On a class of spiral-like functions with respect to a boundary point related to subordination

Abstract

For $\mu \in \mathbb{C}$, φ a starlike univalent function, the class of functions f that are spiral-like with respect to a boundary point satisfying the subordination

$$\frac{2}{\mu }\frac{z{f}^{\prime }(z)}{f(z)}+\frac{1+z}{1-z}\prec \phi (z),z\in \mathbb{D},$$

is investigated. The integral representation, growth and distortion theorem are proved by relating these functions with Ma and Minda starlike functions. Some earlier results are shown to be a special case of the results obtained.

MSC:30C80, 30C45.

Dedication

Dedicated to Professor Hari M Srivastava

1 Introduction and motivation

Let $\mathbb{D}=\{z:|z|<1\}$ be an open unit disk of the complex plane ℂ and let $\mathcal{A}$ be a class of analytic functions f normalized by $f(0)=0$ and $f^{\prime }(0)=1$. Let $w_0$ be an interior or a boundary point of a set $\mathcal{D}$ in ℂ. The set $\mathcal{D}$ is starlike with respect to $w_0$ if the line segment joining $w_0$ to every other point in $\mathcal{D}$ lies in the interior of $\mathcal{D}$. If a function $f\in \mathcal{A}$ maps $\mathbb{D}$ onto a starlike domain with respect to origin, then f is a starlike function. The class of starlike functions with respect to origin is denoted by ${\mathcal{S}}^{\ast }$. Analytically,

$${\mathcal{S}}^{\ast }:=\{f\in \mathcal{A}:Re\frac{z{f}^{\prime }(z)}{f(z)}>0\}.$$

Robertson [1] took a leap forward with the characterization of the class ${\mathcal{S}}^{\ast }$ and defined the class ${\mathcal{S}}_b^{\ast }$ of starlike functions with respect to a boundary point. Geometrically, it is the characterization of a function $f\in {\mathcal{S}}_b=\{f(z)=1+d_{1}z+d_2{z}^2+\cdots |f\text{ univalent}\}$ such that $f(\mathbb{D})$ is starlike with respect to the boundary point $f(1):={lim}_{r\to 1^{-}}f(r)=0$ and lies in a half-plane. The analytic description given by Robertson was

$${\mathcal{S}}_b^{\ast }:=\{f\in {\mathcal{S}}_b:Re(2\frac{z{f}^{\prime }(z)}{f(z)}+\frac{1+z}{1-z})>0\}.$$

This was partially proved in [1]. It was only in 1984 that the characterization was validated by Lyzzaik [2]. Todorov [3] associated this class with a functional $f(z)/(1-z)$ and obtained a structured formula and coefficient estimates in the year 1986. Later, Silverman and Silvia [4] gave a full description of the class of univalent functions on $\mathbb{D}$, the image of which is star-shaped, with respect to a boundary point. Since then, this class of starlike functions with respect to a boundary point has gained notable interest among geometric function theorist and also other researchers. Among them, Abdullah et al. [5] studied the properties of functions in this class. The distortion results for starlike functions with respect to a boundary point were obtained in [6, 7]. The dynamical characterizations of functions starlike with respect to a boundary point can be found in [8]. In the year 2001, Lecko [9] gave another representation of starlike functions with respect to a boundary point. Also, Lecko and Lyzzaik obtained different characterizations of this class in [10].

Following the studies on the class of starlike functions, many authors extensively studied the class of spiral-like functions. For recent work on the class of spiral-like functions, see [11]. Later, there was interest towards the class of spiral-like functions with respect to a boundary point. See [12–15]. Aharonov et al. [16] gave a comprehensive definition for spiral-shaped domains with respect to a boundary point.

Definition 1.1 A simply connected domain $\mathrm{\Omega }\subset \mathbb{C}$, $0\in \partial \mathrm{\Omega }$, is called a spiral-shaped domain with respect to a boundary point if there is a number $\mu \in \mathbb{C}$ with $Re\mu >0$ such that, for any point $\omega \in \mathrm{\Omega }$, the curve $e^{-t\mu }\omega$, $t\ge 0$, is contained in Ω.

It was also showed in [16] (see also [17]) that each spiral-like function with respect to a boundary point is a complex power of starlike function with respect to a boundary point. In particular, if $\mu \in \mathbb{R}$ in Definition 1.1, then Ω is called a star-shaped domain with respect to a boundary point. The following was proved in the same.

Theorem 1.1 Let f be an analytic function with $f(0)=1$, $f(1)=0$, and let it be a spiral-like function with respect to a boundary point. Then there exists a number $\mu \in \mathrm{\Omega }:=\{\lambda \in \mathbb{C}:|\lambda -1|\le 1,\lambda \ne 0\}$ such that

$$Re(\frac{2}{\mu }\frac{z{f}^{\prime }(z)}{f(z)}+\frac{1+z}{1-z})>0.$$

(1.1)

Conversely, if f is a univalent function with $f(0)=1$ and $f(1)=0$ satisfies (1.1) for some $\mu \in \mathrm{\Omega }$, then f is a spiral-like function with respect to a boundary point.

Elin [18] then considered the class of spiral-like functions of order β ($0<\beta \le 1$) with respect to a boundary point and obtained interesting results including the distortion and covering theorems.

On the other hand, Ma and Minda [19] gave a unified presentation of the class starlike using the method of subordination. For two functions h and g in $\mathcal{A}$, the function h is subordinate to g, written

$$h(z)\prec g(z),z\in \mathbb{D},$$

if there exists a function $w\in \mathcal{A}$, with $w(0)=0$ and $|w(z)|<1$, such that $h(z)=g(w(z))$. In particular, if the function g is univalent in $\mathbb{D}$, then $h(z)\prec g(z)$ is equivalent to $h(0)=g(0)$ and $h(\mathbb{D})\subset g(\mathbb{D})$. A function $h\in \mathcal{A}$ is starlike if $z{h}^{\prime }(z)/h(z)$ is subordinated to $(1+z)/(1-z)$. Ma and Minda [19] introduced the class

$${\mathcal{S}}^{\ast }(\phi )=\{h\in \mathcal{A}:\frac{z{h}^{\prime }(z)}{h(z)}\prec \phi (z)\},$$

where φ is an analytic function with a positive real part in $\mathbb{D}$, $\phi (\mathbb{D})$ is symmetric with respect to the real axis and starlike with respect to $\phi (0)=1$ and ${\phi }^{\prime }(0)>0$. A function $f\in {\mathcal{S}}^{\ast }(\phi )$ is called Ma and Minda starlike (with respect to φ). The class ${\mathcal{S}}^{\ast }(\beta )$ consisting of starlike functions of order β, $0\le \beta <1$ and the class ${\mathcal{S}}^{\ast }(A,B)$ of Janowski starlike functions are special cases of ${\mathcal{S}}^{\ast }(\phi )$ when $\phi (z):=(1+(1-2\beta )z)/(1-z)$ and $\phi (z):=(1+Az)/(1+Bz)$ for $-1\le B<A\le 1$, respectively.

In the same direction and motivated mainly by [18] and [19], we consider the following class.

Definition 1.2 Let $f\in {\mathcal{S}}_b$, $f(0)=1$ and $\mu \in \mathrm{\Omega }:=\{\lambda \in \mathbb{C}:|\lambda -1|\le 1,\lambda \ne 0\}$. Also, let φ be an analytic function with a positive real part $\mathbb{D}$, let $\phi (\mathbb{D})$ be symmetric with respect to the real axis and starlike with respect to $\phi (0)=1$ and ${\phi }^{\prime }(0)>0$. The function $f\in {\mathcal{S}}_b^{\ast }(\mu ,\phi )$ if the subordination

$$\frac{2}{\mu }\frac{z{f}^{\prime }(z)}{f(z)}+\frac{1+z}{1-z}\prec \phi (z),z\in \mathbb{D},$$

(1.2)

holds.

For $\phi (z)=(1+Az)/(1+Bz)$ ($-1\le B<A\le 1$), denote the class ${\mathcal{S}}_b^{\ast }(\mu ,\phi )$ by ${\mathcal{S}}_b^{\ast }(\mu ,A,B)$. For $0\le \beta <1$, $A=1-2\beta$ and $B=-1$, denote ${\mathcal{S}}_b^{\ast }(\mu ,A,B)$ by ${\mathcal{S}}_b^{\ast }(\mu ,\beta )$.

The class ${\mathcal{S}}_b^{\ast }(\mu ,\phi )$ defined by subordination is investigated to obtain representation, estimates for f and $f^{\prime }$ and subordination conditions. We obtained some interesting result in a wider context and our approach is mainly based on [19].

2 Representation for the class ${\mathcal{S}}_b^{\ast }(\mu ,\phi )$

The following result provides an integral representation of functions belonging to the class ${\mathcal{S}}_b^{\ast }(\mu ,\phi )$.

Theorem 2.1 The function $f\in {\mathcal{S}}_b^{\ast }(\mu ,\phi )$ if and only if there exists p satisfying $p\prec \phi$ such that

$$f(z)={(1-z)}^{\mu }exp\left(\frac{\mu }{2}{\int }_0^z\frac{p(\zeta )-1}{\zeta }d\zeta \right).$$

Proof Let $f\in {\mathcal{S}}_b^{\ast }(\mu ,\phi )$. Then define $p:\mathbb{D}\to \mathbb{C}$ by

$$p(z)=\frac{2}{\mu }\frac{z{f}^{\prime }(z)}{f(z)}+\frac{1+z}{1-z}.$$

Then $f\in {\mathcal{S}}_b^{\ast }(\mu ,\phi )$ implies that $p\prec \phi$. Rewriting the above equation as

$$\frac{2}{\mu }\frac{f^{\prime }(z)}{f(z)}+\frac{2}{1-z}=\frac{p(z)-1}{z}$$

and integrating from 0 to z, it follows that

$$log\left(\frac{f{(z)}^{\frac{2}{\mu }}}{{(1-z)}^2}\right)={\int }_0^z\frac{p(\zeta )-1}{\zeta }d\zeta .$$

An exponentiation gives

$$f{(z)}^{\frac{2}{\mu }}={(1-z)}^{2}exp\left({\int }_0^z\frac{p(\zeta )-1}{\zeta }d\zeta \right).$$

The desired result follows from this. The converse follows easily. □

3 Estimates for f and $f^{\prime }$in the class ${\mathcal{S}}_b^{\ast }(\mu ,\phi )$

Theorem 3.1 Let $h_{\phi }$ be an analytic function with $h_{\phi }(0)=0$, $h_{\phi }^{\prime }(0)=1$ satisfying the equation $z{h}_{\phi }^{\prime }(z)/h_{\phi }(z)=\phi (z)$. If $f\in {\mathcal{S}}_b^{\ast }(\mu ,\phi )$, then

$$\frac{-h_{\phi }(-r)}{r}{|1-z|}^2\le \left|f{(z)}^{\frac{2}{\mu }}\right|\le \frac{h_{\phi }(r)}{r}{|1-z|}^2,|z|=r.$$

(3.1)

Proof Define the function $h\in \mathcal{A}$ by

$$h(z)=\frac{z}{{(1-z)}^2}f{(z)}^{\frac{2}{\mu }},z\in \mathbb{D}.$$

(3.2)

Since f is univalent and $f(1):={lim}_{r\to 1^{-}}f(r)=0$, it is clear that $f(z)\ne 0$ in $\mathbb{D}$. Therefore, the function h is well defined and analytic in $\mathbb{D}$. A computation shows that

$$\frac{z{h}^{\prime }(z)}{h(z)}=\frac{2}{\mu }\frac{z{f}^{\prime }(z)}{f(z)}+\frac{1+z}{1-z}.$$

(3.3)

Hence we have the relation $f\in {\mathcal{S}}_b^{\ast }(\mu ,\phi )$ if and only if $h\in {\mathcal{S}}^{\ast }(\phi )$. Ma and Minda [[19], Corollary 1′] have shown that for $h\in {\mathcal{S}}^{\ast }(\phi )$,

$$-h_{\phi }(-r)\le |h(z)|\le h_{\phi }(r),|z|=r.$$

Using this inequality for h in (3.2) gives

$$-h_{\phi }(-r)\le \left|\frac{z}{{(1-z)}^2}f{(z)}^{\frac{2}{\mu }}\right|\le h_{\phi }(r),|z|=r$$

and hence the desired result follows. □

If ${\mathcal{S}}_b^{\ast }(\mu ,A,B)$ and hence

$$h_{\phi }(z)=\{\begin{array}{cc}z{(1+Bz)}^{\frac{A-B}{B}},\hfill & B\ne 0,\hfill \\ zexp(Az),\hfill & B=0,\hfill \end{array}$$

then

$$\begin{array}{r}{|1-z|}^2{(1-Br)}^{\frac{A-B}{B}}\le \left|f{(z)}^{\frac{2}{\mu }}\right|\le {|1-z|}^2{(1+Br)}^{\frac{A-B}{B}}\text{for }B\ne 0,\\ {|1-z|}^{2}exp(-Ar)\le \left|f{(z)}^{\frac{2}{\mu }}\right|\le {|1-z|}^{2}exp(Ar)\text{for }B=0.\end{array}$$

If ${\mathcal{S}}_b^{\ast }(\mu ,\beta )$ and

$$h_{\phi }(z)=\frac{z}{{(1-z)}^{2-2\beta }},$$

then

$$\frac{{|1-z|}^2}{{(1+r)}^{2-2\beta }}\le \left|f{(z)}^{\frac{2}{\mu }}\right|\le \frac{{|1-z|}^2}{{(1-r)}^{2-2\beta }}.$$

In particular, for $0\ne \mu \in \mathbb{R}$, the inequality reduces to the following inequality [18]:

$$\frac{{|1-z|}^{\mu }}{{(1+r)}^{\mu (1-\beta )}}\le |f(z)|\le \frac{{|1-z|}^{\mu }}{{(1-r)}^{\mu (1-\beta )}}.$$

Theorem 3.2 Let $\phi (z)=z{h}_{\phi }^{\prime }(z)/h_{\phi }(z)$ and $f\in {\mathcal{S}}_b^{\ast }(\mu ,\phi )$. Then, for $|z|=r$,

$$|arg\frac{f{(z)}^{\frac{1}{\mu }}}{(1-z)}|\le \frac{1}{2}\underset{|z|=r}{max}arg\frac{h_{\phi }(z)}{z}.$$

For $0\ne \mu \in \mathbb{R}$,

$$|arg\frac{f(z)}{{(1-z)}^{\mu }}|\le \frac{|\mu |}{2}\underset{|z|=r}{max}arg\frac{h_{\phi }(z)}{z}.$$

Proof For a function $h\in {\mathcal{S}}^{\ast }(\phi )$, in the paper [[19], Corollary 3′] it is shown that

$$|arg\frac{h(z)}{z}|\le \underset{|z|=r}{max}arg\frac{h_{\phi }(z)}{z},|z|=r.$$

The result then follows easily as the relation (3.3) holds. □

Corollary 3.1 If $f\in {\mathcal{S}}_b^{\ast }(\mu ,A,B)$, then for $|z|=r$,

$$|arg\frac{f{(z)}^{\frac{1}{\mu }}}{(1-z)}|\le \frac{A-B}{2B}\underset{|z|=r}{max}arg(1+Bz)\mathit{\text{for}}B\ne 0$$

and

$$|arg\frac{f{(z)}^{\frac{1}{\mu }}}{(1-z)}|\le \frac{1}{2}\underset{|z|=r}{max}argexp(Az)\mathit{\text{for}}B=0.$$

Corollary 3.2 If $f\in {\mathcal{S}}_b^{\ast }(\mu ,\beta )$, then for $|z|=r$

$$|arg\frac{f{(z)}^{\frac{1}{\mu }}}{(1-z)}|\le (1-\beta )\underset{|z|=r}{max}arg\frac{1}{(1-z)}.$$

Theorem 3.3 Let $\phi (z)=z{h}_{\phi }^{\prime }(z)/h_{\phi }(z)$ and

$$\underset{|z|=r}{min}|\phi (z)|=\phi (-r)\mathit{\text{and}}\underset{|z|=r}{max}|\phi (z)|=\phi (r).$$

(3.4)

Also, let

$$H_{\phi 1}=\frac{|\mu |{|1-z|}^{\mu }}{2r}{\left(\frac{h_{\phi }(-r)}{-r}\right)}^{\frac{\mu }{2}}(-|\frac{1+z}{1-z}|+\phi (-r))$$

and

$$H_{\phi 2}=\frac{|\mu |{|1-z|}^{\mu }}{2r}{\left(\frac{h_{\phi }(r)}{r}\right)}^{\frac{\mu }{2}}\left(\right|\frac{1+z}{1-z}|+\phi (r)).$$

For $\mu \in \mathbb{R}$, if $f\in {\mathcal{S}}_b^{\ast }(\mu ,\phi )$ then

$$H_{\phi 1}\le |f^{\prime }(z)|\le H_{\phi 2}.$$

Proof By Definition 1.2, for $f\in {\mathcal{S}}_b^{\ast }(\mu ,\phi )$, we have

$$\frac{2}{\mu }\frac{z{f}^{\prime }(z)}{f(z)}+\frac{1+z}{1-z}\prec \phi (z),z\in \mathbb{D}.$$

When (3.4) holds, the above subordination indicates that

$$\phi (-r)\le |\frac{2}{\mu }\frac{z{f}^{\prime }(z)}{f(z)}+\frac{1+z}{1-z}|\le \phi (r),|z|=r.$$

This shows that

$$-\left|\frac{1+z}{1-z}\right|+\phi (-r)\le \left|\frac{2}{\mu }\frac{z{f}^{\prime }(z)}{f(z)}\right|\le \left|\frac{1+z}{1-z}\right|+\phi (r)$$

or

$$\frac{|\mu |}{2r}(-|\frac{1+z}{1-z}|+\phi (-r))\le \left|\frac{f^{\prime }(z)}{f(z)}\right|\le \frac{|\mu |}{2r}\left(\right|\frac{1+z}{1-z}|+\phi (r)).$$

(3.5)

For $\mu \in \mathbb{R}$, Theorem 3.1 gives

$${|1-z|}^{\mu }{\left(\frac{h_{\phi }(-r)}{-r}\right)}^{\frac{\mu }{2}}\le |f(z)|\le {|1-z|}^{\mu }{\left(\frac{h_{\phi }(r)}{r}\right)}^{\frac{\mu }{2}}.$$

(3.6)

Combining (3.5) and (3.6), the desired results follows. □

We have the following corollaries as (3.4) holds.

Corollary 3.3 Let $\phi (z)=z{h}_{\phi }^{\prime }(z)/h_{\phi }(z)$. For $B\ne 0$, let

$$H_{\phi 1}=\frac{|\mu |{|1-z|}^{\mu }}{2r}{(1-Br)}^{\frac{\mu (A-B)}{2B}}(-|\frac{1+z}{1-z}|+\frac{1-Ar}{1-Br})$$

and

$$H_{\phi 2}=\frac{|\mu |{|1-z|}^{\mu }}{2r}{(1+Br)}^{\frac{\mu (A-B)}{2B}}\left(\right|\frac{1+z}{1-z}|+\frac{1+Ar}{1+Br}).$$

For $B=0$, let

$$H_{\phi 1}=\frac{|\mu |{|1-z|}^{\mu }}{2r}exp\left(\frac{-\mu Ar}{2}\right)(-|\frac{1+z}{1-z}|-rexp(-Ar))$$

and

$$H_{\phi 2}=\frac{|\mu |{|1-z|}^{\mu }}{2r}exp\left(\frac{\mu Ar}{2}\right)\left(\right|\frac{1+z}{1-z}|+rexp(Ar)).$$

For $\mu \in \mathbb{R}$, if $f\in {\mathcal{S}}_b^{\ast }(\mu ,A,B)$ then

$$H_{\phi 1}\le |f^{\prime }(z)|\le H_{\phi 2}.$$

Corollary 3.4 Let $\phi (z)=z{h}_{\phi }^{\prime }(z)/h_{\phi }(z)$,

$$H_{\phi 1}=\frac{|\mu |{|1-z|}^{\mu }}{2r{(1+r)}^{\mu (1-\beta )}}(-|\frac{1+z}{1-z}|+\frac{1-(1-2\beta )r}{1+r})$$

and

$$H_{\phi 2}=\frac{|\mu |{|1-z|}^{\mu }}{2r{(1-r)}^{\mu (1-\beta )}}\left(\right|\frac{1+z}{1-z}|+\frac{1+(1-2\beta )r}{1-r}).$$

For $\mu \in \mathbb{R}$, if $f\in {\mathcal{S}}_b^{\ast }(\mu ,\beta )$ then

$$H_{\phi 1}\le |f^{\prime }(z)|\le H_{\phi 2}.$$

4 Necessary and sufficient condition

Theorem 4.1 Let φ be a convex univalent function defined on $\mathbb{D}$. The function $f\in {\mathcal{S}}_b^{\ast }(\mu ,\phi )$ if and only if for all $|s|\le 1$, $|t|\le 1$,

$$\frac{s}{t}{\left(\frac{1-tz}{1-sz}\right)}^2{\left(\frac{f(sz)}{f(tz)}\right)}^{\frac{2}{\mu }}\prec \frac{h_{\phi }(sz)}{h_{\phi }(tz)},$$

where $h_{\phi }(z)=zexp({\int }_0^z((\phi (\zeta )-1)/\zeta )d\zeta )$.

Proof Ruscheweyh [[20], Theorem 1] showed that for φ a convex univalent function, F as in the hypothesis and $h\in \mathcal{A}$

$$\frac{z{h}^{\prime }(z)}{h(z)}\prec \phi (z)$$

if and only if for all $|s|\le 1$, $|t|\le 1$,

$$\frac{h(sz)}{h(tz)}\prec \frac{h_{\phi }(sz)}{h_{\phi }(tz)}.$$

(4.1)

From the relation (3.3), we know that $f\in {\mathcal{S}}_b^{\ast }(\mu ,\phi )$ if and only if $h\in {\mathcal{S}}^{\ast }(\phi )$. Substituting (3.2) in (4.1), we have

$$\frac{\frac{sz}{{(1-sz)}^2}f{(sz)}^{\frac{2}{\mu }}}{\frac{tz}{{(1-tz)}^2}f{(tz)}^{\frac{2}{\mu }}}\prec \frac{h_{\phi }(sz)}{h_{\phi }(tz)}$$

and hence the desired result follows. □

The following corollaries hold for $\phi (z)=\frac{1+Az}{1+Bz}$ is convex univalent on $\mathbb{D}$.

Corollary 4.1 The function $f\in {\mathcal{S}}_b^{\ast }(\mu ,A,B)$ if and only if for all $|s|\le 1$, $|t|\le 1$,

$$\begin{array}{c}{\left(\frac{1-tz}{1-sz}\right)}^{\mu }\left(\frac{f(sz)}{f(tz)}\right)\prec {\left(\frac{1+Bsz}{1+Btz}\right)}^{\frac{\mu (A-B)}{2B}}\mathit{\text{for}}B\ne 0,\hfill \\ {\left(\frac{1-tz}{1-sz}\right)}^{\mu }\left(\frac{f(sz)}{f(tz)}\right)\prec exp\left(\frac{\mu Az(s-t)}{2}\right)\mathit{\text{for}}B=0.\hfill \end{array}$$

Let $0\le \beta <1$, $A=1-2\beta$ and $B=-1$ in Corollary 4.1 and hence we have the result.

Corollary 4.2 [18]

The function $f\in {\mathcal{S}}_b^{\ast }(\mu ,\beta )$ if and only if for all $|s|\le 1$, $|t|\le 1$,

$${\left(\frac{1-tz}{1-sz}\right)}^{\mu }\frac{f(sz)}{f(tz)}\prec {\left(\frac{1-tz}{1-sz}\right)}^{\mu (1-\beta )}.$$

Theorem 4.2 as well as Corollaries 4.3 and 4.4 below are respectively special cases of Theorem 4.1 and Corollaries 4.1 and 4.2 when $s=1$ and $t=0$. However, we prove the below without the convexity assumption on φ.

Theorem 4.2 If $f\in {\mathcal{S}}_b^{\ast }(\mu ,\phi )$, then

$$\frac{f{(z)}^{\frac{2}{\mu }}}{{(1-z)}^2}\prec \frac{h_{\phi }(z)}{z},$$

where $h_{\phi }(z)=zexp({\int }_0^z((\phi (\zeta )-1)/\zeta )d\zeta )$.

Proof Clearly $z{h}_{\phi }^{\prime }(z)/h_{\phi }(z)=\phi (z)$. If $h\in {\mathcal{S}}^{\ast }(\phi )$, then

$$\frac{z{h}^{\prime }(z)}{h(z)}\prec \frac{z{h}_{\phi }^{\prime }(z)}{h_{\phi }(z)}.$$

Therefore by [[19], Theorem 1′]

$$\frac{h(z)}{z}\prec \frac{h_{\phi }(z)}{z}.$$

Let $h(z)$ be defined as in (3.2) and hence we arrive at the desired conclusion. □

Corollary 4.3 If $f\in {\mathcal{S}}_b^{\ast }(\mu ,A,B)$ then

$$\frac{f(z)}{{(1-z)}^{\mu }}\prec {(1+Bz)}^{\frac{\mu (A-B)}{2B}}\mathit{\text{for}}B\ne 0$$

and

$$\frac{f(z)}{{(1-z)}^{\mu }}\prec exp\left(\frac{\mu Az}{2}\right)\mathit{\text{for}}B=0.$$

When $0\le \beta <1$, $A=1-2\beta$ and $B=-1$, the above corollary reduces to the following result.

Corollary 4.4 [18]

If $f\in {\mathcal{S}}_b^{\ast }(\mu ,\beta )$ then

$$\frac{f(z)}{{(1-z)}^{\mu }}\prec \frac{1}{{(1-z)}^{\mu (1-\beta )}}.$$

5 Coefficient estimate for $f\in {\mathcal{S}}_b^{\ast }(\phi )$

In particular, when $\mu =1$, (1.2) becomes

$$2\frac{z{f}^{\prime }(z)}{f(z)}+\frac{1+z}{1-z}\prec \phi (z),z\in \mathbb{D}.$$

We denote the class satisfying the above subordination as ${\mathcal{S}}_b^{\ast }(\phi )$.

Theorem 5.1 Let $\phi (z)=1+B_{1}z+B_2{z}^2+\cdots$ . If $f\in {\mathcal{S}}_b^{\ast }(\phi )$, then the coefficients $d_1$, $d_2$, $d_3$ satisfy the following inequalities:

$$\begin{array}{c}|d_1|\le \frac{B_1}{2}+1,\hfill \\ |d_2|\le \frac{B_1}{4}max\{1,|\frac{B_2}{B_1}+\frac{B_1}{2}\left|\right\}+\frac{B_1}{2},\hfill \\ |d_3|\le \frac{B_1}{6}H(\frac{6B_1^2+16B_2}{8B_1},\frac{B_1^3+6B_1{B}_2+8B_3}{8B_1})+\frac{B_1}{4}max\{1,|\frac{B_2}{B_1}+\frac{B_1}{2}\left|\right\},\hfill \end{array}$$

where $H(q_1,q_2)$^a is as defined in [21] (see also [[22], Lemma 3]) and

$$|d_2-\nu d_1^2|\le \{\begin{array}{cc}\frac{B_1}{4}(\frac{B_2}{B_1}-(2\nu -1)\frac{B_1}{2})+(2\nu +1)\frac{B_1}{2}+2\nu ,\hfill & \nu \le {\sigma }_1,\hfill \\ \frac{B_1}{4}+(2\nu +1)\frac{B_1}{2}+2\nu ,\hfill & {\sigma }_1\le \nu \le {\sigma }_2,\hfill \\ \frac{B_1}{4}((2\nu -1)\frac{B_1}{2}-\frac{B_2}{B_1})+(2\nu +1)\frac{B_1}{2}+2\nu ,\hfill & \nu \ge {\sigma }_2,\hfill \end{array}$$

where

$${\sigma }_1=\frac{1}{B_1}(\frac{B_2}{B_1}-1)+\frac{1}{2},{\sigma }_2=\frac{1}{B_1}(\frac{B_2}{B_1}+1)+\frac{1}{2}.$$

Proof Define the function $g(z)=1+g_{1}z+g_2{z}^2+\cdots$ by

$$g(z)=\frac{f(z)}{(1-z)},z\in \mathbb{D}.$$

Then a computation shows that

$$2\frac{z{g}^{\prime }(z)}{g(z)}+1=2\frac{z{f}^{\prime }(z)}{f(z)}+\frac{1+z}{1-z}.$$

Since $f\in {\mathcal{S}}_b^{\ast }(\phi )$, we have

$$2\frac{z{g}^{\prime }(z)}{g(z)}+1\prec \phi (z),$$

or there is an analytic function $w(z)=w_{1}z+w_2{z}^2+\cdots$ such that

$$2\frac{z{g}^{\prime }(z)}{g(z)}+1=\phi (w(z)).$$

Since

$$2\frac{z{g}^{\prime }(z)}{g(z)}+1=1+2g_{1}z+(-2g_1^2+4g_2)z^2+(2g_1^3-6g_1{g}_2+6g_3)z^3+\cdots$$

and

$$\phi (w(z))=1+B_1{w}_{1}z+(B_2{w}_1^2+B_1{w}_2)z^2+(B_3{w}_1^3+2B_2{w}_1{w}_2+b_1{w}_3)z^3+\cdots ,$$

we see that

$$\begin{array}{c}{g}_1=\frac{B_1{w}_1}{2},\hfill \\ g_2=\frac{B_1}{4}(w_2+(\frac{B_2}{B_1}+\frac{B_1}{2})w_1^2),\hfill \\ g_3=\frac{B_1}{6}(w_3+\left(\frac{6B_1^2+16B_2}{8B_1}\right)w_1{w}_2+\left(\frac{B_1^3+6B_1{B}_2+8B_3}{8B_1}\right)w_1^3).\hfill \end{array}$$

In view of the well-known inequality $|w_1|\le 1$, we have

$$|g_1|\le \frac{B_1}{2}.$$

Applying [[23], inequality 7, p.10] and [[22], Lemma 3] (see also [21]), we get

$$|g_2|\le \frac{B_1}{4}max\{1,|\frac{B_2}{B_1}+\frac{B_1}{2}\left|\right\}$$

and

$$|g_3|\le \frac{B_1}{6}H(\frac{6B_1^2+16B_2}{8B_1},\frac{B_1^3+6B_1{B}_2+8B_3}{8B_1}),$$

respectively. Also, we see that applying [[22], Lemma 1] (see also [19]) to inequality

$$g_2-\nu g_1^2=\frac{B_1}{4}(w_2-((2\nu -1)\frac{B_1}{2}-\frac{B_2}{B_1})w_1^2)$$

yields

$$|g_2-\nu g_1^2|\le \{\begin{array}{cc}\frac{B_1}{4}(\frac{B_2}{B_1}-(2\nu -1)\frac{B_1}{2}),\hfill & \nu \le {\sigma }_1,\hfill \\ \frac{B_1}{4},\hfill & {\sigma }_1\le \nu \le {\sigma }_2,\hfill \\ \frac{B_1}{4}((2\nu -1)\frac{B_1}{2}-\frac{B_2}{B_1}),\hfill & \nu \ge {\sigma }_2\hfill \end{array}$$

for ${\sigma }_1$ and ${\sigma }_2$ as in the hypothesis. Todorov in [3] shows that for

$$g(z)=1+\sum _1^{\mathrm{\infty }}{g}_n{z}^n,$$

the coefficient

$$g_n=1+d_1+d_2+\cdots +d_n,$$

and hence from the above relation the desired results are obtained. □

Corollary 5.1 When $\phi (z)=(1+z)/(1-z)$, our results coincide with [[3], Corollary 2.3].

Remark 5.1 All the results for the special case when $\mu =1$ or the class starlike with respect to a boundary point defined by subordination were presented at the 8th International Symposium on GFTA, 27-31 August 2012, Ohrid, Republic of Macedonia and thereafter published as [24].

Endnote

The expression for H is too lengthy to be reproduced here. See [21] or [22] for the full expression.