t+1}^{\mathrm{\infty }}\left(1+\mu \left(j-1\right)\right)\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}{a}_{j}{z}^{j-1}|\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}-\beta |2\nu \left(1-\alpha \right)-\left(2\nu -1\right)\sum _{j=t+1}^{\mathrm{\infty }}\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}×\left[1+\mu \left(j-1\right)\right]{a}_{j}{z}^{j-1}|\hfill \\ \phantom{\rule{1em}{0ex}}\le \sum _{j=t+1}^{\mathrm{\infty }}\left[1+\mu \left(j-1\right)\right]\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}{a}_{k}-2\beta \nu \left(1-\alpha \right)\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\sum _{j=t+1}^{\mathrm{\infty }}\beta \left(2\nu -1\right)\left(1+\mu \left(j-1\right)\right)\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}{a}_{j}\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{j=t+1}^{\mathrm{\infty }}\left[1+\mu \left(j-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}{a}_{j}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}-2\beta \nu \left(1-\alpha \right)<0.\hfill \end{array}$

Hence, by using the maximum modulus Theorem and (1.1), $f\in \mathcal{RI}\left(\gamma ,\lambda ,l,\alpha ,\beta \right)$. Conversely, assume that

$\begin{array}{c}|\frac{R{I}_{n,\lambda ,l}^{\mu ,\gamma }f\left(z\right)-1}{2\nu \left(R{I}_{n,\lambda ,l}^{\mu ,\gamma }f\left(z\right)-\alpha \right)-\left(R{I}_{n,\lambda ,l}^{\mu ,\gamma }f\left(z\right)-1\right)}|\hfill \\ \phantom{\rule{1em}{0ex}}=|\frac{-{\sum }_{j=t+1}^{\mathrm{\infty }}\left[1+\mu \left(j-1\right)\right]\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}{a}_{j}{z}^{j-1}}{2\nu \left(1-\alpha \right)-{\sum }_{j=t+1}^{\mathrm{\infty }}\left[1+\mu \left(j-1\right)\right]\left(2\nu -1\right)\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}{a}_{j}{z}^{j-1}}|\hfill \\ \phantom{\rule{1em}{0ex}}<\beta ,\phantom{\rule{1em}{0ex}}z\in U.\hfill \end{array}$

Since $Re\left(z\right)\le |z|$ for all $z\in U$, we have

$\begin{array}{c}Re\left\{\frac{{\sum }_{j=t+1}^{\mathrm{\infty }}\left[1+\mu \left(j-1\right)\right]\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}{a}_{j}{z}^{j-1}}{2\nu \left(1-\alpha \right)-{\sum }_{j=t+1}^{\mathrm{\infty }}\left[1+\mu \left(j-1\right)\right]\left(2\nu -1\right)\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}{a}_{j}{z}^{j-1}}\right\}\hfill \\ \phantom{\rule{1em}{0ex}}<\beta .\hfill \end{array}$
(2.2)

By choosing choose values of z on the real axis so that $R{I}_{n,\lambda ,l}^{\mu ,\gamma }f\left(z\right)$ is real and letting $z\to 1$ through real values, we obtain the desired inequality (2.1). □

Corollary 2.2 If $f\in \mathcal{T}$ is in $\mathcal{RI}\left(\gamma ,\lambda ,l,\alpha ,\beta \right)$, then

${a}_{j}\le \frac{2\beta \nu \left(1-\alpha \right)}{\left[1+\mu \left(j-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}},\phantom{\rule{1em}{0ex}}j\ge t+1,$
(2.3)

with equality only for functions of the form $F\left(z\right)$.

Theorem 2.3 Let ${f}_{1}\left(z\right)=z$ and

$\begin{array}{c}{f}_{j}\left(z\right)=z-\frac{2\beta \nu \left(1-\alpha \right)}{\left[1+\mu \left(j-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}}{z}^{j},\hfill \\ \phantom{\rule{1em}{0ex}}j\ge t+1,\hfill \end{array}$
(2.4)

for $0\le \alpha <1$, $0<\beta \le 1$, $\gamma ,\lambda ,l\ge 0$ and $0<\nu \le 1$. Then $f\left(z\right)$ is in the class $\mathcal{RI}\left(\gamma ,\lambda ,l,\alpha ,\beta \right)$ if and only if it can be expressed in the form

$f\left(z\right)=\sum _{j=t}^{\mathrm{\infty }}{\omega }_{j}{f}_{j}\left(z\right),$
(2.5)

where ${\omega }_{j}\ge 0$ and ${\sum }_{j=1}^{\mathrm{\infty }}{\omega }_{j}=1$.

Proof Suppose $f\left(z\right)$ can be written as in (2.5). Then

$f\left(z\right)=z-\sum _{j=t+1}^{\mathrm{\infty }}{\omega }_{j}\frac{2\beta \nu \left(1-\alpha \right)}{\left[1+\mu \left(j-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}}{z}^{j}.$

Now,

$\begin{array}{c}\sum _{j=t+1}^{\mathrm{\infty }}\frac{\left[1+\mu \left(j-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}}{2\beta \nu \left(1-\alpha \right)}{\omega }_{j}\hfill \\ \phantom{\rule{1em}{0ex}}×\frac{2\beta \nu \left(1-\alpha \right)}{\left[1+\mu \left(j-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}}=\sum _{j=t+1}^{\mathrm{\infty }}{\omega }_{j}=1-{\omega }_{1}\le 1.\hfill \end{array}$

Thus $f\in \mathcal{RI}\left(\gamma ,\lambda ,l,\alpha ,\beta \right)$.

Conversely, let $f\in \mathcal{RI}\left(\gamma ,\lambda ,l,\alpha ,\beta \right)$. Then by using (2.3), setting

${\omega }_{j}=\frac{\left[1+\mu \left(j-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}}{2\beta \nu \left(1-\alpha \right)}{a}_{j},\phantom{\rule{1em}{0ex}}j\ge t+1,$

and ${\omega }_{1}=1-{\sum }_{j=2}^{\mathrm{\infty }}{\omega }_{j}$, we have $f\left(z\right)={\sum }_{j=t}^{\mathrm{\infty }}{\omega }_{j}{f}_{j}\left(z\right)$. This completes the proof of Theorem 2.3. □

3 Distortion bounds

In this section we obtain distortion bounds for the class $\mathcal{RI}\left(\gamma ,\lambda ,l,\alpha ,\beta \right)$.

Theorem 3.1 If $f\in \mathcal{RI}\left(\gamma ,\lambda ,l,\alpha ,\beta \right)$, then

$\begin{array}{c}r-\frac{2\beta \nu \left(1-\alpha \right)}{\left(1+\mu t\right)\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda t+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+t\right)!}{n!t!}\right\}}{r}^{t+1}\hfill \\ \phantom{\rule{1em}{0ex}}\le |f\left(z\right)|\le r+\frac{2\beta \nu \left(1-\alpha \right)}{\left(1+\mu t\right)\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda t+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+t\right)!}{n!t!}\right\}}{r}^{t+1}\hfill \end{array}$
(3.1)

holds if the sequence ${\left\{{\sigma }_{j}\left(\gamma ,\lambda ,l,\beta ,\nu \right)\right\}}_{j=t+1}^{\mathrm{\infty }}$ is non-decreasing, and

$\begin{array}{c}1-\frac{2\beta \nu \left(1-\alpha \right)\left(t+1\right)}{\left(1+\mu t\right)\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda t+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+t\right)!}{n!t!}\right\}}{r}^{t}\hfill \\ \le |{f}^{\prime }\left(z\right)|\le 1+\frac{2\beta \nu \left(1-\alpha \right)\left(t+1\right)}{\left(1+\mu t\right)\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda t+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+t\right)!}{n!t!}\right\}}{r}^{t}\hfill \end{array}$
(3.2)

holds if the sequence ${\left\{\frac{{\sigma }_{j}\left(\gamma ,\lambda ,l,\beta ,\nu \right)}{j}\right\}}_{j=t+1}^{\mathrm{\infty }}$ is non-decreasing, where

${\sigma }_{j}\left(\gamma ,\beta ,\nu \right)=\left[1+\mu \left(j-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}.$

The bounds in (3.1) and (3.2) are sharp, for $f\left(z\right)$ given by

$f\left(z\right)=z-\frac{2\beta \nu \left(1-\alpha \right)}{\left(1+\mu t\right)\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda t+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+t\right)!}{n!t!}\right\}}{z}^{t+1},\phantom{\rule{1em}{0ex}}z=±r.$
(3.3)

Proof In view of Theorem 2.1, we have

$\sum _{j=t+1}^{\mathrm{\infty }}{a}_{j}\le \frac{2\beta \nu \left(1-\alpha \right)}{\left(1+\mu t\right)\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda t+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+t\right)!}{n!t!}\right\}}.$
(3.4)

We obtain

$|z|-|z{|}^{t+1}\sum _{j=t+1}^{\mathrm{\infty }}{a}_{j}\le |f\left(z\right)|\le |z|+|z{|}^{t+1}\sum _{j=t+1}^{\mathrm{\infty }}{a}_{j}.$

Thus

$\begin{array}{c}r-\frac{2\beta \nu \left(1-\alpha \right)}{\left(1+\mu t\right)\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda t+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+t\right)!}{n!t!}\right\}}{r}^{t+1}\hfill \\ \phantom{\rule{1em}{0ex}}\le |f\left(z\right)|\le r+\frac{2\beta \nu \left(1-\alpha \right)}{\left(1+\mu t\right)\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda t+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+t\right)!}{n!t!}\right\}}{r}^{t+1}.\hfill \end{array}$
(3.5)

Hence (3.1) follows from (3.5). Further,

$\sum _{j=t+1}^{\mathrm{\infty }}j{a}_{j}\le \frac{2\beta \nu \left(1-\alpha \right)}{\left(1+\mu t\right)\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda t+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+t\right)!}{n!t!}\right\}}.$

Hence (3.2) follows from

$1-{r}^{t}\sum _{j=t+1}^{\mathrm{\infty }}j{a}_{j}\le |{f}^{\prime }\left(z\right)|\le 1+{r}^{t}\sum _{j=t+1}^{\mathrm{\infty }}j{a}_{j}.$

□

4 Radius of starlikeness and convexity

The radii of close-to-convexity, starlikeness, and convexity for the class $\mathcal{RI}\left(\gamma ,\lambda ,l,\alpha ,\beta \right)$ are given in this section.

Theorem 4.1 Let the function $f\in \mathcal{T}$ belong to the class $\mathcal{RI}\left(\gamma ,\lambda ,l,\alpha ,\beta \right)$, Then $f\left(z\right)$ is close-to-convex of order δ, $0\le \delta <1$, in the disc $|z|, where

$r:=\underset{j\ge t+1}{inf}{\left[\frac{\left(1-\delta \right)\left[1+\mu \left(j-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}}{2\beta \nu j\left(1-\alpha \right)}\right]}^{\frac{1}{t}}.$
(4.1)

The result is sharp, with extremal function $f\left(z\right)$ given by (2.3).

Proof For given $f\in \mathcal{T}$ we must show that

$|{f}^{\prime }\left(z\right)-1|<1-\delta .$
(4.2)

By a simple calculation we have

$|{f}^{\prime }\left(z\right)-1|\le \sum _{j=t+1}^{\mathrm{\infty }}j{a}_{j}|z{|}^{t}.$

The last expression is less than $1-\delta$ if

$\sum _{j=t+1}^{\mathrm{\infty }}\frac{j}{1-\delta }{a}_{j}|z{|}^{t}<1.$

We use the fact that $f\in \mathcal{RI}\left(\gamma ,\lambda ,l,\alpha ,\beta \right)$ if and only if

$\sum _{j=t+1}^{\mathrm{\infty }}\frac{\left[1+\mu \left(j-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}}{2\beta \nu \left(1-\alpha \right)}{a}_{j}\le 1.$

Equation (4.2) holds true if

$\frac{j}{1-\delta }|z{|}^{t}\le \sum _{j=t+1}^{\mathrm{\infty }}\frac{\left[1+\mu \left(j-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}}{2\beta \nu \left(1-\alpha \right)}.$

Or, equivalently,

$|z{|}^{t}\le \sum _{j=t+1}^{\mathrm{\infty }}\frac{\left(1-\delta \right)\left[1+\mu \left(j-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}}{2\beta \nu j\left(1-\alpha \right)},$

which completes the proof. □

Theorem 4.2 Let $f\in \mathcal{RI}\left(\gamma ,\lambda ,l,\alpha ,\beta \right)$. Then

1. 1.

f is starlike of order δ, $0\le \delta <1$, in the disc $|z|<{r}_{1}$, where

${r}_{1}=\underset{j\ge t+1}{inf}{\left\{\frac{\left(1-\delta \right)\left[1+\mu \left(j-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}}{2\beta \nu \left(1-\alpha \right)\left(j-\delta \right)}\right\}}^{\frac{1}{t}}.$
2. 2.

f is convex of order δ, $0\le \delta <1$, in the disc $|z|<{r}_{2}$ where,

${r}_{2}=\underset{j\ge t+1}{inf}{\left\{\frac{\left(1-\delta \right)\left[1+\mu \left(j-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}}{2\beta \nu j\left(j-1\right)\left(1-\alpha \right)}\right\}}^{\frac{1}{t}}.$

Each of these results is sharp for the extremal function $f\left(z\right)$ given by (2.5).

Proof 1. For $0\le \delta <1$ we need to show that

$|\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}-1|<1-\delta .$
(4.3)

We have

$|\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}-1|\le |\frac{{\sum }_{j=t+1}^{\mathrm{\infty }}\left(j-1\right){a}_{j}|z{|}^{t}}{1-{\sum }_{j=t+1}^{\mathrm{\infty }}{a}_{j}|z{|}^{t}}|.$

The last expression is less than $1-\delta$ if

$\sum _{j=t+1}^{\mathrm{\infty }}\frac{\left(j-\delta \right)}{1-\delta }{a}_{j}|z{|}^{t}<1.$

We use the fact that $f\in \mathcal{RI}\left(\gamma ,\lambda ,l,\alpha ,\beta \right)$ if and only if

$\sum _{j=t+1}^{\mathrm{\infty }}\frac{\left[1+\mu \left(j-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}}{2\beta \nu \left(1-\alpha \right)}{a}_{j}<1.$

Equation (4.3) holds true if

$\frac{j-\delta }{1-\delta }|z{|}^{t}<\frac{\left[1+\mu \left(j-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}}{2\beta \nu \left(1-\alpha \right)}.$

Or, equivalently,

$|z{|}^{t}<\frac{\left(1-\delta \right)\left[1+\mu \left(j-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}}{2\beta \nu \left(1-\alpha \right)\left(j-\delta \right)},$

which yields the starlikeness of the family.

1. 2.

Using the fact that f is convex if and only $z{f}^{\prime }$ is starlike, we can prove (2) with a similar way of the proof of (1). The function f is convex if and only if

$|z{f}^{″}\left(z\right)|<1-\delta .$
(4.4)

We have

$\begin{array}{c}|z{f}^{″}\left(z\right)|\le |\sum _{j=t+1}^{\mathrm{\infty }}j\left(j-1\right){a}_{j}|z{|}^{t-1}|<1-\delta ,\hfill \\ \sum _{j=t+1}^{\mathrm{\infty }}\frac{j\left(j-1\right)}{1-\delta }{a}_{j}|z{|}^{t-1}<1.\hfill \end{array}$

We use the fact that $f\in \mathcal{RI}\left(\gamma ,\lambda ,l,\alpha ,\beta \right)$ if and only if

$\sum _{j=t+1}^{\mathrm{\infty }}\frac{\left[1+\mu \left(j-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}}{2\beta \nu \left(1-\alpha \right)}{a}_{j}<1.$

Equation (4.4) holds true if

$\frac{j\left(j-1\right)}{1-\delta }|z{|}^{t-1}<\frac{\left[1+\mu \left(j-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}}{2\beta \nu \left(1-\alpha \right)},$

or, equivalently,

$|z{|}^{t-1}<\frac{\left(1-\delta \right)\left[1+\mu \left(j-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}}{2\beta \nu j\left(j-1\right)\left(1-\alpha \right)},$

which yields the convexity of the family. □

5 Neighborhood property

In this section we study neighborhood property for functions in the class $\mathcal{RI}\left(\gamma ,\lambda ,l,\alpha ,\beta \right)$.

Definition 5.1 For functions f belong to of the form and $\epsilon \ge 0$, we define $\left(\eta -\epsilon \right)$-neighborhood of f by

${N}_{\epsilon }^{\eta }\left(f\right)=\left\{g\left(z\right)\in \mathcal{A}:g\left(z\right)=z+\sum _{j=2}^{\mathrm{\infty }}{b}_{j}{z}^{j},\sum _{j=2}^{\mathrm{\infty }}{j}^{\eta +1}|{a}_{j}-{b}_{j}|\le \epsilon \right\},$

where η is a fixed positive integer.

By using the following lemmas we will investigate the $\left(\eta -\epsilon \right)$-neighborhood of function in $\mathcal{RI}\left(\gamma ,\lambda ,l,\alpha ,\beta \right)$.

Lemma 5.1 Let $-1\le \beta <1$, if $g\left(z\right)=z+{\sum }_{j=2}^{\mathrm{\infty }}{b}_{j}{z}^{j}$ satisfies

$\sum _{j=2}^{\mathrm{\infty }}{j}^{\rho +1}|{b}_{j}|\le \frac{2\beta \nu \left(1-\alpha \right)}{1+\beta \left(2\nu -1\right)}$

then $g\left(z\right)\in \mathcal{RI}\left(\gamma ,\lambda ,l,\alpha ,\beta \right)$.

Proof By using of Theorem 2.1, it is sufficient to show that

$\frac{\left[1+\mu \left(j-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{\rho }+\left(1-\gamma \right)\frac{\left(\rho +j-1\right)!}{\rho !\left(j-1\right)!}\right\}}{2\beta \nu \left(1-\alpha \right)}=\frac{{j}^{\rho +1}}{2\beta \nu \left(1-\alpha \right)}\left[1+\beta \left(2\nu -1\right)\right].$

But

$\frac{\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{\rho }+\left(1-\gamma \right)\frac{\left(\rho +j-1\right)!}{\rho !\left(j-1\right)!}\right\}}{2\beta \nu \left(1-\alpha \right)}\le \frac{{j}^{\rho +1}}{2\beta \nu \left(1-\alpha \right)}\left[1+\beta \left(2\nu -1\right)\right].$

Therefore it is enough to prove that

$Q\left(j,\rho \right)=\frac{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{\rho }+\left(1-\gamma \right)\frac{\left(\rho +j-1\right)!}{\rho !\left(j-1\right)!}}{{j}^{\rho +1}}\le 1,$

the result follows because the last inequality holds for all $j\ge t+1$. □

Lemma 5.2 Let $f\left(z\right)=z-{\sum }_{k=2}^{\mathrm{\infty }}{a}_{k}{z}^{k}\in \mathcal{T}$, $\gamma ,\lambda ,l\ge 0$, $0\le \alpha <1$, $0<\beta \le 1$ and $\epsilon \ge 0$. If $\frac{f\left(z\right)+ϵz}{1+ϵ}\in \mathcal{RI}\left(\gamma ,\lambda ,l,\alpha ,\beta \right)$, then

$\sum _{j=t+1}^{\mathrm{\infty }}{j}^{\rho +1}{a}_{j}\le \frac{2\beta \nu \left(1-\alpha \right)\left(1+ϵ\right){\left(t+1\right)}^{\rho +1}}{\left[1+\mu \left(j-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda t+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+t\right)!}{n!t!}\right\}},$

where either $\rho =0$ or $\rho =1$. The result is sharp with the extremal function

$f\left(z\right)=z-\frac{2\beta \nu \left(1-\alpha \right)\left(1+ϵ\right)}{\left[1+\mu \left(j-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda t+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+t\right)!}{n!t!}\right\}}{z}^{t+1},\phantom{\rule{1em}{0ex}}z\in U.$

Proof Letting $g\left(z\right)=\frac{f\left(z\right)+ϵz}{1+ϵ}$ we have

$g\left(z\right)=z-\sum _{j=t+1}^{\mathrm{\infty }}\frac{{a}_{j}}{1+ϵ}{z}^{j},\phantom{\rule{1em}{0ex}}z\in U.$

In view of Theorem 2.3, $g\left(z\right)={\sum }_{j=1}^{\mathrm{\infty }}{\eta }_{j}{g}_{j}\left(z\right)$ where ${\eta }_{j}\ge 0$, ${\sum }_{j=1}^{\mathrm{\infty }}{\eta }_{j}=1$,

${g}_{1}\left(z\right)=z$

and

${g}_{j}\left(z\right)=z-\frac{2\beta \nu \left(1-\alpha \right)\left(1+ϵ\right)}{\left[1+\mu \left(j-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}}{z}^{j},\phantom{\rule{1em}{0ex}}j\ge t+1.$

So we obtain

$\begin{array}{rcl}g\left(z\right)& =& {\eta }_{1}z+\sum _{j=t+1}^{\mathrm{\infty }}{\eta }_{j}\left[z-\frac{2\beta \nu \left(1-\alpha \right)\left(1+ϵ\right)}{\left[1+\mu \left(j-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}}{z}^{j}\right]\\ =& z-\sum _{j=t+1}^{\mathrm{\infty }}{\eta }_{k}\frac{2\beta \nu \left(1-\alpha \right)\left(1+ϵ\right)}{\left[1+\mu \left(j-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}}{z}^{j}.\end{array}$

Since ${\eta }_{j}\ge 0$ and ${\sum }_{j=2}^{\mathrm{\infty }}{\eta }_{j}\le 1$, it follows that

$\sum _{j=t+1}^{\mathrm{\infty }}{a}_{k}\le \underset{j\ge t+1}{sup}{j}^{\rho +1}\frac{2\beta \nu \left(1-\alpha \right)\left(1+ϵ\right)}{\left[1+\mu \left(j-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}}.$

Since whenever $\rho =0$ or $\rho =1$ we conclude that

$W\left(j,\rho ,\gamma ,\alpha ,\beta ,ϵ\right)={j}^{\rho +1}\frac{2\beta \nu \left(1-\alpha \right)\left(1+ϵ\right)}{\left[1+\mu \left(j-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda \left(j-1\right)+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}\right\}}$

is a decreasing function of j, the result will follow. The proof is complete. □

Theorem 5.1 Let $\rho =0$ or $\rho =1$ and suppose $0\le \beta <1$ and

$\begin{array}{rcl}-1& \le & \theta \\ <& \frac{\left[1+\mu \left(t-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda t+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+t\right)!}{n!t!}\right\}-2\beta \nu \left(1-\alpha \right)\left(1+ϵ\right){\left(t+1\right)}^{\eta +1}}{\left[1+\mu \left(t-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda t+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+t\right)!}{n!t!}\right\}},\end{array}$

$f\left(z\right)\in \mathcal{T}$ and $\frac{f\left(z\right)+ϵz}{1+ϵ}\in \mathcal{RI}\left(\gamma ,\lambda ,l,\alpha ,\beta \right)$, then the $\left(\eta -\epsilon \right)$-neighborhood of f is the subset of $\mathcal{RI}\left(\lambda ,\lambda ,l,\alpha ,\beta \right)$, where

$\begin{array}{rcl}\epsilon & \le & 2\left(1-\alpha \right)\left\{\theta \gamma \left[1+\mu \left(t-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda t+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+t\right)!}{n!t!}\right\}\\ -\beta \gamma \left[1+\theta \left(2\nu -1\right)\right]\left(1+ϵ\right){\left(t+1\right)}^{\eta +1}\right\}/\left(\left[1+\theta \left(2\nu -1\right)\right]\left[1+\mu \left(t-1\right)\right]\\ ×\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda t+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+t\right)!}{n!t!}\right\}\right).\end{array}$

The result is sharp.

Proof For $f\left(z\right)=z-{\sum }_{j=2}^{\mathrm{\infty }}|{a}_{j}|{z}^{j}$, let $g\left(z\right)=z+{\sum }_{j=2}^{\mathrm{\infty }}{b}_{j}{z}^{j}$ be in ${N}_{\epsilon }^{\eta }\left(f\right)$. So by Lemma 5.2, we have

$\begin{array}{rcl}\sum _{j=2}^{\mathrm{\infty }}{j}^{\eta +1}|{b}_{j}|& =& \sum _{j=2}^{\mathrm{\infty }}{j}^{\eta +1}|{a}_{j}-{b}_{j}-{a}_{j}|\\ \le & \epsilon +\frac{2\beta \nu \left(1-\alpha \right)\left(1+ϵ\right){\left(t+1\right)}^{\eta +1}}{\left[1+\mu \left(t-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda t+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+t\right)!}{n!t!}\right\}}.\end{array}$

By using Lemma 5.1, $g\left(z\right)\in \mathcal{L}\left(\gamma ,\alpha ,\beta \right)$ if

$\epsilon +\frac{2\beta \nu \left(1-\alpha \right)\left(1+ϵ\right){\left(t+1\right)}^{\eta +1}}{\left[1+\mu \left(t-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda t+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+t\right)!}{n!t!}\right\}}\le \frac{2\theta \nu \left(1-\alpha \right)}{1+\theta \left(2\nu -1\right)},$

that is, $\epsilon \le \frac{2\left(1-\alpha \right)\left\{\theta \gamma \left[1+\mu \left(t-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda t+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+t\right)!}{n!t!}\right\}-\beta \gamma \left[1+\theta \left(2\nu -1\right)\right]\left(1+ϵ\right){\left(t+1\right)}^{\eta +1}\right\}}{\left[1+\theta \left(2\nu -1\right)\right]\left[1+\mu \left(t-1\right)\right]\left[1+\beta \left(2\nu -1\right)\right]\left\{\gamma {\left(\frac{1+\lambda t+l}{l+1}\right)}^{n}+\left(1-\gamma \right)\frac{\left(n+t\right)!}{n!t!}\right\}}$, and the proof is complete. □

Author’s contributions

The author drafted the manuscript, read and approved the final manuscript.

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Alb Lupaş, A. Aspects of univalent holomorphic functions involving multiplier transformation and Ruscheweyh derivative. Adv Differ Equ 2014, 117 (2014). https://doi.org/10.1186/1687-1847-2014-117

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Keywords

• analytic functions
• univalent functions
• radii of starlikeness and convexity
• neighborhood property
• Salagean operator
• Ruscheweyh operator