A note on starlike functions associated with symmetric points

Abstract

The objective of the present paper is to study results that are defined using the notions of generalization of Janowski classes and k-symmetrical functions. A representation theorem, coefficients inequality, distortion properties and the result on radius of starlikeness are discussed.

1 Introduction and preliminaries

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

$$\begin{aligned} f(z)=z+\sum _{n=2}^{\infty }a_{n}z^{n}, \end{aligned}$$

(1.1)

which are analytic in the open unit disk $\mathcal {U}=\left\{ z\in \mathbb {C} :|z|<1\right\}$, and $\mathcal {S}$ denote the subclass of $\mathcal {A}$ consisting of all function which are univalent in $\mathcal {U}$.

For two functions f and g,  analytic in $\mathcal {U}$, we say that the function f is subordinate to g in $\mathcal {U}$, if there exists an analytic function w in $\mathcal {U}$ such that $|w(z)|<1$ with $w(0)=0$, and $f(z)=g(w(z))$, and we denote this by $f(z)\prec g(z)$. If g is univalent in $\mathcal {U}$, then the subordination is equivalent to $f(0)=g(0)$ and $f(\mathcal {U})\subset g(\mathcal {U})$.

Using the principle of the subordination we define the class $\mathcal {P}$ of functions with positive real part.

Definition 1.1

[6] Let $\mathcal {P}$ denote the class of analytic functions of the form $p(z)=1+\sum \nolimits _{n=1}^{\infty }p_{n}z^{n}$ defined on $\mathcal {U}$ and satisfying $p(0)=1$, $\mathfrak {R}p(z)>0$, $z\in \mathcal {U}$.

Any function p in $\mathcal {P}$ has the representation $p(z)=\dfrac{1+w(z) }{1-w(z)}$, where $w\in \Omega$ and

$$\begin{aligned} \Omega =\{w\in \mathcal {A}: w(0)=0,|w(z)|<1\}. \end{aligned}$$

(1.2)

The class of functions with positive real part $\mathcal {P}$ plays a crucial role in geometric function theory. Its significance can be seen from the fact that simple subclasses like class of starlike $\mathcal {S}^{*}$, class of convex functions $\mathcal {C}$, class of starlike functions with respect to symmetric points $\mathcal {S}_{s}^{*}$ have been defined by using the concept of class of functions with positive real part.

Let $\mathcal {P}[A,B]$, with $-1\le B<A\le 1$, denote the class of analytic function p defined on $\mathcal {U}$ with the representation $p(z)= \dfrac{1+Aw(z)}{1+Bw(z)}$, $z\in \mathcal {U}$, where $w\in \Omega$.

we note that

$$\begin{aligned} p\in \mathcal {P}[A,B]\text { if and only if }p(z)\prec \dfrac{1+Az}{1+Bz} (\text {see} 7). \end{aligned}$$

The class $\mathcal {P}[A,B,\alpha ]$ of generalized Janowski functions was introduced in [9]. For arbitrary numbers $A,B,\alpha ,$ with $-1\le B<A\le 1$, $0\le \alpha <1,$ a function p analytic in $\mathcal {U}$ with $p(0)=1$ is in the class $\mathcal {P}[A,B,\alpha ]$ if and only if

$$\begin{aligned} p(z)\prec \frac{1+[(1-\alpha )A+\alpha B]z}{1+Bz}\Leftrightarrow p(z)=\frac{ 1+[(1-\alpha )A+\alpha B]w(z)}{1+Bw(z)},\quad w\in \Omega . \end{aligned}$$

The definition of starlike functions with respect to k-symmetric points is as follows.

Definition 1.2

For a positive integer k, let $\varepsilon =\exp \left( \frac{2\pi i}{k} \right)$ denote the kth root of unity for $f\in \mathcal {A}$, let

$$\begin{aligned} M_{f,k}(z)= \sum _{v=1}^{k-1}\varepsilon ^{-v}f(\varepsilon ^v z). \frac{1}{ \sum _{v=1}^{k-1}\varepsilon ^{-v}}, \end{aligned}$$

(1.3)

be its k-weighted mean function.

A function f in $\mathcal {A}$ is said to belong to the class $\mathcal {S} _{k}^{*}$ if functions starlike with respect to k-symmetric points if for every r close to 1 , $r<1$, the angular velocity of f about the point $M_{f_{k}(z_{0})}$ positive at $z=z_{0}$ as z traverses the circle $|z|=r$ in the positive direction, that is

$$\begin{aligned} \mathfrak {R}\left\{ \frac{zf^{\prime }(z)}{f(z)-M_{f,k}(z_{0})}\right\} >0 \end{aligned}$$

for $z=z_{0}$, $|z_{0}|=r$.

Definition 1.3

[11] A function f in $\mathcal {A}$ is univalent and starlike with respect to k-symmetric points, or briefly k-starlike if and only if

$$\begin{aligned} \mathfrak {R}\left\{ \frac{zf^{\prime }(z)}{f_{k}(z)}\right\} >0,\quad z\in \mathcal {U}, \end{aligned}$$

(1.4)

where

$$\begin{aligned} f_{k}(z)=\frac{1}{k}(f(z)-M_{f,k}(z)). \end{aligned}$$

(1.5)

If f(z) defined by (1.1) then,

$$\begin{aligned} f_{k}(z)=z+\sum _{n=2}^{\infty }\chi _{n}a_{n}z^{n},\quad (k=2,3,\dots ), \end{aligned}$$

(1.6)

where

$$\begin{aligned} \chi _{n}= {\left\{ \begin{array}{ll} 1,\quad n=lk+1, \quad l\in \mathbb {N}_{0}, \\ 0,\quad n\ne lk+1. \end{array}\right. } \end{aligned}$$

(1.7)

Al-Sarari and Latha in [1,2,3] (see also, [4]) studied some classes which related to Janowski type functions and symmetric points.

Now using the generalization of Janowski functions and the concept of k-symmetrical functions we define the following:

Definition 1.4

A function f in $\mathcal {A}$ is said to belong to the class $\mathcal {S}^{k}(A,B,\alpha )$, $(-1\le B<A\le 1),0\le \alpha <1$ if

$$\begin{aligned} \frac{zf^{\prime }(z)}{f_{k}(z)}\prec \frac{1+[(1-\alpha )A+\alpha B]z}{1+Bz} ,\quad z\in \mathcal {U}, \end{aligned}$$

where $f_{k}(z)$ defined by (1.6).

We note that for special values of $k,\alpha ,A$ and B yield the following classes:

(i):

$\mathcal {S}^{1}(A,B,\alpha )$${=}$ $\mathcal {S}^{*}(A,B,\alpha )$ the class introduced by Polatoglu et al. [9];

(ii):

$\mathcal {S}^{k}(A,B,0)$${=}$ $\mathcal {S}^{(k)}(A,B)$ is the class studied by Kwon and Sim [8];

(iii):

$\mathcal {S}^{k}(1,-1,0)$${=}$ $\mathcal {S}_{k}^{*}$${=}$ $\mathcal {S}_{k}^{*}(1,-1)$, the class is studied by Sakaguchi [11] and etc. We need the following lemmas to prove our main results.

Lemma 1.5

[5] Let $p(z)=1+\sum _{n=1}^{\infty }p_{n}z^{n}\in \mathcal { P}[A,B,\alpha ]$, then for $n\ge 1$,

$$\begin{aligned} |p_{n}|\le (1-\alpha )(A-B). \end{aligned}$$

Lemma 1.6

pol Any function $f\in \mathcal {S}^{*}(A,B,\alpha )$ can be written in the form

$$\begin{aligned} f(z)=\left\{ \begin{array}{lll} z\left( 1+Bw(z)\right) ^{\frac{(1-\alpha )(A-B)}{B}}, &{} \text {if} &{} B\ne 0, \\ z\,\exp \left[ (1-\alpha )Aw(z)\right] , &{} \text {if} &{} B=0, \end{array} \right. \end{aligned}$$

where $w\in \Omega$, and $\Omega$ was defined by (1.2).

Lemma 1.7

[10] Let $\phi$ be convex and g starlike Then for F analytic in $\mathcal {U}$ with $F(0)=1,$

$$\begin{aligned} \frac{\phi *Fg}{\phi *g}(\mathcal {U})\subset \overline{CO}(F(\mathcal {U})) \end{aligned}$$

where $\overline{CO}(F(\mathcal {U}))$ denotes the closed convex hull of $F( \mathcal {U})$.

Lemma 1.8

[9] Let $p\in \mathcal {P}[A,B,\alpha ]$, then the set of the values of p is in the closed disc with center at C(r) and having the radius $\rho (r)$, where

$$\begin{aligned} \left\{ \begin{array}{ll} C(r)=\left( \frac{1-B[(1-\alpha )A+\alpha B]r^{2}}{1-B^{2}r^{2}},0\right) ,\quad \rho (r)=\frac{(1-\alpha )(A-B)r}{1-B^{2}r^{2}} &{} \text {if } B\ne 0, \\ C(r)=(1,0),\quad \rho (r)=(1-\alpha )|A|r &{} \text {if } B=0. \end{array} \right. \end{aligned}$$

2 Main results

Lemma 2.1

Let $p\in \mathcal {P}[A,B,\alpha ]$. Then

$$\begin{aligned}&\left. \begin{array}{ll} \dfrac{1-(1-\alpha )(A-B)r-B[(1-\alpha )A+\alpha B]r^{2}}{1-B^{2}r^{2}}, &{} \text {if } B\ne 0, \\ 1-(1-\alpha )Ar, &{} \text {if } B=0 \end{array} \right\} \le |p(z)| \\&\quad \le \left\{ \begin{array}{ll} \dfrac{1+(1-\alpha )(A-B)r-B[(1-\alpha )A+\alpha B]r^{2}}{1-B^{2}r^{2}}, &{} \text {if } B\ne 0, \\ 1+(1-\alpha )Ar, &{} \text {if } B=0. \end{array} \right. \end{aligned}$$

Proof

The set of the values of p is in the closed disc with center at $C(r)= \frac{1-B[(1-\alpha )A+\alpha B]r^2}{1-B^2r^2}$ and having the radius $\rho (r)=\frac{(1-\alpha )(A-B)r}{1-B^2r^2}$ using Lemma 1.8, that is

$$\begin{aligned} \left| p-\frac{1-B[(1-\alpha )A+\alpha B]r^2}{1-B^2r^2}\right| \le \frac{ (1-\alpha )(A-B)r}{1-B^2r^2}. \end{aligned}$$

(2.1)

Simplifying (2.1), we get the required result . $\square$

Theorem 2.2

If $f\in \mathcal {S}^{k}(A,B,\alpha )$, then $f_{k}\in \mathcal {S} (A,B,\alpha )$, where $f_{k}$ is defined by (1.6).

Proof

Supposing that $f\in \mathcal {S}^{k}(A,B,\alpha )$, we can get

$$\begin{aligned} \frac{zf^{\prime }(z)}{f_{k}(z)}\prec \frac{1+[(1-\alpha )A+\alpha B]z}{1+Bz} . \end{aligned}$$

(2.2)

Substituting z by $\varepsilon ^{\nu }z$ in (2.2), it follows

$$\begin{aligned} \frac{\varepsilon ^{\nu }zf^{\prime }(\varepsilon ^{v}z)}{f_{k}(\varepsilon ^{\nu }z)}\prec \frac{1+[(1-\alpha )A+\alpha B]\varepsilon ^{\nu }z}{ 1+B\varepsilon ^{\nu }z}\prec \frac{1+[(1-\alpha )A+\alpha B]z}{1+Bz}, \end{aligned}$$

hence

$$\begin{aligned} \frac{\varepsilon ^{\nu -\nu j}zf^{\prime }(\varepsilon ^{\nu }z)}{ f_{k}^{\prime }(z)}\prec \frac{1+[(1-\alpha )A+\alpha B]z}{1+Bz}. \end{aligned}$$

(2.3)

Letting $\nu =0,1,2,\dots ,k-1$ in (2.3) and using the fact that $\mathcal {P}[A,B,\alpha ]$ is a convex set, we deduce that

$$\begin{aligned} \frac{z\frac{1}{k}\sum _{\nu =0}^{k-1}\varepsilon ^{\nu -\nu j}f^{\prime }(\varepsilon ^{\nu }z)}{f_{k}(z)}\prec \frac{1+[(1-\alpha )A+\alpha B]z}{ 1+Bz}, \end{aligned}$$

or equivalently

$$\begin{aligned} \frac{zf_{k}^{\prime }(z)}{f_{k}(z)}\prec \frac{1+[(1-\alpha )A+\alpha B]z}{ 1+Bz}, \end{aligned}$$

that is $f_{k}\in \mathcal {S}(A,B,\alpha )$. $\square$

Theorem 2.3

Let $f\in \mathcal {S}^{k}(A,B,\alpha )$, with $-1\le B<A\le 1$ and $0\le \alpha <1$. Then,

$$\begin{aligned} f(z)=\left\{ \begin{array}{ll} \int _{0}^{z}\frac{1+[(1-\alpha )A+\alpha B]\widetilde{w}(\zeta )}{1+B \widetilde{w}(\zeta )}\left( 1+Bw(\zeta )\right) ^{\frac{(1-\alpha )(A-B)}{B} }\,d\zeta , &{} \text {if } B\ne 0, \\ \int _{0}^{z}[1+A(1-\alpha )\widetilde{w}(\zeta )]\exp [(1-\alpha )Aw(\zeta )]\,d\zeta , &{} \text {if } B=0, \end{array} \right. \end{aligned}$$

for some $w,\widetilde{w}\in \Omega$.

Proof

Supposing that $f\in \mathcal {S}^{k}(A,B,\alpha )$, it follows that there exists a function $\widetilde{w}\in \Omega$ such that

$$\begin{aligned} \frac{zf^{\prime }(z)}{f_{k}(z)}=\frac{1+[(1-\alpha )A+\alpha B]\widetilde{w} (z)}{1+B\widetilde{w}(z)},\;z\in \mathcal {U}. \end{aligned}$$

Using Theorem 2.2 and Lemma 1.6, we have

$$\begin{aligned} f^{\prime }(z)=\left\{ \begin{array}{ll} \frac{1+[(1-\alpha )A+\alpha B]\widetilde{w}(z)}{1+B\widetilde{w}(z)}\left( 1+Bw(z)\right) ^{\frac{(1-\alpha )(A-B)}{B}}, &{} \text {if } B\ne 0, \\ {[} 1+A(1-\alpha )\widetilde{w}(z)]\exp [(1-\alpha )Aw(z)], &{} \text {if } B=0, \end{array} \right. \end{aligned}$$

and integrating the above relations along the line connecting the origin with $z\in \mathcal {U}$ we obtain our result. $\square$

Theorem 2.4

Let $f(z)\in \mathcal {S}^{k}(A,B,\alpha )$ and is of the form (1.1). Then for $n\ge 2$, $-1\le B<A\le 1,0\le \alpha <1$.

$$\begin{aligned} |a_{n}|\le \prod _{m=1}^{n-1}\frac{\chi _{m}\left[ (1-\alpha )(A-B)-1\right] +m}{m+1-\chi _{m+1}}, \end{aligned}$$

(2.4)

where $\chi _{n}$ is defined in (1.7).

Proof

By Definition 1.4, we have

$$\begin{aligned} \frac{zf^{\prime }(z)}{f_{k}(z)}=p(z),\quad p\in \mathcal {P}[A,B,\alpha ], \end{aligned}$$

then we have

$$\begin{aligned} zf^{\prime }(z)=[1+\sum _{n=1}^{\infty }p_{n}z^{n}]f_{k}(z) \end{aligned}$$

by (1.1) and (1.6), we have

$$\begin{aligned} (1-\chi _{1})z+\sum _{n=2}^{\infty }[n-\chi _{n}]a_{n}z^{n}=\left[ \sum _{n=1}^{\infty }p_{n}z^{n}\right] \left[ \sum _{n=1}^{\infty }\chi _{n}a_{n}z^{n}\right] . \end{aligned}$$

Equating coefficients of $z^{n}$ on both sides, we have

$$\begin{aligned} a_{n}=\frac{1}{[n-\chi _{n}]}\sum _{m=1}^{n-1}p_{m}\chi _{n-m}a_{n-m},\quad \chi _{1}=1, \end{aligned}$$

(2.5)

by Lemma 1.5, we have

$$\begin{aligned} |a_{n}|\le \frac{(A-B)(1-\alpha )}{[n-\chi _{n}]}\sum _{m=1}^{n-1}\chi _{m}|a_{m}| \end{aligned}$$

(2.6)

Now we want to prove that

$$\begin{aligned} \frac{(A-B)(1-\alpha )}{[n-\chi _{n}]}\sum _{m=1}^{n-1}\chi _{m}|a_{m}|\le \prod _{m=1}^{n-1}\frac{\chi _{m}[(1-\alpha )(A-B)-1]+m}{[m+1-\chi _{m+1}]}. \end{aligned}$$

(2.7)

For this, we use the induction method.

The inequality (2.7) is true for $n=2$ and 3.

Let the hypothesis be true for $n=m$, we have

$$\begin{aligned} \frac{(A-B)(1-\alpha )}{[m-\chi _{m}]}\sum _{r=1}^{m-1}\chi _{r}|a_{r}|\le \prod _{r=1}^{m-1}\frac{\chi _{r}[(1-\alpha )(A-B)-1]+r}{[r+1-\chi _{r+1}]}, \end{aligned}$$

Multiplying both sides by $\frac{\chi _{m}[(A-B)(1-\alpha )-1]+m}{[m+1-\chi _{m+1}]},$ we get

$$\begin{aligned} \prod _{r=1}^{m}\frac{\chi _{r}[(1-\alpha )(A-B)-1]+r}{[r+1-\chi _{r+1}]}\ge \frac{\chi _{m}[(A-B)(1-\alpha )-1]+m}{[m+1-\chi _{m+1}]}.\frac{ (A-B)(1-\alpha )}{[m-\chi _{m}]}\sum _{r=1}^{m-1}\chi _{r}|a_{r}|, \end{aligned}$$

since

$$\begin{aligned}&\frac{\chi _{m}[(A-B)(1-\alpha )-1]+m}{[m+1-\chi _{m+1}]}.\frac{ (A-B)(1-\alpha )}{[m-\chi _{m}]}\sum _{r=1}^{m-1}\chi _{r}|a_{r}| \\&\quad =\frac{(A-B)(1-\alpha )}{[m+1-\chi _{m+1}]}. \left[ 1+\frac{\chi _{m}(A-B)(1-\alpha )}{[m-\chi _{m}]}\right] \sum _{r=1}^{m-1}\chi _{r}|a_{r}|, \\&\quad \ge \frac{(A-B)(1-\alpha )}{[m+1-\chi _{m+1}]}.\left[ \sum _{r=1}^{m-1} \chi _{r}|a_{r}|+\chi _{m}|a_{m}|\right] , \\&\quad =\frac{(A-B)(1-\alpha )}{[m+1-\chi _{m+1}]}.\left[ \sum _{r=1}^{m}\chi _{r}|a_{r}|\right] . \end{aligned}$$

That is

$$\begin{aligned} |a_{m+1}|\le \frac{(A-B)(1-\alpha )}{[m-\chi _{m}]}\sum _{r=1}^{m}\chi _{r}|a_{r}|\le \prod _{r=1}^{m}\frac{\chi _{r}[(1-\alpha )(A-B)-1]+r}{ [r+1-\chi _{r+1}]}, \end{aligned}$$

which shows that inequality (2.7) is true for $n=m+1$. This completes the proof. $\square$

We now prove the distortion theorem for the class $\mathcal {S} ^{k}(A,B,\alpha )$.

Theorem 2.5

If $f\in \mathcal {S}^{k}(A,B,\alpha )$, then

$$\begin{aligned}&\left. \begin{array}{ll} \dfrac{1-(1-\alpha )(A-B)r-B[(1-\alpha )A+\alpha B]r^2}{1-B^2r^2}(1-Br)^{\frac{ (1-\alpha )(A-B)}{B}}, &{} \text {if } B\ne 0, \\ \left[ 1-(1-\alpha )Ar\right] \exp \left[ -(1-\alpha )Ar\right] , &{} \text {if } B=0 \end{array} \right\} \le |f^{\prime }(z)| \\&\quad \le \left\{ \begin{array}{ll} \dfrac{1+(1-\alpha )(A-B)r-B[(1-\alpha )A+\alpha B]r^2}{1-B^2r^2} (1+Br)^{ \frac{(1-\alpha )(A-B)}{B}}, &{} \text {if } B\ne 0, \\ \left[ 1+(1-\alpha )Ar\right] \exp \left[ (1-\alpha )Ar\right] , &{} \text {if } B=0, \end{array} \right. \end{aligned}$$

where $|z|\le r<1$.

Proof

For an arbitrary function $f\in \mathcal {S}^{k}(A,B,\alpha )$, according to Theorem 2.2 and Lemma 1.6 we need to study the following:

• (i) If $B\ne 0$, then there exists a function $w\in \Omega$, such that

  $f_{k}(z)=z\left( 1+Bw(z)\right) ^{\frac{(1-\alpha )(A-B)}{B}}$, by using Lemma 2.1 and therefore

  $$\begin{aligned}&\dfrac{1-(1-\alpha )(A-B)r-B[(1-\alpha )A+\alpha B]r^2}{1-B^2r^2} \left| 1+Bw(z)\right| ^{\frac{(1-\alpha )(A-B)}{B}}\le |f^{\prime }(z)| \nonumber \\&\quad \le \dfrac{1+(1-\alpha )(A-B)r-B[(1-\alpha )A+\alpha B]r^2}{1-B^2r^2} \left| 1+Bw(z)\right| ^{\frac{(1-\alpha )(A-B)}{B}},\nonumber \\&\qquad |z|\le r<1. \end{aligned}$$

  (2.8)

  Since $w\in \Omega$, we have

  $$\begin{aligned} 1-|B|r\le \left| 1+Bw(z)\right| \le 1+|B|r,\quad |z|\le r<1. \end{aligned}$$

  Case 1 If $B>0$, using the fact that $-1\le B<A\le 1$ and $0\le \alpha <1$, we have

  $$\begin{aligned} (1-|B|r)^{\frac{(1-\alpha )(A-B)}{B}}\le \left| 1+Bw(z)\right| ^{\frac{ (1-\alpha )(A-B)}{B}}\le (1+|B|r)^{\frac{(1-\alpha )(A-B)}{B}},\quad |z|\le r<1, \end{aligned}$$

  and from (2.8) we obtain

  $$\begin{aligned}&\dfrac{1-(1-\alpha )(A-B)r-B[(1-\alpha )A+\alpha B]r^2}{1-B^2r^2}(1-|B|r)^{ \frac{(1-\alpha )(A-B)}{B}}\le |f^{\prime }(z)| \nonumber \\&\quad \le \dfrac{1+(1-\alpha )(A-B)r-B[(1-\alpha )A+\alpha B]r^2}{1-B^2r^2}(1+|B|r)^{ \frac{(1-\alpha )(A-B)}{B}},\nonumber \\&\qquad |z| \le r<1. \end{aligned}$$

  (2.9)

  Case 2 If $B<0$, from the fact that $-1\le B<A\le 1$ and $0\le \alpha <1$, we have

  $$\begin{aligned} (1-|B|r)^{\frac{(1-\alpha )(A-B)}{B}}\ge \left| 1+Bw(z)\right| ^{\frac{ (1-\alpha )(A-B)}{B}}\ge (1+|B|r)^{\frac{(1-\alpha )(A-B)}{B}},\quad |z|\le r<1, \end{aligned}$$

  and from (2.8) we obtain

  $$\begin{aligned}&\dfrac{1-(1-\alpha )(A-B)r-B[(1-\alpha )A+\alpha B]r^2}{1-B^2r^2}(1-|B|r)^{ \frac{(1-\alpha )(A-B)}{B}}\ge |f^{\prime }(z)| \nonumber \\&\quad \ge \dfrac{1+(1-\alpha )(A-B)r-B[(1-\alpha )A+\alpha B]r^2}{1-B^2r^2}(1+|B|r)^{ \frac{(1-\alpha )(A-B)}{B}},\nonumber \\&\qquad |z|\le r<1. \end{aligned}$$

  (2.10)

  Now, combining the inequalities (2.9) and (2.10), we finally conclude that

  $$\begin{aligned}&\dfrac{1-(1-\alpha )(A-B)r-B[(1-\alpha )A+\alpha B]r^2}{1-B^2r^2}(1-Br)^{ \frac{(1-\alpha )(A-B)}{B}}\le |f^{\prime }(z)| \nonumber \\&\quad \le \dfrac{1+(1-\alpha )(A-B)r-B[(1-\alpha )A+\alpha B]r^2}{1-B^2r^2}(1+Br)^{ \frac{(1-\alpha )(A-B)}{B}},\nonumber \\&\qquad |z|\le r<1. \end{aligned}$$

  (2.11)

• (ii) If $B=0$, there exists a function $w\in \Omega$, such that $f_{k}(z)=z\exp \left[ (1-\alpha )Aw(z)\right]$, and therefore

  $$\begin{aligned}&\left[ 1-(1-\alpha )Ar\right] \left| \exp \left[ (1-\alpha )Aw(z)\right] \right| \le |f^{\prime }(z)| \le \left[ 1+(1-\alpha )Ar\right] \nonumber \\&\quad \times \left| \exp \left[ (1-\alpha )Aw(z)\right] \right| , \;|z|\le r<1. \end{aligned}$$

  (2.12)

  Since $\left| \exp \left[ (1-\alpha )Aw(z)\right] \right| =\exp \left[ (1-\alpha )A {\mathrm{Re}}w(z)\right]$, $z\in \mathcal {U}$, using a similar computation as in the previous case, we deduce

  $$\begin{aligned} \exp \left[ -(1-\alpha )Ar\right] \le \left| \exp \left[ (1-\alpha )Aw(z)\right] \right| \le \exp \left[ (1-\alpha )Ar\right] ,\;|z|\le r<1. \end{aligned}$$

  Thus, (2.12) yield to

  $$\begin{aligned}&\left[ 1-(1-\alpha )Ar\right] \exp \left[ -(1-\alpha )Ar\right] \le |f^{\prime }(z)| \nonumber \\&\quad \le \left[ 1+(1-\alpha )Ar\right] \exp \left[ (1-\alpha )Ar\right] ,\;|z|\le r<1, \end{aligned}$$

  (2.13)

  which completes the proof of our theorem.

$\square$

Theorem 2.6

Let $f\in \mathcal {S}^{k}(A,B,\alpha )$ and let $\phi$ be convex. Then $(f*\phi )\in \mathcal {S}^{k}(A,B,\alpha )$.

Proof

To prove that $(f*\phi )\in \mathcal {S}^{k}(A,B,\alpha )$ it is sufficient to show that

$$\begin{aligned} \frac{z(f*\phi )^{\prime }(z)}{(f*\phi )_{k}(z)}\subset \overline{CO} (F(\mathcal {U})), \end{aligned}$$

where $F(z)=\frac{zf^{\prime }(z)}{f_{k}(z)}$. Now

$$\begin{aligned} \frac{z(f*\phi )^{\prime }(z)}{(f*\phi )_{k}(z)}= & {} \frac{zf^{\prime }(z)*\phi (z)}{(f_{k}(z)*\phi (z)} \\= & {} \frac{\phi (z)*\frac{zf^{\prime }(z)}{f_{k}(z)}.f_{k}(z)}{\phi (z)*f_{k}(z)}, \end{aligned}$$

by using Lemma 1.7 with $f_{k}(z)\in \mathcal {S}(A,B,\alpha ),F\in \mathcal {P}[A,B,\alpha ]$, that complete the proof. $\square$

Corollary 2.7

Let $f\in \mathcal {S}^{k}(A,B,\alpha )$. Then

$$\begin{aligned} F_i(z)\in \mathcal {S}^{k}(A,B,\alpha ),\quad (i=1,2,3,4), \end{aligned}$$

where

$$\begin{aligned} F_{1}(z)= & {} \int _{0}^{z}\frac{f(t)}{t}dt, \\ \qquad F_{2}(z)= & {} \int _{0}^{z}\frac{f(t)-f(xt)}{t-xt}dt,\quad |x|\le 1,x\ne 1, \\ F_{3}(z)= & {} \frac{2}{z}\int _{0}^{z}{f(t)}dt,\\ \qquad F_{4}(z)= & {} \frac{m+1}{m}\int _{0}^{z}t^{m-1}{f(t)}dt,\quad \mathfrak {R}m>0. \end{aligned}$$

Proof

Since

$$\begin{aligned} F_{1}(z)= & {} \phi _{1}(z)*f(z),\quad \phi _{1}(z)=\sum _{0}^{\infty } \frac{1}{n}z^{n}=\log (1-z)^{-1}, \\ \qquad \qquad F_{2}(z)= & {} \phi _{2}(z)*f(z), \\ \phi _{2}(z)= & {} \sum _{0}^{\infty }\frac{1-x^{n}}{n(1-x)}z^{n}= \frac{1}{1-x}\log \left( \frac{1-xz}{1-z}\right) ,\quad |x|\le 1,x\ne 1, \\ \qquad F_{3}(z)= & {} \phi _{3}(z)*f(z),\quad \phi _{3}(z)=\sum _{0}^{\infty }\frac{2}{n+1}z^{n}=\frac{-2[z+\log (1-z)]}{z}, \\ F_{4}(z)= & {} \phi _{4}(z)*f(z),\quad \phi _{4}(z)=\sum _{0}^{\infty }\frac{1+m}{n+m}z^{n},\quad \mathfrak {R}m>0. \end{aligned}$$

We note that $\phi _{i},i=1,2,3,4$ are convex. Now using Theorem 2.6. $\square$

Corollary 2.8

The radius of starlikeness of the class $\mathcal {S}^{k}(A,B,\alpha )$ is

$$\begin{aligned} r_{*}=\frac{2}{(1-\alpha )(A-B)+\sqrt{[(1-\alpha )(A-B)]^{2}+4B[(1-\alpha )A+\alpha B]}}. \end{aligned}$$

(2.14)

Proof

From Lemma 2.1

$$\begin{aligned} \mathfrak {R}\left( \frac{zf^{\prime }(z)}{f_{k}(z)}\right) \ge \dfrac{1-(1-\alpha )(A-B)r-B[(1-\alpha )A+\alpha B]r^{2}}{1-B^{2}r^{2}}. \end{aligned}$$

Hence for $r<r_{*}$ the first hand side of the preceding inequality is positive this implies (2.14). $\square$