Study of simply connected convex domain and its geometric properties (II)

Abstract

In this paper, making use of a new integral operator defined in the open unit disk we introduce and study a new class of simply connected convex domain of analytic and univalent functions. We derive some inclusion results and its geometric properties. Moreover, we discuss analytic criteria for a member of the simply connected convex domain to be a member of the class of functions with positive real part greater than μ.

Keywords:

• Analytic functions
• differential operator
• meromorphic convex functions
• open unit disk

2010 Mathematics Subject Classification:

• Primary 30C45

1. Introduction

We study simply connected domain including starlike and convex domain in geometric function theory. Miller and Mocanu (cf. [1]) not only enlarged the scope of a simply connected domain but also worked to spread out in the area. In recent years, more and more researchers are interested to study the simply connected domain including starlike and convex domain (cf. [2–10]). They not only introduced certain new simply connected domain including starlike and convex domain but also used them for the study of geometric properties of analytic and univalent functions. We are motivated by the research works based on analytic and univalent functions (cf. [11] and [12]). The articles provided an idea to introduce new subclasses of analytic and univalent functions in the open unit disk.

The functions analytic and univalent in $\mathbb{U}=\{z\in \mathbb{C}:|z|<1\}$ of the form (1(1) $f\left(z\right)=z^p+\sum _{k=0}^{\mathrm{\infty }}{a}_k{z}^k.$(1) ) with $p\in \mathbb{N}$ are said to form the class $A_p$. (1) $f\left(z\right)=z^p+\sum _{k=0}^{\mathrm{\infty }}{a}_k{z}^k.$(1) For $0\le \rho <1$ and $k\ge 2$, let $P_k\left(\rho \right)$ (cf. [13]) denote the class of analytic function $h\left(z\right)$ with positive real part such that $h\left(0\right)=1$ and satisfy the analytic criterion (2) ${\int }_0^{2\pi }\left|\frac{\mathrm{\Re }\left(p\right(z\left)\right)-\rho }{1-\rho }\right|\mathrm{d}\theta \le k\pi ,z=r{e}^{i\theta }.$(2) Moreover, $P_k\left(0\right)=P_k$. For the classes $P_k\left(\rho \right)$ and $P_k$ (cf.[14].) Note that $P_2\left(\beta \right)=P\left(\beta \right)$, where $P\left(\beta \right)$ is the class of functions with positive real part greater than β and $P_2\left(0\right)=P$ is the class of functions with positive real part. We can write (2(2) ${\int }_0^{2\pi }\left|\frac{\mathrm{\Re }\left(p\right(z\left)\right)-\rho }{1-\rho }\right|\mathrm{d}\theta \le k\pi ,z=r{e}^{i\theta }.$(2) ) in Riemann–Stieltjes sense as (3) $p\left(z\right)=\frac{1}{2}{\int }_0^{2\pi }\frac{1+(1-\rho )z{e}^{-it}}{1-z{e}^{-it}}\mathrm{d}\mu \left(t\right),$(3) where $\mu \left(t\right)$ is a function with bounded variation on $[0,2\pi ]$ such that (4) ${\int }_0^{2\pi }\mathrm{d}\mu \left(t\right)=2\mathrm{\&}{\int }_0^{2\pi }\left|\mathrm{d}\mu \left(t\right)\right|\le k.$(4) By using (2(2) ${\int }_0^{2\pi }\left|\frac{\mathrm{\Re }\left(p\right(z\left)\right)-\rho }{1-\rho }\right|\mathrm{d}\theta \le k\pi ,z=r{e}^{i\theta }.$(2) ), we deduce that $h\in P_k\left(\rho \right)$ if and only if there exist $h_1,h_2\in P\left(\rho \right)$ such that (5) $\begin{array}{rl}h\left(z\right)& =\left(\frac{k}{4}+\frac{1}{2}\right)h_1\left(z\right)-\left(\frac{k}{4}-\frac{1}{2}\right)h_2\left(z\right),\\ \mathrm{\forall }z\in \mathbb{U}& =\left\{z\in \mathbb{C}:\left|z\right|<1\right\}.\end{array}$(5)

2. Materials and methods

For f given by (1(1) $f\left(z\right)=z^p+\sum _{k=0}^{\mathrm{\infty }}{a}_k{z}^k.$(1) ), $\gamma \ge 0$ and $\beta \ge 0$, we introduce a new integral operator as follows: (6) $\begin{array}{rl}{\mathrm{\Psi }}_p^0(\beta ,\gamma )f\left(z\right)& =f\left(z\right);\\ {\mathrm{\Psi }}_p^1(\beta ,\gamma )f\left(z\right)& =\left(p+\frac{\beta }{1+\gamma }\right)z^{-\frac{\beta }{1+\gamma }}{\int }_o^z{t}^{\frac{\beta }{1+\gamma }-1}f\left(t\right)\mathrm{d}t;\\ {\mathrm{\Psi }}_p^2(\beta ,\gamma )f\left(z\right)& =\left(p+\frac{\beta }{1+\gamma }\right)z^{-\frac{\beta }{1+\gamma }}\\ & \times {\int }_o^z{t}^{\frac{\beta }{1+\gamma }-1}{\mathrm{\Psi }}_p^1(\beta ,\gamma )f\left(t\right)\mathrm{d}t;\\ & ⋮\\ {\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)& =\left(p+\frac{\beta }{1+\gamma }\right)z^{-\frac{\beta }{1+\gamma }}\\ & \times {\int }_o^z{t}^{\frac{\beta }{1+\gamma }-1}{\mathrm{\Psi }}_p^{n-1}(\beta ,\gamma )f\left(t\right)\mathrm{d}t.\end{array}$(6) By using (6(6) $\begin{array}{rl}{\mathrm{\Psi }}_p^0(\beta ,\gamma )f\left(z\right)& =f\left(z\right);\\ {\mathrm{\Psi }}_p^1(\beta ,\gamma )f\left(z\right)& =\left(p+\frac{\beta }{1+\gamma }\right)z^{-\frac{\beta }{1+\gamma }}{\int }_o^z{t}^{\frac{\beta }{1+\gamma }-1}f\left(t\right)\mathrm{d}t;\\ {\mathrm{\Psi }}_p^2(\beta ,\gamma )f\left(z\right)& =\left(p+\frac{\beta }{1+\gamma }\right)z^{-\frac{\beta }{1+\gamma }}\\ & \times {\int }_o^z{t}^{\frac{\beta }{1+\gamma }-1}{\mathrm{\Psi }}_p^1(\beta ,\gamma )f\left(t\right)\mathrm{d}t;\\ & ⋮\\ {\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)& =\left(p+\frac{\beta }{1+\gamma }\right)z^{-\frac{\beta }{1+\gamma }}\\ & \times {\int }_o^z{t}^{\frac{\beta }{1+\gamma }-1}{\mathrm{\Psi }}_p^{n-1}(\beta ,\gamma )f\left(t\right)\mathrm{d}t.\end{array}$(6) ), for the function f given in (1(1) $f\left(z\right)=z^p+\sum _{k=0}^{\mathrm{\infty }}{a}_k{z}^k.$(1) ), we get (7) $\begin{array}{rl}& {\mathrm{\Psi }}_p^m(\beta ,\gamma )f\left(z\right)\\ & =z^p+\sum _{k=2}^{\mathrm{\infty }}{\left(\frac{(1+\gamma )p+\beta }{(1+\gamma )k+\beta }\right)}^m{a}_k{z}^k,n\in \mathbb{N}\cup \left\{0\right\}.\end{array}$(7) Further, a straightforward calculation reveals that many differential operators introduced in other papers are special cases of the differential operator defined by (7(7) $\begin{array}{rl}& {\mathrm{\Psi }}_p^m(\beta ,\gamma )f\left(z\right)\\ & =z^p+\sum _{k=2}^{\mathrm{\infty }}{\left(\frac{(1+\gamma )p+\beta }{(1+\gamma )k+\beta }\right)}^m{a}_k{z}^k,n\in \mathbb{N}\cup \left\{0\right\}.\end{array}$(7) ). By specializing the parameters β and γ, we obtain the following operators studied by various authors:

• ${\mathrm{\Psi }}_p^{m+1}(l,0)f\left(z\right)=I_p^{m+1}\left(l\right)f\left(z\right)$[15];

• ${\mathrm{\Psi }}_1^{\alpha }(1,0)f\left(z\right)=I^{\alpha }f\left(z\right)$[16];

• ${\mathrm{\Psi }}_p^{\alpha }(1,0)f\left(z\right)=I_p^{\alpha }f\left(z\right)$[17];

• ${\mathrm{\Psi }}_p^m(1,0)f\left(z\right)=D^{m}f\left(z\right)$[18];

• ${\mathrm{\Psi }}_1^m(1,0)f\left(z\right)=I^{m}f\left(z\right)$[19];

• ${\mathrm{\Psi }}_1^m(0,0)f\left(z\right)=I^{m}f\left(z\right)$[20].

For $\lambda \ge 0$ and $\omega >0$, let ${\mathrm{\Psi }}^p(\lambda ,\omega ,\beta ,\gamma )$ denote the class of functions f defined by (1(1) $f\left(z\right)=z^p+\sum _{k=0}^{\mathrm{\infty }}{a}_k{z}^k.$(1) ) and satisfy (8) $\begin{array}{rl}& \left[(1-\lambda ){\left(\frac{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}{z^p}\right)}^{\omega }\right.\\ & +\lambda \left(\frac{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}\right)\\ & \times \left.{\left(\frac{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}{z^p}\right)}^{\omega }\right]\in P_k\left(\rho \right).\end{array}$(8) The analytic function $g\left(z\right)=z-\left(\beta /1+\gamma +\beta \right)z^2/(1-z{)}^2$, for $z\in \mathbb{U}$ belongs to the class ${\mathrm{\Psi }}^p(\lambda ,\omega ,\beta ,\gamma )$.

Next, we introduce definitions which will be used in our main results.

Definition 1

Let H be the set of complex valued functions $h(r,s,t):{\mathbb{C}}^3\to \mathbb{C}$ such that

1. $h(r,s,t)$ is continuous in a domain $D\subset {\mathbb{C}}^3$,

2. $(1,1,1)\subset D$ and $|h(1,1,1)|<1$,

then $\left|h\left(e^{i\theta },e^{i\theta }+m,e^{i\theta }+2m+\frac{\gamma +1}{\beta +p(\gamma +1)}m+L\right)\right|\ge 1,$ whenever $\left(e^{i\theta },e^{i\theta }+m,e^{i\theta }+2m+\frac{\gamma +1}{\beta +p(\gamma +1)}m+L\right)\in D,$ with $\mathrm{\Re }\left(L\right)\ge m(m-1)$ for real θ and $m\ge 1$ where $L=\left(\gamma +1/\beta +p(\gamma +1){)}^2{z}^2{w}^{\mathrm{\prime \prime }}\right(z)$.

Definition 2

Let H be the set of complex valued functions $h(r,s,t):{\mathbb{C}}^3\to \mathbb{C}$ such that

1. $h(r,s,t)$ is continuous in a domain $D\subset {\mathbb{C}}^3$,

2. $(1,1,1)\subset D$ and $|h(1,1,1)|<1$,

then $\begin{array}{rl}& \left|h\left(\phantom{\frac{m{e}^{i\theta }+\frac{\gamma +1}{\beta +p(\gamma +1)}m-m^2+L}{m+e^{i\theta }}}{e}^{i\theta },m+e^{i\theta },m+e^{i\theta }\right.\right.\\ & +\left.\left.\frac{m{e}^{i\theta }+\frac{\gamma +1}{\beta +p(\gamma +1)}m-m^2+L}{m+e^{i\theta }}\right)\right|\ge 1,\end{array}$ whenever $\begin{array}{rl}& \left(\phantom{\frac{m{e}^{i\theta }+\frac{\gamma +1}{\beta +p(\gamma +1)}m-m^2+L}{m+e^{i\theta }}}{e}^{i\theta },m+e^{i\theta },m+e^{i\theta }\right.\\ & +\left.\frac{m{e}^{i\theta }+\frac{\gamma +1}{\beta +p(\gamma +1)}m-m^2+L}{m+e^{i\theta }}\right)\in D,\end{array}$ with $\mathrm{\Re }\left(L\right)\ge m(m-1)$ for real θ and $m\ge 1$ where $L=(\gamma +1{)}^2{z}^2/(\beta +p(\gamma +1){)}^2{w}^{″}\left(z\right)/w\left(z\right)$.

Lemma 1

[1] Let $w\left(z\right)=a+w_k{z}^k+\cdots$ be analytic in $\mathbb{U}=\{z:|z|<1\}$ with $w\left(z\right)\ne a$ and $k\ge 1$. If $z_0=r_0{e}^{i\theta }$ $(0<r_0<1)$ and $|w\left(z_0\right)|=\underset{|z|\le r_0}{max}|w\left(z\right)|$. Then $z_0{w}^{\mathrm{\prime }}\left(z_0\right)=mw\left(z_0\right)$ and $\mathrm{\Re }\left\{1+\frac{z_0{w}^{″}\left(z_0\right)}{w^{\prime }\left(z_0\right)}\right\}\ge m,$ where m is real number and $m\ge k\frac{{\left|w\left(z_0\right)-a\right|}^2}{{\left|w\left(z_0\right)\right|}^2-{\left|a\right|}^2}\ge k\frac{\left|w\left(z_0\right)\right|-\left|a\right|}{\left|w\left(z_0\right)\right|+\left|a\right|}.$

Lemma 2

[21] Let $u=u_1+i{u}_2$, $v=v_1+i{v}_2$ and $\mathrm{\Phi }(u,v)$ be a complex valued function satisfy the conditions:

1. $\mathrm{\Phi }(u,v)$ is continuous in a domain $D\subseteq {\mathbb{C}}^2$,

2. $(1,0)\in D$ and $\mathrm{\Phi }(1,0)>0$,

3. $\mathrm{\Re }\left\{\mathrm{\Phi }\right(i{u}_2,v_1\left)\right\}>0$ whenever $(i{u}_2,v_1)\in D$ and $v_1\le -1/2(1+u_2^2)$.

If $h\left(z\right)=1+c_n{z}^n+c_{n+1}{z}^{n+1}+\cdots$ is analytic in $\mathbb{U}$ such that $\left(h\right(z),z{h}^{\prime }(z\left)\right)\in D$ and $\mathrm{\Re }\left(h\right(z),z{h}^{\prime }(z\left)\right)>0$ then $\mathrm{\Re }\left\{h\right(z\left)\right\}>0$ in $\mathbb{U}$.

3. Results and discussion

Theorem 3.1

If the function f given by (1(1) $f\left(z\right)=z^p+\sum _{k=0}^{\mathrm{\infty }}{a}_k{z}^k.$(1) ) belongs to the class ${\mathrm{\Psi }}^p(\lambda ,\omega ,\beta ,\gamma )$, then $({\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)/z^p{)}^{\omega }\in P_k(\mu )$, where $\mu =\frac{2\omega \rho \left\{\beta +p(1+\gamma )\right\}+\lambda (1+\gamma )}{2\omega \left\{\beta +p(1+\gamma )\right\}+\lambda (1+\gamma )}.$

Proof.

By using (2(2) ${\int }_0^{2\pi }\left|\frac{\mathrm{\Re }\left(p\right(z\left)\right)-\rho }{1-\rho }\right|\mathrm{d}\theta \le k\pi ,z=r{e}^{i\theta }.$(2) ), if $({\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)/z^p{)}^{\omega }\in P_k(\mu )$, then it must satisfy the condition given by (5(5) $\begin{array}{rl}h\left(z\right)& =\left(\frac{k}{4}+\frac{1}{2}\right)h_1\left(z\right)-\left(\frac{k}{4}-\frac{1}{2}\right)h_2\left(z\right),\\ \mathrm{\forall }z\in \mathbb{U}& =\left\{z\in \mathbb{C}:\left|z\right|<1\right\}.\end{array}$(5) ). Let (9) $\begin{array}{rl}& {\left(\frac{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}{z^p}\right)}^{\omega }\\ & =G\left(z\right)=(1-\mu )h\left(z\right)+\mu \\ & =\left(\frac{k}{4}+\frac{1}{2}\right)\left\{(1-\mu )h_1\left(z\right)+\mu \right\}\\ & +\left(\frac{k}{4}+\frac{1}{2}\right)\left\{(1-\mu )h_2\left(z\right)+\mu \right\},\end{array}$(9) where $h_1,h_2\in P\left(\mu \right)$. By using (9(9) $\begin{array}{rl}& {\left(\frac{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}{z^p}\right)}^{\omega }\\ & =G\left(z\right)=(1-\mu )h\left(z\right)+\mu \\ & =\left(\frac{k}{4}+\frac{1}{2}\right)\left\{(1-\mu )h_1\left(z\right)+\mu \right\}\\ & +\left(\frac{k}{4}+\frac{1}{2}\right)\left\{(1-\mu )h_2\left(z\right)+\mu \right\},\end{array}$(9) ) and (7(7) $\begin{array}{rl}& {\mathrm{\Psi }}_p^m(\beta ,\gamma )f\left(z\right)\\ & =z^p+\sum _{k=2}^{\mathrm{\infty }}{\left(\frac{(1+\gamma )p+\beta }{(1+\gamma )k+\beta }\right)}^m{a}_k{z}^k,n\in \mathbb{N}\cup \left\{0\right\}.\end{array}$(7) ), we get $\begin{array}{rl}& \frac{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}\\ & =\frac{(1+\gamma )(1-\mu )z{h}^{\prime }\left(z\right)}{\omega \left\{(1-\mu )h\left(z\right)+\mu \right\}\left\{\beta +p(\gamma +1)\right\}}+1,\end{array}$ this implies $\begin{array}{rl}& (1-\lambda ){\left(\frac{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}{z^p}\right)}^{\omega }\\ & +\lambda \left(\frac{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}\right){\left(\frac{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}{z^p}\right)}^{\omega }\\ & =\left\{\phantom{\frac{\lambda (1+\gamma )(1-\mu )z{h}^{\prime }\left(z\right)}{\omega \left\{\beta +p(\gamma +1)\right\}}}(1-\mu )h\left(z\right)+\mu \right.\\ & +\left.\frac{\lambda (1+\gamma )(1-\mu )z{h}^{\prime }\left(z\right)}{\omega \left\{\beta +p(\gamma +1)\right\}}\right\}\in P_k\left(\mu \right).\end{array}$ By using (2(2) ${\int }_0^{2\pi }\left|\frac{\mathrm{\Re }\left(p\right(z\left)\right)-\rho }{1-\rho }\right|\mathrm{d}\theta \le k\pi ,z=r{e}^{i\theta }.$(2) ), we can see $\begin{array}{rl}& \frac{1}{1-\rho }\left\{\phantom{\frac{\lambda (1+\gamma )(1-\mu )z{h}_i^{\prime }\left(z\right)}{\omega \left\{\beta +p(\gamma +1)\right\}}}(1-\mu )h_i\left(z\right)+\mu -\rho \right.\\ & +\left.\frac{\lambda (1+\gamma )(1-\mu )z{h}_i^{\prime }\left(z\right)}{\omega \left\{\beta +p(\gamma +1)\right\}}\right\}\in P\left(\mu \right),i=1,2.\end{array}$ For $u=h_i\left(z\right)$ and $v=z{h}_i\left(z\right)$, we define $\mathrm{\Phi }(u,v)$ such that $\mathrm{\Phi }(u,v)=(1-\mu )u+\mu -\rho +\frac{\lambda (1+\gamma )(1-\mu )v}{\omega \left\{\beta +p(\gamma +1)\right\}}.$ Since first two conditions of Lemma 1.2. are obvious so we proceed for third condition as follows: $\begin{array}{rl}\mathrm{\Re }\left\{\mathrm{\Phi }(i{u}_2,v_1)\right\}& =\mu -\rho +\mathrm{\Re }\left\{\frac{\lambda (1+\gamma )(1-\mu )v_1}{\omega \left\{\beta +p(\gamma +1)\right\}}\right\},\\ & \le \mu -\rho -\frac{\lambda (1+\gamma )(1-\mu )(1+u_2^2)}{2\omega \left\{\beta +p(\gamma +1)\right\}}\\ & =\frac{A+B{u}_2^2}{C},\end{array}$ where $\begin{array}{rl}A& =2\omega (\mu -\rho )\left\{\beta +p(\gamma +1)\right\}-\lambda (1+\gamma )(1-\mu ),\\ B& =-\lambda (1+\gamma )(1-\mu ),\\ C& =2\omega \left\{\beta +p(\gamma +1)\right\}.\end{array}$ We concluded that $\mathrm{\Re }\left\{\mathrm{\Phi }\right(i{u}_2,v_1\left)\right\}<0$, ⇔ A=0, B<0 and C>0. Hence by using Lemma 2, we get $({\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)/z^p{)}^{\omega }\in P_k(\mu )$.

Theorem 3.2

If the function f given by (1(1) $f\left(z\right)=z^p+\sum _{k=0}^{\mathrm{\infty }}{a}_k{z}^k.$(1) ) belongs to the class ${\mathrm{\Psi }}^p(\lambda ,\omega ,\beta ,\gamma )$, then $({\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)/z^p{)}^{\omega /2}\in P_k(\mu )$, where $\mu =\frac{\lambda (1+\gamma )+\sqrt{\begin{array}{c}{\lambda }^2(1+\gamma {)}^2+4\omega \rho \left[\beta +p(1+\gamma )\right]\\ \left[\omega (\beta +p(1+\gamma \left)\right)+\lambda (1+\gamma )\right]\end{array}}}{2\left[\omega (\beta +p(1+\gamma \left)\right)+\lambda (1+\gamma )\right]}.$

Proof.

By using (2(2) ${\int }_0^{2\pi }\left|\frac{\mathrm{\Re }\left(p\right(z\left)\right)-\rho }{1-\rho }\right|\mathrm{d}\theta \le k\pi ,z=r{e}^{i\theta }.$(2) ), if $({\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)/z^p{)}^{\omega /2}\in P_k(\mu )$ implies satisfy the condition given by (5(5) $\begin{array}{rl}h\left(z\right)& =\left(\frac{k}{4}+\frac{1}{2}\right)h_1\left(z\right)-\left(\frac{k}{4}-\frac{1}{2}\right)h_2\left(z\right),\\ \mathrm{\forall }z\in \mathbb{U}& =\left\{z\in \mathbb{C}:\left|z\right|<1\right\}.\end{array}$(5) ). We consider (10) $\begin{array}{rl}& {\left(\frac{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}{z^p}\right)}^{\omega }\\ & =G\left(z\right)=\left[\right(1-\mu \left)h\right(z)+\mu {]}^2\\ & =\left(\frac{k}{4}+\frac{1}{2}\right){\left\{(1-\mu )h_1\left(z\right)+\mu \right\}}^2\\ & +\left(\frac{k}{4}+\frac{1}{2}\right){\left\{(1-\mu )h_2\left(z\right)+\mu \right\}}^2,\end{array}$(10) where $h_1,h_2\in P\left(\mu \right)$. By using (10(10) $\begin{array}{rl}& {\left(\frac{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}{z^p}\right)}^{\omega }\\ & =G\left(z\right)=\left[\right(1-\mu \left)h\right(z)+\mu {]}^2\\ & =\left(\frac{k}{4}+\frac{1}{2}\right){\left\{(1-\mu )h_1\left(z\right)+\mu \right\}}^2\\ & +\left(\frac{k}{4}+\frac{1}{2}\right){\left\{(1-\mu )h_2\left(z\right)+\mu \right\}}^2,\end{array}$(10) ) and (7(7) $\begin{array}{rl}& {\mathrm{\Psi }}_p^m(\beta ,\gamma )f\left(z\right)\\ & =z^p+\sum _{k=2}^{\mathrm{\infty }}{\left(\frac{(1+\gamma )p+\beta }{(1+\gamma )k+\beta }\right)}^m{a}_k{z}^k,n\in \mathbb{N}\cup \left\{0\right\}.\end{array}$(7) ), we have $\begin{array}{rl}& \frac{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}\\ & =\frac{2(1+\gamma )(1-\mu )z{h}^{\prime }\left(z\right)}{\omega \left\{(1-\mu )h\left(z\right)+\mu \right\}\left\{\beta +p(\gamma +1)\right\}}+1,\end{array}$ this implies $\begin{array}{rl}& (1-\lambda ){\left(\frac{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}{z^p}\right)}^{\omega }\\ & +\lambda \left(\frac{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}\right){\left(\frac{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}{z^p}\right)}^{\omega }\\ & =\left\{\phantom{\frac{2\lambda (1+\gamma )(1-\mu )z{h}^{\prime }\left(z\right)}{\omega \left\{\beta +p(\gamma +1)\right\}}}{\left[(1-\mu )h\left(z\right)+\mu \right]}^2+\left[(1-\mu )h\left(z\right)+\mu \right]\right.\\ & \times \left.\frac{2\lambda (1+\gamma )(1-\mu )z{h}^{\prime }\left(z\right)}{\omega \left\{\beta +p(\gamma +1)\right\}}\right\}\in P_k\left(\mu \right).\end{array}$ By using (2(2) ${\int }_0^{2\pi }\left|\frac{\mathrm{\Re }\left(p\right(z\left)\right)-\rho }{1-\rho }\right|\mathrm{d}\theta \le k\pi ,z=r{e}^{i\theta }.$(2) ) for i=1,2, we have $\begin{array}{rl}& \frac{1}{1-\rho }\left\{\phantom{\frac{2\lambda (1+\gamma )(1-\mu )z{h}_i^{\prime }\left(z\right)}{\omega \left\{\beta +p(\gamma +1)\right\}}}\left[\right(1-\mu \left)h_i\right(z)+\mu {]}^2-\rho +[(1-\mu )h\left(z\right)+\mu ]\right.\\ & \times \left.\frac{2\lambda (1+\gamma )(1-\mu )z{h}_i^{\prime }\left(z\right)}{\omega \left\{\beta +p(\gamma +1)\right\}}\right\}\in P\left(\mu \right).\end{array}$ For $u=h_i\left(z\right)$ and $v=z{h}_i\left(z\right)$, we define $\mathrm{\Phi }(u,v)$ such that $\begin{array}{rl}\mathrm{\Phi }(u,v)& =\left[\right(1-\mu \left)h_i\right(z)+\mu {]}^2-\rho \\ & +\left[\right(1-\mu \left)h\right(z)+\mu ]\frac{2\lambda (1+\gamma )(1-\mu )z{h}_i^{\prime }\left(z\right)}{\omega \left\{\beta +p(\gamma +1)\right\}}.\end{array}$

Since first two conditions of Lemma 1.2. are obvious so we proceed for third condition as follows: $\begin{array}{rl}\mathrm{\Re }\left\{\mathrm{\Phi }(i{u}_2,v_1)\right\}& =[\mu {]}^2-(1-\mu {)}^2{u}_2^2-\rho \\ & +\mathrm{\Re }\left\{\frac{2\lambda \mu (1+\gamma )(1-\mu )v_1}{\omega \left\{\beta +p(\gamma +1)\right\}}\right\},\\ & \le [\mu {]}^2-(1-\mu {)}^2{u}_2^2-\rho \\ & -\frac{\lambda \mu (1+\gamma )(1-\mu )(1+u_2^2)}{\omega \left\{\beta +p(\gamma +1)\right\}}\\ & =\frac{A+B{u}_2^2}{C},\end{array}$ where $\begin{array}{rl}A& =\omega ({\mu }^2-\rho )\left\{\beta +p(\gamma +1)\right\}-\lambda \mu (1+\gamma )(1-\mu ),\\ B& =-(1-\mu )\left[\omega \left\{\beta +p(\gamma +1)\right\}\right(1-\mu )+\lambda \mu (1+\gamma \left)\right],\\ C& =\omega \left\{\beta +p(\gamma +1)\right\}.\end{array}$ We concluded that $\mathrm{\Re }\left\{\mathrm{\Phi }\right(i{u}_2,v_1\left)\right\}<0$, ⇔ A=0, B<0 and C>0. Hence by using Lemma 2, we get $({\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)/z^p{)}^{\omega /2}\in P_k(\mu )$.

Theorem 3.3

Let $h(r,s,t)\in H$ and f belongs to $A_p$ satisfy the following criterion: $\left(\frac{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}{z^p},\frac{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)}{z^p},\frac{{\mathrm{\Psi }}_p^{n-1}(\beta ,\gamma )f\left(z\right)}{z^p}\right)\in D$ and $\begin{array}{rl}& \left|h\left(\frac{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}{z^p},\frac{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)}{z^p},\frac{{\mathrm{\Psi }}_p^{n-1}(\beta ,\gamma )f\left(z\right)}{z^p}\right)\right|\\ & <1.\end{array}$ Then $\left|\frac{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}{z^p}\right|<1.$

Proof.

Let (11) $\frac{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}{z^p}=w\left(z\right).$(11) Clearly $w\left(0\right)=1$ and $w\left(z\right)\ne 1$. By using (11(11) $\frac{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}{z^p}=w\left(z\right).$(11) ) and (7(7) $\begin{array}{rl}& {\mathrm{\Psi }}_p^m(\beta ,\gamma )f\left(z\right)\\ & =z^p+\sum _{k=2}^{\mathrm{\infty }}{\left(\frac{(1+\gamma )p+\beta }{(1+\gamma )k+\beta }\right)}^m{a}_k{z}^k,n\in \mathbb{N}\cup \left\{0\right\}.\end{array}$(7) ), we get (12) $\frac{z{\mathrm{\Psi }}_p^{n+1^{\prime }}(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}-p=\frac{z{w}^{\prime }\left(z\right)}{w\left(z\right)}.$(12) By using (7(7) $\begin{array}{rl}& {\mathrm{\Psi }}_p^m(\beta ,\gamma )f\left(z\right)\\ & =z^p+\sum _{k=2}^{\mathrm{\infty }}{\left(\frac{(1+\gamma )p+\beta }{(1+\gamma )k+\beta }\right)}^m{a}_k{z}^k,n\in \mathbb{N}\cup \left\{0\right\}.\end{array}$(7) ) and doing some calculation, we have (13) $\begin{array}{rl}& \frac{z{\mathrm{\Psi }}_p^{n+1^{\prime }}(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}\\ & =(\frac{\beta }{\gamma +1}+p)\frac{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}-\frac{\beta }{1+\gamma }.\end{array}$(13) After using (12(12) $\frac{z{\mathrm{\Psi }}_p^{n+1^{\prime }}(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}-p=\frac{z{w}^{\prime }\left(z\right)}{w\left(z\right)}.$(12) ) and (13(13) $\begin{array}{rl}& \frac{z{\mathrm{\Psi }}_p^{n+1^{\prime }}(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}\\ & =(\frac{\beta }{\gamma +1}+p)\frac{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}-\frac{\beta }{1+\gamma }.\end{array}$(13) ) simultaneously, we get (14) $\frac{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)}{z^p}=w\left(z\right)+\frac{\gamma +1}{\beta +p(\gamma +1)}z{w}^{\prime }\left(z\right).$(14) After applying the same steps on (14(14) $\frac{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)}{z^p}=w\left(z\right)+\frac{\gamma +1}{\beta +p(\gamma +1)}z{w}^{\prime }\left(z\right).$(14) ) and (7(7) $\begin{array}{rl}& {\mathrm{\Psi }}_p^m(\beta ,\gamma )f\left(z\right)\\ & =z^p+\sum _{k=2}^{\mathrm{\infty }}{\left(\frac{(1+\gamma )p+\beta }{(1+\gamma )k+\beta }\right)}^m{a}_k{z}^k,n\in \mathbb{N}\cup \left\{0\right\}.\end{array}$(7) ), we get (15) $\begin{array}{rl}& \frac{{\mathrm{\Psi }}_p^{n-1}(\beta ,\gamma )f\left(z\right)}{z^p}\\ & =w\left(z\right)+2\frac{\gamma +1}{\beta +p(\gamma +1)}z{w}^{\mathrm{\prime }}\left(z\right)\\ & +{\left(\frac{\gamma +1}{\beta +p(\gamma +1)}\right)}^2\left(z^2{w}^{\mathrm{\prime \prime }}\left(z\right)+z{w}^{\mathrm{\prime }}\left(z\right)\right).\end{array}$(15) We claim that $|w\left(z\right)|<1$. On contrary, we suppose that for a=k=1, the $w\left(z\right)=1+w_1\left(z\right)$, this implies there exist $z_0\in E$ such that $max|w\left(z\right)|=|w\left(z_0\right)|=1$, hence $w\left(z_0\right)=e^{i\theta }$ and $m=\gamma +1/\beta +p(\gamma +1)z{w}^{\mathrm{\prime }}\left(z\right)\left(say\right)$, then $\begin{array}{rl}\frac{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}{z^p}& ={\mathrm{e}}^{i\theta }\mathrm{\&}\frac{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)}{z^p}={\mathrm{e}}^{i\theta }+m,\\ \frac{{\mathrm{\Psi }}_p^{n-1}(\beta ,\gamma )f\left(z\right)}{z^p}& ={\mathrm{e}}^{i\theta }+2m+\frac{\gamma +1}{\beta +p(\gamma +1)}m+L,\end{array}$ where $L=\left(\gamma +1/\beta +p(\gamma +1){)}^2{z}^2{w}^{\mathrm{\prime \prime }}\right(z)$. After doing calculation, we have $\begin{array}{rl}& \mathrm{\Re }\left\{1+\frac{\gamma +1}{\beta +p(\gamma +1)}\frac{z{w}^{″}\left(z\right)}{w^{\prime }\left(z\right)}\right\}\ge m\\ & \mathrm{\&}\mathrm{\Re }\left\{m+m\frac{\gamma +1}{\beta +(\gamma +1)}\frac{z{w}^{″}\left(z\right)}{w^{\prime }\left(z\right)}\right\}\ge m^2\end{array}$ and $\begin{array}{rl}& \mathrm{\Re }\left\{m+\frac{\gamma +1}{\beta +p(\gamma +1)}\frac{z{w}^{″}\left(z\right)}{w^{\prime }\left(z\right)}\frac{(\gamma +1)}{(\beta +p(\gamma +1\left)\right)}z{w}^{\prime }\left(z\right)\right\}\\ & \ge m^2.\end{array}$ This implies $\begin{array}{rl}& \mathrm{\Re }\left\{m+\frac{(\gamma +1{)}^2{z}^2{w}^{″}(z)}{(\beta +p(\gamma +1){)}^2}\right\}\ge m^2\\ & \mathrm{\&}\mathrm{\Re }\left\{L\right\}\ge m(m-1).\end{array}$ Since $h(r,s,t)\in H$, then by using definition of $h(r,s,t)$, we have $\begin{array}{rl}& \left|h\left(\frac{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}{z^p},\frac{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)}{z^p},\frac{{\mathrm{\Psi }}_p^{n-1}(\beta ,\gamma )f\left(z\right)}{z^p}\right)\right|\\ & =\left|h\left(\phantom{\frac{\gamma +1}{\beta +p(\gamma +1)}}{\mathrm{e}}^{i\theta },{\mathrm{e}}^{i\theta }+m,e^{i\theta }+2m\right.\right.\\ & +\left.\left.\frac{\gamma +1}{\beta +p(\gamma +1)}m+L\right)\right|\ge 1,\end{array}$ which is contradiction and hence $\left|\frac{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}{z^p}\right|<1.$

Theorem 3.4

Let $h(r,s,t)\in H$ and f belongs to $A_p$ satisfy the following criterion: $\begin{array}{rl}& \left(\frac{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)},\frac{{\mathrm{\Psi }}_p^{n-1}(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)},\right.\\ & \left.\frac{{\mathrm{\Psi }}_p^{n-2}(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^{n-1}(\beta ,\gamma )f\left(z\right)}\right)\in D\end{array}$ and $\begin{array}{rl}& \left|h\left(\frac{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)},\frac{{\mathrm{\Psi }}_p^{n-1}(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)},\right.\right.\\ & \left.\left.\frac{{\mathrm{\Psi }}_p^{n-2}(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^{n-1}(\beta ,\gamma )f\left(z\right)}\right)\right|<1.\end{array}$ Then $\left|\frac{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}\right|<1.$

Proof.

Since, (16) $\frac{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}=w\left(z\right),$(16) then clearly $w\left(0\right)=1$ and $w\left(z\right)\ne 1$. By using (16(16) $\frac{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}=w\left(z\right),$(16) ) and (7(7) $\begin{array}{rl}& {\mathrm{\Psi }}_p^m(\beta ,\gamma )f\left(z\right)\\ & =z^p+\sum _{k=2}^{\mathrm{\infty }}{\left(\frac{(1+\gamma )p+\beta }{(1+\gamma )k+\beta }\right)}^m{a}_k{z}^k,n\in \mathbb{N}\cup \left\{0\right\}.\end{array}$(7) ), we get (17) $\begin{array}{rl}& \frac{(\gamma +1)z{\mathrm{\Psi }}_p^{n^{\prime }}(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)}-\frac{(\gamma +1)z{\mathrm{\Psi }}_p^{n+1^{\prime }}(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}\\ & =\frac{(\gamma +1)z{w}^{\prime }\left(z\right)}{w\left(z\right)}.\end{array}$(17) By using (7(7) $\begin{array}{rl}& {\mathrm{\Psi }}_p^m(\beta ,\gamma )f\left(z\right)\\ & =z^p+\sum _{k=2}^{\mathrm{\infty }}{\left(\frac{(1+\gamma )p+\beta }{(1+\gamma )k+\beta }\right)}^m{a}_k{z}^k,n\in \mathbb{N}\cup \left\{0\right\}.\end{array}$(7) ) and doing some calculation, we have (18) $\begin{array}{rl}& \frac{z{\mathrm{\Psi }}_p^{n^{\prime }}(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)}\\ & =\left(\frac{\beta }{\gamma +1}+p\right)\frac{{\mathrm{\Psi }}_p^{n-1}(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)}-\frac{\beta }{\gamma +1}\end{array}$(18) and (19) $\begin{array}{rl}& \frac{z{\mathrm{\Psi }}_p^{n+1^{\prime }}(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}\\ & =\left(\frac{\beta }{\gamma +1}+p\right)\frac{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^{n+11}(\beta ,\gamma )f\left(z\right)}-\frac{\beta }{\gamma +1}.\end{array}$(19) Simultaneously using (17(17) $\begin{array}{rl}& \frac{(\gamma +1)z{\mathrm{\Psi }}_p^{n^{\prime }}(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)}-\frac{(\gamma +1)z{\mathrm{\Psi }}_p^{n+1^{\prime }}(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}\\ & =\frac{(\gamma +1)z{w}^{\prime }\left(z\right)}{w\left(z\right)}.\end{array}$(17) ), (18(18) $\begin{array}{rl}& \frac{z{\mathrm{\Psi }}_p^{n^{\prime }}(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)}\\ & =\left(\frac{\beta }{\gamma +1}+p\right)\frac{{\mathrm{\Psi }}_p^{n-1}(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)}-\frac{\beta }{\gamma +1}\end{array}$(18) ) and (19(19) $\begin{array}{rl}& \frac{z{\mathrm{\Psi }}_p^{n+1^{\prime }}(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}\\ & =\left(\frac{\beta }{\gamma +1}+p\right)\frac{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^{n+11}(\beta ,\gamma )f\left(z\right)}-\frac{\beta }{\gamma +1}.\end{array}$(19) ) implies $\begin{array}{rl}& \frac{(\beta +p(\gamma +1\left)\right){\mathrm{\Psi }}_p^{n-1}(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)}\\ & =\frac{(1+\gamma )z{w}^{\prime }\left(z\right)}{w\left(z\right)}+(\beta +p(\gamma +1\left)\right)w\left(z\right)\end{array}$ or (20) $\frac{{\mathrm{\Psi }}_p^{n-1}(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)}=\frac{(1+\gamma )z{w}^{\prime }\left(z\right)}{(\beta +p(\gamma +1\left)\right)w\left(z\right)}+w\left(z\right).$(20)

After differentiating (20(20) $\frac{{\mathrm{\Psi }}_p^{n-1}(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)}=\frac{(1+\gamma )z{w}^{\prime }\left(z\right)}{(\beta +p(\gamma +1\left)\right)w\left(z\right)}+w\left(z\right).$(20) ) and simplification, we get (21) $\begin{array}{rl}& \frac{z{\mathrm{\Psi }}_p^{n-1^{\prime }}(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^{n-1}(\beta ,\gamma )f\left(z\right)}-\frac{z{\mathrm{\Psi }}_p^{n^{\prime }}(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)}\\ & =\frac{z{w}^{\prime }\left(z\right)+\frac{\gamma +1}{(\beta +p(\gamma +1\left)\right)}\left(\frac{z^2{w}^{″}\left(z\right)}{w\left(z\right)}+\frac{z{w}^{\prime }\left(z\right)}{w\left(z\right)}-\frac{\left(z{w}^{\prime }\right(z){)}^2}{\left(w\right(z){)}^2}\right)}{w\left(z\right)+\frac{\gamma +1}{(\beta +p(\gamma +1\left)\right)}\frac{z{w}^{\prime }\left(z\right)}{w\left(z\right)}}.\end{array}$(21) Now by using (7(7) $\begin{array}{rl}& {\mathrm{\Psi }}_p^m(\beta ,\gamma )f\left(z\right)\\ & =z^p+\sum _{k=2}^{\mathrm{\infty }}{\left(\frac{(1+\gamma )p+\beta }{(1+\gamma )k+\beta }\right)}^m{a}_k{z}^k,n\in \mathbb{N}\cup \left\{0\right\}.\end{array}$(7) ) and (20(20) $\frac{{\mathrm{\Psi }}_p^{n-1}(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)}=\frac{(1+\gamma )z{w}^{\prime }\left(z\right)}{(\beta +p(\gamma +1\left)\right)w\left(z\right)}+w\left(z\right).$(20) ) for right-hand side of (21(21) $\begin{array}{rl}& \frac{z{\mathrm{\Psi }}_p^{n-1^{\prime }}(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^{n-1}(\beta ,\gamma )f\left(z\right)}-\frac{z{\mathrm{\Psi }}_p^{n^{\prime }}(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)}\\ & =\frac{z{w}^{\prime }\left(z\right)+\frac{\gamma +1}{(\beta +p(\gamma +1\left)\right)}\left(\frac{z^2{w}^{″}\left(z\right)}{w\left(z\right)}+\frac{z{w}^{\prime }\left(z\right)}{w\left(z\right)}-\frac{\left(z{w}^{\prime }\right(z){)}^2}{\left(w\right(z){)}^2}\right)}{w\left(z\right)+\frac{\gamma +1}{(\beta +p(\gamma +1\left)\right)}\frac{z{w}^{\prime }\left(z\right)}{w\left(z\right)}}.\end{array}$(21) ), we get $\begin{array}{rl}& \frac{{\mathrm{\Psi }}_p^{n-2}(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^{n-1}(\beta ,\gamma )f\left(z\right)}\\ & =w\left(z\right)+\frac{(\gamma +1)z{w}^{\prime }\left(z\right)}{(\beta +p(\gamma +1\left)\right)w\left(z\right)}\\ & +\frac{\begin{array}{c}\frac{\gamma +1}{(\beta +p(\gamma +1\left)\right)}z{w}^{\prime }\left(z\right)+{\left(\frac{\gamma +1}{(\beta +p(\gamma +1\left)\right)}\right)}^2\\ \left(\frac{z^2{w}^{″}\left(z\right)}{w\left(z\right)}+\frac{z{w}^{\prime }\left(z\right)}{w\left(z\right)}-\frac{\left(z{w}^{\prime }\right(z){)}^2}{\left(w\right(z){)}^2}\right)\end{array}}{w\left(z\right)+\frac{\gamma +1}{(\beta +p(\gamma +1\left)\right)}\frac{z{w}^{\prime }\left(z\right)}{w\left(z\right)}}.\end{array}$ We claim that $|w\left(z\right)|<1$. On contrary, we suppose that for a=k=1, $w\left(z\right)=1+w_1\left(z\right)$, this implies there exist $z_0\in E$ such that $max|w\left(z\right)|=|w\left(z_0\right)|=1$, hence $w\left(z_0\right)=e^{i\theta }$ and $m=\frac{(\gamma +1)z_0{w}^{\prime }\left(z_0\right)}{(\beta +p(\gamma +1\left)\right)w\left(z_0\right)}\left(say\right)$, then $m(\beta +p(\gamma +1\left)\right)e^{i\theta }=(\gamma +1)z_0{w}^{\prime }\left(z_0\right)$. This implies that $\begin{array}{rl}\frac{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z_0\right)}{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z_0\right)}& ={\mathrm{e}}^{i\theta }\mathrm{\&}\frac{{\mathrm{\Psi }}_p^{n-1}(\beta ,\gamma )f\left(z_0\right)}{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z_0\right)}\\ & =m+{\mathrm{e}}^{i\theta },\\ \frac{{\mathrm{\Psi }}_p^{n-2}(\beta ,\gamma )f\left(z_0\right)}{{\mathrm{\Psi }}_p^{n-1}(\beta ,\gamma )f\left(z_0\right)}& =m+{\mathrm{e}}^{i\theta }\\ & +\frac{m{\mathrm{e}}^{i\theta }+\frac{\gamma +1}{\beta +p(\gamma +1)}m-m^2+L}{m+{\mathrm{e}}^{i\theta }},\end{array}$ where $L=(\gamma +1{)}^2{z}^2{w}^{″}(z)/(\beta +p(\gamma +1\left){)}^{2}w\right(z)$. After doing calculation, we have $\begin{array}{rl}& \mathrm{\Re }\left\{1+\frac{\gamma +1}{\beta +p(\gamma +1)}\frac{z{w}^{″}\left(z\right)}{w^{\prime }\left(z\right)}\right\}\ge m\\ & \mathrm{\&}\mathrm{\Re }\left\{m+m\frac{\gamma +1}{\beta +p(\gamma +1)}\frac{z{w}^{″}\left(z\right)}{w^{\prime }\left(z\right)}\right\}\ge m^2\end{array}$ and $\begin{array}{rl}& \mathrm{\Re }\left\{m+\frac{\gamma +1}{\beta +p(\gamma +1)}\frac{z{w}^{″}\left(z\right)}{w^{\prime }\left(z\right)}\frac{(\gamma +1)}{(\beta +p(\gamma +1\left)\right)}\frac{z{w}^{\prime }\left(z\right)}{w\left(z\right)}\right\}\\ & \ge m^2.\end{array}$ This implies $\begin{array}{rl}& \mathrm{\Re }\left\{m+\frac{(\gamma +1{)}^2{z}^2}{(\beta +p(\gamma +1){)}^2}\frac{w^{″}\left(z\right)}{w\left(z\right)}\right\}\ge m^2\\ & \mathrm{\&}\mathrm{\Re }\left\{L\right\}\ge m(m-1).\end{array}$ Since $h(r,s,t)\in H$, then by using definition of $h(r,s,t)$, we have $\begin{array}{rl}& \left|h\left(\frac{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)},\frac{{\mathrm{\Psi }}_p^{n-1}(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)},\right.\right.\\ & \left.\left.\frac{{\mathrm{\Psi }}_p^{n-2}(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^{n-1}(\beta ,\gamma )f\left(z\right)}\right)\right|\\ & =\left|h\left(\phantom{\frac{m{e}^{i\theta }+\frac{\gamma +1}{\beta +p(\gamma +1)}m-m^2+L}{m+e^{i\theta }}}{e}^{i\theta },m+e^{i\theta },m+e^{i\theta }\right.\right.\\ & +\left.\left.\frac{m{e}^{i\theta }+\frac{\gamma +1}{\beta +p(\gamma +1)}m-m^2+L}{m+e^{i\theta }}\right)\right|\ge 1,\end{array}$ which is contradiction and therefore $\left|\frac{{\mathrm{\Psi }}_p^n(\beta ,\gamma )f\left(z\right)}{{\mathrm{\Psi }}_p^{n+1}(\beta ,\gamma )f\left(z\right)}\right|<1.$

Example 1

Since we have $\begin{array}{rl}g\left(z\right)& =\frac{z-\left(\frac{\beta }{1+\gamma +\beta }\right)z^2}{{\left(1-z\right)}^2},\mathrm{\forall }z\in \mathbb{U},\\ & =z(1-z{)}^{-1}+z^2\left(\frac{1+\gamma }{1+\gamma +\beta }\right){\left(1-z\right)}^{-2},\\ & =z+z^2+z^3+\cdots +\left(\frac{1+\gamma }{1+\gamma +\beta }\right)z^2\\ & +\left(\frac{2(1+\gamma )}{1+\gamma +\beta }\right)z^3+\cdots ,\\ g\left(z\right)& =z+\left(\frac{(1+\gamma )+\beta }{1+\gamma +\beta }\right)z^2\\ & +\left(\frac{3(1+\gamma )+\beta }{1+\gamma +\beta }\right)z^3+\cdots ,\\ g\left(z\right)& =z+\sum _{k=2}^{\mathrm{\infty }}\left(\frac{(1+\gamma )k+\beta }{1+\gamma +\beta }\right)z^k.\end{array}$ After taking n-times hadamard product of g, we have $g\left(z\right)=z+\sum _{k=2}^{\mathrm{\infty }}{\left(\frac{(1+\gamma )k+\beta }{1+\gamma +\beta }\right)}^n{z}^k.$ We define $\begin{array}{rl}g\left(z\right)\ast f\left(z\right)& =\frac{z}{1-z},\text{implies }f\left(z\right)\\ & =z+\sum _{k=2}^{\mathrm{\infty }}{\left(\frac{1+\gamma +\beta }{(1+\gamma )k+\beta }\right)}^n{z}^k\mathrm{for}p=1,\end{array}$ which clearly satisfy our result.

4. Conclusion

In this paper, an integral operator is used to introduce new simply connected convex domain of analytic functions in the open unit disk $\mathbb{U}$. Further, analytic criteria for p-valent analytic function belongs to simply connected convex domain to be a member of class of functions with positive real part greater than μ have been discussed.

Acknowledgements

The author is deeply indebted to the referees for providing constructive comments and helps in improving the content of this article.

Disclosure statement

No potential conflict of interest was reported by the author.

ORCID

M. I. Faisal http://orcid.org/0000-0002-4169-4366