Bi-univalent properties for certain class of Bazilevič functions defined by convolution and with bounded boundary rotation

Abstract

In this paper, we obtain bi-univalent properties for certain class of Bazilevič functions defined by convolution and with bounded boundary rotation. We will find coefficient bounds for |a₂| and |a₃| for the class $\mathcal {M}_{\alpha,\lambda,\rho,k,\beta }(f * h)\mathsf {.}$

Introduction

Let $\mathcal {A}$ denote the class of analytic functions of the form:

$$ f(z)=z+\sum_{n=2}^{\infty}a_{n}z^{n}\ \ (z\in\mathbb{U}:\mathbb{U} =\{z\in\mathbb{C}:\left\vert z\right\vert <1\}). $$

(1)

For $h(z)\in \mathcal {A}$, given by $h(z)=z+\sum \limits _{n=2}^{\infty }h_{n}z^{n},$ the Hadamard product (or convolution) of f(z) and h(z) is defined by:

$$ (f * h)(z)=z+\sum_{n=2}^{\infty} a_{n} h_{n} z^{n}=(h\times f)(z). $$

(2)

Definition 1

([1, 2], and [3] with p = 1). Let $\mathcal {P}_{k}^{\lambda }(\rho)\, \left (0\leq \rho <1,\ k\geq 2 \text { and } \left \vert \lambda \right \vert <\frac {\pi }{2}\right)$ denote the class of functions $p(z)=1+\sum \limits _{n=1}^{\infty }c_{n}z^{n},$ which are analytic in $\mathbb {U}$ and satisfy the conditions:

$$(i)\ p(0)=1, $$

$$ (ii)\int\limits_{0}^{2\pi}\left\vert \frac{\mathfrak{R}\left\{ e^{i\lambda }p(z)\right\} -\rho\cos\lambda}{1-\rho}\right\vert \leq k\pi\cos \lambda\ \left(r<1,z=re^{i\theta}\in\mathbb{U}\right). $$

(3)

We note that:

(i) $\mathcal {P}_{k}^{\lambda }(0)=\mathcal {P}_{k}^{\lambda }\ (\ k\geq 2\$and $\left \vert \lambda \right \vert <\frac {\pi }{2})\$ is the class of functions introduced by Robertson (see [4]), and he derived a variational formula for functions in this class.

(ii) $\mathcal {P}_{k}^{0}(\rho)=\mathcal {P}_{k}(\rho)\ (0\leq \rho <1,\ k\geq 2)\$is the class of functions introduced by Padmanabhan and Parvatham [5] (see also Umarani and Aouf [6]).

(iii) $\mathcal {P}_{k}^{0}(0)=\mathcal {P}_{k}(k\geq 2)\$ is the class of functions having their real parts bounded in the mean on $\mathbb {U}$, introduced by Robertson [4] and studied by Pinchuk [7].

(iv) $\mathcal {P}_{2}^{0}(\rho)=\mathcal {P}\left (\rho \right)\ (0\leq \rho <1)\$is the class of functions with positive real part of order ρ, 0≤ρ<1.

(v) $\mathcal {P}_{2}^{0}(0)=\mathcal {P}$ is the class of functions having positive real part for $z\in \mathbb {U}$.

By the Koebe one-quarter theorem [8], we know that the image of $\mathbb {U\ }$under every univalent function $f\in \mathcal {A}$ contains the disk with center in the origin and radius 1/4. Therefore, every univalent function f has an inverse f⁻¹ satisfies:

$$ f^{-1}(f(z))=z\ (z\in\mathbb{U})\ \text{and}\ f(f^{-1}(w))=w\ (|w|< r_{0} (f),\ r_{0}(f)\geq1/4). $$

(4)

It is easy to see that the inverse function has the form:

$$ f^{-1}(w)=w-a_{2}w^{2}+\left(2a_{2}^{2}-a_{3}\right)w^{3}-\left(5a_{2}^{3}-5a_{2}a_{3} +a_{4}\right)w^{4}+....\ \ \ . $$

(5)

A function $f\in \mathcal {A}$ is said to be bi-univalent in $\mathbb {U}$ if both f and its inverse map g=f⁻¹are univalent in $\mathbb {U}$.

Let $\sum$ denote the class of bi-univalent functions in $\mathbb {U}$ in the form (1). For interesting examples about the class $\sum$, see [9].

The object of this paper is to introduce new subclass of Bazilevič functions [10] for the class $\sum$ with bounded boundary rotation and defined by using convolution as follows:

Definition 2

Let $f,h\in \sum,\ \alpha \in \mathbb {C}^{\ast },\ \beta \geq 0,\ 0\leq \rho <1,\ k\geq 2\ {and}\ \left \vert \lambda \right \vert <\frac {\pi }{2},\$then $(f * h)(z)\in \sum$ is said to be in the class $\mathcal {M}_{\alpha,\lambda,\rho,k,\beta }(f * h)\$if it satisfies the following conditions:

$$ \left\{ (1-\alpha)\left(\frac{(f * h)(z)}{z}\right)^{\beta}+\alpha \frac{z(f * h)^{\prime}(z)}{(f * h)(z)}\left(\frac{(f * h)(z)} {z}\right)^{\beta}\right\} \in\mathcal{P}_{k}^{\lambda}(\rho)\ (z\in \mathbb{U)} $$

(6)

and

$$ {} \left\{ (1-\alpha)\left(\frac{(f * h)^{-1}(w)}{w}\right)^{\beta} \,+\,\alpha\frac{w((f * h)^{-1}(w))^{\prime}}{(f * h)^{-1}(w)}\left(\frac{(f * h)^{-1}(w)}{w}\right)^{\beta}\right\} \in\mathcal{P}_{k}^{\lambda}(\rho)\ (w\in\mathbb{U)}. $$

(7)

We note that by putting different values for h, α, β, k, λ, and ρ, in the above definition, we have:

(1) $\mathcal {M}_{1,0,\rho,k,\beta }\left (f\times \frac {z}{1-z}\right)=R_{\sum } (\rho,k,\beta)\ (f\in \sum,\ \beta \geq 0,\ 0\leq \rho <1,\ k\geq 2)\$(see [11], with γ=1);

(2) $\mathcal {M}_{\alpha,0,\rho,k,1}(f * h)=\mathcal {L} _{\alpha,\rho,k}(f * h)\ \left (\ f,h\in \sum,\ \alpha \in \mathbb {C} ^{\ast },\ 0\leq \rho <1,\ k\geq 2\right)\$(see [12]);

(3) $\mathcal {M}_{\eta,0,\rho,2,1}(f * h)=\mathcal {L}_{\eta,\rho }(f * h)\ \left (\ f,h\in \sum,\ \eta \geq 0,\ 0\leq \rho <1\right)\$(see [13] and [14]);

(4) $\mathcal {M}_{\eta,0,\rho,2,1}\left (f\times \frac {z}{1-z}\right)=\mathcal {L}_{\eta,\rho }(f)(z)\ \left (\ f\in \sum,\ \eta \geq 0,\ 0\leq \rho <1\right)\$(see [15]);

(5) $\mathcal {M}_{1,0,\rho,2,\beta }\left (f\times \frac {z}{1-z}\right)=\mathcal {L} _{\rho,\beta }(f)(z)\left (\ f\in \sum,\ \beta \geq 0,\ 0\leq \rho <1\right)\$(see [16]);

(6) $\mathcal {M}_{1,0,\rho,2,1}\left (f\times \frac {z}{1-z}\right)=\mathcal {L}_{\rho }(f)(z)\left (\ f\in \sum,\ 0\leq \rho <1\right)\$(see [9]);

(7) $\mathcal {M}_{\alpha,0,\rho,2,\beta }\left (f\times \frac {z}{1-z} \right)=\mathcal {NP}_{\sum }^{\beta,\alpha }(0,\rho)\ \left (f\in \sum,\ \beta,\alpha \geq 0,\ 0\leq \rho <1\right)\$(see [[17], with β=0]);

(8) $\mathcal {M}_{1,0,\rho,2,\beta }\left (f\times \frac {z}{1-z}\right)=\mathcal {R}_{\sum }(\beta,\rho)\ \left (\ f\in \sum,\ \beta \geq 0,\ 0\leq \rho <1\ \right)\$(see [18]).

Also, we can obtain the following subclasses:

(i) $\mathcal {M}_{\alpha,\lambda,\rho,k,\beta }\left (f\times \frac {z} {1-z\ }\right)=\mathcal {\digamma }_{\alpha,\lambda,\rho,k,\beta }(f)$

$$\begin{array}{*{20}l} & =\left\{ f\in\sum:(1-\alpha)\left(\frac{f(z)}{z}\right)^{\beta} +\alpha\frac{zf^{\prime}(z)}{f(z)}\left(\frac{f(z)}{z}\right)^{\beta} \in\mathcal{P}_{k}^{\lambda}(\rho)\right. \\ & \text{and }\left. (1-\alpha)\left(\frac{f^{-1}(w)}{w}\right)^{\beta }+\alpha\frac{w\left((f^{-1}(w)\right)^{\prime}}{f^{-1}(w)}\left(\frac{f^{-1}(w)} {w}\right) ^{\beta}\in\mathcal{P}_{k}^{\lambda}(\rho)\right\} ; \end{array} $$

(ii) $\mathcal {M}_{\alpha,0,\rho,k,\beta }(f\ast h)=\mathcal {F}_{\alpha,\rho,k,\beta }(f * h)$

$$\begin{array}{*{20}l} & =\left\{ f,h\in\sum:(1-\alpha)\left(\frac{(f * h)(z)}{z}\right)^{\beta}+\alpha\frac{z(f * h)^{\prime}(z)}{(f * h)(z)}\left(\frac{(f * h)(z)}{z}\right)^{\beta}\in\mathcal{P}_{k}(\rho)\right. \\ & \text{and }\left. (1-\alpha)\left(\frac{(f * h)^{-1}(w)}{w}\right)^{\beta}+\alpha\frac{w((f * h)^{-1}(w))^{\prime}}{(f * h)^{-1}(w)}\left(\frac{(f * h)^{-1}(w)}{w}\right) ^{\beta}\in\mathcal{P}_{k}(\rho)\right\} ; \end{array} $$

(iii) $\mathcal {M}_{\alpha,0,\rho,2,\beta }(f\ast h)=\mathcal {F}_{\alpha,\rho,\beta }(f * h)$

$$\begin{array}{*{20}l} & =\left\{ f,h\in\sum:\mathfrak{R}\left[ (1-\alpha)\left(\frac{(f * h)(z)} {z}\right)^{\beta}+\alpha\frac{z(f * h)^{\prime}(z)}{(f * h)(z)}\left(\frac{(f * h)(z)}{z}\right)^{\beta}\right] >\rho\right. \\ & \text{and }\left. \mathfrak{R}\left[ (1-\alpha)\left(\frac{(f * h)^{-1}(w)} {w}\right)^{\beta}\,+\,\alpha\frac{w\left((f * h)^{-1}(w)\right)^{\prime}}{(f * h)^{-1}(w)}\left(\frac{(f * h)^{-1}(w)}{w}\right)^{\beta}\right] >\rho\right\} ; \end{array} $$

(iv) $\mathcal {M}_{\alpha,\lambda,0,k,\beta }(f * h)=\mathcal {M}_{\alpha,\lambda,k,\beta }(f * h)$

$$\begin{array}{*{20}l} & =\left\{ f,h\in\sum:(1-\alpha)\left(\frac{(f * h)(z)}{z}\right)^{\beta}+\alpha\frac{z(f * h)^{\prime}(z)}{(f * h)(z)}\left(\frac{(f * h)(z)}{z}\right)^{\beta}\in\mathcal{P}_{k}^{\lambda}\right. \\ & \text{and }\left. (1-\alpha)\left(\frac{(f * h)^{-1}(w)}{w}\right)^{\beta}+\alpha\frac{w\left((f * h)^{-1}(w)\right)^{\prime}}{(f * h)^{-1}(w)}\left(\frac{(f * h)^{-1}(w)}{w}\right)^{\beta}\in\mathcal{P}_{k}^{\lambda }\right\} ; \end{array} $$

(v) $\mathcal {M}_{\alpha,0,0,k,\beta }(f * h)=\mathcal {M}_{\alpha,k,\beta }(f * h)$

$$\begin{array}{*{20}l} & =\left\{ f,h\in\sum:(1-\alpha)\left(\frac{(f * h)(z)}{z}\right)^{\beta}+\alpha\frac{z(f * h)^{\prime}(z)}{(f * h)(z)}\left(\frac{(f * h)(z)}{z}\right)^{\beta}\in\mathcal{P}_{k}\right. \\ & \text{and }\left. (1-\alpha)\left(\frac{(f * h)^{-1}(w)}{w}\right)^{\beta}+\alpha\frac{w\left((f * h)^{-1}(w)\right)^{\prime}}{(f * h)^{-1}(w)}\left(\frac{(f * h)^{-1}(w)}{w}\right)^{\beta}\in\mathcal{P}_{k}\right\} ; \end{array} $$

(vi) $\mathcal {M}_{\alpha,0,0,2,\beta }(f * h)=\mathcal {M}_{\alpha,\beta }(f * h)$

$$\begin{array}{*{20}l} & =\left\{ f,h\in\sum:(1-\alpha)\left(\frac{(f * h)(z)}{z}\right)^{\beta}+\alpha\frac{z(f * h)^{\prime}(z)}{(f\ast h)(z)}\left(\frac{(f * h)(z)}{z}\right)^{\beta}\in\mathcal{P}\right. \\ & \text{and }\left. (1-\alpha)\left(\frac{(f * h)^{-1}(w)}{w}\right)^{\beta}+\alpha\frac{w((f * h)^{-1}(w))^{\prime}}{(f * h)^{-1}(w)}\left(\frac{(f * h)^{-1}(w)}{w}\right)^{\beta}\in\mathcal{P}\right\} ; \end{array} $$

(vii) $\mathcal {M}_{1,\lambda,\rho,k,\beta }(f * h)=\mathbb {F}_{\lambda,\rho,k,\beta }(f * h)$

$${\begin{aligned} &=\left\{ f,h\in\sum:\frac{z(f * h)^{\prime}(z)}{(f * h)(z)}\left(\frac{(f * h)(z)}{z}\right) ^{\beta}\in\mathcal{P}_{k}^{\lambda} (\rho)\right. \text{and}\\ &\quad \left. \frac{w((f * h)^{-1}(w))^{\prime}}{(f * h)^{-1}(w)}\left(\frac{(f * h)^{-1}(w)}{w}\right)^{\beta}\in \mathcal{P}_{k}^{\lambda}(\rho)\right\} ; \end{aligned}} $$

or

$$\begin{array}{*{20}l} & =\left\{ f\in\sum:\frac{e^{i\lambda}\left[ \frac{z(f * h)^{\prime} (z)}{(f * h)(z)}\left(\frac{(f * h)(z)}{z}\right)^{\beta}\right] -\rho\cos\lambda-i\sin\lambda}{\left(1-\rho\right) \cos\lambda} \in\mathcal{P}_{k}\right. \\ & \text{and\ }\left. \frac{e^{i\lambda}\left[ \frac{z(f * h)^{\prime} (z)}{(f * h)(z)}\left(\frac{(f * h)(z)}{z}\right)^{\beta}\right] -\rho\cos\lambda-i\sin\lambda}{\left(1-\rho\right) \cos\lambda} \in\mathcal{P}_{k}\right\} ; \end{array} $$

(viii) $\mathcal {M}_{1,0,\rho,2,\beta }(f * h)=\mathbb {F}_{\rho,\beta }(f * h)$

$$=\left\{ f,h\in\sum:\mathfrak{R}\left[ \frac{z(f * h)^{\prime}(z)}{(f * h)(z)}\left(\frac{(f * h)(z)}{z}\right) ^{\beta}\right] >\rho \right.\\ \left. and\ \mathfrak{R}\left[ \frac{w((f * h)^{-1}(w))^{\prime}}{(f * h)^{-1}(w)}\left(\frac{(f * h)^{-1}(w)}{w}\right)^{\beta}\right] >\rho\right\}. $$

In order to obtain our main results, we have to recall here the following lemma.

Lemma 1

([3] with p = 1). If $p(z)=1+\sum \limits _{n=1}^{\infty }c_{n}z^{n}\in \mathcal {P}_{k}^{\lambda }(\rho),$ then

$$ \left\vert c_{n}\right\vert \leq(1-\rho)\ k\ \cos\lambda. $$

(8)

The result is sharp. Equality is attained for the odd coefficients and even coefficients respectively for the functions:

$$p_{1}\left(z\right) =1+\left(1-\rho\right) \cos\lambda\ e^{-i\lambda} \left[ \left(\frac{k+2}{4}\right) \left(\frac{1-z}{1+z}\right) -\left(\frac{k-2}{4}\right) \left(\frac{1+z}{1-z}\right) -1\right], $$

$$p_{2}\left(z\right) =1+\left(1-\rho\right) \cos\lambda\ e^{-i\lambda} \left[ \left(\frac{k+2}{4}\right) \left(\frac{1-z^{2}}{1+z^{2}}\right) -\left(\frac{k-2}{4}\right) \left(\frac{1+z^{2}}{1-z^{2}}\right) -1\right]. $$

We note that for λ=0 in Lemma 1, we obtain the result obtained by Goswami et al. [19] [Lemma 2.1] for the class $\mathcal {P}_{k}(\rho).$

In this paper, we will obtain the coefficients bounds |a₂| and |a₃| for the class $\mathcal {M}_{\alpha,\lambda,\rho,k,\beta }(f * h)$, which defined in Definition 2.

Coefficient estimates for functions in the class $\mathcal {M}_{\alpha,\lambda,\rho,k,\beta }(f * h)$

Theorem 1

Let $f,h\in \sum,\ \alpha \in \mathbb {C} ^{\ast }\backslash \{-1,\frac {-1}{2}\},\ \beta \geq 0,\ 0\leq \rho <1,\ k\geq 2,\ \left \vert \lambda \right \vert <\frac {\pi }{2},$ f∗h given by (2) and h₂, h₃≠0. If f∗h belongs to $\mathcal {M}_{\alpha,\lambda,\rho,k,\beta }(f * h)$, then:

$$ \left\vert a_{2}\right\vert \leq\min\left\{ \sqrt[\ ]{\frac{2k(1-\rho)\cos\lambda}{\left\vert 2\alpha+\beta\right\vert \left(\beta+1\right) \left\vert h_{2}\right\vert^{2}}};\ \frac{k(1-\rho)\cos\lambda}{\left\vert \alpha+\beta\right\vert \left\vert h_{2}\right\vert }\right\} $$

(9)

and

$$ \left\vert a_{3}\right\vert \leq\frac{k(1-\rho)\cos\lambda}{\left\vert 2\alpha+\beta\right\vert \left\vert h_{3}\right\vert }+\frac{\left[ k(1-\rho)\cos\lambda\right]^{2}}{\left\vert \alpha+\beta\right\vert^{2}\left\vert h_{3}\right\vert }. $$

(10)

The result is sharp.

Proof 1 If $(f * h)\in \mathcal {M}_{\alpha,\lambda,\rho,k,\beta }(f * h)$, then from Definition 2, we have:

$$ (1-\alpha)\left(\frac{(f * h)(z)}{z}\right)^{\beta}+\alpha\frac{z(f * h)^{\prime}(z)}{(f * h)(z)}\left(\frac{(f * h)(z)}{z}\right)^{\beta }=p(z),\ p\in\mathcal{P}_{k}^{\lambda}(\rho) $$

(11)

and

$$ (1-\alpha)\left(\frac{(f * h)^{-1}(w)}{w}\right)^{\beta}\,+\,\alpha \frac{w\left((f * h)^{-1}(w)\right)^{\prime}}{(f * h)^{-1}(w)}\left(\frac{(f * h)^{-1}(w)}{w}\right)^{\beta}\,=\,q(w),\ q\in\mathcal{P}_{k}^{\lambda} (\rho), $$

(12)

where p and q have Taylor expansions as follows:

$$ p(z)=1+p_{1}z+p_{2}z^{2}+p_{3}z^{3}+....,z\in\mathbb{U}, $$

(13)

$$ q(w)=1+q_{1}w+q_{2}w^{2}+q_{3}w^{3}+....,w\in\mathbb{U}. $$

(14)

By comparing the coefficients in (11) with (13) and coefficients in (12) with (14), we obtain:

$$ p_{1}=\left(\beta+\alpha\right) \ a_{2}h_{2}, $$

(15)

$$ p_{2}=\left(\beta+2\alpha\right) \ a_{3}h_{3}+\frac{\left(\beta +2\alpha\right) \left(\beta-1\right) }{2}\ a_{2}^{2}h_{2}^{2}, $$

(16)

$$ q_{1}=-\left(\beta+\alpha\right) \ a_{2}h_{2} $$

(17)

and

$$ q_{2}=\left(\beta+2\alpha\right) \ \left(2a_{2}^{2}h_{2}^{2}-a_{3}h_{3} \right)+\frac{\left(\beta+2\alpha\right) \left(\beta-1\right) }{2}\ a_{2}^{2}h_{2}^{2}. $$

(18)

Since $p,q\in \mathcal {P}_{k}^{\lambda }(\rho)\ {and}$ by applying Lemma 1, we have:

$$ \left\vert p_{n}\right\vert \leq k(1-\rho)\cos\lambda\ (n\geq1) $$

(19)

and

$$ \left\vert q_{n}\right\vert \leq k(1-\rho)\cos\lambda\ (n\geq1). $$

(20)

From (16) and (18) and using inequalities (19) and (20), we obtain:

$$ \left\vert a_{2}\right\vert^{2}\leq\frac{1}{\left\vert 2\alpha+\beta \right\vert \left\vert \beta+1\right\vert }\frac{\left\vert p_{2}\right\vert +\left\vert q_{2}\right\vert }{\left\vert h_{2}\right\vert^{2}}\leq \frac{2k(1-\rho)\cos\lambda}{\left\vert 2\alpha+\beta\right\vert \left(\beta+1\right) \left\vert h_{2}\right\vert^{2}}\ . $$

(21)

Also, from (15) and (19), we obtain:

$$ \left\vert a_{2}\right\vert \leq\frac{k(1-\rho)\cos\lambda}{\left\vert \alpha+\beta\right\vert \left\vert h_{2}\right\vert }\ . $$

(22)

Subtracting (18) from (16), we have:

$$ p_{2}-q_{2}=2\left(2\alpha+\beta\right) \ \left(a_{3}h_{3}-a_{2}^{2}h_{2}^{2}\right). $$

(23)

Also, we have:

$$ p_{1}^{2}+q_{1}^{2}=2\left(\alpha+\beta\right)^{2}a_{2}^{2}h_{2}^{2}. $$

(24)

After using (23), (24), (19), and (20), and some easily calculations, we obtain:

$$ \left\vert a_{3}\right\vert \leq\frac{k(1-\rho)\cos\lambda}{\left\vert 2\alpha+\beta\right\vert \left\vert h_{3}\right\vert }+\frac{\left[ k(1-\rho)\cos\lambda\right]^{2}}{\left\vert \alpha+\beta\right\vert^{2}\left\vert h_{3}\right\vert }, $$

(25)

which completes the proof of Theorem 1. The result is sharp in view of the fact that assertion (8) of Lemma 1 is sharp.

Remark 1

For $h(z)=\frac {z}{1-z\ },\ \beta =\alpha =1,\ k=2,\$and λ=0 in Theorem 1, we obtain the result obtained by Srivastava et al. [9] [Theorem 2].

Putting $h(z)=\frac {z}{1-z\ }\$in Theorem 1, we obtain the following corollary.

Corollary 1

Let $\ f\in \sum,\ \alpha \in \mathbb {C}^{\ast }\backslash \left \{-1,\frac {-1}{2}\right \},\ \beta \geq 0,\ 0\leq \rho <1,\ k\geq 2\ and\ \left \vert \lambda \right \vert <\frac {\pi }{2}.\$If $f\in \mathcal {\digamma }_{\alpha,\lambda,\rho,k,\beta }(f)$, then:

$$\left\vert a_{2}\right\vert \leq\min\left\{ \sqrt[\ ]{\frac{2k(1-\rho)\cos\lambda}{\left\vert 2\alpha+\beta\right\vert \left(\beta+1\right)} };\ \frac{k(1-\rho)\cos\lambda}{\left\vert \alpha+\beta\right\vert }\right\} $$

and

$$\left\vert a_{3}\right\vert \leq\frac{k(1-\rho)\cos\lambda}{\left\vert 2\alpha+\beta\right\vert }+\frac{\left[ k(1-\rho)\cos\lambda\right]^{2} }{\left\vert \alpha+\beta\right\vert^{2}}. $$

The result is sharp.

Putting λ=0 in Theorem 1, we obtain the following corollary.

Corollary 2

Let $\ f,h\in \sum,\ \alpha \in \mathbb {C}^{\ast }\backslash \left \{-1,\frac {-1}{2}\right \},\ \beta \geq 0,\ 0\leq \rho <1,\ k\geq 2,$ f∗h given by (2) and h₂, h₃≠0. If $f * h\in \mathcal {F}_{\alpha,\rho,k,\beta }(f * h)$, then:

$$\left\vert a_{2}\right\vert \leq\min\left\{ \sqrt[\ ]{\frac{2k(1-\rho)}{\left\vert 2\alpha+\beta\right\vert \left(\beta+1\right) \left\vert h_{2}\right\vert^{2}}};\ \frac{k(1-\rho)}{\left\vert \alpha+\beta\right\vert \left\vert h_{2}\right\vert }\right\} $$

and

$$\left\vert a_{3}\right\vert \leq\frac{k(1-\rho)}{\left\vert 2\alpha +\beta\right\vert \left\vert h_{3}\right\vert }+\frac{\left[ k(1-\rho)\right]^{2}}{\left\vert \alpha+\beta\right\vert^{2}\left\vert h_{3}\right\vert }. $$

The result is sharp.

Putting λ=0 and k=2 in Theorem 1, we obtain the following corollary.

Corollary 3

Let $\ f,h\in \sum,\ \alpha \in \mathbb {C}^{\ast }\backslash \ \left \{-1,\frac {-1}{2}\right \},\ \beta \geq 0,\ 0\leq \rho <1,$ f∗h given by (2) and h₂, h₃≠0. If $f * h\in \mathcal {F}_{\alpha,\rho,\beta }(f * h)$, then:

$$\left\vert a_{2}\right\vert \leq\min\left\{ \sqrt[\ ]{\frac{4(1-\rho)}{\left\vert 2\alpha+\beta\right\vert \left(\beta+1\right) \left\vert h_{2}\right\vert^{2}}};\ \frac{2(1-\rho)}{\left\vert \alpha+\beta\right\vert \left\vert h_{2}\right\vert }\right\} $$

and

$$\left\vert a_{3}\right\vert \leq\frac{2(1-\rho)}{\left\vert 2\alpha +\beta\right\vert \left\vert h_{3}\right\vert }+\frac{\left[ 2(1-\rho)\right]^{2}}{\left\vert \alpha+\beta\right\vert^{2}\left\vert h_{3}\right\vert }. $$

The result is sharp.

Putting α=1 in Theorem 1, we obtain the following corollary.

Corollary 4

Let $\ f,h\in \sum,\ \beta \geq 0,\ 0\leq \rho <1,\ k\geq 2,\ \left \vert \lambda \right \vert <\frac {\pi }{2},$ f∗h given by (2) and h₂, h₃≠0. If $f * h\in \mathbb {F} _{\lambda,\rho,k,\beta }(f * h)$, then:

$$\left\vert a_{2}\right\vert \leq\min\left\{ \sqrt[\ ]{\frac{2k(1-\rho)\cos\lambda}{(2+\beta)\left(\beta+1\right) \left\vert h_{2}\right\vert^{2}}};\ \frac{k(1-\rho)\cos\lambda}{(1+\beta)h_{2}}\right\} $$

and

$$\left\vert a_{3}\right\vert \leq\frac{k(1-\rho)\cos\lambda}{(2+\beta)\left\vert h_{3}\right\vert }+\frac{\left[ k(1-\rho)\cos\lambda\right]^{2}}{(1+\beta)^{2}\left\vert h_{3}\right\vert }. $$

The result is sharp.

Putting α=1, k=2, and λ=0 in Theorem 1, we obtain the following corollary.

Corollary 5

Let $\ f,h\in \sum,\ \beta \geq 0,\ 0\leq \rho <1,$ f∗h given by (2) and h₂, h₃≠0. If $f * h\in \mathbb {F}_{\rho,\beta }(f * h)$, then:

$$\left\vert a_{2}\right\vert \leq\min\left\{ \sqrt[\ ]{\frac{4(1-\rho)}{(2+\beta)\left(\beta+1\right) \left\vert h_{2}\right\vert^{2}}} ;\ \frac{2(1-\rho)}{(1+\beta)h_{2}}\right\} $$

and

$$\left\vert a_{3}\right\vert \leq\frac{2(1-\rho)}{(2+\beta)\left\vert h_{3}\right\vert }+\frac{\left[ 2(1-\rho)\right] ^{2}}{(1+\beta)^{2}\left\vert h_{3}\right\vert }. $$

The result is sharp.

Putting ρ=0 in Theorem 1, we obtain the following corollary.

Corollary 6

Let $\ f,h\in \sum,\ \alpha \in \mathbb {C}^{\ast }\backslash \ \left \{-1,\frac {-1}{2}\right \},\ \beta \geq 0,\ \left \vert \lambda \right \vert <\frac {\pi }{2},\ k\geq 2,$ f∗h given by (2) and h₂, h₃≠0. If $f * h\in \mathcal {M}_{\alpha,\lambda,k,\beta }(f * h)$, then:

$$\left\vert a_{2}\right\vert \leq\min\left\{ \sqrt[\ ]{\frac{2k\cos\lambda }{\left\vert 2\alpha+\beta\right\vert \left(\beta+1\right) \left\vert h_{2}\right\vert^{2}}};\ \frac{k\cos\lambda}{\left\vert \alpha+\beta \right\vert \left\vert h_{2}\right\vert }\right\} $$

and

$$\left\vert a_{3}\right\vert \leq\frac{k\cos\lambda}{\left\vert 2\alpha +\beta\right\vert \left\vert h_{3}\right\vert }+\frac{\left[ k\cos \lambda\right]^{2}}{\left\vert \alpha+\beta\right\vert^{2}\left\vert h_{3}\right\vert }. $$

The result is sharp.

Putting ρ=λ=0 in Theorem 1, we obtain the following corollary.

Corollary 7

Let $\ f,h\in \sum,\ \alpha \in \mathbb {C}^{\ast }\backslash \left \{-1,\frac {-1}{2}\right \},\ \beta \geq 0,\ k\geq 2,$ f∗h given by (2) and h₂, h₃≠0. If $f * h\in \mathcal {M}_{\alpha,k,\beta }(f * h)$, then:

$$\left\vert a_{2}\right\vert \leq\min\left\{ \sqrt[\ ]{\frac{2k}{\left\vert 2\alpha+\beta\right\vert \left(\beta+1\right) \left\vert h_{2}\right\vert^{2}}};\ \frac{k}{\left\vert \alpha+\beta\right\vert \left\vert h_{2} \right\vert }\right\} $$

and

$$\left\vert a_{3}\right\vert \leq\frac{k}{\left\vert 2\alpha+\beta\right\vert \left\vert h_{3}\right\vert }+\frac{k^{2}}{\left\vert \alpha+\beta\right\vert^{2}\left\vert h_{3}\right\vert }. $$

The result is sharp.

Putting ρ=λ=0 and k=2 in Theorem 1, we obtain the following corollary.

Corollary 8

Let $\ f,h\in \sum,\ \alpha \in \mathbb {C} ^{\ast }\backslash \left \{-1,\frac {-1}{2}\right \},\ \beta \geq 0,$ f∗h given by (2) and h₂, h₃≠0. If $f * h\in \mathcal {M}_{\alpha,\beta }(f * h)$, then:

$$\left\vert a_{2}\right\vert \leq\min\left\{ \sqrt[\ ]{\frac{4}{\left\vert 2\alpha+\beta\right\vert \left(\beta+1\right) \left\vert h_{2}\right\vert^{2}}};\ \frac{2}{\left\vert \alpha+\beta\right\vert \left\vert h_{2} \right\vert }\right\} $$

and

$$\left\vert a_{3}\right\vert \leq\frac{2}{\left\vert 2\alpha+\beta\right\vert \left\vert h_{3}\right\vert} +\frac{4}{\left\vert \alpha+\beta\right\vert^{2}\left\vert h_{3}\right\vert }. $$

The result is sharp.

Putting λ=0, α=1 and $h(z)=\frac {z}{1-z}\$in Theorem 1, we obtain the following corollary.

Corollary 9

Let $\ f\in \sum,\ 0\leq \rho <1\$ and β≥0. If $f\in R_{\sum }(\rho,k,\beta)$, then:

$$\left\vert a_{2}\right\vert \leq\min\left\{ \sqrt[\ ]{\frac{2k(1-\rho)}{\left(2+\beta\right) \left(\beta+1\right) }};\ \frac{k(1-\rho)}{\left(1+\beta\right) }\right\} $$

and

$$\left\vert a_{3}\right\vert \leq\frac{k(1-\rho)}{\left(2+\beta\right) }+\frac{\left[ k(1-\rho)\right]^{2}}{\left(1+\beta\right)^{2}}. $$

The result is sharp.

Remark 2

The results in Corollary 9 correct the results obtained by Orhan et al. [11] [Theorem 2.11, with γ=1. ].