New subclass of analytic functions defined by $ q $-analogue of $ p $-valent Noor integral operator

Ibtisam Aldawish; Mohamed Aouf; Basem Frasin; Tariq Al-Hawary; Ibtisam Aldawish; Mohamed Aouf; Basem Frasin; Tariq Al-Hawary

doi:10.3934/math.2021607

AIMS Mathematics

2021, Volume 6, Issue 10: 10466-10484. doi: 10.3934/math.2021607

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Research article

New subclass of analytic functions defined by $q$ -analogue of $p$ -valent Noor integral operator

1.
Mathematics and Statistics Department, College of Science, IMSIU (Imam Mohammad Ibn Saud Islamic University), Riyadh 11623, Saudi Arabia
2.
Department of Mathematics, Faculty of Science, Mansoura University, 35516, Egypt
3.
Faculty of Science, Department of Mathematics, AL AL-Bayt University Mafraq, Jordon
4.
Department of Applied Science, Ajloun College, Al-Balqa Applied University, Ajloun 26816, Jordan

Received: 29 March 2021 Accepted: 07 July 2021 Published: 19 July 2021
MSC : 30C45

In this paper, we introduce a certain subclass of analytic functions associated with $q$ -analogue of $p$ -valent Noor integral operator in the open unit disc. A variety of useful properties for this subclass are investigated including coefficient estimates and the familiar Fekete-Szeg ö type inequalities. Several known sequences of the main results are also highlighted.

Keywords:

Citation: Ibtisam Aldawish, Mohamed Aouf, Basem Frasin, Tariq Al-Hawary. New subclass of analytic functions defined by $q$ -analogue of $p$ -valent Noor integral operator[J]. AIMS Mathematics, 2021, 6(10): 10466-10484. doi: 10.3934/math.2021607

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Abstract

1. Introduction and definitions

Let $\mathcal{A(}p\mathcal{)}$ denote the class of functions of the form:

$\begin{equation} f(z) = z^{p}+\sum \limits_{n = p+1}^{\infty }a_{n}z^{n}{\rm{ \ \ }}(p\in \mathbb{N} = \left \{ 1,2,\cdots \right \} ), \end{equation}$

(1.1)

which are $p$ -valent and analytic in the open unit disc $\mathcal{U} = \left \{ z:\left \vert z\right \vert < 1\right \}.$ We note that $\mathcal{ A(}1\mathcal{) = A}.$ For functions $f(z)$ given by (1.1) and $g(z)$ defined by

$\begin{equation} g(z) = z^{p}+\sum \limits_{n = p+1}^{\infty }b_{n}z^{n}{\rm{ \ \ }}(p\in \mathbb{N} ), \end{equation}$

(1.2)

the convolution of $f(z)$ and $g(z)$ is defined by

$\begin{equation} (f\ast g)(z) = z^{p}+\sum \limits_{n = p+1}^{\infty }a_{n}b_{n}z^{n} = (g\ast f)(z) {\rm{.}} \end{equation}$

(1.3)

For $f\in \mathcal{A(}p\mathcal{)}$ given by (1.1) and $0 < q < 1,$ the $q$ -derivative of a function $f(z)$ is given by (see ^[1,6,7])

$\begin{equation} \mathcal{D}_{q,p}f(z) = \left \{ \begin{array}{lcl} \dfrac{f(z)-f(qz)}{(1-q)z} & for & z\neq 0, \\ {\rm{ \ \ \ \ \ }}f^{\prime }(0) & for & z = 0, \end{array} \right. \end{equation}$

(1.4)

provided that $f^{\prime }(z)$ exists. From (1.1) and (1.4), we deduce that

$\begin{equation} \mathcal{D}_{q,p}f(z) = [p]_{q}z^{p-1}+\sum \limits_{n = p+1}^{\infty }[n]_{q}a_{n}z^{n-1}, \end{equation}$

(1.5)

where

$\begin{equation} \lbrack n]_{q} = \frac{1-q^{n}}{1-q} = 1+q+\cdots +q^{n-1},{\rm{ \ }}[0]_{q} = 0, {\rm{ \ }}0 < q < 1. \end{equation}$

(1.6)

We note that

$\begin{equation*} \underset{q\longrightarrow 1^{-}}{\lim }\mathcal{D}_{q,p}f(z) = \underset{ q\longrightarrow 1^{-}}{\lim }\dfrac{f(z)-f(qz)}{(1-q)z} = f^{\prime }(z) \end{equation*}$

for a function $f$ which is differentiable in a given subset of $\mathbb{C}.$ Further, for $p = 1$ , we have $\mathcal{D}_{q, 1}f(z) = \mathcal{D}_{q}f(z)$ (see ^[20]).

The $q$ -number shift factorial for any non-negative integer $n$ is defined by

$\begin{equation*} \lbrack n]_{q}! = \left \{ \begin{array}{lcl} 1 & for & n = 0 \\ \lbrack 1]_{q}[2]_{q}\cdots \lbrack n]_{q} & for & n\in \mathbb{N} . \end{array} \right. \end{equation*}$

The Pochhammer $q$ -generalized symbol for $x > 0$ and $n\in \mathbb{N}$ is also

$\begin{equation*} \lbrack x,q]_{n} = \left \{ \begin{array}{lcl} 1 & for & n = 0 \\ \lbrack x]_{q}[x+1]_{q}\cdots \lbrack x+n-1]_{q} & for & n\in \mathbb{N} , \end{array} \right. \end{equation*}$

and for $x > 0$ , the $q$ -gamma function is defined by

$\begin{equation*} \Gamma _{q}(x+1) = [x]_{q}\Gamma _{q}(x){\rm{ and }}\Gamma _{q}(1) = 1. \end{equation*}$

For $\lambda > -p \; (p\in \mathbb{N}),$ we define the function $f_{\lambda +p-1, q}^{-1}(z)$ by

$\begin{equation} f_{\lambda +p-1,q}(z)\ast f_{\lambda +p-1,q}^{-1}(z) = z^{p}+\sum \limits_{n = p+1}^{\infty }\frac{[p+1,q]_{n-p}}{[1,q]_{n-p}}z^{n}, \end{equation}$

(1.7)

where the function $f_{\lambda +p-1, q}(z)$ is given by

$\begin{equation} f_{\lambda +p-1,q}(z) = z^{p}+\sum \limits_{n = p+1}^{\infty }\frac{[\lambda +p,q]_{n-p}}{[1,q]_{n-p}}z^{n}. \end{equation}$

(1.8)

It is clear that the function defined in (1.8) converges absolutely in $\mathcal{U}$ . Using the idea of convolution we define the $q$ - $p$ -valent Noor integral operator $\mathcal{I}_{q}^{\lambda +p-1}:\mathcal{A(}p\mathcal{)\longrightarrow A(}p\mathcal{)}$ as follows:

$\begin{equation} \mathcal{I}_{q}^{\lambda +p-1}f(z) = f_{\lambda +p-1,q}^{-1}(z)\ast f(z) = z^{p}+\sum \limits_{n = p+1}^{\infty }\Phi _{q}(\lambda ,p,n)a_{n}z^{n}, \end{equation}$

(1.9)

where

$\begin{equation} \Phi _{q}(\lambda ,p,n) = \frac{[p+1,q]_{n-p}}{[\lambda +p,q]_{n-p}}{\rm{ \ }} (\lambda > -p,{\rm{ }}p\in \mathbb{N} ). \end{equation}$

(1.10)

From (1.9), we can easily get the identity

$\begin{equation} q^{\lambda }z\mathcal{D}_{q,p}(\mathcal{I}_{q}^{\lambda +p}f(z)) = [\lambda +p]_{q}\mathcal{I}_{q}^{\lambda +p-1}f(z)-[\lambda ]_{q}\mathcal{I} _{q}^{\lambda +p}f(z). \end{equation}$

(1.11)

We note that:

(i) For $p = 1$ , we have the $q$ -Noor integral operator $\mathcal{I} _{q}^{\lambda }f(z)$ ( $f\in \mathcal{A}$ ) which was introduced and studied by Arif et al. ^[4];

(ii) $\underset{q\longrightarrow 1^{-}}{\lim }\mathcal{I}_{q}^{\lambda +p-1}f(z) = \mathcal{I}^{\lambda +p-1}f(z)$ which is the $p$ -valent Noor integral operator (see ^[11]);

(iii) Taking $p = 1$ and letting $q\longrightarrow 1^{-}$ in (1.9), we obtain Noor integral operator for univalent functions (see ^[13,14]);

(iv) For $\lambda = 1,$ we have $\mathcal{I}_{q}^{p}f(z) = f(z)$ and for $\lambda = 0,$ we have

$\begin{equation*} \mathcal{I}_{q}^{p-1}f(z) = z^{p}+\sum \limits_{n = p+1}^{\infty }\frac{ [p+1,q]_{n-p}}{[1,q]_{n-p}}a_{n}z^{n} = z^{p}+\sum \limits_{n = p+1}^{\infty } \frac{[n]_{q}}{[p]_{q}}a_{n}z^{n} = \frac{z\mathcal{D}_{q,p}f(z)}{[p]_{q}}, \end{equation*}$

$\begin{equation*} \underset{q\longrightarrow 1^{-}}{\lim }\mathcal{I}_{q}^{p-1}f(z) = \mathcal{I} ^{p-1}f(z) = z+\sum \limits_{n = p+1}^{\infty }\left( \frac{n}{p}\right) a_{n}z^{n} = \frac{zf^{\prime }(z)}{p}. \end{equation*}$

By using the operator $\mathcal{I}_{q}^{\lambda +p-1}f(z)$ we define the subclass $\mathcal{ST}_{q}(\lambda, p, k, b)$ of $\mathcal{A(}p\mathcal{)}$ as follows:

Definition 1.1. Let $k\geq 0, \; \lambda > -p, \; p\in \mathbb{N}, \; b\in \mathbb{C} ^{\ast } = \mathbb{C} \backslash \{0\}$ and $0 < q < 1.$ A function $f\in \mathcal{A(}p\mathcal{)}$ is said to be in the class $\mathcal{ST}_{q}(\lambda, p, k, b)$ if it satisfies

$\begin{eqnarray} &&{{\rm{Re}}}\left \{ 1+\frac{1}{b}\left( \frac{1}{[p]_{q}}\frac{z\mathcal{D} _{q,p}\left( \mathcal{I}_{q}^{\lambda +p-1}f(z)\right) }{\mathcal{I} _{q}^{\lambda +p-1}f(z)}-1\right) \right \} \\ & > &k\left \vert \frac{1}{b}\left( \frac{1}{[p]_{q}}\frac{z\mathcal{D} _{q,p}\left( \mathcal{I}_{q}^{\lambda +p-1}f(z)\right) }{\mathcal{I} _{q}^{\lambda +p-1}f(z)}-1\right) \right \vert ,\;(z\in \mathcal{U}). \end{eqnarray}$

(1.12)

We note that: (1) $\underset{q\longrightarrow 1^{-}}{\lim }\mathcal{ST}_{q}(1, p, k, 1-\frac{ \alpha }{p}) = \mathcal{ST}(p, k, \alpha) =$

$\left \{ f\in \mathcal{A(}p\mathcal{)}:{{\rm{Re}}}\left(\frac{zf^{\prime }(z)}{f(z)}-\alpha \right) > k\left \vert \frac{zf^{\prime }(z)}{f(z)} -p\right \vert, {\rm{ }}0\leq \alpha < p, z\in \mathcal{U}\right \}$ (see ^[19]) $;$

(2) $\underset{q\longrightarrow 1^{-}}{\lim }\mathcal{ST}_{q}(0, p, k, 1-\frac{ \alpha }{p}) = \mathcal{UST}(p, k, \alpha) =$

$\left \{ f\in \mathcal{A(}p\mathcal{)}:{{\rm{Re}}}\left(1+\frac{zf^{\prime \prime }(z)}{f^{\prime }(z)}-\alpha \right) > k\left \vert 1+\frac{zf^{\prime \prime }(z)}{f^{\prime }(z)}-p\right \vert, {\rm{ }}0\leq \alpha < p, z\in \mathcal{U}\right \}$ (see ^[19]).

2. Geometric interpretation

A functions $f\in \mathcal{A(}p\mathcal{)}$ is in the class $\mathcal{ST} _{q}(\lambda, p, k, b)$ if

$\begin{equation*} 1+\frac{1}{b}\left( \frac{1}{[p]_{q}}\frac{z\mathcal{D}_{q,p}\left( \mathcal{ I}_{q}^{\lambda +p-1}f(z)\right) }{\mathcal{I}_{q}^{\lambda +p-1}f(z)} -1\right) \end{equation*}$

takes all the values in the conic domain $\Omega _{k} = p_{k}(\mathcal{U}),$ where

$\begin{equation*} \Omega _{k} = \left \{ u+iv:u > k\sqrt{(u-1)^{2}+v^{2}}\right \} , \end{equation*}$

or, equivalently,

$\begin{equation} 1+\frac{1}{b}\left( \frac{1}{[p]_{q}}\frac{z\mathcal{D}_{q,p}\left( \mathcal{ I}_{q}^{\lambda +p-1}f(z)\right) }{\mathcal{I}_{q}^{\lambda +p-1}f(z)} -1\right) \prec p_{k}(z),{\rm{ }}\Omega _{k} = p_{k}(\mathcal{U}). \end{equation}$

(2.1)

The boundary $\partial \Omega _{k}$ of the above set when $k = 0$ becomes the imaginary axis, when $0 < k < 1$ a hyperbola $,$ when $k = 1$ a parabola and an ellipse when $1 < k < \infty.$ The functions $p_{k}(z)$ are defined by

$\begin{equation} p_{k}(z) = \left \{ \begin{array}{c} {\rm{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }}\frac{1+z}{1-z},{\rm{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }} k = 0, \\ {\rm{ \ \ \ \ \ \ \ \ \ \ \ }}1+\frac{2}{\pi ^{2}}\log \left( \frac{1+\sqrt{z }}{1-\sqrt{z}}\right) ^{2},{\rm{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }}k = 1, \\ {\rm{ \ }}1+\frac{1}{1-k^{2}}\cos \left( \frac{2}{\pi }(\cos ^{-1}k)i\log \frac{1+\sqrt{z}}{1-\sqrt{z}}\right) -\frac{k^{2}}{1-k^{2}},{\rm{ \ \ \ \ }}0 < k < 1, \\ 1+\frac{1}{k^{2}-1}\sin \left( \frac{\pi }{2R(t)}\underset{}{\underset{0}{ \overset{u(z)/\sqrt{t}}{\int }}\frac{dx}{\sqrt{1-x^{2}}\sqrt{1-t^{2}x^{2}}}} \right) +\frac{k^{2}}{k^{2}-1},{\rm{ \ \ \ \ }}1 < k < \infty , \end{array} \right. \end{equation}$

(2.2)

where $u(z) = \frac{z-\sqrt{t}}{1-\sqrt{t}z} \; (0 < t < 1, \; z\in \mathcal{U}), \; t$ is chosen such that $k = \cosh \left(\frac{\pi R^{\prime }(t)}{4R(t)} \right), \; R(t)$ is the Legendre's complete elliptic integral of the first kind, and $R^{\prime }(t)$ is complementary integral of $R(t)$ (see ^[9,10,18]).

By giving a specific value to the parameters $q, \lambda, p, k, \; \$ and $b$ in the class $\mathcal{ST}_{q}(\lambda, p, k, b),$ we get a lot of new and known subclasses studied by various others, for example,

(1) $\mathcal{ST}_{q}(\lambda, 1, k, b) = \mathcal{ST}_{q}(\lambda, k, b) = \; \left \{ f\in \mathcal{A}:1+\frac{1}{b}\left(\frac{z\mathcal{D}_{q}\left(\mathcal{I}_{q}^{\lambda }f(z)\right) }{\mathcal{I}_{q}^{\lambda }f(z)} -1\right) \prec p_{k}(z), {\rm{ }}z\in \mathcal{U}\right \};$

(2) $\mathcal{ST}_{q}(\lambda, 1, k, 1) = \mathcal{ST}_{q}(\lambda, k) = \; \left \{ f\in \mathcal{A}:\frac{z\mathcal{D}_{q}\left(\mathcal{I} _{q}^{\lambda }f(z)\right) }{\mathcal{I}_{q}^{\lambda }f(z)}\prec p_{k}(z), {\rm{ }}z\in \mathcal{U}\right \};$

(3) $\mathcal{ST}_{q}(\lambda, p, k, 1-\frac{\alpha }{[p]_{q}}) = \mathcal{ST} _{q}(\lambda, p, k, \alpha) =$

$\left \{ f\in \mathcal{A(}p\mathcal{)}:\frac{1}{([p]_{q}-\alpha)}\left(\frac{z\mathcal{D}_{q, p}\left(\mathcal{I}_{q}^{\lambda +p-1}f(z)\right) }{ \mathcal{I}_{q}^{\lambda +p-1}f(z)}-\alpha \right) \prec p_{k}(z), {\rm{ }} 0\leq \alpha < [p]_{q}, {\rm{ }}z\in \mathcal{U}\right \};$

(4) $\mathcal{ST}_{q}\left(\lambda, p, k, \left(1-\frac{\alpha }{[p]_{q}} \right) \cos \gamma e^{-i\gamma }\right) = \; \mathcal{ST}_{q}^{\gamma }(\lambda, p, k, \alpha) =$

$\left \{ f\in \mathcal{A(}p\mathcal{)}:e^{i\gamma }\frac{z\mathcal{D} _{q, p}\left(\mathcal{I}_{q}^{\lambda +p-1}f(z)\right) }{\mathcal{I} _{q}^{\lambda +p-1}f(z)}\prec ([p]_{q}-\alpha)\cos \gamma p_{k}(z)+\alpha \cos \gamma +i[p]_{q}\sin \gamma, {\rm{ }}\right.$

$\left. 0\leq \alpha < [p]_{q}, \left \vert \gamma \right \vert < \frac{\pi }{2} {\rm{, }}z\in \mathcal{U}\right \};$

(5) $\mathcal{ST}_{q}(1, p, k, b) = \mathcal{ST}_{q}(p, k, b) = \; \left \{ f\in \mathcal{A(}p\mathcal{)}:1+\frac{1}{b}\left(\frac{1}{[p]_{q}}\frac{z \mathcal{D}_{q, p}f(z)}{f(z)}-1\right) \prec p_{k}(z), {\rm{ }}z\in \mathcal{U} \right \};$

(6) $\mathcal{ST}_{q}(1, p, k, 1-\frac{\alpha }{[p]_{q}}) = \mathcal{ST} _{q}(p, k, \alpha) =$

$\left \{ f\in \mathcal{A(}p\mathcal{)}:\frac{1}{([p]_{q}-\alpha)}\left(\frac{z\mathcal{D}_{q, p}f(z)}{f(z)}-\alpha \right) \prec p_{k}(z), {\rm{ }} 0\leq \alpha < [p]_{q}, {\rm{ }}z\in \mathcal{U}\right \};$

(7) $\mathcal{ST}_{q}\left(1, p, k, \left(1-\frac{\alpha }{[p]_{q}} \right) \cos \gamma e^{-i\gamma }\right) = \mathcal{ST}_{q}^{\gamma }(p, k, \alpha) =$

$\left \{ f\in \mathcal{A(}p\mathcal{)}:e^{i\gamma }\frac{z\mathcal{D} _{q, p}f(z)}{f(z)}\prec ([p]_{q}-\alpha)\cos \gamma p_{k}(z)+\alpha \cos \gamma +i[p]_{q}\sin \gamma, {\rm{ }}\right.$

$\left. 0\leq \alpha < [p]_{q}, \left \vert \gamma \right \vert < \frac{\pi }{2} {\rm{, }}z\in \mathcal{U}\right \}.$

Also we note that:

(8) $\mathcal{ST}_{q}(\lambda, p, 0, b) = \mathcal{S}_{q}(\lambda, p, b) =$

$\left \{ f\in \mathcal{A(}p\mathcal{)}:{{\rm{Re}}}\left \{ [p]_{q}+\frac{1}{ b}\left(\frac{z\mathcal{D}_{q, p}\left(\mathcal{I}_{q}^{\lambda +p-1}f(z)\right) }{\mathcal{I}_{q}^{\lambda +p-1}f(z)}-[p]_{q}\right) \right \} > 0, {\rm{ }}z\in \mathcal{U}\right \},$

$\mathcal{S}_{q}(\lambda, p, 1-\frac{\alpha }{[p]_{q}}) = \mathcal{S} _{q}(\lambda, p, \alpha) =$

$\left \{ f\in \mathcal{A(}p\mathcal{)}:{{\rm{Re}}}\left \{ \frac{z\mathcal{D }_{q, p}\left(\mathcal{I}_{q}^{\lambda +p-1}f(z)\right) }{\mathcal{I} _{q}^{\lambda +p-1}f(z)}\right \} > \alpha, 0\leq \alpha < [p]_{q}{\rm{, }} z\in \mathcal{U}\right \},$

$\mathcal{S}_{q}(1, p, \alpha) = \mathcal{S}_{q}(p, \alpha) =$

$\left \{ f\in \mathcal{A(}p\mathcal{)}:{{\rm{Re}}}\left \{ \frac{z\mathcal{D }_{q, p}f(z)}{f(z)}\right \} > \alpha, 0\leq \alpha < [p]_{q}{\rm{, }}z\in \mathcal{U}\right \}$ , $\mathcal{S}_{q}(1, \alpha) = \mathcal{S}_{q}(\alpha)$ (see ^[20]);

(9) $\mathcal{ST}_{q}\left(\lambda, p, 0, \left(1-\frac{\alpha }{[p]_{q}} \right) \cos \gamma e^{-i\gamma }\right) = \mathcal{S}_{q}^{\gamma }(\lambda, p, \alpha) =$

$\left \{ f\in \mathcal{A(}p\mathcal{)}:{{\rm{Re}}}\left \{ e^{i\gamma } \frac{z\mathcal{D}_{q, p}\left(\mathcal{I}_{q}^{\lambda +p-1}f(z)\right) }{ \mathcal{I}_{q}^{\lambda +p-1}f(z)}\right \} > \alpha \cos \gamma, 0\leq \alpha < [p]_{q}{\rm{, }}\left \vert \gamma \right \vert < \frac{\pi }{2}, {\rm{ }}z\in \mathcal{U}\right \},$

$\mathcal{S}_{q}^{\gamma }(1, p, \alpha) = \mathcal{S}_{q}^{\gamma }(p, \alpha) =$

$\left \{ f\in \mathcal{A(}p\mathcal{)}:{{\rm{Re}}}\left \{ e^{i\gamma } \frac{z\mathcal{D}_{q, p}f(z)}{f(z)}\right \} > \alpha \cos \gamma, 0\leq \alpha < [p]_{q}{\rm{, }}\left \vert \gamma \right \vert < \frac{\pi }{2}, {\rm{ }}z\in \mathcal{U}\right \};$

(10) $\underset{q\longrightarrow 1^{-}}{\lim }\mathcal{ST}_{q}(\lambda, p, 0, b) = \mathcal{S}(\lambda, p, b) =$

$\left \{ f\in \mathcal{A(}p\mathcal{)}:{{\rm{Re}}}\left \{ p+\frac{1}{b} \left(\frac{z\left(\mathcal{I}_{q}^{\lambda +p-1}f(z)\right) ^{\prime }}{ \mathcal{I}_{q}^{\lambda +p-1}f(z)}-p\right) \right \} > 0, z\in \mathcal{U} \right \},$

$\mathcal{S}\left(\lambda, p, \left(1-\frac{\alpha }{p}\right) \cos \gamma e^{-i\gamma }\right) = \mathcal{S}^{\gamma }(\lambda, p, \alpha) =$

$\left \{ f\in \mathcal{A(}p\mathcal{)}:{{\rm{Re}}}\left \{ e^{i\gamma } \frac{z\left(\mathcal{I}_{q}^{\lambda +p-1}f(z)\right) ^{\prime }}{\mathcal{ I}_{q}^{\lambda +p-1}f(z)}\right \} > \alpha \cos \gamma, {\rm{ }}0\leq \alpha < p, {\rm{ }}\left \vert \gamma \right \vert < \frac{\pi }{2}, {\rm{ }} z\in \mathcal{U}\right \},$

$\mathcal{S}^{\gamma }\left(1, p, \left(1-\frac{\alpha }{p}\right) \cos \gamma e^{-i\gamma }\right) = \mathcal{S}^{\gamma }(p, \alpha) =$

$\left \{ f\in \mathcal{A(}p\mathcal{)}:{{\rm{Re}}}\left \{ e^{i\gamma } \frac{zf^{\prime }(z)}{f(z)}\right \} > \alpha \cos \gamma, 0\leq \alpha < p {\rm{, }}\left \vert \gamma \right \vert < \frac{\pi }{2}, {\rm{ }}z\in \mathcal{U}\right \}$ (see ^[22]),

$\mathcal{S}(1, p, b) = \mathcal{S}(p, b) = \left \{ f\in \mathcal{A(}p\mathcal{)}:{ {\rm{Re}}}\left \{ p+\frac{1}{b}\left(\frac{zf^{\prime }(z)}{f(z)}-p\right) \right \} > 0, {\rm{ }}z\in \mathcal{U}\right \}$ (see ^[23]),

$\mathcal{S}(0, p, b) = \mathcal{C}(p, b) = \left \{ f\in \mathcal{A(}p\mathcal{)}:{ {\rm{Re}}}\left \{ p+\frac{1}{b}\left(1+\frac{zf^{\prime \prime }(z)}{ f^{\prime }(z)}-p\right) \right \} > 0, {\rm{ }}z\in \mathcal{U}\right \}$ (see ^[2,3,21,23]), $\mathcal{S}(1, b) = \mathcal{S}(b)$ and $\mathcal{C}(1, b) = \mathcal{C}(b)$ (see ^[15,16,17]);

(11) $\underset{q\longrightarrow 1^{-}}{\lim }\mathcal{ST} _{q}(1, 1, k, 1-\alpha) = \mathcal{ST}(k, \alpha) =$

$\left \{ f\in \mathcal{A}:{{\rm{Re}}}\left(\frac{zf^{\prime }(z)}{f(z)} -\alpha \right) > k\left \vert \frac{zf^{\prime }(z)}{f(z)}-1\right \vert, {\rm{ }}0\leq \alpha < 1{\rm{, }}z\in \mathcal{U}\right \}$ (see ^[5]) $;$

(12) $\underset{q\longrightarrow 1^{-}}{\lim }\mathcal{ST}_{q}\left(1, p, k, \left(1-\frac{\alpha }{p}\right) \cos \gamma e^{-i\gamma }\right) = \mathcal{ST}^{\gamma }(p, k, \alpha) =$

$\left \{ f\in \mathcal{A(}p\mathcal{)}:{{\rm{Re}}}\left(e^{i\gamma }\frac{ zf^{\prime }(z)}{f(z)}-\alpha \cos \gamma \right) > k\left \vert \frac{ zf^{\prime }(z)}{f(z)}-p\right \vert, {\rm{ }}0\leq \alpha < p{\rm{, }}\left \vert \gamma \right \vert < \frac{\pi }{2}, z\in \mathcal{U}\right \},$

$\underset{q\longrightarrow 1^{-}}{\lim }\mathcal{ST}_{q}\left(0, p, k, \left(1-\frac{\alpha }{p}\right) \cos \gamma e^{-i\gamma }\right) = \mathcal{UST} ^{\gamma }(p, k, \alpha) =$

$\left \{ f\in \mathcal{A(}p\mathcal{)}:{{\rm{Re}}}\left \{ e^{i\gamma }\left(1+\frac{zf^{\prime \prime }(z)}{f^{\prime }(z)}\right) -\alpha \cos \gamma \right \} > k\left \vert 1+\frac{zf^{\prime \prime }(z)}{f^{\prime }(z) }-p\right \vert, \right.$

$\left. 0\leq \alpha < p{\rm{, }}\left \vert \gamma \right \vert < \frac{\pi }{ 2}, z\in \mathcal{U}\right \}.$

We need the following lemmas in order to establish our main results.

Lemma 2.1. ^[8] Let $0\leq k < \infty$ be fixed and let $p_{k}$ be defined by (2.2). If $p_{k}(z) = 1+Q_{1}z+Q_{2}z^{2}+\cdots,$ then

$Q_{1} = \left\{ \begin{array}{l} \frac{2A^{2}}{1-k^{2}},&&0\leq k < 1, \\ \\ \frac{8}{\pi ^{2}},&&k = 1, \\ \\ \frac{\pi ^{2}}{4\sqrt{t}(k^{2}-1)R^{2}(t)(1+t)},&&1 < k < \infty ,\end{array} \right.$

(2.3)

and

$Q_{2} = \left\{ \begin{array}{l} \frac{(A^{2}+2)}{3}Q_{1},&&0\leq k < 1, \\ \\ \frac{2}{3}Q_{1},&&k = 1, \\ \\ \frac{4R^{2}(t)(t^{2}+6t+1)-\pi ^{2}}{24\sqrt{t}R^{2}(t)(1+t)}Q_{1},&&1 < k < \infty , \end{array} \right.$

(2.4)

where $A = \frac{2\cos ^{-1}k}{\pi },$ and $t\in (0, 1)$ is chosen such that $k = \cosh \left(\frac{\pi R^{\prime }(t)}{R(t)}\right),$ where $R(t)$ is the Legendre's complete elliptic integral of the first kind.

Lemma 2.2. ^[12] Let $h(z) = 1+\sum \limits_{n = 1}^{\infty }c_{n}z^{n}\in P,$ i.e., let $h$ be analytic in $\mathcal{U}$ and satisfies ${\mathit{{\rm{Re}}}}\left(h(z)\right) > 0 \; (z\in \mathcal{U)}$ , then

$\begin{equation} \left \vert c_{2}-vc_{1}^{2}\right \vert \leq 2\max \left \{ 1,\left \vert 2v-1\right \vert \right \} \mathit{{\rm{}}}(v\in \mathbb{C} )\mathit{{\rm{.}}} \end{equation}$

(2.5)

The result is sharp for a function given by

$\begin{equation*} g(z) = \frac{1+z^{2}}{1-z^{2}}\mathit{{\rm{or}}}g(z) = \frac{1+z}{1-z}. \end{equation*}$

Lemma 2.3. ^[12] If $h(z) = 1+\sum \limits_{n = 1}^{\infty }c_{n}z^{n}\in P,$ then

$\begin{equation} | c_{2}-vc_{1}^{2}| \leq \left \{ \begin{array}{c} 2-4v&&if&&v\leq 0, \\ 2&&if&&0\leq v\leq 1, \\ 4v-2&&if&&v\geq 1, \end{array} \right. \end{equation}$

(2.6)

where $v < 0$ or $v > 1,$ the equality holds iff $h(z) = \frac{1+z}{1-z}$ or one of its rotations. If $0 < v < 1,$ then he equality holds iff $h(z) = \frac{1+z^{2} }{1-z^{2}}$ or one of its rotations. If $v = 0,$ then he equality holds iff $h(z) = \left(\frac{1+\lambda }{2}\right) \frac{1+z}{1-z}+\left(\frac{ 1-\lambda }{2}\right) \frac{1-z}{1+z}(0\leq \lambda \leq 1)$ or one of its rotations. If $v = 1,$ then he equality holds if and only if $g$ is reciprocal of one of the function such that the equality holds in the case of $v = 0.$

Also the above upper bound is sharp, and it can improved as follows when $0 < v < 1:$

$\begin{equation*} \left \vert c_{2}-vc_{1}^{2}\right \vert +v\left \vert c_{1}\right \vert ^{2}\leq 2{\rm{ \ }}(0\leq v\leq \frac{1}{2}), \end{equation*}$

and

$\begin{equation*} \left \vert c_{2}-vc_{1}^{2}\right \vert +(1-v)\left \vert c_{1}\right \vert ^{2}\leq 2{\rm{ \ }}(\frac{1}{2}\leq v\leq 1). \end{equation*}$

3. Main results

We shall assume throughout this paper, unless otherwise stated, that $0\leq k < \infty, \; p\in \mathbb{N}, \; \lambda > -p, \; b\in \mathbb{C} ^{\ast }, \; 0 < q < 1, \; Q_{1}$ is given by (2.3) and $Q_{2}$ is given by (2.4), $\Phi _{q}(\lambda, p, n)$ is given by (1.10) and $z\in \mathcal{U}$ .

Theorem 3.1. Let $f\in \mathcal{A(}p\mathcal{)}$ be given by (1.1) $.$ If the inequality

$\begin{equation} \sum \limits_{n = p+1}^{\infty }\left \{ (k+1)\left( [n]_{q}-[p]_{q}\right) +[p]_{q}\left \vert b\right \vert \right \} \Phi _{q}(\lambda ,p,n)\left \vert a_{n}\right \vert \leq \lbrack p]_{q}\left \vert b\right \vert , \end{equation}$

(3.1)

holds, then $f\in \mathcal{ST}_{q}(\lambda, p, k, b)$ .

Proof. Assume the inequality (3.1) holds. Let us assume that

$\begin{equation*} {H(z) = }1+\frac{1}{b}\left( \frac{1}{[p]_{q}}\frac{z\mathcal{D}_{q,p}\left( \mathcal{I}_{q}^{\lambda +p-1}f(z)\right) }{\mathcal{I}_{q}^{\lambda +p-1}f(z)}-1\right) . \end{equation*}$

We have

$\begin{eqnarray*} \left \vert {H(z)-1}\right \vert &{ = }&\frac{1}{[p]_{q}\left \vert b\right \vert }\left \vert \frac{\sum \limits_{n = p+1}^{\infty }\left( [n]_{q}-[p]_{q}\right) \Phi _{q}(\lambda ,p,n)a_{n}z^{n-p}}{1+\sum \limits_{n = p+1}^{\infty }\Phi _{q}(\lambda ,p,n)a_{n}z^{n-p}}\right \vert \\ &\leq &\frac{1}{[p]_{q}\left \vert b\right \vert }\frac{\sum \limits_{n = p+1}^{\infty }\left( [n]_{q}-[p]_{q}\right) \Phi _{q}(\lambda ,p,n)\left \vert a_{n}\right \vert }{1-\sum \limits_{n = p+1}^{\infty }\Phi _{q}(\lambda ,p,n)\left \vert a_{n}\right \vert }. \end{eqnarray*}$

Now consider

$\begin{eqnarray*} k\left \vert {H(z)-1}\right \vert -{{\rm{Re}}}\left( {H(z)-1}\right) &\leq &(k+1)\left \vert {H(z)-1}\right \vert \\ & < &\frac{(k+1)\sum \limits_{n = p+1}^{\infty }\left( [n]_{q}-[p]_{q}\right) \Phi _{q}(\lambda ,p,n)\left \vert a_{n}\right \vert }{[p]_{q}\left \vert b\right \vert \left( 1-\sum \limits_{n = p+1}^{\infty }\Phi _{q}(\lambda ,p,n)\left \vert a_{n}\right \vert \right) }. \end{eqnarray*}$

The last inequality is bounded by $1$ if (3.1) holds.

Corollary 3.2. If $f\in \mathcal{ST}_{q}(\lambda, p, k, b),$ then

$\begin{equation} \left \vert a_{n}\right \vert \leq \frac{\lbrack p]_{q}\left \vert b\right \vert }{\left \{ (k+1)\left( [n]_{q}-[p]_{q}\right) +[p]_{q}\left \vert b\right \vert \right \} \Phi _{q}(\lambda ,p,n)}(n\geq p+1). \end{equation}$

(3.2)

The inequality (3.2) is sharp for the function

$\begin{equation} f(z) = z^{p}+\frac{[p]_{q}\left \vert b\right \vert }{\left \{ (k+1)\left( [n]_{q}-[p]_{q}\right) +[p]_{q}\left \vert b\right \vert \right \} \Phi _{q}(\lambda ,p,n)}z^{n}(n\geq p+1). \end{equation}$

(3.3)

Choosing $p = 1$ and $b = 1-\alpha, \; 0\leq \alpha < 1,$ in Theorem 3.1, we obtain the following corollary.

Corollary 3.3. Let $f\in \mathcal{A}$ be given by (1.1) $\$ with $p = 1$ and satisfy

$\begin{equation*} \sum \limits_{n = 2}^{\infty }\left \{ (k+1)\left( [n]_{q}-1\right) +(1-\alpha )\right \} \Phi _{q}(\lambda ,1,n)\left \vert a_{n}\right \vert \leq 1-\alpha . \end{equation*}$

Then $f\in \mathcal{ST}_{q}(\lambda, k, \alpha)$ .

Taking $b = 1-\frac{\alpha }{[p]_{q}} \; (0\leq \alpha < [p]_{q})$ in Theorem 3.1, we obtain the following consequence.

Corollary 3.4. Let $f\in \mathcal{A(}p\mathcal{)}$ be given by (1.1) $\$ and satisfy

$\begin{equation*} \sum \limits_{n = p+1}^{\infty }\left \{ (k+1)\left( [n]_{q}-[p]_{q}\right) +([p]_{q}-\alpha )\right \} \Phi _{q}(\lambda ,p,n)\left \vert a_{n}\right \vert \leq \lbrack p]_{q}-\alpha . \end{equation*}$

Then $f\in \mathcal{ST}_{q}(\lambda, p, k, \alpha)$ .

Letting $q\longrightarrow 1^{-}$ in Theorem 3.1, we obtain the following corollary.

Corollary 3.5. Let $f\in \mathcal{A(}p\mathcal{)}$ be given by (1.1) $\$ and satisfy

$\begin{equation*} \sum \limits_{n = p+1}^{\infty }\left \{ (k+1)\left( n-p\right) +p\left \vert b\right \vert \right \} \Phi _{q}(\lambda ,p,n)\left \vert a_{n}\right \vert \leq p\left \vert b\right \vert . \end{equation*}$

Then $f\in \mathcal{ST}(\lambda, p, k, b)$ .

Putting $b = \left(1-\frac{\alpha }{[p]_{q}}\right) \cos \gamma e^{-i\gamma } \; (0\leq \alpha < [p]_{q}, \; \left \vert \gamma \right \vert < \frac{\pi }{2})$ in Theorem 3.1, we obtain the following consequence.

Corollary 3.6. Let $f\in \mathcal{A(}p\mathcal{)}$ be given by (1.1) $\$ and satisfy

$\begin{equation*} \sum \limits_{n = p+1}^{\infty }\left \{ (k+1)\left( [n]_{q}-[p]_{q}\right) +([p]_{q}-\alpha )\cos \gamma \right \} \Phi _{q}(\lambda ,p,n)\left \vert a_{n}\right \vert \leq \left( \lbrack p]_{q}-\alpha \right) \cos \gamma . \end{equation*}$

Then $f\in \mathcal{ST}_{q}^{\gamma }(\lambda, p, k, \alpha)$ .

Letting $q\longrightarrow 1^{-}$ and putting $b = 1-\frac{\alpha }{p} \; (0\leq \alpha < p)$ and $\lambda = 1$ in Theorem 3.1, we obtain the following corollary (see also ^[19], Theorem 1, with $n = 0$ ).

Corollary 3.7. Let $f\in \mathcal{A(}p\mathcal{)}$ be given by (1.1) $\$ and satisfy

$\begin{equation*} \sum \limits_{n = p+1}^{\infty }\left \{ (k+1)\left( n-p\right) +(p-\alpha )\right \} \left \vert a_{n}\right \vert \leq p-\alpha . \end{equation*}$

Then $f\in \mathcal{ST}(p, k, \alpha)$ .

Letting $q\longrightarrow 1^{-}$ and putting $b = 1-\frac{\alpha }{p} \; (0\leq \alpha < p)$ and $\lambda = 0$ in Theorem 3.1, we obtain the following corollary.

Corollary 3.8. Let $f\in \mathcal{A(}p\mathcal{)}$ be given by (1.1) $\$ and satisfy

$\begin{equation*} \sum \limits_{n = p+1}^{\infty }\left( \frac{n}{p}\right) \left \{ (k+1)\left( n-p\right) +(p-\alpha )\right \} \left \vert a_{n}\right \vert \leq p-\alpha . \end{equation*}$

Then $f\in \mathcal{UST}(p, k, \alpha)$ .

Taking $p = 1$ in Theorem 3.1, we obtain the following corollary.

Corollary 3.9. If a function $f\in \mathcal{A}$ has the form (1.1) $($ with $p = 1)$ and satisfy

$\begin{equation*} \sum \limits_{n = 2}^{\infty }\left \{ (k+1)\left( [n]_{q}-1\right) +\left \vert b\right \vert \right \} \Phi _{q}(\lambda ,n)\left \vert a_{n}\right \vert \leq \left \vert b\right \vert . \end{equation*}$

Then $f\in \mathcal{ST}_{q}(\lambda, k, b)$ .

Theorem 3.10. If $f\in \mathcal{ST}_{q}(\lambda, p, k, b).$ Then

$\begin{equation} \left \vert a_{p+1}\right \vert \leq \frac{\lbrack p]_{q}\left \vert b\right \vert Q_{1}}{q^{p}\Phi _{q}(\lambda ,p,p+1)} = \frac{[p]_{q}\left \vert b\right \vert Q_{1}[\lambda +p]_{q}}{q^{p}[p+1]_{q}}, \end{equation}$

(3.4)

and for all $n\geq 3$

$\left|a_{n+p-1}\right| \leq \frac{[p]_{q}|b| Q_{1}}{q^{p}[n-1]_{q} \Phi_{q}(\lambda, p, n+p-1)} \prod\limits_{j=1}^{n-2}\left(1+\frac{[p]_{q}|b| Q_{1}}{q^{p}[j]_{q}}\right),$

(3.5)

where $Q_{1}$ is given by (2.3).

Proof. Let

$\begin{equation*} {p(z) = }1+\frac{1}{b}\left( \frac{1}{[p]_{q}}\frac{z\mathcal{D}_{q,p}\left( \mathcal{I}_{q}^{\lambda +p-1}f(z)\right) }{\mathcal{I}_{q}^{\lambda +p-1}f(z)}-1\right) , \end{equation*}$

where $p(z) = 1+\sum \limits_{n = 1}^{\infty }c_{n}z^{n}$ is analytic in $\mathcal{U}$ and it can be written as

$\begin{equation} \sum \limits_{n = p+1}^{\infty }\left( [n]_{q}-[p]_{q}\right) \Phi _{q}(\lambda ,p,n)a_{n}z^{n}\leq \lbrack p]_{q}b\left( \mathcal{I} _{q}^{\lambda +p-1}f(z)\right) \left( \sum \limits_{n = 1}^{\infty }c_{n}z^{n}\right) . \end{equation}$

(3.6)

Comparing the coefficients of $z^{n+p-1}$ on both sides of (3.6), we obtain

$\begin{eqnarray*} &&\left( [n+p-1]_{q}-[p]_{q}\right) \Phi _{q}(\lambda ,p,n+p-1)a_{n+p-1} \\ & = &[p]_{q}b\left \{ c_{1}\Phi _{q}(\lambda ,p,n+p-1)a_{n+p-2}+\cdots +c_{n-1}\right \} . \end{eqnarray*}$

Taking the absolute value on both sides and using $\left \vert c_{n}\right \vert \leq Q_{1} \; (n\geq 1)$ (see ^[18]), we obtain

$\begin{equation*} \left \vert a_{n+p-1}\right \vert \leq \frac{\lbrack p]_{q}\left \vert b\right \vert Q_{1}}{q^{p}[n-1]_{q}\Phi _{q}(\lambda ,p,n+p-1)} \end{equation*}$

$\begin{equation} \times \left \{ 1+\Phi _{q}(\lambda ,p,p+1)\left \vert a_{p+1}\right \vert +\cdots +\Phi _{q}(\lambda ,p,n+p-2)\left \vert a_{n+p-2}\right \vert \right \} . \end{equation}$

(3.7)

We apply the mathematical induction on (3.7), so for $n = 2,$ we have

(3.8)

this shows that the result is true for $n = 2$ . Now for $n = 3$ we have

$\begin{equation*} \left \vert a_{p+2}\right \vert \leq \frac{\lbrack p]_{q}\left \vert b\right \vert Q_{1}}{q^{p}[2]_{q}\Phi _{q}(\lambda ,p,p+2)}\left( 1+\Phi _{q}(\lambda ,p,p+1)\left \vert a_{p+1}\right \vert \right) , \end{equation*}$

using (3.8), we obtain

$\begin{equation*} \left \vert a_{p+2}\right \vert \leq \frac{\lbrack p]_{q}\left \vert b\right \vert Q_{1}}{q^{p}[2]_{q}\Phi _{q}(\lambda ,p,p+2)}\left( 1+\frac{ [p]_{q}\left \vert b\right \vert Q_{1}}{q^{p}[1]_{q}}\right) , \end{equation*}$

which is true for $n = 3$ . Let us assume that (3.7) is true for $n\leq t,$ that is

$\begin{equation*} \left \vert a_{t+p-1}\right \vert \leq \frac{\lbrack p]_{q}\left \vert b\right \vert Q_{1}}{q^{p}[t-1]_{q}\Phi _{q}(\lambda ,p,t+p-1)}\prod\limits_{j=1}^{t-2}\left( 1+\frac{[p]_{q}\left \vert b\right \vert Q_{1} }{q^{p}[j]_{q}}\right) . \end{equation*}$

Consider

$\begin{eqnarray*} \left \vert a_{t+p}\right \vert &\leq &\frac{[p]_{q}\left \vert b\right \vert Q_{1}}{q^{p}[t]_{q}\Phi _{q}(\lambda ,p,t+p)} \\ &&\times \left \{ 1+\Phi _{q}(\lambda ,p,p+1)\left \vert a_{p+1}\right \vert +\cdots +\Phi _{q}(\lambda ,p,t+p-1)\left \vert a_{t+p-1}\right \vert \right \} \\ &\leq &\frac{[p]_{q}\left \vert b\right \vert Q_{1}}{q^{p}[t]_{q}\Phi _{q}(\lambda ,p,t+p)}\left \{ 1+\frac{[p]_{q}\left \vert b\right \vert Q_{1} }{q^{p}}+\frac{[p]_{q}\left \vert b\right \vert Q_{1}}{q^{p}[2]_{q}}\left( 1+ \frac{[p]_{q}\left \vert b\right \vert Q_{1}}{q^{p}[1]_{q}}\right) \right. \\ &&+\frac{[p]_{q}\left \vert b\right \vert Q_{1}}{q^{p}[3]_{q}}\left( 1+\frac{ [p]_{q}\left \vert b\right \vert Q_{1}}{q^{p}[1]_{q}}\right) \left( 1+\frac{ [p]_{q}\left \vert b\right \vert Q_{1}}{q^{p}[2]_{q}}\right) +\cdots + \\ &&\left. \frac{[p]_{q}\left \vert b\right \vert Q_{1}}{q^{p}[t-1]_{q}} \overset{t-2}{\underset{j = 1}{\prod }}\left( 1+\frac{[p]_{q}\left \vert b\right \vert Q_{1}}{q^{p}[j]_{q}}\right) \right \} \\ & = &\frac{[p]_{q}\left \vert b\right \vert Q_{1}}{q^{p}[t]_{q}\Phi _{q}(\lambda ,p,t+p)}\overset{t-1}{\underset{j = 1}{\prod }}\left( 1+\frac{ [p]_{q}\left \vert b\right \vert Q_{1}}{q^{p}[j]_{q}}\right) . \end{eqnarray*}$

So, the result is true for $n = t+1$ . Also, we proved that the result true for all $n(n\geq 2)$ using mathematical induction.

Taking $p = 1$ in Theorem 3.10, we obtain the following corollary.

Corollary 3.11. Let $f\in \mathcal{A}$ be given by (1.1) $\ ($ with $p = 1).$ If $f\in \mathcal{ST}_{q}(\lambda, k, b),$ then

$\begin{equation*} \left \vert a_{2}\right \vert \leq \frac{\lbrack \lambda +1]_{q}\left \vert b\right \vert Q_{1}}{q[2]_{q}}, \end{equation*}$

and

$\begin{equation*} \left \vert a_{n}\right \vert \leq \frac{\left \vert b\right \vert Q_{1}}{ q[n-1]_{q}\Phi _{q}(\lambda ,1,n)}\prod\limits_{j=1}^{n-2} \left( 1+\frac{\left \vert b\right \vert Q_{1}}{q[j]_{q}}\right) \mathit{{\rm{}}} (n\geq 3). \end{equation*}$

Taking $b = 1-\alpha \; (0\leq \alpha < 1)$ and $p = 1$ in Theorem 3.10, we obtain the following consequence.

Corollary 3.12. Let $f\in \mathcal{A}$ be given by (1.1) $\ ($ with $p = 1).$ If $f\in \mathcal{ST}_{q}(\lambda, k, \alpha),$ then

$\begin{equation*} \left \vert a_{2}\right \vert \leq \frac{P_{1}[\lambda +1]_{q}}{q[2]_{q}}, \end{equation*}$

and

$\begin{equation*} \left \vert a_{n}\right \vert \leq \frac{\lbrack p]_{q}P_{1}}{q[n-1]_{q}\Phi _{q}(\lambda ,n)}\prod\limits_{j=1}^{n-2}\left( 1+\frac{P_{1}}{ q[j]_{q}}\right) \mathit{{\rm{}}}(n\geq 3), \end{equation*}$

where $P_{1} = (1-\alpha)Q_{1}$ and $Q_{1}$ is given by (2.3).

Taking $b = 1-\frac{\alpha }{[p]_{q}} \; (0\leq \alpha < [p]_{q})$ in Theorem 3.10, we obtain the following result.

Corollary 3.13. Let $f\in \mathcal{A(}p\mathcal{)}$ be given by (1.1) $.$ If $f\in \mathcal{ST}_{q}(\lambda, p, k, \alpha),$ then

$\begin{equation*} \left \vert a_{p+1}\right \vert \leq \frac{\left( \lbrack p]_{q}-\alpha \right) Q_{1}}{q^{p}\Phi _{q}(\lambda ,p,n+p-1)}, \end{equation*}$

and for all $n\geq 3,$

$\begin{equation*} \left \vert a_{n+p-1}\right \vert \leq \frac{\left( \lbrack p]_{q}-\alpha \right) Q_{1}}{q^{p}[n-1]_{q}\Phi _{q}(\lambda ,p,n+p-1)}\prod\limits_{j=1}^{n-2}\left( 1+\frac{\left( [p]_{q}-\alpha \right) Q_{1}}{ q^{p}[j]_{q}}\right) . \end{equation*}$

Putting $b = \left(1-\frac{\alpha }{[p]_{q}}\right) \cos \gamma e^{-i\gamma } \; (0\leq \alpha < [p]_{q}, \; \left \vert \gamma \right \vert < \frac{\pi }{2})$ in Theorem 3.10, we obtain the following consequence.

Corollary 3.14. Let $f\in \mathcal{A(}p\mathcal{)}$ be given by (1.1) $.$ If $f\in \mathcal{ST}_{q}(\lambda, p, k, \alpha),$ then

$\begin{equation*} \left \vert a_{p+1}\right \vert \leq \frac{\left( \lbrack p]_{q}-\alpha \right) \cos \gamma Q_{1}}{q^{p}\Phi _{q}(\lambda ,p,p+1)}, \end{equation*}$

and for all $n\geq 3,$

$\begin{equation*} \left \vert a_{n+p-1}\right \vert \leq \frac{\left( \lbrack p]_{q}-\alpha \right) \cos \gamma Q_{1}}{q^{p}[n-1]_{q}\Phi _{q}(\lambda ,p,n+p-1)}\prod\limits_{j=1}^{n-2}\left( j+\frac{\left( [p]_{q}-\alpha \right) \cos \gamma Q_{1}}{q^{p}[j]_{q}}\right) . \end{equation*}$

Theorem 3.15. Let $f\in \mathcal{ST}_{q}(\lambda, p, k, b).$ Then $f(\mathcal{U})$ contains an open disc

$\begin{equation*} r = \frac{q^{p}[p+1]_{q}}{q^{p}(p+1)[p+1]_{q}+[p]_{q}\left \vert b\right \vert }. \end{equation*}$

Proof. Let $w_{0}\in \mathbb{C}$ and $w_{0}\neq 0$ such that $f(z)\neq w_{0}$ for $z\in \mathcal{U}$ . Then

$\begin{equation*} f_{1}(z) = \frac{w_{0}f(z)}{w_{0}-f(z)} = z^{p}+\left( a_{p+1}+\frac{1}{w_{0}} \right) z^{p+1}+\cdots . \end{equation*}$

Since $f_{1}$ is univalent, so

$\begin{equation*} \left \vert a_{p+1}+\frac{1}{w_{0}}\right \vert \leq p+1. \end{equation*}$

Now using Theorem 3.10, we have

$\begin{equation*} \left \vert \frac{1}{w_{0}}\right \vert \leq p+1+\frac{[p]_{q}\left \vert b\right \vert Q_{1}[\lambda +p]_{q}}{q^{p}[p+1]_{q}}, \end{equation*}$

and hence we have

$\begin{equation*} \left \vert w_{0}\right \vert \geq \frac{q^{p}[p+1]_{q}}{ q^{p}(p+1)[p+1]_{q}+[p]_{q}\left \vert b\right \vert Q_{1}[\lambda +p]_{q}}. \end{equation*}$

This completes the proof of Theorem 3.15

Theorem 3.16. Let $0\leq k < \infty$ be fixed and let $f\in \mathcal{ST} _{q}(\lambda, p, k, b)\$ with the form (1.1). Then for a complex $\mu,$ we have

$\begin{equation} \left \vert a_{p+2}-\mu a_{p+1}^{2}\right \vert \leq \frac{\lbrack p]_{q}\left \vert b\right \vert Q_{1}[\lambda +p,q]_{2}}{ 2[2]_{q}q^{p}[p+1,q]_{2}}\max \left \{ 1,\left \vert 2v-1\right \vert \right \} , \end{equation}$

(3.9)

where

$\begin{equation*} v = \frac{1}{2}\left \{ 1-\frac{Q_{2}}{Q_{1}}-\frac{[p]_{q}bQ_{1}}{q^{p}} \left( 1-\frac{[2]_{q}[\lambda +p]_{q}[p+2]_{q}}{[\lambda +p+1]_{q}[p+1]_{q}} \mu \right) \right \} , \end{equation*}$

where $Q_{1}$ and $Q_{2}$ are given by (2.3) and (2.4), respectively. The result is sharp.

Proof. Let $f\in \mathcal{ST}_{q}(\lambda, p, k, b)$ , then there exist a function $w$ , with $w(0) = 0$ and $\left \vert w(z)\right \vert < 1$ such that

$\begin{equation} 1+\frac{1}{b}\left( \frac{1}{[p]_{q}}\frac{z\mathcal{D}_{q,p}\left( \mathcal{ I}_{q}^{\lambda +p-1}f(z)\right) }{\mathcal{I}_{q}^{\lambda +p-1}f(z)} -1\right) = p_{k}(w(z)){\rm{ \ }}(z\in \mathcal{U)}. \end{equation}$

(3.10)

Let $h\in P$ be a function defined by

$\begin{equation*} h(z) = \frac{1+w(z)}{1-w(z)} = 1+c_{1}z+c_{2}z^{2}+\cdots {\rm{ \ }}(z\in \mathcal{U)}. \end{equation*}$

This gives

$\begin{equation*} w(z) = \frac{c_{1}}{2}z+\frac{1}{2}(c_{2}-\frac{c_{1}^{2}}{2})z^{2}+\cdots , \end{equation*}$

and

$\begin{equation} p_{k}(w(z)) = 1+\frac{1}{2}c_{1}Q_{1}z+\frac{1}{2}\left \{ \frac{c_{1}^{2}Q_{2} }{2}+(c_{2}-\frac{c_{1}^{2}}{2})Q_{1}\right \} z^{2}+\cdots . \end{equation}$

(3.11)

Using (3.11) in (3.10) along with (1.9), we obtain

$\begin{equation*} a_{p+1} = \frac{[p]_{q}bc_{1}Q_{1}[\lambda +1]_{q}}{2q^{p}[p+1]_{q}}, \end{equation*}$

and

$\begin{equation*} a_{p+2} = \frac{[p]_{q}b[\lambda +p,q]_{2}}{[2]_{q}q^{p}[p+1,q]_{2}}\left \{ \frac{c_{1}^{2}Q_{2}}{4}+\frac{1}{2}(c_{2}-\frac{c_{1}^{2}}{2})Q_{1}+\frac{ [p]_{q}bQ_{1}^{2}c_{1}^{2}}{4q^{p}}\right \} . \end{equation*}$

For any complex number $\mu$ , we have

$\begin{eqnarray} a_{p+2}-\mu a_{p+1}^{2} & = &\frac{[p]_{q}b[\lambda +p,q]_{2}}{ 2[2]_{q}q^{p}[p+1,q]_{2}}\left \{ \frac{c_{1}^{2}Q_{2}}{2}+(c_{2}-\frac{ c_{1}^{2}}{2})Q_{1}+\frac{[p]_{q}bQ_{1}^{2}c_{1}^{2}}{4q^{p}}\right \} \\ &&-\frac{[p]_{q}^{2}b^{2}c_{1}^{2}Q_{1}^{2}}{4q^{2p}}\left( \frac{[\lambda +1]_{q}}{[p+1]_{q}}\right) ^{2}\mu . \end{eqnarray}$

(3.12)

Thus (3.12) can be written as

$\begin{equation} a_{p+2}-\mu a_{p+1}^{2} = \frac{[p]_{q}bQ_{1}[\lambda +p,q]_{2}}{ 2[2]_{q}q^{p}[p+1,q]_{2}}\left \{ c_{2}-vc_{1}^{2}\right \} , \end{equation}$

(3.13)

where

$\begin{equation} v = \frac{1}{2}\left \{ 1-\frac{Q_{2}}{Q_{1}}-\frac{[p]_{q}bQ_{1}}{q^{p}} \left( 1-\frac{[2]_{q}[\lambda +p]_{q}[p+2]_{q}}{[\lambda +p+1]_{q}[p+1]_{q}} \mu \right) \right \} . \end{equation}$

(3.14)

Now, taking absolute value and using Lemma 2.2, we obtain the required result. The sharpness of (3.9) follows from the sharpness of (2.5).

Putting $p = 1$ in Theorem 3.16, we obtain the following consequence.

Corollary 3.17. Let $0\leq k < \infty$ be fixed and let $f\in \mathcal{ST}_{q}(\lambda, k, b)$ with the form (1.1) (with $p = 1$ ). Then for a complex parameter $\mu,$ we have

$\begin{equation*} \left \vert a_{3}-\mu a_{2}^{2}\right \vert \leq \frac{\left \vert b\right \vert Q_{1}[\lambda +1,q]_{2}}{2[2]_{q}q[2,q]_{2}}\max \left \{ 1,\left \vert 2v-1\right \vert \right \} , \end{equation*}$

where

$\begin{equation*} v = \frac{1}{2}\left \{ 1-\frac{Q_{2}}{Q_{1}}-\frac{bQ_{1}}{q}\left( 1-\frac{ [\lambda +1]_{q}[3]_{q}}{[\lambda +2]_{q}}\mu \right) \right \} , \end{equation*}$

where $Q_{1}$ and $Q_{2}$ are given by (2.3) and (2.4), respectively. The result is sharp.

Putting $p = 1$ and $b = 1-\alpha \; (0\leq \alpha < 1)$ in Theorem 3.16, we get the following corollary.

Corollary 3.18. Suppose that the function $f(z)$ given by (1.1) (with $p = 1$ ) is in the class $\mathcal{ST}_{q}(\lambda, k, \alpha).$ Then for a complex parameter $\mu,$ we have

$\begin{equation} \left \vert a_{3}-\mu a_{2}^{2}\right \vert \leq \frac{P_{1}[\lambda +1,q]_{2}}{2q[2]_{q}[2,q]_{2}}\max \left \{ 1,\frac{P_{2}}{P_{1}}-\frac{P_{1} }{q}\left( 1-\frac{[\lambda +1]_{q}[3]_{q}}{[\lambda +2]_{q}}\mu \right) \right \} , \end{equation}$

(3.15)

where $P_{1} = (1-\alpha)Q_{1}$ and $P_{2} = (1-\alpha)Q_{2}, \; Q_{1}$ and $Q_{2}$ are given by (2.3) and (2.4), respectively. The result is sharp.

Putting $b = 1-\frac{\alpha }{[p]_{q}} \; (0\leq \alpha < [p]_{q})$ in Theorem 3.16, we get the following corollary.

Corollary 3.19. Let $0\leq k < \infty$ be fixed and let $f\in \mathcal{ST}_{q}(\lambda, p, k, \alpha)\$ with\ the form (1.1). Then for a complex parameter $\mu,$ we have

$\begin{eqnarray*} \left \vert a_{p+2}-\mu a_{p+1}^{2}\right \vert &\leq &\frac{([p]_{q}-\alpha )Q_{1}[\lambda +p,q]_{2}}{2[2]_{q}q^{p}[p+1,q]_{2}} \\ &&\times \max \left \{ 1,\left \vert \frac{Q_{2}}{Q_{1}}-\frac{ ([p]_{q}-\alpha )Q_{1}}{q^{p}}\left( 1-\frac{[2]_{q}[\lambda +p]_{q}[p+2]_{q} }{[\lambda +p+1]_{q}[p+1]_{q}}\mu \right) \right \vert \right \} , \end{eqnarray*}$

where $Q_{1}$ and $Q_{2}$ are given by (2.3) and (2.4), respectively. The result is sharp.

Putting $b = \left(1-\frac{\alpha }{[p]_{q}}\right) \cos \gamma e^{-i\gamma } \; (0\leq \alpha < [p]_{q}, \; \left \vert \gamma \right \vert < \frac{\pi }{2})$ in Theorem 3.16, we get the following corollary.

Corollary 3.20. Let $0\leq k < \infty$ be fixed and let $f\in \mathcal{ST}_{q}^{\gamma }(\lambda, p, k, \alpha).$ Then for a complex parameter $\mu,$ we have

$\begin{eqnarray*} \left \vert a_{p+2}-\mu a_{p+1}^{2}\right \vert &\leq &\frac{([p]_{q}-\alpha )\cos \gamma Q_{1}[\lambda +p,q]_{2}}{2[2]_{q}q^{p}[p+1,q]_{2}} \\ &&\times \max \left \{ 1,\left \vert \frac{Q_{2}}{Q_{1}}-\frac{ ([p]_{q}-\alpha )\cos \gamma Q_{1}}{q^{p}}\left( 1-\frac{[2]_{q}[\lambda +p]_{q}[p+2]_{q}}{[\lambda +p+1]_{q}[p+1]_{q}}\mu \right) \right \vert \right \} . \end{eqnarray*}$

The result is sharp.

Theorem 3.21. Let

$\begin{eqnarray*} \sigma _{1} & = &\frac{\left[ [p]_{q}bQ_{1}^{2}+q^{p}(Q_{2}-Q_{1})\right] [\lambda +p+1]_{q}[p+1]_{q}}{[2]_{q}[p]_{q}bQ_{1}^{2}[\lambda +p]_{q}[p+2]_{q}}, \\ \sigma _{2} & = &\frac{\left[ [p]_{q}bQ_{1}^{2}+q^{p}(Q_{2}+Q_{1})\right] [\lambda +p+1]_{q}[p+1]_{q}}{[2]_{q}[p]_{q}bQ_{1}^{2}[\lambda +p]_{q}[p+2]_{q}}, \\ \sigma _{3} & = &\frac{\left[ [p]_{q}bQ_{1}^{2}+q^{p}Q_{2}\right] [\lambda +p+1]_{q}[p+1]_{q}}{[2]_{q}[p]_{q}bQ_{1}^{2}[\lambda +p]_{q}[p+2]_{q}}. \end{eqnarray*}$

If $f$ given by (1.1) belong to the class $\mathcal{ST}_{q}(\lambda, p, k, b) \; (b > 0),$ then

$\begin{equation*} \left \vert a_{p+2}-\mu a_{p+1}^{2}\right \vert \leq \left \{ \begin{array}{c} \frac{\lbrack p]_{q}bQ_{1}[\lambda +p,q]_{2}}{q^{p}[2]_{q}[p+1,q]_{2}}\left \{ \frac{Q_{2}}{Q_{1}}+\frac{[p]_{q}bQ_{1}}{q^{p}}\left( 1-\frac{ [2]_{q}[\lambda +p]_{q}[p+2]_{q}}{[\lambda +p+1]_{q}[p+1]_{q}}\mu \right) \right \} ,\mu \leq \sigma _{1}, \\ \\ \frac{[p]_{q}bQ_{1}[\lambda +p,q]_{2}}{q^{p}[2]_{q}[p+1,q]_{2}}, \sigma _{1}\leq \mu \leq \sigma _{2}, \\ \\ -\frac{[p]_{q}bQ_{1}[\lambda +p,q]_{2}}{q^{p}[2]_{q}[p+1,q]_{2}}\left \{ \frac{Q_{2}}{Q_{1}}+\frac{[p]_{q}bQ_{1}}{q^{p}}\left( 1-\frac{ [2]_{q}[\lambda +p]_{q}[p+2]_{q}}{[\lambda +p+1]_{q}[p+1]_{q}}\mu \right) \right \} ,\mu \geq \sigma _{2}. \end{array} \right. \end{equation*}$

Further, if $\sigma _{1}\leq \mu \leq \sigma _{3}$ , then \newpage

$\begin{eqnarray*} &&\left \vert a_{p+2}-\mu a_{p+1}^{2}\right \vert +\frac{q^{p}\left( [p+1]_{q}\right) ^{2}[\lambda +p,q]_{2}}{[2]_{q}[p]_{q}bQ_{1}\left( [\lambda +p]_{q}\right) ^{2}[p+1,q]_{2}} \\ &&\times \left \{ 1-\frac{Q_{2}}{Q_{1}}-\frac{bQ_{1}}{q^{p}}\left( 1-\frac{ [2]_{q}[\lambda +p]_{q}[p+2]_{q}}{[\lambda +p+1]_{q}[p+1]_{q}}\mu \right) \right \} \left \vert a_{p+1}\right \vert ^{2}\leq \frac{\lbrack p]_{q}bQ_{1}[\lambda +p,q]_{2}}{q^{p}[2]_{q}[p+1,q]_{2}}. \end{eqnarray*}$

If $\sigma _{3}\leq \mu \leq \sigma _{2},$ then

$\begin{eqnarray*} &&\left \vert a_{p+2}-\mu a_{p+1}^{2}\right \vert +\frac{q^{p}\left( [p+1]_{q}\right) ^{2}[\lambda +p,q]_{2}}{[2]_{q}bQ_{1}\left( [\lambda +p]_{q}\right) ^{2}[p+1,q]_{2}} \\ &&\times \left \{ 1+\frac{Q_{2}}{Q_{1}}+\frac{[p]_{q}bQ_{1}}{q^{p}}\left( 1- \frac{[2]_{q}[\lambda +p]_{q}[p+2]_{q}}{[\lambda +p+1]_{q}[p+1]_{q}}\mu \right) \right \} \left \vert a_{p+1}\right \vert ^{2}\leq \frac{\lbrack p]_{q}bQ_{1}[\lambda +p,q]_{2}}{q^{p}[2]_{q}[p+1,q]_{2}}. \end{eqnarray*}$

The result is sharp.

Proof. Applying Lemma 2.3 to (3.12) and (3.13), we can obtain our results asserted by Theorem 3.21.

Putting $p = 1$ in Theorem 3.21, we obtain the following corollary.

Corollary 3.22. Let

$\begin{eqnarray*} \varsigma _{1} & = &\frac{\left[ bQ_{1}^{2}+q(Q_{2}-Q_{1})\right] [\lambda +2]_{q}}{bQ_{1}^{2}[\lambda +1]_{q}[3]_{q}}, \\ \varsigma _{2} & = &\frac{\left[ bQ_{1}^{2}+q(Q_{2}+Q_{1})\right] [\lambda +2]_{q}}{bQ_{1}^{2}[\lambda +1]_{q}[3]_{q}}, \\ \varsigma _{3} & = &\frac{\left[ bQ_{1}^{2}+qQ_{2}\right] [\lambda +2]_{q}}{ bQ_{1}^{2}[\lambda +1]_{q}[3]_{q}}. \end{eqnarray*}$

If $f$ given by (1.1) (with $p = 1$ ) belong to the class $\mathcal{ST} _{q}(\lambda, k, b)$ with $b > 0,$ then

$\begin{equation*} \left \vert a_{3}-\mu a_{2}^{2}\right \vert \leq \left \{ \begin{array}{c} \frac{bQ_{1}[\lambda +1,q]_{2}}{q[2]_{q}[2,q]_{2}}\left \{ \frac{Q_{2}}{Q_{1} }+\frac{bQ_{1}}{q}\left( 1-\frac{[\lambda +1]_{q}[3]_{q}}{[\lambda +2]_{q}} \mu \right) \right \} ,&&\mu \leq \varsigma _{1}, \\ \frac{bQ_{1}[\lambda +1,q]_{2}}{ q[2]_{q}[2,q]_{2}},&&\varsigma _{1}\leq \mu \leq \varsigma _{2}, \\ \\ -\frac{bQ_{1}[\lambda +1,q]_{2}}{q[2]_{q}[2,q]_{2}}\left \{ \frac{Q_{2}}{ Q_{1}}+\frac{bQ_{1}}{q}\left( 1-\frac{[\lambda +1]_{q}[3]_{q}}{[\lambda +2]_{q}}\mu \right) \right \} ,&&\mu \geq \varsigma _{2}. \end{array} \right. \end{equation*}$

Further, if $\varsigma _{1}\leq \mu \leq \varsigma _{3}$ , then

$\begin{eqnarray*} &&\left \vert a_{3}-\mu a_{2}^{2}\right \vert +\frac{q[2]_{q}[\lambda +1,q]_{2}}{bQ_{1}\left( [\lambda +1]_{q}\right) ^{2}[2,q]_{2}} \\ &&\times \left \{ 1-\frac{Q_{2}}{Q_{1}}-\frac{bQ_{1}}{q}\left( 1-\frac{ [\lambda +1]_{q}[3]_{q}}{[\lambda +2]_{q}}\mu \right) \right \} \left \vert a_{2}\right \vert ^{2}\leq \frac{bQ_{1}[\lambda +1,q]_{2}}{q[2]_{q}[2,q]_{2}} . \end{eqnarray*}$

If $\varsigma _{3}\leq \mu \leq \varsigma _{2},$ then

$\begin{eqnarray*} &&\left \vert a_{p+2}-\mu a_{p+1}^{2}\right \vert +\frac{q[2]_{q}[\lambda +1,q]_{2}}{qbQ_{1}\left( [\lambda +1]_{q}\right) ^{2}[2,q]_{2}} \\ &&\times \left \{ 1+\frac{Q_{2}}{Q_{1}}+\frac{bQ_{1}}{q}\left( 1-\frac{ [\lambda +1]_{q}[3]_{q}}{[\lambda +2]_{q}}\mu \right) \right \} \left \vert a_{2}\right \vert ^{2}\leq \frac{bQ_{1}[\lambda +1,q]_{2}}{q[2]_{q}[2,q]_{2}} . \end{eqnarray*}$

The result is sharp.

Putting $p = 1$ and $b = 1-\alpha \; (0\leq \alpha < 1)$ in Theorem 3.21, we obtain the following corollary.

Corollary 3.23. Let

$\begin{eqnarray*} \vartheta _{1} & = &\frac{\left[ P_{1}^{2}+q(P_{2}-P_{1})\right] [\lambda +2]_{q}}{P_{1}^{2}[\lambda +1]_{q}[3]_{q}}, \\ \vartheta _{2} & = &\frac{\left[ P_{1}^{2}+q(P_{2}+P_{1})\right] [\lambda +2]_{q}}{P_{1}^{2}[\lambda +1]_{q}[3]_{q}}, \\ \vartheta _{3} & = &\frac{\left[ P_{1}^{2}+qP_{2}\right] [\lambda +2]_{q}}{ P_{1}^{2}[\lambda +1]_{q}[3]_{q}}. \end{eqnarray*}$

If $f$ given by (1.1) (with $p = 1$ ) belong to the class $\mathcal{ST} _{q}(\lambda, k, \alpha),$ then

$\begin{equation*} \left \vert a_{3}-\mu a_{2}^{2}\right \vert \leq \left \{ \begin{array}{c} \frac{P_{1}[\lambda +1,q]_{2}}{q[2]_{q}[2,q]_{2}}\left \{ \frac{P_{2}}{P_{1}} +\frac{P_{1}}{q}\left( 1-\frac{[\lambda +1]_{q}[3]_{q}}{[\lambda +2]_{q}}\mu \right) \right \} ,\mu \leq \vartheta _{1}, \\ \\ \frac{P_{1}[\lambda +1,q]_{2}}{q[2]_{q}[2,q]_{2}}, \vartheta _{1}\leq \mu \leq \vartheta _{2}, \\ \\ -\frac{P_{1}[\lambda +1,q]_{2}}{q[2]_{q}[2,q]_{2}}\left \{ \frac{P_{2}}{P_{1} }+\frac{P_{1}}{q}\left( 1-\frac{[\lambda +1]_{q}[3]_{q}}{[\lambda +2]_{q}} \mu \right) \right \} ,\mu \geq \vartheta _{2}. \end{array} \right. \end{equation*}$

Further, if $\vartheta _{1}\leq \mu \leq \vartheta _{3}$ , then

$\begin{eqnarray*} &&\left \vert a_{3}-\mu a_{2}^{2}\right \vert +\frac{q[2]_{q}[\lambda +1,q]_{2}}{P_{1}\left( [\lambda +1]_{q}\right) ^{2}[2,q]_{2}} \\ &&\times \left \{ 1-\frac{P_{2}}{P_{1}}-\frac{P_{1}}{q}\left( 1-\frac{ [\lambda +1]_{q}[3]_{q}}{[\lambda +2]_{q}}\mu \right) \right \} \left \vert a_{2}\right \vert ^{2}\leq \frac{P_{1}[\lambda +1,q]_{2}}{q[2]_{q}[2,q]_{2}}. \end{eqnarray*}$

If $\vartheta _{3}\leq \mu \leq \vartheta _{2},$ then

$\begin{eqnarray*} &&\left \vert a_{3}-\mu a_{2}^{2}\right \vert +\frac{q[2]_{q}[\lambda +1,q]_{2}}{qP_{1}\left( [\lambda +1]_{q}\right) ^{2}[2,q]_{2}} \\ &&\times \left \{ 1+\frac{P_{2}}{P_{1}}+\frac{P_{1}}{q}\left( 1-\frac{ [\lambda +1]_{q}[3]_{q}}{[\lambda +2]_{q}}\mu \right) \right \} \left \vert a_{2}\right \vert ^{2}\leq \frac{P_{1}[\lambda +1,q]_{2}}{q[2]_{q}[2,q]_{2}}. \end{eqnarray*}$

The result is sharp.

Putting $b = \left(1-\frac{\alpha }{[p]_{q}}\right) \; (0\leq \alpha < [p]_{q})$ in Theorem 3.21, we obtain the following corollary.

Corollary 3.24. Let

$\begin{eqnarray*} \epsilon _{1} & = &\frac{\left[ \left( [p]_{q}-\alpha \right) Q_{1}^{2}+q^{P}(Q_{2}-Q_{1})\right] [\lambda +P+1]_{q}[P+1]_{q}}{ [2]_{q}\left( [p]_{q}-\alpha \right) Q_{1}^{2}[\lambda +P]_{q}[P+2]_{q}}, \\ \epsilon _{2} & = &\frac{\left[ \left( [p]_{q}-\alpha \right) Q_{1}^{2}+q^{P}(Q_{2}+Q_{1})\right] [\lambda +P+1]_{q}[P+1]_{q}}{ [2]_{q}\left( [p]_{q}-\alpha \right) Q_{1}^{2}[\lambda +P]_{q}[P+2]_{q}}, \\ \epsilon _{3} & = &\frac{\left[ \left( [p]_{q}-\alpha \right) Q_{1}^{2}+q^{P}Q_{2}\right] [\lambda +P+1]_{q}[P+1]_{q}}{[2]_{q}\left( [p]_{q}-\alpha \right) Q_{1}^{2}[\lambda +P]_{q}[P+2]_{q}}. \end{eqnarray*}$

If $f$ given by (1.1) belong to the class $\mathcal{ST}_{q}(\lambda, P, k, b)$ with $b > 0,$ then

$\begin{equation*} \left \vert a_{p+2}-\mu a_{p+1}^{2}\right \vert \leq \left \{ \begin{array}{c} \frac{\left( \lbrack p]_{q}-\alpha \right) Q_{1}[\lambda +p,q]_{2}}{ q^{p}[2]_{q}[p+1,q]_{2}}\left \{ \frac{Q_{2}}{Q_{1}}+\frac{\left( [p]_{q}-\alpha \right) Q_{1}}{q}\left( 1-\frac{[2]_{q}[\lambda +p]_{q}[p+2]_{q}}{[\lambda +p+1]_{q}[p+1]_{q}}\mu \right) \right \} ,\mu \leq \epsilon _{1}, \\ \frac{\left( \lbrack p]_{q}-\alpha \right) Q_{1}[\lambda +p,q]_{2}}{ q^{p}[2]_{q}[p+1,q]_{2}},\epsilon _{1}\leq \mu \leq \epsilon _{2}, \\ -\frac{\left( [p]_{q}-\alpha \right) Q_{1}[\lambda +p,q]_{2}}{ q^{p}[2]_{q}[p+1,q]_{2}}\left \{ \frac{Q_{2}}{Q_{1}}+\frac{\left( [p]_{q}-\alpha \right) Q_{1}}{q}\left( 1-\frac{[2]_{q}[\lambda +p]_{q}[p+2]_{q}}{[\lambda +p+1]_{q}[p+1]_{q}}\mu \right) \right \} ,\mu \geq \epsilon _{2}. \end{array} \right. \end{equation*}$

Further, if $\epsilon _{1}\leq \mu \leq \epsilon _{3}$ , then

$\begin{eqnarray*} &&\left \vert a_{p+2}-\mu a_{p+1}^{2}\right \vert +\frac{q^{p}\left( [p+1]_{q}\right) ^{2}[\lambda +p,q]_{2}}{[2]_{q}\left( [p]_{q}-\alpha \right) Q_{1}\left( [\lambda +p]_{q}\right) ^{2}[p+1,q]_{2}} \\ &&\times \left \{ 1-\frac{Q_{2}}{Q_{1}}-\frac{\left( [p]_{q}-\alpha \right) Q_{1}}{q^{p}}\left( 1-\frac{[2]_{q}[\lambda +p]_{q}[p+2]_{q}}{[\lambda +p+1]_{q}[p+1]_{q}}\mu \right) \right \} \left \vert a_{p+1}\right \vert ^{2} \\ &\leq &\frac{\left( [p]_{q}-\alpha \right) Q_{1}[\lambda +p,q]_{2}}{ q^{p}[2]_{q}[p+1,q]_{2}}. \end{eqnarray*}$

If $\epsilon _{3}\leq \mu \leq \epsilon _{2},$ then

$\begin{eqnarray*} &&\left \vert a_{p+2}-\mu a_{p+1}^{2}\right \vert +\frac{q^{p}\left( [p+1]_{q}\right) ^{2}[\lambda +p,q]_{2}}{[2]_{q}\left( [p]_{q}-\alpha \right) \left( [\lambda +p]_{q}\right) ^{2}[p+1,q]_{2}} \\ &&\times \left \{ 1+\frac{Q_{2}}{Q_{1}}+\frac{\left( [p]_{q}-\alpha \right) Q_{1}}{q^{p}}\left( 1-\frac{[2]_{q}[\lambda +p]_{q}[p+2]_{q}}{[\lambda +p+1]_{q}[p+1]_{q}}\mu \right) \right \} \left \vert a_{p+1}\right \vert ^{2} \\ &\leq &\frac{\left( [p]_{q}-\alpha \right) Q_{1}[\lambda +p,q]_{2}}{ q^{p}[2]_{q}[p+1,q]_{2}}. \end{eqnarray*}$

The result is sharp.

4. Conclusions

Studies of the coefficient problems including the Fekete-Szegö problems continue to motivate researchers in Geometric Function Theory of Complex Analysis. In our present investigation, we have introduced and studied a new class $\mathcal{ST}_{q}(\lambda, p, k, b)$ of analytic functions associated with $q$ -analogue of $p$ -valent Noor integral operator in the open unit disc $\mathcal{U}$ . For functions in this class, we have derived the coefficient estimates of the coefficients $\left \vert a_{p+1}\right \vert$ and $\left \vert a_{n+p+1}\right \vert$ for $n\geq 3$ , and Fekete-Szegö functional problems for functions belonging to this new class. Several of new results are shown to follow upon specializing the parameters involved in our main results.

Acknowledgments

The authors would like to thank the referees for their helpful comments and suggestions.

Conflict of interest

The authors declare that they have no competing interests.

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AIMS Mathematics

New subclass of analytic functions defined by $q$ -analogue of $p$ -valent Noor integral operator

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Abstract

1. Introduction and definitions

2. Geometric interpretation

3. Main results

4. Conclusions

Acknowledgments

Conflict of interest

References

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通讯作者: 陈斌, bchen63@163.com

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AIMS Mathematics

New subclass of analytic functions defined by q q -analogue of p p -valent Noor integral operator

Related Papers:

Abstract

1. Introduction and definitions

2. Geometric interpretation

3. Main results

4. Conclusions

Acknowledgments

Conflict of interest

References

This article has been cited by:

Reader Comments

通讯作者: 陈斌, bchen63@163.com

Metrics

Other Articles By Authors

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New subclass of analytic functions defined by $q$ -analogue of $p$ -valent Noor integral operator