On some inequalities of a certain class of analytic functions

Abstract

The aim of this paper is to study the properties of a subclass of analytic functions related to p-valent Bazilevic functions by using the concept of differential subordination. We investigate some results concerned with coefficient bounds, inclusion results, radius problem, covering theorem, angular estimation of a certain integral operator, and some other interesting properties.

MSC:30C45, 30C50.

1 Introduction and preliminaries

Let $A_p$ be the class of analytic functions

$$f(z)=z^p+\sum _{n=p+1}^{\mathrm{\infty }}{a}_n{z}^n(p\in \mathbb{N}=\{1,2,3,\dots \}),$$

(1.1)

defined in the open unit disc $E=\{z:|z|<1\}$. A function $f\in A_p$ is a p-valent starlike function of order ρ if and only if

$$Re\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}>\rho ,0\le \rho <p,z\in E.$$

This class of functions is denoted by $S_p^{\ast }(\rho )$. It is noted that $S_p^{\ast }(0)=S_p^{\ast }$. If $f(z)\in A_p$ satisfies

$$|arg\frac{z{f}^{\mathrm{\prime }}(z)}{pf(z)}|<\frac{\pi \eta }{2}$$

(1.2)

for some $\eta \in (0,1]$ and for all $z\in E$, then the function f is called strongly starlike p-valent of order η in E. We denote this class by ${\tilde{S}}_p^{\ast }(\eta )$. Let $f_1(z)$ and $f_2(z)$ be analytic in E. We say $f_1(z)$ is subordinate to $f_2(z)$, written $f_1\prec f_2$ or $f_1(z)\prec f_2(z)$, if there exists a Schwarz function $w(z)$, $w(0)=0$, and $|w(z)|<1$ in E, then $f_1(z)=f_2(w(z))$. A function f in $A_p$ is said to belong to the class $S_p^{\ast }[A,B]$, $-1\le B<A\le 1$, if and only if

$$\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}\prec p\frac{1+Az}{1+Bz},z\in E.$$

For $p=1$, we obtain the class $S^{\ast }[A,B]$ of Janowski starlike functions. Janowski functions have extensively been studied by several researchers; see for example [1–3]. It is clear that $f\in S_p^{\ast }[A,B]$ if and only if

$$|\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}-p\frac{1-AB}{1-B^2}|<p\frac{A-B}{1-B^2}(-1<B<A\le 1,z\in E),$$

(1.3)

and

$$Re\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}>p\frac{1-A}{2}(B=-1,z\in E).$$

A function $f\in A_p$ is p-valent Bazilevic function of type $(\alpha ,\beta )$ and order ρ if and only if

$$Re\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}{\left(\frac{f(z)}{g(z)}\right)}^{\alpha }{\left(\frac{f(z)}{z^p}\right)}^{i\beta }>\rho ,0\le \rho <p,z\in E,$$

where $\alpha \ge 0$, $\beta \in \mathbb{R}$ and $g\in S_p^{\ast }$. For $p=1$ and $\rho =0$, this class was introduced by Bazilevic and these functions are univalent for $\alpha \ge 0$, $\beta \in \mathbb{R}$. This class of functions is studied by many authors; for some details, see [4–11].

Using this concept, we generalize and define a subclass of p-valent Bazilevic functions of type $(\alpha ,\beta )$ as follows.

Definition 1.1 A function $f\in M_p(\alpha ,\beta ,\mu ,A,B)$ if it satisfies the condition

where $\alpha \ge 0$, $\mu >0$, $g\in S_p^{\ast }$, $-1\le B<A\le 1$ and β is any real.

We have the following special cases.

• (i) For $\beta =0$, we have the subclass of Bazilevic functions defined by Patel [12].

• (ii) For $\beta =0$, $p=1$, $g(z)=z$, $A=1-2\rho$, $B=-1$, we obtain the subclass of Bazilevic functions defined in [13].

For $A=1-2\rho$, $B=-1$, we have the following subclass of analytic functions.

Definition 1.2 A function $f\in B_p(\alpha ,\beta ,\rho )$ if it satisfies the condition

where $\alpha \ge 0$, $\mu >0$, $g\in S_p^{\ast }$, $0\le \rho <1$, β is any real and $z\in E$. In other words, a function $f\in B_p(\alpha ,\beta ,\rho )$ if it satisfies the condition

We need the following definition and lemmas which will be used in our main results.

Definition 1.3 Let $\mathrm{\Psi }:{\mathbb{C}}^2\times E\to \mathbb{C}$ be analytic in a domain D and h be univalent in E. If p is analytic in E with $(p(z),z{p}^{\mathrm{\prime }}(z);z)\in D$ when $z\in E$, then we say that p satisfies a first-order differential subordination if

$$\mathrm{\Psi }(p(z),z{p}^{\mathrm{\prime }}(z);z)\prec h(z),z\in E.$$

(1.4)

The univalent function q is called dominant of the differential subordination (1.4) if $p\prec q$ for all p satisfies (1.4). If $\tilde{q}\prec q$ for all dominants of (1.4), then we say that $\tilde{q}$ is the best dominant of (1.4).

Lemma 1.4 ([14])

If $-1\le B<A\le 1$, $\lambda >0$ and the complex number γ satisfies $Re\{\gamma \}\ge -\lambda (1-A)/(1-B)$, then the differential equation

$$q(z)+\frac{z{q}^{\mathrm{\prime }}(z)}{\lambda q(z)+\gamma }=\frac{1+Az}{1+Bz},z\in E,$$

has a univalent solution in E given by

$$q(z)=\{\begin{array}{cc}\frac{z^{\lambda +\gamma }{(1+Bz)}^{\lambda (A-B)/B}}{\lambda {\int }_0^z{t}^{\lambda +\gamma -1}{(1+Bt)}^{\lambda (A-B)/B}dt}-\frac{\gamma }{\lambda },\hfill & B\ne 0,\hfill \\ \frac{z^{\lambda +\gamma }{e}^{\lambda Az}}{\lambda {\int }_0^z{t}^{\lambda +\gamma -1}{e}^{\lambda At}dt}-\frac{\gamma }{\lambda },\hfill & B=0.\hfill \end{array}$$

If $h(z)=1+c_{1}z+c_2{z}^2+\cdots$ is analytic in E and satisfies

$$h(z)+\frac{z{h}^{\mathrm{\prime }}(z)}{\lambda h(z)+\gamma }\prec \frac{1+Az}{1+Bz},z\in E,$$

then

$$h(z)\prec q(z)\prec \frac{1+Az}{1+Bz},$$

and $q(z)$ is the best dominant.

Lemma 1.5 ([15])

Let ε be a positive measure on $[0,1]$. Let g be a complex-valued function defined on $E\times [0,1]$ such that $g(\cdot ,t)$ is analytic in E for each $t\in [0,1]$ and $g(z,\cdot )$ is ε-integrable on $[0,1]$ for all $z\in E$. In addition, suppose that $Reg(z,t)>0$, $g(-r,t)$ is real and $Re\{1/g(z,t)\}\ge 1/g(-r,t)$ for $|z|\le r<1$ and $t\in [0,1]$. If $g(z)={\int }_0^{1}g(z,t)d\epsilon (t)$, then $Re\{1/g(z)\}\ge 1/g(-r)$.

Lemma 1.6 ([[16], Chapter 14])

Let $a_1$, $b_1$ and $c_1\ne 0,-1,-2,\dots$ be complex numbers. Then, for $Re{c}_1>Re{b}_1>0$,

Lemma 1.7 ([17])

Let $-1\le B_1\le B_2<A_2\le A_1\le 1$. Then

$$\frac{1+A_{2}z}{1+B_{2}z}\prec \frac{1+A_{1}z}{1+B_{1}z}.$$

Lemma 1.8 ([18])

Let F be analytic and convex in E. If $f,g\in A_p$ and $f,g\prec F$, then

$$\mu f+(1-\mu )g\prec F,0\le \mu \le 1.$$

Lemma 1.9 ([19])

Let $f(z)={\sum }_{k=0}^{\mathrm{\infty }}{a}_k{z}^k$ be analytic in E and $F(z)={\sum }_{k=0}^{\mathrm{\infty }}{b}_k{z}^k$ be analytic and convex in E. If $f\prec F$, then

$$|a_k|\le |b_1|(k\in \mathbb{N}).$$

Lemma 1.10 ([20])

Let $h(z)=1+d_{1}z+d_2{z}^2+\cdots$ be analytic in E and $h(z)\ne 0$ in E. If there exists a point $z_0\in E$ such that $|argh(z)|<\frac{\pi }{2}\eta$ ($|z|<|z_0|$) and $|argh(z_0)|=\frac{\pi }{2}\eta$ ($0<\eta \le 1$), then we have $\frac{z_0{h}^{\mathrm{\prime }}(z_0)}{h(z_0)}=ik\eta$, where

$$\{\begin{array}{cc}k\ge \frac{1}{2}(x+\frac{1}{x}),\hfill & \mathit{\text{when}}argh(z_0)=\frac{\pi }{2}\eta ,\hfill \\ k\le -\frac{1}{2}(x+\frac{1}{x}),\hfill & \mathit{\text{when}}argh(z_0)=-\frac{\pi }{2}\eta ,\hfill \end{array}$$

and ${(h(z_0))}^{1/\eta }=\pm ix$ ($x>0$).

Lemma 1.11 Let $g\in S^{\ast }[A,B]$. Then the function

$$G(z)={[\frac{c+\alpha +i\beta }{z^{c+i\beta }}{\int }_0^z{t}^{c+i\beta -1}{g}^{\alpha }(t)dt]}^{\frac{1}{\alpha }}$$

(1.5)

belongs to $S^{\ast }[A,B]$ for $c\ge -\alpha \frac{1-A}{1-B}$.

Proof is straightforward by using Lemma 1.4.

Throughout this paper, $\alpha \ge 0$, $\beta \in \mathbb{R}$, $\mu >0$, and $-1\le B<A\le 1$ unless otherwise stated.

2 Main results

Theorem 2.1 If $f\in M_p(\alpha ,\beta ,\mu ,A,B)$, then

$$\frac{z{f}^{\mathrm{\prime }}(z)}{pf(z)}{\left(\frac{f(z)}{g(z)}\right)}^{\alpha }{\left(\frac{f(z)}{z^p}\right)}^{i\beta }\prec q(z),$$

(2.1)

where $q(z)=\frac{\mu }{pQ(z)}$ and

$$Q(z)=\{\begin{array}{cc}{\int }_0^1{t}^{\frac{p}{\mu }-1}{(\frac{1+Bzt}{1+Bz})}^{\frac{p}{\mu }(A-B)/B}dt,\hfill & B\ne 0,\hfill \\ {\int }_0^1{t}^{\frac{p}{\mu }-1}{e}^{\frac{p}{\mu }(t-1)Az}dt,\hfill & B=0.\hfill \end{array}$$

(2.2)

In hypergeometric function form,

$$q(z)=\{\begin{array}{cc}{{[}_2{F}_1(1,\frac{p}{\mu }(1-\frac{A}{B});\frac{p}{\mu }+1;\frac{Bz}{Bz+1})]}^{-1},\hfill & B\ne 0,\hfill \\ {{[}_1{F}_1(1,\frac{p}{\mu }+1;-\frac{p}{\mu }Az)]}^{-1},\hfill & B=0,\hfill \end{array}$$

(2.3)

and if $A<-\frac{\mu B}{p}$, $-1\le B<0$, then $M_p(\alpha ,\beta ,\mu ,A,B)\subset B_p(\alpha ,\beta ,\rho )$, where

$$\rho =p{\{}_2{F}_1(1,\frac{p}{\mu }(1-\frac{A}{B});\frac{p}{\mu }+1;\frac{B}{B-1}){\}}^{-1}.$$

(2.4)

This result is best possible.

Proof Let

$$h(z)=\frac{z{f}^{\mathrm{\prime }}(z)}{pf(z)}{\left(\frac{f(z)}{g(z)}\right)}^{\alpha }{\left(\frac{f(z)}{z^p}\right)}^{i\beta },$$

where $h(z)$ is analytic in E with $h(0)=1$. Differentiating logarithmically, we obtain

(2.5)

Using Lemma 1.4 for $\lambda =\frac{p}{\mu }$ and $\gamma =0$, we have

$$h(z)\prec q(z)\prec \frac{1+Az}{1+Bz},$$

where $q(z)$ is given in (2.3) and is the best dominant of (2.5). Next, in order to prove $M_p(\alpha ,\beta ,\mu ,A,B)\subset B_p(\alpha ,\beta ,\rho )$, we show that ${inf}_{|z|<1}\{Req(z)\}=q(-1)$. Now, we set $a=\frac{p}{\mu }(B-A)/B$, $b=\frac{p}{\mu }$ and $c=\frac{p}{\mu }+1$, then it is clear that $c>b>0$; therefore, for $B\ne 0$ it follows from (2.2) by using Lemma 1.6 that

$$\begin{array}{rcl}Q(z)& =& {(1+Bz)}^a{\int }_0^1{t}^{b-1}{(1+Btz)}^{-a}dt\\ =& {\frac{\mathrm{\Gamma }(b)}{\mathrm{\Gamma }(c)}}_2{F}_1(1,a,c;\frac{Bz}{Bz+1}).\end{array}$$

(2.6)

To prove that ${inf}_{|z|<1}\{Req(z)\}=q(-1)$, we need to show that

$$Re\{1/Q(z)\}\ge 1/Q(-1).$$

Since $A<-\frac{\mu B}{p}$ with $-1\le B<0$ implies that $c>a>0$, therefore, by using Lemma 1.6, (2.6) yields

$$Q(z)={\int }_0^{1}g(z,t)d\epsilon (t),$$

where

which is a positive measure on $[0,1]$. For $-1\le B<0$ it is clear that $Reg(z,t)>0$ and $g(-r,t)$ is real for $0\le |z|\le r<1$ and $t\in [0,1]$. Also,

$$Re\left\{\frac{1}{g(z,t)}\right\}=Re\left\{\frac{1+(1-t)Bz}{1+Bz}\right\}\ge \frac{1-(1-t)Br}{1-Br}=\frac{1}{g(-r,t)}$$

for $|z|\le r<1$. Therefore, using Lemma 1.5, we have

$$Re\{1/Q(z)\}\ge 1/Q(-r).$$

Now, letting $r\to 1^{-}$, it follows

$$Re\{1/Q(z)\}\ge 1/Q(-1).$$

Therefore, $M_p(\alpha ,\beta ,\mu ,A,B)\subset B_p(\alpha ,\beta ,\rho )$. □

For $\beta =0$, we have the following result proved in [12].

Corollary 2.2 If $f\in M_p(\alpha ,\mu ,A,B)$, then

$$\frac{z{f}^{\mathrm{\prime }}(z)}{pf(z)}{\left(\frac{f(z)}{g(z)}\right)}^{\alpha }\prec \{\begin{array}{cc}{{[}_2{F}_1(1,\frac{p}{\mu }(1-\frac{A}{B});\frac{p}{\mu }+1;\frac{Bz}{Bz+1})]}^{-1},\hfill & B\ne 0,\hfill \\ {{[}_1{F}_1(1,\frac{p}{\mu }+1;-\frac{p}{\mu }Az)]}^{-1},\hfill & B=0,\hfill \end{array}$$

and if $A<-\frac{\mu B}{p}$, $-1\le B<0$, then $M_p(\alpha ,\mu ,A,B)\subset B_p(\alpha ,\rho )$, where

$$\rho =p{\{}_2{F}_1(1,\frac{p}{\mu }(1-\frac{A}{B});\frac{p}{\mu }+1;\frac{B}{B-1}){\}}^{-1}.$$

This result is best possible.

For $p=1$, we have the class $M_1(\alpha ,\beta ,\mu ,A,B)=M(\alpha ,\beta ,\mu ,A,B)$. We denote the class of functions $f\in A$, having Taylor series representation of the form

$$f(z)=z+\sum _{k=n+1}^{\mathrm{\infty }}{a}_k{z}^k$$

and satisfying the condition

(2.7)

by $M^{\ast }(\alpha ,\beta ,\mu ,A,B)$, where $g(z)=z+{\sum }_{k=n+1}^{\mathrm{\infty }}{b}_k{z}^k$ such that $Re\frac{z{g}^{\mathrm{\prime }}(z)}{g(z)}>0$. Now, we derive the following result for the class $M^{\ast }(\alpha ,\beta ,\mu ,A,B)$.

Theorem 2.3 Let $f\in M^{\ast }(\alpha ,\beta ,\mu ,A,B)$. Then

$$|a_{n+1}|\le \frac{(A-B)+\alpha (n+1)(1+\mu n)}{|\alpha +i\beta +n|(1+\mu n)}.$$

(2.8)

Proof Since $f\in M^{\ast }(\alpha ,\beta ,\mu ,A,B)$, therefore,

Now, using the fact that $f(z)=z+{\sum }_{k=n+1}^{\mathrm{\infty }}{a}_k{z}^k$ and $g(z)=z+{\sum }_{k=n+1}^{\mathrm{\infty }}{b}_k{z}^k$, we obtain

$$1+\{(\alpha +i\beta +n)(1+\mu n)a_{n+1}-\alpha (1+\mu n)b_{n+1}\}{z}^n+\cdots \prec \frac{1+Az}{1+Bz}.$$

By a well-known result due to Janowski and Lemma 1.9, we have

$$|(\alpha +i\beta +n)(1+\mu n)a_{n+1}-\alpha (1+\mu n)b_{n+1}|\le A-B.$$

By the triangle inequality, we obtain

$$|(\alpha +i\beta +n)(1+\mu n)a_{n+1}|-|\alpha (1+\mu n)b_{n+1}|\le A-B.$$

Using the coefficient bound for the class $S^{\ast }$, we have the required result. □

For $\beta =0$ and $g(z)=z$, we have the following result proved in [13].

Corollary 2.4 Let $f\in M(\alpha ,\mu ,A,B)$. Then

$$|a_{n+1}|\le \frac{(A-B)}{(\alpha +n)(1+\mu n)}.$$

For $\beta =0$, $p=1$, $g(z)=z$, $A=1-2\rho$ and $B=-1$, we have the following result proved in [21].

Corollary 2.5 Let f satisfy the condition

$$Re\{\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}{\left(\frac{f(z)}{z}\right)}^{\alpha }+\mu [1+\frac{z{f}^{\mathrm{\prime }\mathrm{\prime }}(z)}{f^{\mathrm{\prime }}(z)}+(1-\alpha )(1-\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)})]\}>\rho .$$

Then

$$|a_{n+1}|\le \frac{2(1-\rho )}{(n+\alpha )(1+\mu n)}.$$

Theorem 2.6 For ${\mu }_2\ge {\mu }_1\ge 0$ and $-1\le B_1\le B_2<A_2\le A_1\le 1$,

$$M_p(\alpha ,\beta ,{\mu }_2,A_2,B_2)\subset M_p(\alpha ,\beta ,{\mu }_1,A_1,B_1).$$

Proof Let $f\in M_p(\alpha ,\beta ,{\mu }_2,A_2,B_2)$. Then

Since $-1\le B_1\le B_2<A_2\le A_1\le 1$, therefore by Lemma 1.7, we have

Hence, we have $f\in M_p(\alpha ,\beta ,{\mu }_2,A_1,B_1)$. For ${\mu }_2={\mu }_1\ge 0$, we have the required result. When ${\mu }_2>{\mu }_1\ge 0$, Theorem 2.1 implies that

$$\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}{\left(\frac{f(z)}{g(z)}\right)}^{\alpha }{\left(\frac{f(z)}{z^p}\right)}^{i\beta }\prec p\frac{1+A_{1}z}{1+B_{1}z}.$$

Now

Using Lemma 1.8, we have the required result. □

For $\beta =0$, we have the following result.

Corollary 2.7 For ${\mu }_2\ge {\mu }_1\ge 0$ and $-1\le B_1\le B_2<A_2\le A_1\le 1$,

$$M_p(\alpha ,{\mu }_2,A_2,B_2)\subset M_p(\alpha ,{\mu }_1,A_1,B_1).$$

This result is proved in [13].

For $\beta =0$, $p=1$, $g(z)=z$, $A=1-2\rho$ and $B=-1$, we have the class $M(\alpha ,\mu ,\rho )$ defined as

$$Re[\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}{\left(\frac{f(z)}{z}\right)}^{\alpha }+\mu \{1+\frac{z{f}^{\mathrm{\prime }\mathrm{\prime }}(z)}{f^{\mathrm{\prime }}(z)}-\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}+\alpha (\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}-1)\}]>\rho$$

(2.9)

for $z\in E$. Now have the following result for the class $M(\alpha ,\mu ,\rho )$ proved in [21].

Corollary 2.8 For $\alpha \ge 0$, ${\mu }_2\ge {\mu }_1\ge 0$ and $1>{\rho }_2\ge {\rho }_1\ge 0$,

$$M(\alpha ,{\mu }_2,{\rho }_2)\subset M(\alpha ,{\mu }_1,{\rho }_1).$$

Theorem 2.9 Let $f\in A_p$ satisfy

$$Re\left\{\frac{f(z)}{z^p}\right\}>0\mathit{\text{and}}|\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}{\left(\frac{f(z)}{g(z)}\right)}^{\alpha }{\left(\frac{f(z)}{z^p}\right)}^{i\beta }-p|<\sigma p,0<\sigma \le 1,$$

for $g\in S_p^{\ast }$. Then f is p-valent convex in $|z|<R_{\alpha ,\beta ,\sigma }$, where

$$\begin{array}{rcl}{R}_{\alpha ,\beta ,\sigma }& =& ((2|1-\alpha -i\beta |+2\alpha p+\sigma )\\ -\sqrt{{(2|1-\alpha -i\beta |+2\alpha p+\sigma )}^2-4p(2\alpha p-p-\sigma )})/(2(2\alpha p-p-\sigma )).\end{array}$$

(2.10)

Proof Let

$$h(z)=\frac{z{f}^{\mathrm{\prime }}(z)}{pf(z)}{\left(\frac{f(z)}{g(z)}\right)}^{\alpha }{\left(\frac{f(z)}{z^p}\right)}^{i\beta }-1,$$

where $h(z)$ is analytic in E with $h(0)=0$ and $|h(z)|<1$. By using the Schwarz lemma, we get

$$h(z)=\sigma z\psi (z),$$

where $\psi (z)$ is analytic in E with $|\psi (z)|<1$. Differentiating logarithmically, we have

$$1+\frac{z{f}^{\mathrm{\prime }\mathrm{\prime }}(z)}{f^{\mathrm{\prime }}(z)}=(1-\alpha -i\beta )\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}+\alpha \frac{z{g}^{\mathrm{\prime }}(z)}{g(z)}+\frac{\sigma z(z{\psi }^{\mathrm{\prime }}(z)+\psi (z))}{1+\sigma z\psi (z)}+ip\beta .$$

Since $Re\{\frac{f(z)}{z^p}\}>0$, therefore,

$$\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}=p+\frac{z{\phi }^{\mathrm{\prime }}(z)}{\phi (z)},Re\phi (z)>0.$$

This implies that

$$Re\{1+\frac{z{f}^{\mathrm{\prime }\mathrm{\prime }}(z)}{f^{\mathrm{\prime }}(z)}\}\ge (1-\alpha )p+\alpha Re\frac{z{g}^{\mathrm{\prime }}(z)}{g(z)}-|1-\alpha -i\beta |\left|\frac{z{\phi }^{\mathrm{\prime }}(z)}{\phi (z)}\right|-\sigma \left|\frac{z(z{\psi }^{\mathrm{\prime }}(z)+\psi (z))}{1+\sigma z\psi (z)}\right|.$$

Now, using the well-known results for classes $S_p^{\ast }$, P and the Schwarz function [22], we have

$$\begin{array}{rcl}Re\{1+\frac{z{f}^{\mathrm{\prime }\mathrm{\prime }}(z)}{f^{\mathrm{\prime }}(z)}\}& \ge & (1-\alpha )p+\alpha p\frac{1-r}{1+r}-|1-\alpha -i\beta |\frac{2r}{(1-r^2)}-\frac{\sigma r}{(1-r)}(0<\sigma \le 1)\\ =& \frac{(2\alpha p-p-\sigma )r^2-(2|1-\alpha -i\beta |+2\alpha p+\sigma )r+p}{(1-r^2)}.\end{array}$$

Let $P(r)=(2\alpha p-p-\sigma )r^2-(2|1-\alpha -i\beta |+2\alpha p+\sigma )r+p$. Since $p\in \mathbb{N}$ and $0<\sigma \le 1$, therefore, $P(0)=p>0$ and $P(1)=-2(|1-\alpha -i\beta |+\sigma )<0$. It follows that the root lies in $(0,1)$. This implies that $Re\{1+\frac{z{f}^{\mathrm{\prime }\mathrm{\prime }}(z)}{f^{\mathrm{\prime }}(z)}\}>0$ if $r<R_{\alpha ,\beta ,\sigma }$, where $R_{\alpha ,\beta ,\sigma }$ is given by (2.10). □

For $\sigma =1$ and $\beta =0$, we have the following result which is proved in [12].

Corollary 2.10 Let $f\in A_p$ satisfy

$$Re\left\{\frac{f(z)}{z^p}\right\}>0\mathit{\text{and}}|\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}{\left(\frac{f(z)}{g(z)}\right)}^{\alpha }-p|<p,$$

$g\in S_p^{\ast }$. Then f is p-valent convex in $|z|<R_{\alpha }$, where

$$R_{\alpha }=\frac{3+2\alpha (p-1)-\sqrt{{(3+2\alpha (p-1))}^2-4p(2\alpha p-p-1)}}{2(2\alpha p-p-1)}.$$

Theorem 2.11 Let $f\in A_p$ satisfy

$$|\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}{\left(\frac{f(z)}{g(z)}\right)}^{\alpha }-p|<\sigma p,$$

for $g\in S_p^{\ast }$. Then, for $\alpha >0$, f is p-valent $\frac{1}{\alpha }$-convex in $|z|<R_{\alpha ,\sigma }$, where

$$R_{\alpha ,\sigma }=\frac{2\alpha p+\sigma -\sqrt{{(2\alpha p+\sigma )}^2-4\alpha p(\alpha p-\sigma )}}{2(\alpha p-\sigma )}.$$

(2.11)

Proof Let

$$h(z)=\frac{z{f}^{\mathrm{\prime }}(z)}{pf(z)}{\left(\frac{f(z)}{g(z)}\right)}^{\alpha }-1,$$

where $h(z)$ is analytic in E with $h(0)=0$ and $|h(z)|<1$. By using the Schwarz lemma, we get

$$h(z)=\sigma z\psi (z),$$

where $\psi (z)$ is analytic in E with $|\psi (z)|<1$. Differentiating logarithmically, we have

$$\frac{1}{\alpha }(1+\frac{z{f}^{\mathrm{\prime }\mathrm{\prime }}(z)}{f^{\mathrm{\prime }}(z)})+(1-\frac{1}{\alpha })\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}=\frac{z{g}^{\mathrm{\prime }}(z)}{g(z)}+\frac{\sigma }{\alpha }\frac{z(z{\psi }^{\mathrm{\prime }}(z)+\psi (z))}{1+\sigma z\psi (z)}.$$

This implies that

$$Re\{\frac{1}{\alpha }(1+\frac{z{f}^{\mathrm{\prime }\mathrm{\prime }}(z)}{f^{\mathrm{\prime }}(z)})+(1-\frac{1}{\alpha })\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}\}\ge Re\frac{z{g}^{\mathrm{\prime }}(z)}{g(z)}-\frac{\sigma }{\alpha }\left|\frac{(z{\psi }^{\mathrm{\prime }}(z)+\psi (z))}{1+\sigma z\psi (z)}\right|.$$

Now, using the well-known results for classes $S_p^{\ast }$, and the Schwarz function, we have

$$\begin{array}{rcl}Re\frac{1}{p}\{\frac{1}{\alpha }(1+\frac{z{f}^{\mathrm{\prime }\mathrm{\prime }}(z)}{f^{\mathrm{\prime }}(z)})+(1-\frac{1}{\alpha })\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}\}& \ge & \frac{1-r}{1+r}-\frac{\sigma r}{\alpha p(1-r)}(0<\sigma \le 1)\\ =& \frac{(\alpha p-\sigma )r^2-(2\alpha p+\sigma )r+\alpha p}{\alpha (1-r^2)}.\end{array}$$

Let $Q(r)=(\alpha p-\sigma )r^2-(2\alpha p+\sigma )r+\alpha p$. Then for $p\in \mathbb{N}$, $\alpha >0$ and $0<\sigma \le 1$, $Q(0)=\alpha p>0$ and $Q(1)=-2\sigma <0$. It shows that the root lies in $(0,1)$. This implies that $Re\{\frac{1}{\alpha }(1+\frac{z{f}^{\mathrm{\prime }\mathrm{\prime }}(z)}{f^{\mathrm{\prime }}(z)})+(1-\frac{1}{\alpha })\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}\}>0$ if $r<R_{\alpha ,\sigma }$, where $R_{\alpha ,\sigma }$ is given by (2.11). □

Theorem 2.12 Let $f\in M(\alpha ,\beta ,\mu ,A,B)$. Then E is mapped by f on a domain that contains the disc $|w|<R_{\alpha ,\beta ,\mu }\in (0,1)$, where

$$R_{\alpha ,\beta ,\mu }=\frac{|\alpha +i\beta +1|(1+\mu )}{2|\alpha +i\beta +1|(1+\mu )+(A-B)+2\alpha (1+\mu )}.$$

(2.12)

Proof Let $w_0$ be any complex number such that $f(z)\ne w_0$. Then

$$\frac{w_{0}f(z)}{w_0-f(z)}=z+(a_2+\frac{1}{w_0})z^2+\cdots ,$$

is univalent in E, so that

$$|a_2+\frac{1}{w_0}|\le 2.$$

Therefore,

$$\left|\frac{1}{w_0}\right|-|a_2|\le 2.$$

Hence,

$$w_0\ge \frac{|\alpha +i\beta +1|(1+\mu )}{2|\alpha +i\beta +1|(1+\mu )+(A-B)+2\alpha (1+\mu )}=R_{\alpha ,\beta ,\mu }.$$

 □

For $\beta =0$, $g(z)=z$, $A=1-2{\rho }_1$, $B=-1$, we have the following result proved in [13].

Corollary 2.13 Let $f\in M(\alpha ,\mu ,\rho )$. Then E is mapped by f on a domain that contains the disc $|w|<R_{\alpha ,\mu }$, where

$$R_{\alpha ,\beta ,\mu }=\frac{(1+\alpha )(1+\mu )}{2(1+\alpha )(1+\mu )+2(1-{\rho }_1)}.$$

Theorem 2.14 Let $\alpha >0$, $c\ge -\alpha \frac{1-A}{1-B}$ and let $f\in A$. If

$$|arg\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}{\left(\frac{f(z)}{g(z)}\right)}^{\alpha }{\left(\frac{f(z)}{z}\right)}^{i\beta }-\rho |<\frac{\pi }{2}\upsilon (0<\upsilon \le 1,0\le \rho <1),$$

(2.13)

for some $g\in S^{\ast }$, then

$$|arg\frac{z{F}^{\mathrm{\prime }}(z)}{F(z)}{\left(\frac{F(z)}{G(z)}\right)}^{\alpha }{\left(\frac{F(z)}{z}\right)}^{i\beta }-\rho |<\frac{\pi }{2}\eta (0<\eta \le 1),$$

(2.14)

where

$$F(z)={[\frac{c+\alpha +i\beta }{z^c}{\int }_0^z{t}^{c-1}{f}^{\alpha +i\beta }(t)dt]}^{\frac{1}{\alpha +i\beta }},$$

(2.15)

with

$$\upsilon =\{\begin{array}{cc}\eta +\frac{2}{\pi }{tan}^{-1}(\frac{(1+B)\eta sin(\pi (1-t(\alpha ,\beta ,c,A,B))/2)}{|c+i\beta |(1+B)+\alpha (1+A)+\eta (1+B)cos(\pi (1-t(\alpha ,\beta ,c,A,B))/2)}),\hfill & B\ne -1,\hfill \\ \eta ,\hfill & B=-1,\hfill \end{array}$$

(2.16)

and

$$t(\alpha ,\beta ,c,A,B)=\frac{2}{\pi }{sin}^{-1}\frac{\alpha (A-B)+\beta (1-B^2)}{|c+i\beta |(1-B^2)+\alpha (1-AB)}.$$

(2.17)

Proof Since

$$F^{\alpha +i\beta }(z)=\frac{c+\alpha +i\beta }{z^c}{\int }_0^z{t}^{c-1}{f}^{\alpha +i\beta }(t)dt,$$

therefore,

$$z^{1-i\beta }{F}^{\alpha +i\beta -1}(z)F^{\mathrm{\prime }}(z)=\frac{1}{\alpha +i\beta }\{(c+\alpha +i\beta )z^{-i\beta }{f}^{\alpha +i\beta }(z)-c{z}^{-i\beta }{F}^{\alpha +i\beta }(z)\}.$$

Now using (1.5), we have

$$\frac{z{F}^{\mathrm{\prime }}(z)}{F(z)}{\left(\frac{F(z)}{G(z)}\right)}^{\alpha }{\left(\frac{F(z)}{z}\right)}^{i\beta }=\frac{\frac{1}{\alpha +i\beta }\{(c+\alpha +i\beta )z^c{f}^{\alpha +i\beta }(z)-c{z}^c{F}^{\alpha +i\beta }(z)\}}{(c+\alpha +i\beta ){\int }_0^z{t}^{c+i\beta -1}{g}^{\alpha }(t)dt}.$$

Let

$$h(z)=\frac{1}{1-\rho }(\frac{z{F}^{\mathrm{\prime }}(z)}{F(z)}{\left(\frac{F(z)}{G(z)}\right)}^{\alpha }{\left(\frac{F(z)}{z}\right)}^{i\beta }-\rho )=\frac{N(z)}{D(z)},$$

where $h(z)$ is analytic with $h(0)=1$. Now

$$\begin{array}{rcl}\frac{N^{\mathrm{\prime }}(z)}{D^{\mathrm{\prime }}(z)}& =& \frac{1}{1-\rho }(\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}{\left(\frac{f(z)}{g(z)}\right)}^{\alpha }{\left(\frac{f(z)}{z}\right)}^{i\beta }-\rho )\\ =& h(z)\{1+\frac{D(z)}{z{D}^{\mathrm{\prime }}(z)}\cdot \frac{z{h}^{\mathrm{\prime }}(z)}{h(z)}\},\end{array}$$

where

$$\begin{array}{rcl}N(z)& =& \frac{1}{1-\rho }(\frac{1}{\alpha +i\beta }\{(c+\alpha +i\beta )z^c{f}^{\alpha +i\beta }(z)-c{z}^c{F}^{\alpha +i\beta }(z)\}-\rho z^{c+i\beta -1}{G}^{\alpha }(z)),\\ D(z)& =& (c+\alpha +i\beta ){\int }_0^z{t}^{c+i\beta -1}{g}^{\alpha }(t)dt.\end{array}$$

Since $G\in S^{\ast }[A,B]$, therefore, we can write

$$\frac{z{D}^{\mathrm{\prime }}(z)}{D(z)}=c+i\beta +\alpha \frac{z{G}^{\mathrm{\prime }}(z)}{G(z)}=r_1{e}^{i\frac{\pi }{2}\theta },$$

where

$$\{\begin{array}{cc}|c+i\beta |+\alpha \frac{1-A}{1-B}<r_1<|c+i\beta |+\alpha \frac{1+A}{1+B},\hfill & B\ne -1,\hfill \\ -t(\alpha ,\beta ,c,A,B)<\theta <t(\alpha ,\beta ,c,A,B),\hfill & B\ne -1,\hfill \end{array}$$

(2.18)

similarly

$$\{\begin{array}{cc}|c+i\beta |+\alpha \frac{1-A}{2}<r_1<\mathrm{\infty },\hfill & B=-1,\hfill \\ -1<\theta <1,\hfill & B=-1.\hfill \end{array}$$

(2.19)

Suppose that $h(z)\ne 0$ in E, there exists a point $z_0\in E$ such that $|argh(z)|<\frac{\pi }{2}\eta$ ($|z|<|z_0|$) and $|argh(z_0)|=\frac{\pi }{2}\eta$. Now, using Lemma 1.10, we have $\frac{z{h}^{\mathrm{\prime }}(z_0)}{h(z_0)}=ik\eta$. At first suppose that $h(z_0)={(ix)}^{\eta }$ ($x>0$) for the case $B\ne -1$, we obtain

where υ and $t(\alpha ,\beta ,c,A,B)$ are given by (2.16) and (2.17) respectively. For $B=-1$, we have

$$arg(\frac{z_0{f}^{\mathrm{\prime }}(z_0)}{f(z_0)}{\left(\frac{f(z_0)}{g(z_0)}\right)}^{\alpha }{\left(\frac{f(z_0)}{z_0}\right)}^{i\beta }-\rho )\ge \frac{\pi }{2}\eta ,$$

which is a contradiction to the assumption of our theorem. Now we suppose that $h(z_0)={(-ix)}^{\eta }$. For the case $B\ne -1$ using a similar method, we obtain

$$arg(\frac{z_0{f}^{\mathrm{\prime }}(z_0)}{f(z_0)}{\left(\frac{f(z_0)}{g(z_0)}\right)}^{\alpha }{\left(\frac{f(z_0)}{z_0}\right)}^{i\beta }-\rho )\le -\frac{\pi }{2}\upsilon ,$$

and for $B=-1$, we have

$$arg(\frac{z_0{f}^{\mathrm{\prime }}(z_0)}{f(z_0)}{\left(\frac{f(z_0)}{g(z_0)}\right)}^{\alpha }{\left(\frac{f(z_0)}{z_0}\right)}^{i\beta }-\rho )\le -\frac{\pi }{2}\eta ,$$

which is a contradiction to the assumption of our theorem. Hence, we have the proof. □

This kind of problem is also considered in [23]. For $\beta =0$, we have the following result proved in [24].

Corollary 2.15 Let $\alpha >0$, $c\ge -\alpha \frac{1-A}{1-B}$ and let $f\in A$. If

$$|arg(\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}{\left(\frac{f(z)}{g(z)}\right)}^{\alpha }-\rho )|<\frac{\pi }{2}\upsilon (0<\upsilon \le 1),$$

then

$$|arg(\frac{z{F}^{\mathrm{\prime }}(z)}{F(z)}{\left(\frac{F(z)}{G(z)}\right)}^{\alpha }-\rho )|<\frac{\pi }{2}\eta (0<\eta \le 1),$$

where

$$F(z)={[\frac{c+\alpha }{z^c}{\int }_0^z{t}^{c-1}{f}^{\alpha }(t)dt]}^{\frac{1}{\alpha }},$$

with

$$\upsilon =\{\begin{array}{cc}\eta +\frac{2}{\pi }{tan}^{-1}(\frac{(1+B)\eta sin(\pi (1-t(\alpha ,\beta ,c,A,B))/2)}{c(1+B)+\alpha (1+A)+\eta (1+B)cos(\pi (1-t(\alpha ,\beta ,c,A,B))/2)}),\hfill & B\ne -1,\hfill \\ \eta ,\hfill & B=-1,\hfill \end{array}$$

and

$$t(\alpha ,\beta ,c,A,B)=\frac{2}{\pi }{sin}^{-1}\frac{\alpha (A-B)}{c(1-B^2)+\alpha (1-AB)}.$$

Remark 2.16 By using the suitable choices of parameters c, α, A and B, we can find many results proved in the literature.