The main results in this section is to define two bounded subclasses of the class Σ, then some estimations of the first two Maclaurin coefficients of functions belonging to those subclasses were calculated.
Definition 1
For \(0\leq \lambda <1,b\in \mathbb {C}^{\ast }\) and \(M>\frac {1}{2},\) let \(\mathcal {S}_{\Sigma,q}^{n}(\lambda,b,M) \) be the subclass of Σ consisting of functions of the form (1) and satisfying the following condition
$$ \left\vert \frac{b-1+\frac{D_{q}^{n+1}f(z)}{\lambda D_{q}^{n+1}f(z)+(1-\lambda)D_{q}^{n}f(z)}}{b}-M\right\vert <M,(z\in U) $$
(9)
and
$$ \left\vert \frac{b-1+\frac{D_{q}^{n+1}g(w)}{\lambda D_{q}^{n+1}g(w)+(1-\lambda)D_{q}^{n}g(w)}}{b}-M\right\vert <M,(w\in U) $$
(10)
where z,w∈U, and g=f−1∈Σ is given by (3). Also, let \(\mathcal {C}_{\Sigma,q}^{n}(\lambda,b,M)\)be the subclass of Σ consisting of functions of the form (1) and satisfying the following condition
$$ \left\vert \frac{b-1+\frac{D_{q}^{n+2}f(z)}{\lambda D_{q}^{n+2}f(z)+(1-\lambda)D_{q}^{n+1}f(z)}}{b}-M\right\vert <M,(z\in U) $$
(11)
and
$$ \left\vert \frac{b-1+\frac{D_{q}^{n+2}g(w)}{\lambda D_{q}^{n+2}g(w)+(1-\lambda)D_{q}^{n+1}g(w)}}{b}-M\right\vert <M,(w\in U) $$
(12)
where z,w∈U, and g=f−1∈Σ is given by (3).
It is clear that
$$f(z)\in\mathcal{C}_{\Sigma,q}^{n}(\lambda,b,M)\Longleftrightarrow z\partial_{q}f(z)\in\mathcal{S}_{\Sigma,q}^{n}(\lambda,b,M) $$
Lemma 1
Let \(m=1-\frac {1}{M}\left (M>\frac {1}{2}\right)\), f defined by (1) and g=f−1, then we have
$$ f(z)\in\mathcal{S}_{\Sigma,q}^{n}(\lambda,b,M)\Longleftrightarrow\left\{ \begin{array} [c]{c} 1+\frac{1}{b}\left(\frac{D_{q}^{n+1}f(z)}{\lambda D_{q}^{n+1}f(z)+(1-\lambda)D_{q}^{n}f(z)}-1\right) \prec\frac{1+z}{1-mz}\\ 1+\frac{1}{b}\left(\frac{D_{q}^{n+1}g(w)}{\lambda D_{q}^{n+1}g(w)+(1-\lambda)D_{q}^{n}g(w)}-1\right) \prec\frac{1+w}{1-mw} \end{array}\right., $$
(13)
and also,
$$ f(z)\in\mathcal{C}_{\Sigma,q}^{n}(\lambda,b,M)\Longleftrightarrow\left\{ \begin{array} [c]{c} 1+\frac{1}{b}\left(\frac{D_{q}^{n+2}f(z)}{\lambda D_{q}^{n+2}f(z)+(1-\lambda)D_{q}^{n+1}f(z)}-1\right) \prec\frac{1+z}{1-mz}\\ 1+\frac{1}{b}\left(\frac{D_{q}^{n+2}g(w)}{\lambda D_{q}^{n+2}g(w)+(1-\lambda)D_{q}^{n+1}g(w)}-1\right) \prec\frac{1+w}{1-mw} \end{array}\right., $$
(14)
where
$$(0\leq\lambda<1,b\in \mathbb{C}^{\ast}\text{{{and}}} z,w\in U) $$
Lemma 2
(see [22]) If \(h\in \mathcal {P}\), then |cn|≤2 for each \(n\in \mathbb {N}\), where \(\mathcal {P}\) is the family of all functions h which is analytic in U for which Re{h(z)}>0, where h(z)=1+c1z+c2z2+... for z∈U.
Remark 1
In Definitions 1, 2 and for special choices of the parameters λ,b,M, also, taking q→1−, then we can obtain the following subclasses:
$$\begin{array}{@{}rcl@{}} \mathcal{S}_{\Sigma}^{0}(\lambda,1-\beta,\infty) &=& \mathcal{M}_{\Sigma}(\beta,\lambda) \\ &=& \left\{f\in \Sigma:\begin{array}{c} \text{Re}\left(\frac{zf^{\prime}(z)}{(1-\lambda)f(z)+\lambda zf^{\prime}(z)}\right)> \beta, \\ \text{Re}\left(\frac{wg^{\prime}(w)}{(1-\lambda)g(w)+\lambda wg^{\prime}(w)}\right)> \beta \end{array}, 0<\beta \leq 1, z,w\in U \right\} \\ \mathcal{S}_{\Sigma}^{0}(\lambda,b, \infty) &=& \mathcal{S}_{\Sigma}(1,\lambda) \\ &=& \left\{f\in \Sigma:\begin{array}{c} \left|\text{arg}\left\{\frac{zf^{\prime}(z)}{(1-\lambda)f(z)+\lambda zf^{\prime}(z)}\right\}\right|<\frac{\pi}{2}, \\ \left|\text{arg}\left\{\frac{wg^{\prime}(w)}{(1-\lambda)g(w)+\lambda wg^{\prime}(w)}\right\}\right|<\frac{\pi}{2} \end{array}, z,w\in U \right\}, \end{array} $$
which were introduced by Murugusundaramoorthy et al. [23].
$$\begin{array}{@{}rcl@{}} \mathcal{S}_{\Sigma}^{0}(\gamma, \tau, \infty) &=& \mathcal{H}_{\Sigma}(\tau,0,1, \gamma,0) \\ &=& \left\{f\in \Sigma:\begin{array}{c} \text{Re}\left(1+\frac{1}{\tau}\left(\frac{zf^{\prime}(z)}{(1-\gamma)f(z)+\gamma zf^{\prime}(z)}-1 \right)\right)> 0, \\ \text{Re}\left(1+\frac{1}{\tau}\left(\frac{zg^{\prime}(w)}{(1-\gamma)g(w)+\gamma wg^{\prime}(w)}-1 \right)\right)> 0 \end{array}, z,w\in U \right\}, \end{array} $$
which was introduced by Srivastava et al. [24].
$$\begin{array}{@{}rcl@{}} \mathcal{S}_{\Sigma}^{0}(0, 1-\beta, \infty) &=& \mathcal{S}_{\Sigma}(\beta) \\ &=& \left\{f\in \Sigma:\begin{array}{c} \text{Re}\left(\frac{zf^{\prime}(z)}{f(z)}\right)> \beta, \\ \text{Re}\left(\frac{wg^{\prime}(w)}{g(w)}\right)> \beta \end{array}, 0\leq\beta<1, z,w\in U \right\},\\ \mathcal{C}_{\Sigma}^{0}(0, 1-\beta, \infty) &=& \mathcal{K}_{\Sigma}(\beta) \\ &=& \left\{f\in \Sigma:\begin{array}{c} \text{Re}\left(1+\frac{zf^{\prime\prime}(z)}{f^{\prime}(z)}\right)> \beta, \\ \text{Re}\left(1+\frac{wg^{\prime\prime}(w)}{g^{\prime}(w)}\right)> \beta \end{array}, 0\leq\beta<1, z,w\in U \right\}, \end{array} $$
which are the classes of bi-starlike and bi-convex functions introduced by Brannan and Taha [8].
Theorem 1
Let f given by (1) be in the subclass \(\mathcal {S}_{\Sigma,q}^{n}(\lambda,b,M)\). Then
$$\left\vert a_{2}\right\vert \leq \sqrt{\frac{\left\vert b\right\vert (m+1)} { (1-\lambda) \left\vert \left[ 3\right]_{q} ^{n}(\left[ 3\right]_{q}-1)-\left[ 2\right]_{q}^{2n}\left(\left[ 2\right]_{q}-1\right) \left(1-\lambda+\lambda\left[ 2\right]_{q} +\frac{(1-\lambda)(m-1)(\left[ 2\right]_{q}-1)}{b(m+1)}\right) \right\vert}}. $$
and
$$\left\vert a_{3}\right\vert \leq\frac{\left\vert b\right\vert (m+1)} {(1-\lambda)}\left\{ \frac{1}{\left[ 3\right]_{q}^{n}(\left[ 3\right]_{q}-1)}+\frac{\left\vert b\right\vert (m+1)}{\left[ 2\right]_{q}^{2n}(1-\lambda)(\left[ 2\right]_{q}-1)^{2}}\right\}, $$
where
$$0\leq\lambda<1,b\in \mathbb{C}^{\ast},z\in U\text{{{and}}}m=1-\frac{1}{M}(M>\frac{1}{2}), $$
Proof
Let \(f\in \mathcal {S}_{\Sigma,q}^{n}(\lambda,b,M)\) and g=f−1. Then, it satisfy the conditions (13). By the definition, there exist two analytic functions u,v:U→U with u(0)=v(0)=0 and |u(z)|<1,|v(w)|<1 for all z,w∈U satisfying
$$ 1+\frac{1}{b}\left(\frac{D_{q}^{n+1}f(z)}{\lambda D_{q}^{n+1}f(z)+(1-\lambda)D_{q}^{n}f(z)}-1\right) =\frac{1+u(z)}{1-mu(z)}, $$
(15)
and
$$ 1+\frac{1}{b}\left(\frac{D_{q}^{n+1}g(w)}{\lambda D_{q}^{n+1}g(w)+(1-\lambda)D_{q}^{n}g(w)}-1\right) =\frac{1+v(w)}{1-mv(w)}, $$
(16)
Now, define the two functions p(z) and q(z) by
$$ p(z):=\frac{1+u(z)}{1-u(z)}=1+p_{1}z+p_{2}z^{2}+..., $$
$$p(z):=\frac{1+v(z)}{1-v(z)}=1+q_{1}z+q_{2}z^{2}+... $$
It is equivalent to
$$ u(z):=\frac{1-p(z)}{1+p(z)}=\frac{1}{2}\left(p_{1}z+\left(p_{2} -\frac{p_{1}^{2}}{2}\right) z^{2}+...\right), $$
(17)
$$ v(z):=\frac{1-q(z)}{1+q(z)}=\frac{1}{2}\left(q_{1}z+\left(q_{2} -\frac{q_{1}^{2}}{2}\right) z^{2}+...\right). $$
(18)
Then p(z) and q(z) are analytic in U with p(0)=1=q(0). In view of Janowski [4], Since u,v:U→U, the functions p(z),q(z)∈P(M) and have a positive real part in U where P(M) is the class of all function ψ(z)=1+δ1z+δ2z2+... which are analytic in U and satisfy the condition
$$| \psi(z)-\rho |< \rho,(\rho \geq 1, z\in U). $$
Therefore, in view of the Lemma 1, we have
$$ \left\vert p_{i}\right\vert \leq2\text{and}\left\vert q_{i}\right\vert \leq2(n\in \mathbb{N}) $$
(19)
By substituting from (7), (8), (17), and (18) into (15)and (16), we obtain
$$ {\begin{aligned} 1+\left[ 2\right]_{q}^{n}(1-\lambda)(\left[ 2\right]_{q}-1)a_{2} z+(1-\lambda) \left(\left[ 3\right]_{q}^{n}(\left[ 3\right]_{q} -1)a_{3}-\left[ 2\right]_{q}^{2n}(\left[ 2\right]_{q}-1).\right. \\ \left. (1-\lambda+\lambda\left[ 2\right]_{q})a_{2}^{2}\right) z^{2}+...=1+\frac{b(m+1)p_{1}}{2}z+\frac{b(m+1)}{2}\left(p_{2}+\frac{p_{1}^{2}(m-1)}{2}\right) z^{2}+... \end{aligned}} $$
(20)
$$ {\begin{aligned} 1-\left[ 2\right]_{q}^{n}(1-\lambda)(\left[ 2\right]_{q}-1)a_{2} w+(1-\lambda) \left(\left[ 3\right]_{q}^{n}(\left[ 3\right]_{q}-1)(2a_{2}^{2}-a_{3})+\left[ 2\right]_{q}^{2n}(\left[ 2\right]_{q}-1).\right. \\ \left. (1-\lambda+\lambda\left[ 2\right]_{q})a_{2}^{2}\right) w^{2}+...=1+\frac{b(m+1)q_{1}}{2}w+\frac{b(m+1)}{2}\left(q_{2}+\frac{q_{1}^{2}(m-1)}{2}\right) w^{2}+... \end{aligned}} $$
(21)
which yields the following relations
$$ \left[ 2\right]_{q}^{n}(1-\lambda)(\left[ 2\right]_{q}-1)a_{2}=\frac{b}{2}(m+1)p_{1} $$
(22)
$$ \begin{aligned} \left[ 3\right]_{q}^{n}(1-\lambda)(\left[ 3\right]_{q}-1)a_{3}- \left[2\right]_{q}^{2n}(1-\lambda) (\left[ 2\right]_{q}-1)(1-\lambda+\lambda\left[ 2\right]_{q})a_{2}^{2}\\ =\frac{b}{2}(m+1)\left(p_{2}+\frac{p_{1}^{2}}{2}(m-1)\right) \end{aligned} $$
(23)
$$ -\left[ 2\right]_{q}^{n}(1-\lambda)(\left[ 2\right]_{q}-1)a_{2}=\frac{b}{2}(m+1)q_{1} $$
(24)
$$ \begin{aligned} \left[ 3\right]_{q}^{n}(1-\lambda)(\left[ 3\right]_{q}-1)(2a_{2}^{2}-a_{3})-\left[ 2\right]_{q}^{2n}(1-\lambda)(\left[ 2\right]_{q}-1)(1-\lambda+\lambda\left[ 2\right]_{q})a_{2}^{2}\\ =\frac{b}{2}(m+1)\left(q_{2}+\frac{q_{1}^{2}}{2}(m-1)\right) \end{aligned} $$
(25)
From (22) and (24), we obtain
and
$$ 2\left[ 2\right]_{q}^{2n}(1-\lambda)^{2}(\left[ 2\right]_{q}-1)^{2} a_{2}^{2}=\frac{b^{2}}{4}(m+1)^{2}\left(p_{1}^{2}+q_{1}^{2}\right). $$
(27)
By adding (23) to (25) then use (27), we obtain
$$ a_{2}^{2}=\frac{b(m+1)(p_{2}+q_{2})}{4(1-\lambda)\left(\left[ 3\right]_{q}^{n}(\left[ 3\right]_{q}-1)-\left[ 2\right]_{q}^{2n}\left(\left[ 2\right]_{q}-1\right) (1-\lambda+\lambda\left[ 2\right]_{q} +\frac{(1-\lambda)(m-1)(\left[ 2\right]_{q}-1)}{b(m+1)})\right) }, $$
(28)
applying Lemma 2 to the coefficients p2 and q2, we conclude
$$\left\vert a_{2}\right\vert \leq \sqrt{\frac{\left\vert b\right\vert (m+1)} {(1-\lambda) \left\vert \left[ 3\right]_{q}^{n}(\left[ 3\right]_{q}-1)-\left[ 2\right]_{q}^{2n}\left(\left[ 2\right]_{q}-1\right) \left(1-\lambda+\lambda\left[ 2\right]_{q} +\frac{(1-\lambda)(m-1)(\left[ 2\right]_{q}-1)}{b(m+1)}\right) \right\vert}}. $$
By subtracting (25) from (23), we have
$$ 2\left[ 3\right]_{q}^{n}(1-\lambda)(\left[ 3\right]_{q}-1)\left(a_{3} -a_{2}^{2}\right)=\frac{b}{2}(m+1)\left(\left(p_{2}-q_{2}\right) +\frac{(m-1)}{2}\left(p_{1}^{2}-q_{1}^{2}\right) \right), $$
(29)
by substituting from (26) and (27) into (29), we conclude
$$ a_{3}=\frac{b(m+1)\left(p_{2}-q_{2}\right) }{4\left[ 3\right]_{q}^{n}(1-\lambda)(\left[ 3\right]_{q}-1)}+\frac{b^{2}(m+1)^{2}\left(p_{1}^{2}+q_{1}^{2}\right) }{8\left[ 2\right]_{q}^{2n}(1-\lambda)^{2}(\left[ 2\right]_{q}-1)^{2}}. $$
(30)
Finally, by applying Lemma 2 to the coefficients p1,p2,q1 and q2, we conclude
$$\left\vert a_{3}\right\vert \leq\frac{\left\vert b\right\vert (m+1)}{\left[ 3\right]_{q}^{n}(1-\lambda)(\left[ 3\right]_{q}-1)}+\frac{\left\vert b\right\vert^{2}(m+1)^{2}}{\left[ 2\right]_{q}^{2n}(1-\lambda)^{2}(\left[2\right]_{q}-1)^{2}} $$
The proof is completed. □
For n=0,b=1−β,m=1, and q→1−, we obtain the bounds corresponding to the class MΣ(β,λ) given by Murugusundaramoorthy et al. [23].
Corollary 1
Let f given by (1) be a function in the class MΣ(β,λ), then
$$|a_{2}|\leq \frac{\sqrt{2(1-\beta)}}{(1-\lambda)} $$
and
$$|a_{3}|\leq \frac{4(1-\beta)^{2}}{(1-\lambda)^{2}}+\frac{1-\beta}{1-\lambda} $$
Additionally, put λ=0, we obtain bounds of the class of bi-starlike function of order β donated by \(\mathcal {S}_{\Sigma }(\beta)\).
Corollary 2
[8] Let f given by (1) be in the class \(\mathcal {S}_{\Sigma }(\beta)\), then
$$|a_{2}|\leq \sqrt{2(1-\beta)} $$
and
$$|a_{3}|\leq (1-\beta)(5-4\beta) $$
Theorem 2
Let f given by (1) be in the subclass \(\mathcal {C}_{\Sigma,q}^{n}(\lambda,b,M)\). Then
$$\left\vert a_{2}\right\vert \leq \sqrt{\frac{\left\vert b\right\vert (m+1)} {(1-\lambda) \left\vert \left[ 3\right]_{q}^{n+1}(\left[ 3\right]_{q}-1)-\left[ 2\right]_{q}^{2n+2}\left(\left[ 2\right]_{q}-1\right) \left(1-\lambda+\lambda\lbrack2\right)_{q} +\frac{(m-1)(1-\lambda)(\left[ 2\right]_{q}-1)}{b(m+1)}\right\vert}} $$
and
$$\left\vert a_{3}\right\vert \leq\frac{\left\vert b\right\vert (m+1)} {(1-\lambda)}\left\{ \frac{1}{\left[ 3\right]_{q}^{n+1}(\left[ 3\right]_{q}-1)}+\frac{\left\vert b\right\vert (m+1)}{\left[ 2\right]_{q}^{2n+2}(1-\lambda)(\left[ 2\right]_{q}-1)^{2}}\right\}, $$
where
$$0\leq\lambda<1,b\in \mathbb{C}^{\ast},z\in U~\text{and}~m=1-\frac{1}{M}\left(M>\frac{1}{2}\right), $$
Proof
Let \(f\in \mathcal {C}_{\Sigma,q}^{n}(\lambda,b,M) \) and g=f−1. Then, it satisfy the conditions (14). By the definition, there exist two analytic functions u,v:U→U with u(0)=v(0)=0 and |u(z)|<1,|v(w)|<1 for all z,w∈U satisfying
$$ 1+\frac{1}{b}\left(\frac{D_{q}^{n+2}f(z)}{\lambda D_{q}^{n+2}f(z)+(1-\lambda)D_{q}^{n+1}f(z)}-1\right) =\frac{1+u(z)}{1-mu(z)}, $$
(31)
and
$$ 1+\frac{1}{b}\left(\frac{D_{q}^{n+2}g(w)}{\lambda D_{q}^{n+2}g(w)+(1-\lambda)D_{q}^{n+1}g(w)}-1\right) =\frac{1+v(w)}{1-mv(w)}. $$
(32)
Now, define the two functions r(z) and s(z) by
$$r(z):=\frac{1+u(z)}{1-u(z)}=1+r_{1}z+r_{2}z^{2}+..., $$
$$s(z):=\frac{1+v(z)}{1-v(z)}=1+s_{1}z+s_{2}z^{2}+... $$
It is equivalent to
$$ u(z):=\frac{1-r(z)}{1+r(z)}=\frac{1}{2}\left(r_{1}z+\left(r_{2} -\frac{r_{1}^{2}}{2}\right) z^{2}+...\right), $$
(33)
$$ v(z):=\frac{1-s(z)}{1+s(z)}=\frac{1}{2}\left(s_{1}z+\left(s_{2} -\frac{s_{1}^{2}}{2}\right) z^{2}+...\right). $$
(34)
Then r(z) and s(z) are analytic in U with p(0)=1=q(0). Since u,v:U→U, the functions r(z) and s(z) have a positive real part in U. Therefore, in view of the Lemma 2, we have
$$ \left\vert r_{i}\right\vert \leq2\text{and}\left\vert s_{i}\right\vert \leq2(n\in \mathbb{N}). $$
(35)
By following the same steps in proving Theorem 1, we can complete the proof of this theorem. □
For n=0,b=1−β,m=1,λ=0 and q→1−, we obtain the bounds corresponding to the class \(\mathcal {K}_{\Sigma }(\beta)\) given by Brannan and Taha [8].
Corollary 3
Let f given by (1) be in the class \(\mathcal {K}_{\Sigma }(\beta)\), then
$$|a_{2}|\leq \sqrt{1-\beta} $$
and
$$|a_{3}|\leq \frac{(4-3\beta)(1-\beta)}{3} $$