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

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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  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 ), 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  (see also Umarani and Aouf ).

(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  and studied by Pinchuk .

(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 , 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 .

The object of this paper is to introduce new subclass of Bazilevič functions  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 , 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 );

(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  and );

(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 );

(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 );

(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 );

(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 [, 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 ).

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

( 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.  [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 $$f,h\in \sum,\ \alpha \in \mathbb {C} ^{\ast }\backslash \{-1,\frac {-1}{2}\},\ \beta \geq 0,\ 0\leq \rho <1,\ k\geq 2,\ \left \vert \lambda \right \vert <\frac {\pi }{2},$$ fh given by (2) and h2, h3≠0. If fh 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.  [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 $$\ f,h\in \sum,\ \alpha \in \mathbb {C}^{\ast }\backslash \left \{-1,\frac {-1}{2}\right \},\ \beta \geq 0,\ 0\leq \rho <1,\ k\geq 2,$$ fh 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 $$\ f,h\in \sum,\ \alpha \in \mathbb {C}^{\ast }\backslash \ \left \{-1,\frac {-1}{2}\right \},\ \beta \geq 0,\ 0\leq \rho <1,$$ fh 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 $$\ f,h\in \sum,\ \beta \geq 0,\ 0\leq \rho <1,\ k\geq 2,\ \left \vert \lambda \right \vert <\frac {\pi }{2},$$ fh 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 $$\ f,h\in \sum,\ \beta \geq 0,\ 0\leq \rho <1,$$ fh 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 $$\ f,h\in \sum,\ \alpha \in \mathbb {C}^{\ast }\backslash \ \left \{-1,\frac {-1}{2}\right \},\ \beta \geq 0,\ \left \vert \lambda \right \vert <\frac {\pi }{2},\ k\geq 2,$$ fh 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 $$\ f,h\in \sum,\ \alpha \in \mathbb {C}^{\ast }\backslash \left \{-1,\frac {-1}{2}\right \},\ \beta \geq 0,\ k\geq 2,$$ fh 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 $$\ f,h\in \sum,\ \alpha \in \mathbb {C} ^{\ast }\backslash \left \{-1,\frac {-1}{2}\right \},\ \beta \geq 0,$$ fh 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.  [Theorem 2.11, with γ=1. ].

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Acknowledgements

The authors are grateful to the referees for their valuable suggestions.

Funding

Higher Institute for Engineering and Technology, New Damietta, Egypt

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All authors jointly worked on the results, and they read and approved the final manuscript.

Correspondence to Mohamed K. Aouf.

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