1. Introduction

Let $A(p)$ denote the class of all functions $f$ defined by

$f(z) = z^{p}+\sum\limits_{n = 1}^{\infty }a_{n+p}z^{n+p}\ \ \ \ (p\in N = 1, 2, 3...)$

(1.1)

which are analytic and $p$-valent in the unit disk

$\mathbb{E} = \{z\in \mathbb{C} :\left \vert z\right \vert <1\}.$

For a real number $\alpha$ $(0\leq \alpha < p)$ the well-known subclasses $S_{p}^{\ast }(\alpha)$, $p$-valently starlike functions of order $\alpha$ and $C_{p}(\alpha),$ $p$-valently convex functions of order $\alpha$ of $A(p)$ are given by

$\begin{eqnarray*} S^{\ast }(\alpha, p) & = &\left \{ f\in A(p):\Re\left( \frac{zf^{\prime }(z)}{f(z)}\right) >\alpha, \quad (z\in \mathbb{E})\right \}, \\ C(\alpha, p) & = &\left \{ f\in A(p):\Re\left( 1+\frac{zf^{\prime \prime }(z)}{f^{^{\prime }}(z)}\right) >\alpha, \quad (z\in \mathbb{E}% )\right \} . \end{eqnarray*}$

For $\left \vert \beta \right \vert$ $< \frac{\pi }{2}$ and $0\leq \alpha < 1,$ a function $f\in A$ is said to be $\beta$-spirallike of order $\alpha$ in $\mathbb{E}$ if

$\Re\left \{ e^{\iota \beta }\frac{zf^{\prime }(z)}{f(z)}\right \} >\alpha \cos \beta \ \ \ \ \ (z\in \mathbb{E}).$

(1.2)

The class of all such functions is denoted by $S_{\beta }(\alpha)$ ^[3], (also see ^[5,14,15]). In recent years many interesting subclasses of analytic univalent, multivalent and spirallike functions and their many special cases were investigated, see for example ^{[1,2,6,7,8,9,10]}.

Motivated and inspired by the above mention work, we here define the following.

Definition 1.1. A function $f\in A(p)$ belongs to the class $S_{\beta }(\alpha, p)$ if it satisfies the inequality

$\left \vert \frac{f^{(p-1)}(z)}{e^{\iota \beta }zf^{(p)}(z)}-\frac{1}{2\alpha }\right \vert <\frac{1}{2\alpha }, \ \ \ \ \ \ \ z\in \mathbb{E},$

for some real $\beta$ and $0 < \alpha < 1,$ where $f^{(p)}(z)$ is the $p^{th}$ derivative of $f(z)$.

Remark 1.2. First of all, it is easily seen that, for

$p = 1\quad \text{and}\quad \beta = 0, S_{0}(\alpha , 1) = M(\alpha ),$

where $M(\alpha)$ is a function class introduced and studied in ^[12]. Secondly, we have

$p = 1, \quad S_{\beta }(\alpha, 1) = S_{\beta }(\alpha ),$

where $S_{\beta }(\alpha)\ $is a function class introduced by Owa and Kamali ^[13].

Definition 1.3. A function $f\in A(p)$ is said to be in the class $C_{\beta }(\alpha, p)$ if it satisfies the inequality

$\left \vert \frac{f^{(p)}(z)}{e^{\iota \beta }\left( zf^{(p)}(z)\right) ^{\prime }}-\frac{1}{2\alpha }\right \vert <\frac{1}{2\alpha }, \ \ \ \ \ \ \ \ z\in \mathbb{E},$

(1.3)

for some real $\beta$ and $0 < \alpha < 1$, where $f^{(p)}(z)$ is the $p$th derivative of $f(z)$.

As a special case, the class $C_{\beta }(\alpha, 1) = K_{\beta }(\alpha),$ is introduced by Owa and Kamali ^[13]. Using essentially their technique, we prove the main results for the classes $S_{\beta }(\alpha, p)$ and $C_{\beta }(\alpha, p)$ which is the main motivation of this paper.

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2. Preliminary results

Lemma 2.1. ^[4]. Let $\phi (u, \upsilon)$ be a complex-valued function such that

$\phi :D\rightarrow \mathbb{C} , \ \ D\subset \mathbb{C} \times \mathbb{C}$

$\mathbb{C}$ being the complex plane and let $u = u_{1}+\iota u_{2\text{ }}$and $v = v_{1}+\iota v_{2}.$ Suppose that the function $\phi (u, \upsilon)$ satisfies each of the following conditions

1. $\phi (u, \upsilon)$ is continuous in $D;$

2. $(1, 0)\in D$ and $\Re\{ \phi (1, 0)\}>0;$

3. $\Re\{ \phi (\iota u_{2}, v_{1})\} \leq 0$ for all $(\iota u_{2}, v_{1})\in D$ such that $v_{1}\leq -\frac{(1+u_{2}^{2})}{2}.$ Let

$p(z) = 1+p_{1}z+p_{2}z^{2}+...$

be analytic (regular) in the unit disk $\mathbb{E}$ such that

$\left( p(z), zp^{\prime }(z)\right) \in D, \;\; \text{ for all }\;\;z\in \mathbb{E }.$

If

$\Re\left \{ \phi \left( p(z), zp^{\prime }(z)\right) \right \} >0, \;\; \text{then}\;\; \Re\{p(z)\}>0\;\; (z\in \mathbb{E}).$

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3. Main results

Theorem 3.1. A function $f\in S_{\beta }(\alpha, p)$ if and only if, $\Re\left(e^{\iota \beta }\frac{zf^{(p)}(z)}{f^{(p-1)}(z)}\right) >\alpha.$

Proof. Let $f(z)\in S_{\beta }(\alpha, p)$, then we can write

$\left \vert \frac{1}{e^{\iota \beta }F(z)}-\frac{1}{2\alpha }\right \vert < \frac{1}{2\alpha }\ \ (z\in E)$

where $F(z) = \frac{zf^{(p)}(z)}{f^{(p-1)}(z)}.$ From above, we have

$\begin{align*} \left \vert \frac{2\alpha -e^{\iota \beta }F(z)}{2\alpha e^{\iota \beta }F(z)% }\right \vert & <\left( \frac{1}{2\alpha }\right) \\ & \Leftrightarrow \left \vert 2\alpha -e^{\iota \beta }F(z)\right \vert ^{2}<\left( e^{\iota \beta }F(z)\right) ^{2}\text{ } \\ & \Leftrightarrow \left[2\alpha-e^{\iota \beta }F(z)\right] \left[ \overline{2\alpha-e^{\iota \beta }F(z)}\right] <\left( e^{\iota \beta }F(z)\right) \left[\overline{e^{\iota \beta }F(z)}\right] \\ & \Leftrightarrow \left[2\alpha-e^{\iota \beta }F(z)\right] \left[ 2\alpha -e^{-\iota \beta }\overline{F(z)}\right] <\left( e^{\iota \beta }F(z)\right) % \left[e^{-\iota \beta }\overline{F(z)}\right] \\ & \Leftrightarrow 4\alpha ^{2}-2\alpha e^{-\iota \beta }\overline{F(z)}% -2\alpha e^{\iota \beta }F(z)+F(z)\overline{F(z)}<F(z)\overline{F(z)} \\ & \Leftrightarrow 4\alpha ^{2}-2\alpha \left( e^{-\iota \beta }\overline{F(z)% }+e^{\iota \beta }F(z)\right) <0 \\ & \Leftrightarrow 2\alpha -2\Re(e^{\iota \beta }F(z))<0 \\ & \Leftrightarrow -2\Re(e^{\iota \beta }F(z))<-2\alpha \\ & \Leftrightarrow \Re\left( e^{\iota \beta }\frac{zf^{(P)}(z)}{% f^{(P-1)}(z)}\right) >\alpha . \end{align*}$

This complete the proof.

When $p = 1$ we have the following known result proved by Owa and Kamali ^[13].

Corollary 3.2. $f(z)\in S_{\beta }(\alpha)$ iff $\Re\left(e^{i\beta }\frac{zf^{\prime }(z)% }{f(z)}\right) >\alpha.$

Theorem 3.3. If $f(z)\in A(p)$ satisfies

$\sum\limits_{n = 1}^{\infty }\frac{(p+n)!}{(n+1)!}\left( n+1+\left \vert \left( n+1\right) -2\alpha e^{-\iota \beta }\right \vert \right) \left \vert a_{n+p}\right \vert \leq 1-\left \vert 1-2\alpha e^{-\iota \beta }\right \vert$

(3.1)

for some $\left \vert \beta \right \vert < \frac{\pi }{2}$ and $0 < \alpha < \cos \beta,$ then $f(z)\in S_{\beta }(\alpha, p).$

Proof. If $f(z)\in S_{\beta }(\alpha, p)$ then, it suffices to show that

$\left \vert \frac{2\alpha -e^{\iota \beta }F(z)}{e^{\iota \beta }F(z)}\right \vert <1$

for some $\left \vert \beta \right \vert < \frac{\pi }{2}$ and $0 < \alpha < \cos \beta,$ where $F(z) = \frac{zf^{(p)}(z)}{f^{(p-1)}(z)}.$

Now we have

$\left \vert \frac{2\alpha -e^{\iota \beta }F(z)}{e^{\iota \beta }F(z)}\right \vert = \left \vert \frac{2\alpha e^{-\iota \beta }f^{(p-1)}(z)-zf^{(p)}(z)}{% zf^{(p)}(z)}\right \vert \notag \\ = \left \vert \frac{2\alpha e^{-\iota \beta }-1+\sum_{n = 1}^{\infty }\frac{ (p+n)!}{(n+1)!}\left( 2\alpha e^{-\iota \beta }-(n+1)\right) a_{n+p}z^{n}}{ 1+\sum_{n = 1}^{\infty }\frac{(p+n)!}{(n+1)!}\left( n+1\right) a_{n+p}z^{n}} \right \vert \notag \\ \leq \frac{\left \vert 2\alpha e^{-\iota \beta }-1\right \vert +\sum_{n = 1}^{\infty }\frac{(p+n)!}{(n+1)!}\left \vert \left( 2\alpha e^{-\iota \beta }-(n+1)\right) \right \vert \left \vert a_{n+p}\right \vert \left \vert z^{n}\right \vert }{1-\sum_{n = 1}^{\infty }\frac{(p+n)!}{(n+1)!} \left( n+1\right) \left \vert a_{n+p}\right \vert \left \vert z^{n}\right \vert } \notag \\ <\frac{\left \vert 1-2\alpha e^{-\iota \beta }\right \vert +\sum_{n = 1}^{\infty }\frac{(p+n)!}{(n+1)!}\left \vert (n+1)-2\alpha e^{-\iota \beta }\right \vert \left \vert a_{n+p}\right \vert }{ 1-\sum_{n = 1}^{\infty }\frac{(p+n)!}{(n+1)!}\left( n+1\right) \left \vert a_{n+p}\right \vert }.$

(3.2)

The last expression in (3.2) is bounded above by 1 if

$\left \vert 1-2\alpha e^{-\iota \beta }\right \vert +\sum\limits_{n = 1}^{\infty } \frac{(p+n)!}{(n+1)!}\left \vert \left( n+1\right) -2\alpha e^{-\iota \beta }\right \vert \left \vert a_{n+p}\right \vert \leq 1-\sum\limits_{n = 1}^{\infty } \frac{(p+n)!}{(n+1)!}\left \vert \left( n+1\right) a_{n+p}\right \vert .$

(3.3)

After simplification of (3.3) we have

$\sum\limits_{n = 1}^{\infty }\frac{(p+n)!}{(n+1)!}\left \{ \left( n+1+\left \vert \left( n+1\right) -2\alpha e^{-\iota \beta }\right \vert \right) \right \} \left \vert a_{n+p}\right \vert \leq 1-\left \vert 1-2\alpha e^{-\iota \beta }\right \vert .$

Therefore, $f(z)\in S_{\beta }(\alpha, p)\ $ for some $\left \vert \beta \right \vert < \frac{\pi }{2}$ and$\ 0 < \alpha < \cos \beta.$

When $\beta = 0$ and $p = 1,$ we have the following result proved by Owa et al. ^[12].

Corollary 3.4. Let $0 < \alpha < 1.$ If $f(z)\in A$ satisfies the following coefficient inequality

$\sum\limits_{n = 2}^{\infty }\left( n-\alpha \right) \left \vert a_{n}\right \vert \leq \frac{1}{2}\left( 1-\left \vert 1-2\alpha \right \vert \right) = \left \{ \begin{array}{c} \alpha ;\quad \left( 0<\alpha \leq \frac{1}{2}\right) \\ \\ 1-\alpha ;\quad \left( \frac{1}{2}<\alpha <1\right)% \end{array} \right.$

then $f(z)\in M(\alpha).$

Taking $\beta = \frac{\pi }{4}$ in above Theorem we have the following result.

Corollary 3.5. If $f(z)\in A(p)$ satisfies

$\sum\limits_{n = 1}^{\infty }\frac{(p+n)!}{(n+1)!}\left( n+1+\sqrt{\left( n+1\right) ^{2}-2\sqrt{2}\alpha (n+1)+4\alpha ^{2}}\right) \left \vert a_{n+p}\right \vert \leq 1-\sqrt{1-2\sqrt{2}\alpha +4\alpha ^{2}}$

for some $0 < \alpha < \frac{\sqrt{2}}{2},$ then $f(z)\in S_{\frac{\pi }{4}% }(\alpha, p).$

Theorem 3.6. Let the function $f(z)$ defined by $(1.1)$ be in the class $S_{\beta }(\alpha, p)$ and let

$0<\lambda \leq \frac{1}{2(\cos \beta -\alpha )}, \quad 0<\alpha <\cos \beta .$

(3.4)

Then we have

$\Re\left \{ \left( \frac{f^{\left( p-1\right) }(z)}{z}\right) ^{\lambda e^{\iota \beta }}\right \} >\frac{\left( p!\right) ^{-\lambda e^{\iota \beta }}}{2\lambda (\cos \beta -\alpha )+1}\ (z\in E).$

(3.5)

Proof. If we put

$A = \frac{1}{2\lambda (\cos \beta -\alpha )+1}$

and

$\left( \frac{f^{\left( p-1\right) }(z)}{p!z}\right) ^{\lambda e^{\iota \beta }} = (1-A)p(z)+A$

(3.6)

where $\lambda$ satisfies (3.4) then $p(z)$ is regular in the unit disk $E$ and $p(z) = 1+p_{1}z+p_{2}z^{2}+...$

Logarithmic differentiation of (3.6) yields

$\lambda e^{\iota \beta }\left[\frac{f^{(p)}(z)}{f^{(p-1)}(z)}-\frac{1}{z} \right] = (1-A)\frac{p^{\prime }(z)}{(1-A)p(z)+A}.$

This can be written as

$e^{\iota \beta }\frac{zf^{(p)}(z)}{f^{(p-1)}(z)}-e^{\iota \beta } = (1-A)\frac{ zp^{\prime }(z)}{\lambda \left \{ (1-A)p(z)+A\right \} },$

equivalently

$e^{\iota \beta }\frac{zf^{(p)}(z)}{f^{p-1}(z)}-\alpha = e^{\iota \beta }-\alpha +(1-A)\frac{zp^{\prime }(z)}{\lambda \left \{ (1-A)p(z)+A\right \} }.$

(3.7)

Since $f(z)\in S_{\beta }(\alpha, p)$ then from (3.7) we have

$\Re\left \{ e^{\iota \beta }-\alpha +(1-A)\frac{zp^{\prime }(z)}{ \lambda \left \{ (1-A)p(z)+A\right \} }\right \} >0, \ \ \ \ (z\in E, 0<\alpha <\cos \beta ).$

Let us consider the functional $\theta (u, v)$ defined by

$\theta (u, v) = e^{\iota \beta }-\alpha +(1-A)\frac{v}{\lambda \left \{ (1-A)u+A\right \} },$

where $u = p(z)$ and $v = zp^{\prime }(z).$ Then $\theta (u, v)$ is continuous in $D = \left(\mathbb{C} -\left \{ \frac{A}{A-1}\right \} \right) \times \mathbb{C}$.

Also, $(1, 0)\in D$ and $\Re\left \{ \theta (1, 0)\right \} = \cos \beta -\alpha >0.$ Furthermore, for all $(\iota u_{2}, v_{1})\in D$ such that $v_{1}\leq -\frac{(1+u_{2}^{2})}{2},$ we have

$\begin{eqnarray*} \Re\left \{ \theta (\iota u_{2}, v_{1})\right \} & = &\cos \beta -\alpha +% \Re\left \{ (1-A)\frac{v_{1}}{\left[\lambda (1-A)\iota u_{2}+A\right] }% \right \} \\ & = &\cos \beta -\alpha +\frac{A(1-A)v_{1}}{\lambda \left[ (1-A)^{2}u_{2}^{2}+A^{2}\right] } \\ &<&\cos \beta -\alpha -\frac{A(1-A)\left( 1+u_{2}^{2}\right) }{2\lambda % \left[(1-A)^{2}u_{2}^{2}+A^{2}\right] } \\ & = &(\cos \beta -\alpha )\frac{A^{2}\left[4\lambda ^{2}(\cos \beta -\alpha )^{2}-1\right] u_{2}^{2}}{\left[(1-A)^{2}u_{2}^{2}+A^{2}\right] } \\ &\leq &0, \end{eqnarray*}$

because $0 < \alpha < \cos \beta$ and $4\lambda ^{2}(\cos \beta -\alpha)^{2}-1\leq 0$ implies that $\lambda \leq \frac{1}{2(\cos \beta -\alpha)}.$

Therefore, the functional $\theta (u, v)$ satisfies all the conditions of Lemma 2.1.This proves that $\Re\left \{ p(z)\right \}$ $>0$, that is from (3.6)

$\begin{eqnarray*} \Re\left( \frac{f^{(p-1)}(z)}{p!z}\right) ^{\lambda e^{\iota \beta }} &>&A \\ \Re\left( \frac{f^{(p-1)}(z)}{z}\right) ^{\lambda e^{\iota \beta }} &>& \frac{\left( p!\right) ^{-\lambda e^{\iota \beta }}}{2\lambda (\cos \beta -\alpha )+1}. \end{eqnarray*}$

This completes the proof.

For $\beta = 0$ and $p = 1,$ in above theorem we have the following known result given by ^[11].

Corollary 3.7. Let $f\in A$ be in the class $S_{0}(\alpha, 1)$ and $0 < \lambda \leq \frac{1}{ 2(1-\alpha)},$ $0 < \alpha < 0$ then

$\Re\left( \frac{f(z)}{z}\right) ^{\lambda }>\frac{1}{2\lambda (1-\alpha )+1}, \;\;z\in E.$

Theorem 3.8. A function $f\in C_{\beta }(\alpha, p)$ if and only if

$\Re\left \{ e^{\iota \beta }\left( 1+\frac{zf^{\left( p+1\right) }(z)}{ f^{(p)}(z)}\right) \right \} >\alpha .$

Proof. Let $f(z)\in C_{\beta }(\alpha, p)$, then we can write

$\left \vert \frac{1}{e^{\iota \beta }G(z)}-\frac{1}{2\alpha }\right \vert < \frac{1}{2\alpha }.$

This can be written as

$\begin{eqnarray*} \left \vert \frac{1}{e^{\iota \beta }G(z)}-\frac{1}{2\alpha }\right \vert &<& \frac{1}{2\alpha }\Leftrightarrow \left \vert \frac{2\alpha -e^{\iota \beta }G(z)}{2\alpha e^{\iota \beta }G(z)}\right \vert <\frac{1}{2\alpha } \\ &\Leftrightarrow &\left \vert 2\alpha -e^{\iota \beta }G(z)\right \vert ^{2}<\left( e^{\iota \beta }G(z)\right) ^{2} \\ &\Leftrightarrow &\left( 2\alpha -e^{\iota \beta }G(z)\right) \overline{ \left( 2\alpha -e^{\iota \beta }G(z)\right) }<\left( e^{\iota \beta }G(z)\right) \overline{\left( e^{\iota \beta }G(z)\right) } \\ &\Leftrightarrow &\left( 2\alpha -e^{\iota \beta }G(z)\right) \left( 2\alpha -e^{-\iota \beta }\overline{G(z)}\right) <\left( e^{\iota \beta }G(z)\right) \left( e^{-\iota \beta }\overline{G(z)}\right) \\ &\Leftrightarrow &4\alpha ^{2}-2\alpha \left[e^{-\iota \beta }\overline{G(z) }+e^{\iota \beta }G(z)\right] +G(z)\overline{G(z)}<G(z)\overline{G(z)} \\ &\Leftrightarrow &4\alpha ^{2}-2\alpha \left[e^{-\iota \beta }\overline{G(z) }+e^{\iota \beta }G(z)\right] <0 \\ &\Leftrightarrow &2\alpha \left[2\alpha-\Re\left( e^{\iota \beta }G(z)\right) \right] <0 \\ &\Leftrightarrow &2\alpha -2\Re\left( e^{\iota \beta }G(z)\right) <0 \\ \ &\Leftrightarrow &\Re\left \{ e^{\iota \beta }\left( 1+\frac{ zf^{\left( p+1\right) }(z)}{f^{(p)}(z)}\right) \right \} >\alpha . \end{eqnarray*}$

This completes the proof.

Theorem 3.9. If $f(z)\in A(p)$ satisfies

$\sum\limits_{n = 1}^{\infty }\frac{(p+n)!}{n!}\left \{ n+1+\left \vert \left( n+1\right) -2\alpha e^{-\iota \beta }\right \vert \right \} \left \vert a_{n+p}\right \vert \leq 1-\left \vert 1-2\alpha e^{-\iota \beta }\right \vert$

(3.8)

for some $\left \vert \beta \right \vert < \frac{\pi }{2}$ and $0 < \alpha < \cos \beta,$ then $f(z)\in C_{\beta }(\alpha, p).$

Proof. To prove that $f(z)\in C_{\beta }(\alpha, p)$ we need to prove that

$\left \vert \frac{2\alpha -e^{\iota \beta }G(z)}{e^{\iota \beta }G(z)} \right \vert <1$

(3.9)

for some $\left \vert \beta \right \vert < \frac{\pi }{2}, 0 < \alpha < 1,$ where $G(z) = 1+\frac{zf^{(p+1)}(z)}{f^{(p)}(z)}.$

For this consider the left hand side of (3.9), we have

$\begin{eqnarray*} \left \vert \frac{2\alpha -e^{\iota \beta }G(z)}{e^{\iota \beta }G(z)} \right \vert & = &\left \vert \frac{2\alpha e^{-\iota \beta }f^{(p)}(z)-\left( f^{(p)}(z)+zf^{\left( p+1\right) }(z)\right) }{\left( f^{(p)}(z)+zf^{\left( p+1\right) }(z)\right) }\right \vert \\ & = &\left \vert \frac{2\alpha e^{-\iota \beta }-1+\sum_{n = 1}^{\infty }\frac{ (p+n)!}{(n+1)!}\left( n+1\right) a_{n+p}2\alpha e^{-\iota \beta }-\left( n+1\right) }{1+\sum_{n = 1}^{\infty }\frac{(p+n)!}{(n+1)!}\left( n+1\right) ^{2}\left \vert a_{n+p}\right \vert }\right \vert \\ & = &\frac{\left \vert 1-2\alpha e^{-\iota \beta }\right \vert +\sum_{n = 1}^{\infty }\frac{(p+n)!}{(n+1)!}\left \vert \left( n+1\right) a_{n+p}\right \vert \left \vert \left( n+1\right) -2\alpha e^{-\iota \beta }\right \vert }{1-\sum_{n = 1}^{\infty }\frac{(p+n)!}{(n+1)!}\left( n+1\right) ^{2}\left \vert a_{n+p}\right \vert }. \end{eqnarray*}$

The last expression is bounded above by 1 if

$\begin{eqnarray} &&\left \vert 1-2\alpha e^{-\iota \beta }\right \vert +\sum\limits_{n = 1}^{\infty } \frac{(p+n)!}{(n+1)!}\left \vert \left( n+1\right) a_{n+p}\right \vert \left \vert \left( n+1\right) -2\alpha e^{-\iota \beta }\right \vert \notag \\ &\leq &1-\sum\limits_{n = 1}^{\infty }\frac{(p+n)!}{(n+1)!}\left( n+1\right) ^{2}\left \vert a_{n+p}\right \vert \label{2a} \end{eqnarray}$

for some $\left \vert \beta \right \vert < \frac{\pi }{2}$ and $0 < \alpha < 1.$ After simplification, inequality (3.10) can be written as

$\sum\limits_{n = 1}^{\infty }\frac{(p+n)!}{n!}\left \{ n+1+\left \vert \left( n+1\right) -2\alpha e^{-\iota \beta }\right \vert \right \} \left \vert a_{n+p}\right \vert \leq 1-\left \vert 1-2\alpha e^{-\iota \beta }\right \vert .$

This completes the proof.

When we take $p = 1,$ we have the following known result given in ^[13].

Corollary 3.10. If $f\in A$ satisfies

$\sum\limits_{n = 2}^{\infty }\left \{ n\left( n+\left \vert n-2\alpha e^{-\iota \beta }\right \vert \right) \right \} \left \vert a_{n}\right \vert \leq 1-\left \vert 1-2\alpha e^{-\iota \beta }\right \vert$

for some $\left \vert \beta \right \vert < \frac{\pi }{2}$ and $0 < \alpha < \cos \beta,$ then $f\in K_{\beta }(\alpha).$

Taking $p = 1$, $\beta = 0$ in above theorem we have the following result given in ^[12].

Corollary 3.11. Let $0 < \alpha < 1.$ If $f\in A$ satisfies the following coefficient inequality

$\sum\limits_{n = 2}^{\infty }n\left( n-\alpha \right) \left \vert a_{n}\right \vert \leq \frac{1}{2}\left( 1-\left \vert 1-2\alpha \right \vert \right) = \left \{ \begin{array}{c} \alpha ;\quad \left( 0<\alpha \leq \frac{1}{2}\right) \\ \\ 1-\alpha ;\quad \left( \frac{1}{2}<\alpha <1\right) \end{array} \right.$

then $f(z)\in N(\alpha).$

Taking $\beta = \frac{\pi }{4}$ in above Theorem we have the following result.

Corollary 3.12. If $f(z)\in A(p)$ satisfies

$\begin{eqnarray*} &&\sum\limits_{n = 1}^{\infty }\frac{(p+n)!}{n!}\left( n+1+\sqrt{\left( n+1\right) ^{2}-2\sqrt{2}\alpha (n+1)+4\alpha ^{2}}\right) \left \vert a_{n+p}\right \vert \\ &\leq &1-\sqrt{1-2\sqrt{2}\alpha +4\alpha ^{2}} \end{eqnarray*}$

for some $0 < \alpha < \frac{\sqrt{2}}{2},$ then $f(z)\in S_{\frac{\pi }{4} }(\alpha, p).$

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Acknowledgments

The authors are grateful to the referees for their helpful comments and suggestions which improved the presentation of the paper. This research is carried out under the HEC projects grant No. 21-1235/SRGP/R & D/HEC/2016 and 21-1193/SRGP/R & D/HEC/2016.

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Conflict of Interest

No potential conflict of interest was reported by the authors.

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