On strongly starlike and strongly convex functions with bounded radius and bounded boundary rotation

Abstract

In this paper, we introduce and investigate new classes of normalized analytic functions in an open unit disk with bounded radius and bounded boundary rotation by using the subordination. We discuss inclusion results, co-efficient bounds, growth and distortion theorems of the classes. Moreover, we compute the radii of strong starlikeness, convexity and starlikeness of the classes. It is interesting to mention that most of our findings are best possible as compared to the existing results in the literature.

1 Introduction and preliminaries

We represent $\mathcal{A}$ as the class of functions $f(z)$, which is analytic in $\mathtt{E}= \{z\in C: \vert z \vert <1 \}$. Then, for each $z\in E$, $f(z)$ has the form

$$ f(z)=z+\sum_{n=2}^{\infty }a _{{n}}z^{ {n}} . $$

(1.1)

Suppose that $\mathtt{g}(z)$ and $\mathtt{G}(z)$ are analytic functions in E. Then $\mathtt{g}(z)$ is said to be subordinate to $\mathtt{G}(z)$ written as $\mathtt{g}(z)\prec \mathtt{G}(z)$ if and only if there exists $w(z)$, which is analytic in E with $w(\mathtt{0})= \mathtt{0}$ and $\vert w(z) \vert <1$. Therefore, $\mathtt{g}(z)= \mathtt{G}(w(z))$ belongs to E. $\mathtt{g}(z)$ is univalent in E follows that $\mathtt{g}\prec \mathtt{G}$ is equivalent to $\mathtt{g}( \mathtt{0})=\mathtt{G}(\mathtt{0})$ and $\mathtt{g}( E)\subset \mathtt{G}(\mathtt{E})$.

The function $F(A,B,z)= ( \frac{1+Az}{1+Bz} )$ is the conformal mapping of the unit disk to circle, which is symmetric with respect to the real axis having center $\frac{1-AB}{1-B^{2}}$ and radius $\frac{A-B}{1-B^{2}}$ for each A, B such that $-1\leq B< A\leq 1$. A function $p(z)$ (with $p(\mathtt{0})=1$) is analytic in E and belongs to the class $P[A,B]$ if $p(z)$ is subordinate to $F(A,B,z)$. In [5], Janowski introduced and investigated the class $P[A,B]$. Later on, Noor and Arif [8] investigated the class $P_{m}[A,B]$. They investigated that a function $p(z)$ (with $p( \mathtt{0})=1$) is analytic in E and belongs to the class $P_{m}[A,B]$ if and only if there exist $p_{1}(z), p_{2}(z)\in \mathrm{P}[A,B]$ such that

$$ p(z)= \biggl( \frac{m}{4}+\frac{1}{2} \biggr) p_{1}(z)- \biggl( \frac{m}{4}-\frac{1}{2} \biggr) p_{2}(z) $$

(1.2)

for ${m}\geq 2$, $-1\leq B< A\leq 1$, and $z\in E$.

In [14], Rajapat et al. used the known family of fractional integral operators (with the Gauss hypergeometric function in the kernel) and defined new subclasses of strongly starlike and strongly convex functions of order β and type α in the open unit disk U. Moreover, they established several inclusion relationships and interesting results associated with the fractional integral operators.

Shiraishi et al. [16] investigated some new sufficient conditions for the class of strong Caratheodory functions in the open unit disk U. Example and several corollaries of the main results were presented.

We consider a function $F_{\beta }(A,B,z)= ( \frac{1+Az}{1+Bz} ) ^{\beta }$ for $-1\leq B\leq 1$, $-1\leq A\leq 1$ ($A\neq B$), and $0<\beta \leq 1$, which is analytic and univalent in E. If $p(z)$ (with $p(\mathtt{0})=1$) is subordinate to $F_{\beta }(A,B,z)$, then the function $p(z)$ is analytic in E and belongs to the class ${\tilde{\mathrm{P}}}^{\beta }[A,B]$.

Lemma 1.1

([10])

Suppose$f\in \mathcal{V}_{m} ( \rho )$for$m \geq 2$and$\mathtt{0} \leq \rho <1$. Then$f ( z ) = ( f_{1} ( z ) ) ^{ ( 1-\rho ) }$for$f_{1} ( z ) \in \mathcal{V}_{m}$.

Lemma 1.2

([6])

Suppose${\xi }={\xi }_{1}+ i{\xi }_{2}$, ${\zeta }={\zeta }_{1}+i {\zeta }_{2}$and$\varPhi :D\subset C^{2}\rightarrow C$is a complex-valued function, which satisfies the following conditions:

• (i) ${\varPsi } ( {\xi },{\zeta } )$is continuous in a domain$D\subset C^{2}$,

• (ii) $( 1,\mathtt{0} ) \in \mathsf{D}$and${\varPhi } ( 1,\mathtt{0} ) >\mathtt{0}$,

• (iii) $\operatorname{Re}{\varPhi } ( i{\xi }_{2},{\zeta }_{1} ) \leq \mathtt{0}$, whenever$( i{\xi }_{2},{\zeta }_{1} ) \in \mathsf{D}$and${\zeta }_{1}\leq -\frac{1}{2} ( 1+{\xi }_{2}^{2} )$.

If$q(z)=1+\sum_{n=1}^{\infty }c_{n}z^{n}$is an analytic function inEwith$( q(z),zq^{\prime }(z) ) \in D$and$\operatorname{Re}\varPhi (q(z), zq^{\prime }(z))$for$z\in E$, then$\operatorname{Re}(q(z))>0$, for$z\in E$.

Lemma 1.3

([3])

Suppose$\mathsf{r}_{a}\leq \operatorname{Re}(a)\sin ( \frac{\alpha \pi }{2} ) -\operatorname{Re}(a) \cos ( \frac{\alpha \pi }{2} )$, $\operatorname{Im}(a)\geq 0$. Then the disk$\vert w-a \vert \leq \mathsf{r} _{a}$is contained in the sector$\vert \arg (w) \vert \leq \frac{\alpha \pi }{2}$for$\mathtt{0}\leq \alpha <1$.

2 Main results

In this section, we discuss the coefficient problem, analytic property, inclusion results, and radius problem. It is interesting to mention that our obtained results are sharp as compared to the existing results in the literature.

First we define the following.

Definition 2.1

A function $q(z)$ (with $q(\mathtt{0})=1$) is analytic in E and belongs to the class $\tilde{P}_{m}^{ \beta }[A,B]$ (for ${m}\geq 2$) if and only if there exist $p_{1}(z), p_{2}(z)\in {\tilde{\mathrm{P}}}^{\beta }[A,B]$ such that

$$ q(z)= \biggl( \frac{m}{4}+\frac{1}{2} \biggr) q_{1}(z)- \biggl( \frac{m}{4}-\frac{1}{2} \biggr) q_{2}(z)\quad \text{for } z\in E. $$

It is easy to see that the set ${\tilde{\mathrm{P}}}_{m}^{\beta }[A,B]$ is convex. For special values of A, B, and β, we obtain several subclasses of analytic functions studied and investigated by several researchers [4, 7–9, 13–15].

Definition 2.2

An analytic function $f(z )\in \tilde{\mathsf{R}}_{m}^{\beta }[A,B]$ if and only if

$$ \frac{zf^{{\prime }}(z)}{f(z)}\in {\tilde{\mathrm{P}}}_{m}^{\beta }[A,B] \quad \text{for }z\in E. $$

(2.1)

Note that, for $\beta =1$, the class $\tilde{\mathsf{R}}_{m}^{\beta }[A,B]$ yields the class $\mathsf{R}_{m}[A,B]$ [8] and $\tilde{\mathsf{R}}_{m}^{\beta }[1,-1]=\mathsf{R}_{m}(\beta )$. For $\beta =\frac{1}{2}$, $B= \mathtt{0}$, and $A=a$ with $a\in (\mathtt{0},1]$, we obtain the class

$$ \tilde{\mathsf{R}}_{m}^{{\frac{1}{2}}}[a,\mathtt{0}]= q_{1}, q_{2}\prec \sqrt{1+az}\quad \text{for }z \in E. $$

(2.2)

Definition 2.3

A function $f\in \mathcal{A}$ belongs to $\tilde{\mathsf{V}}_{m}^{\beta }[A,B]$ if and only if f is locally univalent and

$$ \frac{(zf^{{\prime }}(z))^{{\prime }}}{f^{{\prime }}(z)}\in { \tilde{\mathrm{P}}}_{m}^{\beta }[A,B] \quad \text{for }z\in E. $$

(2.3)

From (2.1) and (2.3), it is considered that

$$ f\in \tilde{\mathsf{V}}_{m}^{\beta }[A,B]\quad \Leftrightarrow\quad zf'(z)\in \tilde{\mathsf{R}}_{m}^{\beta }[A,B]. $$

(2.4)

Note that, for particular cases of A, B, and β, several classes are obtained and investigated in [8]. For $\beta =\frac{1}{2}$, $m=2$, $B=\mathtt{0}$ and $A=a$, $a\in (\mathtt{0},1]$, Aouf [1] obtained the following class:

$$ \tilde{\mathsf{V}}_{m}^{{\frac{1}{2}}}[a,\mathtt{0}]= \biggl\{ f\in A:\frac{(zf^{{\prime }}(z))^{{\prime }}}{f^{{\prime }}(z)}= \biggl( \frac{m}{4}+\frac{1}{2} \biggr) q_{1}(z)- \biggl( \frac{m}{4}- \frac{1}{2} \biggr) q_{2}(z), q_{1}, q_{2}\prec \sqrt{1+az} \biggr\} . $$

Now we prove the following lemmas, which represent the coefficient inequality of functions belonging to the class ${\tilde{\mathrm{P}}}_{m}^{\beta }[A,B]$. By choosing special values of A, B, and β, we may relate our finding with the existing results. We shall assume $m\geq 2$, $-1\leq B< A\leq 1$, $\mathtt{0}<\beta \leq 1$, unless otherwise stated.

Lemma 2.4

Suppose$q(z)=1+\sum_{n=1}^{\infty }a_{n}z^{n}$belongs to${\tilde{\mathrm{P}}}_{m}^{\beta }[A,B]$. Then

$$ \vert a_{n} \vert \leq \frac{\beta m}{2} \vert A-B \vert \quad \textit{for all }n\geq 1. $$

(2.4*)

Proof

Suppose $q(z)$ belongs to ${\tilde{\mathrm{P}}}_{m}^{\beta }[A,B]$ with $q(z)=1+\sum_{n=1}^{\infty }a_{n}z^{n}$. Then

$$ q(z)= \biggl( \frac{m}{4}+\frac{1}{2} \biggr) q_{1}(z)- \biggl( \frac{m}{4}-\frac{1}{2} \biggr) q_{2}(z), \quad q_{1},q_{2}\in { \tilde{\mathrm{P}}}^{ \beta }[A,B]. $$

If $q_{1}(z)=1+\sum_{n=1}^{\infty }{b}_{n}z^{n}$ and $q_{2}(z)=1+\sum_{n=1}^{\infty }{c}_{n}z^{n}$, then

$$ 1+\sum_{n=1}^{\infty }a_{n}z^{n}= \biggl( \frac{m}{4}+ \frac{1}{2} \biggr) \Biggl( 1+\sum_{n=1}^{\infty }{b}_{n}z^{n} \Biggr) - \biggl( \frac{m}{4}-\frac{1}{2} \biggr) \Biggl( 1+ \sum_{n=1}^{\infty }{c}_{n}z^{n} \Biggr) . $$

By comparing the coefficients of $z^{n}$ and using the triangle inequality, we get

$$ \vert a_{n} \vert \leq \biggl( \frac{m}{4}+ \frac{1}{2} \biggr) \vert {b}_{n} \vert + \biggl( \frac{m}{4}- \frac{1}{2} \biggr) \vert {c}_{n} \vert . $$

Since $q_{i}(z)\prec ( \frac{1+Az}{1+Bz} ) ^{\beta }=1+ \beta (A-B)z+\cdots$ for $i=1,2$. Using the well-known result of Rogosinski [15] on subordination, we have $\vert {b}_{n} \vert \leq \beta \vert A-B \vert$ and $\vert c_{n} \vert \leq \beta \vert A-B \vert$ for all $n\geq 1$. This implies that

$$ \vert a_{n} \vert \leq \biggl( \frac{m}{4}+ \frac{1}{2} \biggr) \beta \vert A-B \vert + \biggl( \frac{m}{4}- \frac{1}{2} \biggr) \beta \vert A-B \vert , $$

which is as required. This completes the proof. □

The following lemma extends the inequality bound of functions in the class $\mathrm{P}_{m}[A,B]$.

Lemma 2.5

If$q(z)$belongs to${\tilde{\mathrm{P}}}_{m}^{\beta }[A,B]$and$z=\mathrm{r}e^{i\theta }$, then

$$ \frac{m(k_{1}-k_{2})+2(k_{1}+k_{2})}{4(1-B^{2}r^{2})}\leq \operatorname{Re}\bigl\{ q(z)\bigr\} \leq \frac{m}{2}\biggl(\frac{1+Ar}{1+Br}\biggr)^{\beta }, $$

(2.5)

where$k_{1}= ( 1-(A-B)r-ABr^{2} ) ^{\beta }$, $k_{2}= ( 1+(A-B)r-ABr^{2} ) ^{\beta }$.

Proof

Suppose $q(z) \in {\tilde{\mathrm{P}}}_{m}^{\beta }[A,B]$. Then

$$ q(z)= \biggl( \frac{m}{4}+\frac{1}{2} \biggr) q_{1}(z)- \biggl( \frac{m}{4}-\frac{1}{2} \biggr) q_{2}(z),\quad \text{where }q_{1}(z),q_{2}(z)\text{ in } \tilde{\mathrm{P}}^{\beta }[A,B]. $$

(2.6)

Using the triangle inequality, we have

$$ \operatorname{Re}\bigl(q(z)\bigr)\leq \bigl\vert {q(z)} \bigr\vert \leq \biggl( \frac{m}{4}+\frac{1}{2} \biggr) \bigl\vert {q_{1}(z)} \bigr\vert + \biggl( \frac{m}{4}- \frac{1}{2} \biggr) \bigl\vert {q_{2}(z)} \bigr\vert . $$

(2.7)

Since $q_{1}(z), q_{2}(z)\in {\tilde{\mathrm{P}}}^{\beta }[A,B]$, then

$$ \operatorname{Re}\bigl\{ q_{i}(z)\bigr\} \leq \biggl( \frac{1+A \vert w \vert }{1+B \vert w \vert }\biggr)^{\beta }< \biggl( \frac{1+Ar}{1+Br}\biggr)^{\beta } \quad \text{for }i=1,2, $$

(2.8)

where $w(z)$ is defined in E, which is an analytic function with $w{(0)}=0$ and $\vert w(z) \vert < r$. Using (2.8) in (2.7), we get the right-hand side of (2.5).

Moreover, (2.6) can be written as

$$ {\operatorname{Re}\bigl(q(z)\bigr)}\geq \biggl( \frac{m}{4}+ \frac{1}{2} \biggr) \operatorname{Re}\bigl\{ q_{1}(z)\bigr\} - \biggl( \frac{m}{4}-\frac{1}{2} \biggr) \bigl\vert {q_{2}(z)} \bigr\vert . $$

(2.9)

Since $q_{1}(z), q_{2}(z)\in {\tilde{\mathrm{P}}}^{\beta }[A,B]$ and

$$ \operatorname{Re}\bigl\{ q_{i}(z)\bigr\} \geq \biggl( \frac{1-A \vert w \vert }{1-B \vert w \vert }\biggr)^{\beta }>\biggl( \frac{1-Ar}{1-Br}\biggr)^{\beta } \quad \text{for } i=1,2. $$

(2.10)

Using (2.10) in (2.9), we have

$$ \bigl\{ \operatorname{Re}q(z)\bigr\} \geq \biggl( \frac{m}{4}+ \frac{1}{2} \biggr) \biggl( \frac{1-Ar}{1-Br}\biggr)^{\beta }- \biggl( \frac{m}{4}-\frac{1}{2} \biggr) \biggl(\frac{1+Ar}{1+Br}\biggr)^{ \beta }. $$

(2.11)

This implies that

$$\begin{aligned} {\bigl\{ \operatorname{Re}q(z)\bigr\} }&\geq \biggl( \frac{m}{4}+ \frac{1}{2} \biggr) \biggl( \frac{1-(A-B)r-ABr^{2}}{1-B^{2}r^{2}}\biggr)^{\beta } \\ &\quad {}- \biggl( \frac{m}{4}- \frac{1}{2} \biggr) \biggl( \frac{1+(A-B)r-ABr^{2}}{1-B^{2}r^{2}}\biggr)^{\beta }. \end{aligned}$$

(2.11*)

By simple calculations, we get the left-hand side of (2.5). Hence the proof. □

Lemma 2.6

If a function$q(z)\in \mathcal{A}$belongs to${\tilde{\mathrm{P}}}_{m}^{\beta }[A,B]$, then$q(z)\in \mathrm{P}_{m}(\gamma )$, where$\gamma = ( \frac{1-A}{1-B} ) ^{\beta }$for$z \in E$.

Proof

Let $q(z)$ belong to ${\tilde{\mathrm{P}}}_{m}^{\beta }[A,B]$. From (2.6) and (2.8), we have

$$ \operatorname{Re}\bigl\{ q_{i}(z)\bigr\} \geq \biggl( \frac{1-A \vert w \vert }{1-B \vert w \vert }\biggr)^{\beta }>\biggl( \frac{1-Ar}{1-Br}\biggr)^{\beta },\quad i=1,2,\ z\in E. $$

(2.12)

From this it follows that each $q_{i}(z)\in \mathrm{P}(\gamma )$ for $i=1,2$, where $\gamma = ( \frac{1-A}{1-B} ) ^{\beta }$ and $q(z)\in \mathrm{P}_{m}(\gamma )$ for $\gamma = ( \frac{1-A}{1-B} ) ^{\beta }$. This implies that ${\tilde{\mathrm{P}}}_{m}^{\beta }[A,B]\subseteq \mathrm{P}_{m}(\gamma )$. Hence the proof. □

In the following theorems, we discuss and investigate the coefficient problem, analytic property, inclusion results, and radius problem.

Theorem 2.7

If$f\in \tilde{\mathsf{V}}_{m}^{\beta }[A,B]$, then there exist${\sigma }_{1},{\sigma }_{2}\in \overset{\ast }{S}(\gamma )$with$\gamma = ( \frac{1-A}{1-B} ) ^{\beta }$such that

$$ f^{{\prime }}(z)= \frac{ ( \frac{{\sigma }_{1}(z)}{z} ) ^{(\frac{m}{4}+\frac{1}{2})}}{ ( \frac{{\sigma }_{2}(z)}{z} ) ^{(\frac{m}{4}-\frac{1}{2})}},\quad z\in E. $$

(2.13)

Proof

Let $f(z)$ belong to $\tilde{\mathsf{V}}_{m}^{\beta }[A,B]$. Then $q(z)=\frac{(zf^{{\prime }}(z))^{{\prime }}}{f^{{\prime }}(z)}\in {\tilde{\mathrm{P}}}_{m}^{\beta }[A,B]$. By Lemma 2.6, ${\tilde{\mathrm{P}}} _{m}^{\beta }[A, B]\subseteq P_{m}(\gamma )$. This implies that $\frac{(zf^{{\prime }}(z))^{{\prime }}}{f^{{\prime }}(z)}\in P_{m}( \gamma )$ with $\gamma = ( \frac{1-A}{1-B} ) ^{\beta }$. It follows that $f\in V_{m}(\gamma )$. By Lemma 1.1, it has the form

$$ f'(z)= \bigl( \mathtt{g}(z) \bigr) ^{1-\lambda },\quad \text{where }\lambda =\biggl(\frac{1-A}{1-B}\biggr)^{\beta } \text{ and } \mathtt{g}(z)\in \mathsf{V}_{m}. $$

(2.14)

Brannan [2] showed that each $\mathtt{g}(\mathsf{z})\in \mathsf{V}_{m}$ has the representation of the form

$$ g'(z)= \frac{ ( \frac{\mathsf{s}_{1}(z)}{z} ) ^{(\frac{m}{4}+\frac{1}{2})}}{ ( \frac{\mathsf{s}_{2}(z)}{z} ) ^{(\frac{m}{4}-\frac{1}{2})}},\quad \mathsf{s}_{1}, \mathsf{s}_{2}\in \overset{\ast }{S}. $$

(2.15)

It has been proved in [12] that each ${\sigma }_{i}\in \overset{\ast }{S}(\gamma )$ is

$$ {\sigma }_{i}(z)=z \biggl( \frac{\mathsf{s}_{{i}}(z)}{z} \biggr) ^{1-\gamma },\quad \mathsf{s}_{{i}}(z )\in \overset{\ast }{S},\ i=1,2. $$

(2.16)

Using (2.14), (2.15), and (2.16), we get our required result (2.13). This completes the proof. □

Theorem 2.8

Let$f(z)$belong to$\tilde{\mathsf{R}}_{m}^{\beta }[A,B]$with$f(z)=z+\sum_{n=2}^{\infty }a_{n}z^{n}$for$z\in E$. Then

$$ \vert a_{n} \vert \leq \frac{(\beta \frac{m}{2} \vert A-B \vert )_{n-1}}{(n-1)!}\quad \textit{for all }n\geq 2. $$

(2.17)

Proof

Let $f(z)$ belong to $\tilde{\mathsf{R}}_{m}^{\beta }[A,B]$. Then $q(z)=\frac{zf^{{\prime }}(z)}{f(z)}\in {\tilde{\mathrm{P}}}_{m}^{ \beta }[A,B]$. Suppose $q(z)=\frac{zf^{{\prime }}(z)}{f(z)}=1+\sum_{n=1}^{\infty }c_{n}z^{n}$, where $q(z)$ is defined in E and is analytic with $q( \mathtt{0})=1$. This implies

$$ zf^{{\prime }}(z)=f(z)q(z). $$

(2.18)

The series representation of (2.18) is

$$ z+\sum_{n=2}^{\infty }na_{n}z^{n}= \Biggl( z+\sum_{n=2}^{ \infty }a_{n}z^{n} \Biggr) \Biggl( 1+\sum_{n=1}^{\infty }c_{n}z^{n} \Biggr) =z+\sum_{n=2}^{\infty }\sum _{ {i}=\mathtt{0}}^{n-1}c_{{i}}a_{n-{i}}z^{n}. $$

Comparing the coefficient of $z^{n}$, we obtain $( n-1 ) a_{n}=\sum_{{i}=1}^{n-1}c_{ {i}}a_{n-{i}}$, $c_{\mathtt{0}}=1$ and hence

$$ \vert a_{n} \vert =\frac{1}{n-1}\sum _{ {i}=1}^{n-1} \vert c_{{i}} \vert \vert a_{n-{i}} \vert \quad \text{for }n\geq 2. $$

(2.19)

Since $q(z)\in {\tilde{\mathrm{P}}}_{m}^{\beta }[A,B]$ and using Lemma 2.4, we get $\vert c_{{i}} \vert \leq \frac{\beta m}{2}(A-B)$ for all n ≥1. From (2.19), we have $\vert a_{n} \vert \leq \frac{\beta {m}(A-B)}{2(n-1)}\sum_{ {i}=1}^{n-1} \vert a_{{i}} \vert$. In particular,

$$\begin{aligned} & \vert a_{2} \vert \leq \frac{\beta m}{2}(A-B), \\ & \vert a_{3} \vert \leq \frac{\beta \frac{m}{2}(A-B)}{2} \biggl(\frac{\beta m}{2}(A-B)+1\biggr)= \frac{(\frac{m}{2}(A-B))_{2}}{(2)!}. \end{aligned}$$

For $n=\mathsf{j}$,

$$\begin{aligned} \vert a_{\mathsf{j}} \vert &\leq \frac{\beta \frac{m}{2}(A-B)}{\mathsf{j}-1}\sum _{{i}=1}^{j-1} \vert a_{{i}} \vert = \frac{\beta \frac{m}{2}(A-B)}{\mathsf{j}-1}\prod_{i=1}^{\mathsf{j}-2} \biggl( \frac{\beta \frac{m}{2}(A-B)}{i}+1\biggr) \\ &= \frac{(\beta \frac{m}{2}(A-B))_{\mathsf{j}-1}}{(\mathsf{j}-1)!}\quad \text{for } \mathsf{j}\geq 3, \end{aligned}$$

(2.20)

where we use the Pochhammer notation $( \beta ) _{\mathsf{m}}$ introduced in [11]:

$$ (\beta )_{m}= \textstyle\begin{cases} 1, & m=0,\beta \in \mathbb{C}\setminus \{0\}, \\ \beta (\beta +1)\cdot (\beta +m-1), & m\in \mathbb{N},\beta \in \mathbb{C}. \end{cases} $$

Consider

$$ \vert a_{m+1} \vert \leq \frac{\beta \frac{m}{2}(A-B)}{m}\sum _{{i}=1}^{m-1} \vert a_{ {i}} \vert +\frac{\beta \frac{m}{2}(A-B)}{m} \vert a_{m} \vert . $$

Using (2.20), we obtain the following:

$$\begin{aligned} \vert a_{\mathsf{j}+1} \vert & \leq \frac{\beta \frac{m}{2}(A-B)}{\mathsf{j}}\prod _{i=1}^{\mathsf{j}-2}\biggl( \frac{\beta \frac{m}{2}(A-B)}{i}+1\biggr)+ \frac{\frac{m}{2}(\beta (A-B))^{2}}{\mathsf{j}(\mathsf{j}-1)}\prod_{ i=1}^{\mathsf{j}-2} \biggl(\frac{\beta (A-B)}{i}+1\biggr) \\ & =\frac{\beta \frac{m}{2}(A-B)}{\mathsf{j}}\prod_{ i=1}^{\mathsf{j}-1} \biggl(\frac{\beta \frac{m}{2}(A-B)}{i}+1\biggr)=\frac{(\beta \frac{m}{2}(A-B))_{m}}{(\mathsf{j})!}. \end{aligned}$$

By induction, we get (2.17). Hence the proof. □

Theorem 2.9

Let$f(z)$belong to$\tilde{\mathsf{V}}_{m}^{\beta }[A,B]$with$f(z)=z+\sum_{n=2}^{\infty }a_{n}z^{n}$, $z\in E$. Then

$$ \vert a_{n} \vert \leq \frac{(\beta \frac{m}{2}(A-B))_{n-1}}{(n)!}\quad \textit{for }n \geq 2. $$

Proof

Let $f(z)$ belong to $\tilde{\mathsf{V}}_{m}^{\beta }[A,B]$ with $f(z)=z+\sum_{n=2}^{\infty }a_{n}z^{n}$. Then $zf^{{\prime }}(z)\in \tilde{\mathsf{R}}_{m}^{\beta }[A,B]$. By using Theorem 2.8, we get

$$ \vert a_{n} \vert \leq \frac{(\beta \frac{m}{2}(A-B))_{n-1}}{(n)!}\quad \text{for }n\geq 2. $$

 □

Corollary 2.10

For$A=1$, $B=-1$, $f\in \tilde{\mathsf{V}}_{m}^{\beta }[1,-1]\equiv \tilde{\mathsf{V}}_{m}(\beta )$and using Theorem 2.9, we obtain

$$ \vert a_{n} \vert \leq \frac{(2m\beta )_{n-1}}{(n)!}\quad \textit{for } n\geq 2. $$

Theorem 2.11

If$f\in \tilde{\mathsf{V}}_{m}^{\beta }[A,B]$, then$f\in \tilde{\mathsf{R}}_{m}^{\beta }(\rho )$, where

$$ \rho = \frac{ ( 2\gamma -1 ) +\sqrt{ ( 1-2\gamma ) ^{2}+8}}{4},\qquad \gamma = \biggl( \frac{1-A}{1-B} \biggr) ^{ \beta }. $$

(2.21)

Proof

Suppose that

$$\begin{aligned} \frac{zf^{{\prime }}(z)}{f(z)}& = ( 1-\rho ) q(z)+\rho \\ & = ( 1-\rho ) \biggl\{ \biggl( \frac{m}{4}+\frac{1}{2} \biggr) q_{1}(z)- \biggl( \frac{m}{4}-\frac{1}{2} \biggr) q_{2}(z) \biggr\} +\rho , \end{aligned}$$

(2.22)

where $q(z)$ is analytic in E with $q(\mathtt{0})=1$. From (2.21), we have

$$ \frac{(zf^{{\prime }}(z))^{{\prime }}}{f^{{\prime }}(z)}= ( 1- \rho ) \biggl[ q(z)+ \frac{zq^{\prime }(z)}{ ( 1-\rho ) q(z)+\rho. } \biggr] +\rho . $$

(2.23)

$f\in \tilde{\mathsf{V}}_{m}^{\beta }[A,B]$ implies that $\frac{(zf^{{\prime }}(z))^{{\prime }}}{f^{{\prime }}(z)}\in {\tilde{\mathrm{P}}}_{m}^{\beta }[A,B]$ and ${\tilde{\mathrm{P}}}_{m}^{\beta }[A,B]\subseteq \mathrm{P}_{m} (\gamma )$, where $\gamma =(\frac{1-A}{1-B})^{\beta }$. From this it follows that

$$ \frac{1}{1-\gamma }\biggl( \frac{(zf^{{\prime }}(z))^{{\prime }}}{f^{{\prime }}(z)}-\gamma \biggr)\in P_{m}. $$

(2.24)

Using (2.23) and (2.24), we get

$$ \frac{ ( 1-\rho ) }{1-\gamma } \biggl[ q(z)+ \frac{\frac{1}{1-\rho }z\mathsf{q}'(z)}{q(z)+\frac{\rho }{1-\rho }.} \biggr] + \frac{(\rho -\gamma )}{1-\gamma }\in P_{m}. $$

(2.25)

Now we define

$$ \varPhi _{\eta _{1},\eta _{2}}(z)=\frac{1}{\eta _{2}+1} \frac{z}{(1-z)^{\eta _{1}}}+ \frac{\eta _{2}}{\eta _{2}+1} \frac{z}{(1-z)^{\eta _{1}+1}}, $$

with $\eta _{1}=\frac{1}{1-\rho }$, $\eta _{2}=\frac{\rho }{1-\rho }$. Using (2.22) with the convolution techniques given by Noor [7], we have

$$\begin{aligned} q(z)+\frac{\eta _{1}z\mathsf{q}(z)}{q(z)+\eta _{2}} =& ( 1- \rho ) \biggl[ \biggl( \frac{m}{4}+ \frac{1}{2} \biggr) \biggl\{ q_{1}(z)+ \frac{\eta _{1}z\mathsf{q}'_{1}(z)}{q_{1}(z)+\eta _{2}.} \biggr\} \\ &{}- \biggl( \frac{m}{4}- \frac{1}{2} \biggr) \biggl\{ q_{2}(z)+ \frac{\eta _{1}z\mathsf{q}'_{2}(z)}{q_{2}(z)+\eta _{2}.} \biggr\} \biggr] +\rho . \end{aligned}$$

(2.26)

Thus, using (2.25) and (2.26), we have

$$ \frac{ ( 1-\rho ) }{1-\gamma } \biggl\{ q_{i}(z)+ \frac{\eta _{1}z\mathsf{q}'_{i}(z)}{q_{i}(z)+\eta _{2}.} \biggr\} +\frac{(\rho -\gamma )}{1-\gamma }\in \mathrm{P}\quad \text{for }i=1,2\text{ and }z \in E. $$

(2.27)

Since the first two conditions of Lemma 1.2 are obviously satisfied, we satisfy condition (iii) as follows:

$$\begin{aligned} \operatorname{Re}\varPhi (i\xi _{2},\zeta _{1}) =& \frac{\rho -\gamma }{1-\gamma }+ \frac{1-\rho }{1-\gamma }\operatorname{Re}\biggl\{ i\xi _{2}+ \frac{\rho \zeta _{1}}{(1-\rho )i\xi _{2}+\rho }\biggr\} \\ & \leq \frac{\rho -\gamma }{1-\gamma }-\frac{1-\rho }{2(1-\gamma )} \biggl\{ \frac{\rho (1+\zeta _{2}^{2}}{(1-\rho )^{2}\xi _{2}^{2}+\rho ^{2}}\biggr\} , \quad \text{for } \xi _{1}\leq -\frac{1+\xi _{2}^{2}}{2} \\ & =\frac{A+B\xi _{2}^{2}}{D}, \end{aligned}$$

(2.28)

where

$$\begin{aligned} &A =2(\rho -\gamma )\rho ^{2}-\rho ( 1-\rho ) , \\ &B =2(\rho -\gamma ) ( 1-\rho ) ^{2}-\rho ( 1-\rho ) , \\ &D =2(1-\gamma ) \bigl( ( 1-\rho ) ^{2}{u}_{2}^{2}+ \rho ^{2} \bigr) . \end{aligned}$$

It is observed that (2.28) gives negative value for ${A}\leq \mathtt{0}$ and $B\leq {0}$. Therefore, using $A\leq {0}$, we have

$$ \rho = \frac{ ( 2\gamma -1 ) +\sqrt{ ( 1-2\gamma ) ^{2}+8}}{4}, \qquad \gamma = \biggl( \frac{1-A}{1-B} \biggr) ^{ \beta }, $$

(2.29)

and from $B\leq \mathtt{0}$, we obtain $\mathtt{0} \leq \rho <1$. Hence all three conditions of Lemma 1.2 are satisfied. From this it follows that $q_{i}(z)\in \mathrm{P}$ for $i=1,2$ and $z\in E$. Therefore $q(z)\in \mathrm{P}_{m}$ and $f\in R_{m}(\rho )$, where ρ is given in (2.29). This completes the proof. □

Corollary 2.12

If$\beta =\frac{1}{2}$, $B=0$, $A=a$, $a\in (0,1]$and$f\in \tilde{\mathsf{V}}_{m}^{\frac{1}{2}}[a,0]$. Then, using Theorem 2.11, we have$f\in\tilde{\mathsf{R}}_{m}^{\frac{1}{2}}(\rho )$, where

$$ \rho = \frac{ ( 2\gamma -1 ) +\sqrt{ ( 1-2\gamma ) ^{2}+8}}{4},\qquad \gamma =\sqrt{1-a}. $$

(2.30)

Theorem 2.13

If$f(z)$belongs to$\tilde{\mathsf{R}}_{m}^{\beta }[A,B]$, then$f(z)\in \mathsf{R}_{m}(\gamma )$, $\gamma = ( \frac{1-A}{1-B} ) ^{\beta }$for$z \in E$.

Proof

$f(z)$ belongs to $\tilde{\mathsf{R}}_{m}^{\beta }[A,B]$ implies $q(z)=\frac{zf^{{\prime }}(z)}{f(z)}$ for some $q\in\tilde{\mathrm{P}}_{m}^{\beta }[A,B]$. Using Lemma 2.6, we get $\frac{zf^{{\prime }}(z)}{f(z)}\in \mathrm{P}_{m}(\gamma )$, where $\gamma = ( \frac{1-A}{1-B} ) ^{\beta }$. Thus $f(z)\in \mathsf{R}_{m}(\gamma )$ for $z\in E$. This completes the proof. □

Corollary 2.14

If$\beta =1$, $m=2$and$f\in S^{\ast }[A,B]$. Then, using Theorem 2.13, we have$f(z)\in S^{\ast }(\gamma )$, where$\gamma =(\frac{1-A}{1-B})$.

3 Conclusion

We introduced some new classes of analytic functions with bounded radius and bounded boundary rotation by using the subordination. We discussed inclusion results, coefficient bounds, growth and distortion theorems of the classes. The radii of starlikeness and strong starlikeness of the classes have been computed. It is observed that most of the results are best possible.