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

$f(z) = z + \sum\limits_{n = 2}^\infty {{a_n}{z^n}} ,$

(1.1)

which are analytic in the unit disk $\mathbb{U}:=\{z:{|z|}<1\}$.

Let $S, S^{*}(\gamma), C(\gamma), K(\gamma)$ be the subclasses of $\mathcal{A}$ whose members are univalent, starlike of order $\gamma$, convex of order $\gamma$, and close-to-convex of order $\gamma$, respectively, where $0\leq \gamma<1$.

Let $f$ and $g$ be analytic in $\mathbb{U}$. Then $f$ is said to be subordinate to $g$, written $f\prec g$, if there exists an analytic function $\omega(z)$, with $\omega(0)=0$ and ${|w(z)|}<1$ such tat $f(z)=g(\omega(z))$. Indeed, it is known that

$f(z)\prec g(z)(z\in\mathbb{U})\Longrightarrow f(0)=g(0)\quad{\rm{and}}\quad f(\mathbb{U})\subset g(\mathbb{U}).$

Furthermore, if the function $g$ is univalent in $\mathbb{U}$, then we have the following equivalence:

$f(z)\prec g(z)(z\in\mathbb{U})\Longleftrightarrow f(0)=g(0) \quad{\rm and}\quad f(\mathbb{U})\subset g(\mathbb{U}).$

Sakaguchi [13] introduced a class $S_{s}^{*}$ of starlike functions with respect to symmetric points which satisfy the inequality

${\rm{Re}}\left(\frac{zf^{\prime}(z)}{f(z)-f(-z)}\right)>0\qquad(z\in\mathbb{U}).$

Since then, many authors discussed this class and its subclasses. Also, a function $f(z)\in\mathcal{A}$ is in the class $C_{s}$ if and only if $zf^{\prime}(z)\in S_{s}^{*}$.

Let $P_{m}(\gamma)$ be the class of functions $p$ analytic in $\mathbb{U}$ satisfying the conditions $p(0)=1$ and

$\int_{0}^{2\pi}|{\frac{{\rm{Re}}(p(z))-\gamma}{1-\gamma}}|d\theta\leq m\pi\qquad(0\leq\gamma<1;\, m\geq2;\, z=re^{i\theta}).$

This class was introduced in [11]. We note that $P_{m}(0)\equiv P_{m}$ is introduced in [12] and $P_{2}(\gamma)\equiv P(\gamma)$ is the class of functions with positive real part of order $\gamma$. With $m=2$, $\gamma=0$, we have the class $P$ of functions with positive real part.

The classes $V_{m}(\gamma)$ of functions of bounded boundary rotation of order $\gamma$ and $R_{m}(\gamma)$ of functions of bounded radius rotation of order $\gamma$ are closely related with $P_{m}(\gamma)$. A function $f\in\mathcal{A}$ is said to be in the class $V_{m}(\gamma)$ if and only if

$\frac{(zf^{\prime}(z))^{\prime}}{f^{\prime}(z)}\in P_{m}(\gamma)\qquad(z\in\mathbb{U}).$

Moreover, we know that

$f\in R_{m}(\gamma)\Longleftrightarrow\frac{zf^{\prime}(z)}{f(z)}\in P_{m}(\gamma)\qquad(z\in\mathbb{U}).$

Motivated essentially by the above work, we introduce and study the following classes $R_{m}^{s}(b, k, \lambda)$ and $K_{m}^{s}(\alpha, b, k, \lambda, \delta)$ with respect to $k$-symmetric points.

Definition 1.1  Suppose that $b\in\mathbb{C}\setminus\{0\}$, $0\leq\lambda\leq1$, $m\geq2$ and $k$ is a fixed positive integer. A function $f\in\mathcal{A}$ is said to be in the class $R_{m}^{s}(b, k, \lambda)$ if and only if

$\begin{equation}\label{12} 1+\frac{1}{b}\left\{\frac{zf^{\prime}(z)+\lambda z^{2}f^{\prime\prime}(z)}{(1-\lambda)f_{k}(z)+\lambda zf_{k}^{\prime}(z)}-1\right\}\in P_{m}\qquad(z\in\mathbb{U}), \end{equation}$

(1.2)

where $f_{k}(z)$ is defined by

${f_k}(z) = \frac{1}{k}\sum\limits_{\nu = 0}^{k - 1} {{\varepsilon ^{ - \nu }}} f({\varepsilon ^\nu }z)\qquad \left( {z \in\mathbb{U};\;\varepsilon = {\rm{exp}}\left( {\frac{{2\pi i}}{k}} \right)} \right).$

(1.3)

Remark 1.1  For some recent investigations on analytic functions involving $k$-symmetric points, one can refer to [6, 14-17].

Definition 1.2  Let $\alpha>0$, $m\geq2$ and $0\leq\delta<1$. A function $f\in\mathcal{A}$ is said to be in the class $K_{m}^{s}(\alpha, b, k, \lambda, \delta)$ if and only if

$\begin{equation}\label{14} \frac{zf^{\prime}(z)}{f(z)}\left(\frac{f(z)}{g_{k}(z)}\right)^{\alpha}\in P_{m}(\delta)\qquad(z\in\mathbb{U}) \end{equation}$

(1.4)

for some $g\in R_{2}^{s}(b, k, \lambda)$.

Remark 1.2  For special choices of $\alpha, b, k, \lambda, m$ and $\delta$, several related function classes have been studied extensively, see for example [1, 2, 8-10].

In the present paper, we aim at proving some basic properties of the classes $R_{m}^{s}(b,k,\lambda)$ and $K_{m}^{s}(\alpha,b,k,\lambda,\delta)$. Such results as integral representations, coefficient inequalities, covering theorems and arc-length estimates are derived. The results presented here would provide extensions of those given in earlier works.

In order to prove our main results, we need the following lemmas.

Lemma 2.1  (see [3]) Let $h$ be convex in $\mathbb{U}$ with ${\rm{Re}}(\beta h(z)+\gamma)>0$. If $q$ is analytic in $\mathbb{U}$ with $q(0)=h(0)$, then

$\begin{equation}\label{21} q(z)+\frac{zq^{\prime}(z)}{\beta q(z)+\gamma}\prec h(z)\Longrightarrow q(z)\prec h(z). \end{equation}$

(2.1)

Lemma 2.2  (see [5]) If $f\in S^{*}(\alpha)$, $0\leq\alpha<1$ and ${|z|}=r<1$, then

$\begin{equation}\label{23} \frac{r}{(1+r)^{2(1-\alpha)}}\leq{f(z)}\leq\frac{r}{(1-r)^{2(1-\alpha)}}. \end{equation}$

(2.2)

Lemma 2.3 (see [7]) Let $p\in P_{m}(\gamma)$ and ${|z|}=r<1$. Then

$\begin{equation}\label{24} \frac{1}{2\pi}\int_{0}^{2\pi}{p(z)}^{2}d\theta\leq\frac{1+[m^{2}(1-\gamma)^{2}-1]r^{2}}{1-r^{2}}. \end{equation}$

(2.3)

Lemma 2.4  (see [4]) Let $q$ be univalent in $\mathbb{U}$. Then there exists a point $\xi$ with ${|\xi|}=r$ such that for all $z$, ${|z|}=r$,

$\begin{equation}\label{25} {z-\xi}{q(z)}\leq\frac{2r^{2}}{1-r^{2}}. \end{equation}$

(2.4)

We begin by stating the following result which involved the connections between $R_{m}^{s}(b, k, \lambda)$ and $R_{m}(1-b)$.

Theorem 3.1 Let $f\in R_{m}^{s}(b, k, \lambda)$. Then

$\begin{equation}\label{31} (1-\lambda)f_{k}(z)+{\lambda}zf_{k}^{\prime}(z)\in R_{m}(1-b)\qquad(z\in\mathbb{U}). \end{equation}$

(3.1)

Proof  Let

$\begin{equation}\label{32} F(z)=(1-\lambda)f(z)+{\lambda}zf^{\prime}(z) \end{equation}$

(3.2)

and

$\begin{equation}\label{33} F_{k}(z)=(1-\lambda)f_{k}(z)+\lambda zf_{k}^{\prime}(z). \end{equation}$

(3.3)

Then condition (1.2) can be written as

$\begin{equation}\label{34} 1+\frac{1}{b}\left(\frac{zF^{\prime}(z)}{F_{k}(z)}-1\right)=p(z) \end{equation}$

(3.4)

for some $p\in P_{m}$. Substituting $z$ by $\varepsilon^{\mu}z\, (\mu=0, 1, 2, \cdots, k-1)$ in (3.4) gives

$\begin{equation}\label{35} 1+\frac{1}{b}\left(\frac{\varepsilon^{\mu}zf^{\prime}(\varepsilon^{\mu}z)+\lambda(\varepsilon^{\mu}z)^{2}f^{\prime\prime}(\varepsilon^{\mu}z)}{(1-\lambda)f_{k}(\varepsilon^{\mu}z) +\lambda\varepsilon^{\mu}zf_{k}^{\prime}(\varepsilon^{\mu}z)}-1\right)=p(\varepsilon^{\mu}z) . \end{equation}$

(3.5)

We note that $f_{k}(\varepsilon^{\nu}z)=\varepsilon^{\nu}f_{k}(z)$, $f_{k}^{\prime}(\varepsilon^{\nu}z)=f_{k}^{\prime}(z)$ and $\varepsilon^{\mu}f_{k}^{\prime\prime}(\varepsilon^{\nu}z)=f_{k}^{\prime\prime}(z)$. Thus taking $\mu=0, 1, 2, \cdots, k-1$ in (3.5), respectively, and summing the resulting equations, we get

$1 + \frac{1}{b}\left( {\frac{{zF_k^\prime (z)}}{{{F_k}(z)}} - 1} \right) = \frac{1}{k}\sum\limits_{\mu = 0}^{k - 1} p ({\varepsilon ^\mu }z).$

(3.6)

Since $P_{m}$ is a convex set, it is clear that

$\begin{equation}\label{37} 1+\frac{1}{b}\left(\frac{zF_{k}^{\prime}(z)}{F_{k}(z)}-1\right)\in P_{m}, \end{equation}$

(3.7)

which implies that

$\begin{equation}\label{38} \frac{zF_{k}^{\prime}(z)}{F_{k}(z)}\in P_{m}(1-b) \end{equation}$

(3.8)

and hence $F_{k}(z)\in R_{m}(1-b)$.

Next, we give the integral representations of functions belonging to the class $R_{m}^{s}(b, k, \lambda)$.

Theorem 3.2 Let $f\in R_{m}^{s}(b, k, \lambda)$ with $0<\lambda\leq1$. Then

${f_k}(z) = \frac{1}{\lambda }{z^{1 - \frac{1}{\lambda }}}\int_0^z {{\rm{exp}}} \left( {\frac{b}{k}\sum\limits_{\mu = 0}^{k - 1} {\int_0^{{\varepsilon ^\mu }u} {\frac{{p(\xi ) - 1}}{\xi }} } d\xi } \right){u^{\frac{1}{\lambda } - 1}}du$

(3.9)

for some $p\in P_{m}$.

Proof  Suppose that $f\in R_{m}^{s}(b, k, \lambda)$. From (1.2), we get

$\begin{equation}\label{40} \frac{zf^{\prime}(z)+\lambda z^{2}f^{\prime\prime}(z)}{(1-\lambda)f_{k}(z)+\lambda zf^{\prime}_{k}(z)}=b\left(p(z)-1\right)+1 \end{equation}$

(3.10)

for some $p\in P_{m}$. Substituting $z$ by $\varepsilon^{\mu}z\ (\mu=0, 1, 2, \cdots, k-1)$ in (3.10), we have

$\begin{equation}\label{41} \frac{\varepsilon^{\mu}zf^{\prime}(\varepsilon^{\mu}z)+\lambda(\varepsilon^{\mu}z)^{2}f^{\prime\prime}(\varepsilon^{\mu}z)}{(1-\lambda)f_{k}(\varepsilon^{\mu}z)+\lambda \varepsilon^{\mu}zf^{\prime}_{k}(\varepsilon^{\mu}z)}=b\left(p(\varepsilon^{\mu}z)-1\right)+1. \end{equation}$

(3.11)

By observing that $f_{k}(\varepsilon^{\mu}z)=\varepsilon^{\mu}f_{k}(z)$ and $f_{k}^{\prime}(\varepsilon^{\mu}z)=f_{k}^{\prime}(z)$, we know that (3.11) can be written as

$\begin{equation}\label{42} \frac{zf^{\prime}(\varepsilon^{\mu}z)+\lambda \varepsilon^{\mu}z^{2}f^{\prime\prime}(\varepsilon^{\mu}z)}{(1-\lambda)f_{k}(z)+\lambda zf^{\prime}_{k}(z)}=b\left(p(\varepsilon^{\mu}z)-1\right)+1. \end{equation}$

(3.12)

Taking $\mu=0, 1, 2, \cdots, k-1$ in (3.12), respectively, and summing the resulting equations, we obtain

$\frac{{zf_k^\prime (z) + \lambda {z^2}f_k^{\prime \prime }(z)}}{{(1 - \lambda ){f_k}(z) + \lambda zf_k^\prime (z)}} = \frac{1}{k}\sum\limits_{\mu = 0}^{k - 1} {\left( {b\left( {p({\varepsilon ^\mu }z) - 1} \right) + 1} \right)} ,$

(3.13)

which follows that

$\frac{{(1 - \lambda )f_k^\prime (z) + \lambda {{(zf_k^\prime (z))}^\prime }}}{{(1 - \lambda ){f_k}(z) + \lambda zf_k^\prime (z)}} - \frac{1}{z} = \frac{b}{k}\sum\limits_{\mu = 0}^{k - 1} {\frac{{p({\varepsilon ^\mu }z) - 1}}{z}} .$

(3.14)

Integrating (3.14), we get

${\rm{log}}\left( {\frac{{(1 - \lambda ){f_k}(z) + \lambda zf_k^\prime (z)}}{z}} \right) = \frac{b}{k}\sum\limits_{\mu = 0}^{k - 1} {\int_0^{{\varepsilon ^\mu }z} {\frac{{p(\xi ) - 1}}{\xi }} } d\xi $

(3.15)

or equivalently

$(1 - \lambda ){f_k}(z) + \lambda zf_k^\prime (z) = z \cdot {\rm{exp}}\left( {\frac{b}{k}\sum\limits_{\mu = 0}^{k - 1} {\int_0^{{\varepsilon ^\mu }z} {\frac{{p(\xi ) - 1}}{\xi }} } d\xi } \right).$

(3.16)

The assertion of Theorem 3.2 can now be derived from (3.16).

Theorem 3.3 Let $f\in R_{m}^{s}(b, k, \lambda)$ with $0<\lambda\leq1$. Then

$f(z) = \frac{1}{\lambda }{z^{1 - \frac{1}{\lambda }}}\int_0^z {\int_0^u {{\rm{exp}}} } \left( {\frac{b}{k}\sum\limits_{\mu = 0}^{k - 1} {\int_0^{{\varepsilon ^\mu }\xi } {\frac{{p(t) - 1}}{t}} } dt} \right) \cdot \left( {b\left( {p(\xi ) - 1} \right) + 1} \right)d\xi {u^{\frac{1}{\lambda } - 2}}du$

(3.17)

for some $p\in P_{m}$.

Proof  Suppose that $f\in R_{m}^{s}(b, k, \lambda)$. From (1.2) and (3.16), we have

$\begin{align}\label{48}\begin{split} (1-\lambda)f^{\prime}(z)+\lambda(zf^{\prime}(z))^{\prime}&=\frac{(1-\lambda)f_{k}(z)+\lambda zf_{k}^{\prime}(z)}{z}\cdot\left(b\left(p(z)-1\right)+1\right)\\ &={\rm{exp}}\left(\frac{b}{k}\sum_{\mu=0}^{k-1}\int_{0}^{\varepsilon^{\mu }z}\frac{p(t)-1}{t}dt\right)\cdot\left(b\left(p(z)-1\right)+1\right).\end{split} \end{align}$

(3.18)

Integrating (3.18) yields

$(1 - \lambda )f(z) + \lambda z{f^\prime }(z) = \int_0^z {{\rm{exp}}} \left( {\frac{b}{k}\sum\limits_{\mu = 0}^{k - 1} {\int_0^{{\varepsilon ^\mu }\xi } {\frac{{p(t) - 1}}{t}} } dt} \right) \cdot \left( {b\left( {p(\xi ) - 1} \right) + 1} \right)d\xi .$

(3.19)

From (3.19), we can get (3.17) easily.

In what follows, we provide some coefficient inequalities and covering theorems for functions in the class $R_{m}^{s}(b, k, \lambda)$.

Theorem 3.4 Let $f\in R_{m}^{s}(b, k, \lambda)$ with $k\geq2$. Then

$\begin{equation}\label{50} |{a_{2}}|\leq\frac{m{b}}{2(1+\lambda)}. \end{equation}$

(3.20)

Proof Suppose that $f\in R_{m}^{s}(b, k, \lambda)$. In view of Theorem 3.1, there exists a function $\phi\in R_{m}(1-b)$, $\phi(z)=(1-\lambda)f_{k}(z)+\lambda zf_{k}^{\prime}(z)$ such that

$\begin{equation}\label{52} zf^{\prime}(z)+\lambda z^{2}f^{\prime\prime}(z)=\phi(z) p(z) \end{equation}$

(3.21)

for some $p\in P_{m}(1-b)$. Using the fact that

$\begin{equation}\label{53}\begin{split} f_{k}(z)=\frac{1}{k}\sum_{\nu=0}^{k-1}\varepsilon^{-\nu}f(\varepsilon^{\nu}z)=\frac{1}{k}\sum_{\nu=0}^{k-1}\left(\varepsilon^{\nu}z+\sum_{n=2}^{\infty} a_{n}(\varepsilon^{\nu}z)^{n}\right)=z+\sum_{l=2}^{\infty}a_{(l-1)k+1}z^{(l-1)k+1}, \end{split} \end{equation}$

(3.22)

we have

$\begin{equation}\phi(z)=z+\mathop \sum \limits_{l = 2}^\infty [1+\lambda(l-1)k]a_{(l-1)k+1}z^{(l-1)k+1}. \end{equation}$

(3.23)

Let

$p(z) = 1 + \sum\limits_{n = 1}^\infty {{c_n}} {z^n}.$

(3.24)

Then we find from (3.21) that

$\begin{equation}\label{56} z+\mathop \sum \limits_{n = 2}^\infty n[1+\lambda(n-1)]a_{n}z^{n}=\left(z+\mathop \sum \limits_{l = 2}^\infty [1+\lambda(l-1)k]a_{(l-1)k+1}z^{(l-1)k+1}\right)\left(1+\mathop \sum \limits_{n = 1}^\infty c_{n}z^{n}\right). \end{equation}$

(3.25)

Comparing the coefficients $z^{2}$ in both sides of (3.25), we get $2(1+\lambda)a_{2}=c_{1}$, which follows that

$|{a_2}| \le \frac{{|{c_1}|}}{{2(1 + \lambda )}}.$

(3.26)

Since ${|c_{1}|}\leq m{b}$ for $p\in P_{m}(1-b)$, we get the desired assertion of Theorem 3.4.

Theorem 3.5 Let $f\in R_{m}^{s}(b, k, \lambda)$ with $k\geq2$. Then the unit disk $\mathbb{U}$ is mapped by every univalent function $f$ onto a domain that contains the disk ${|\omega|}<r_{1}$, where

$\begin{equation}\label{58} r_{1}=\frac{2(1+\lambda)}{4(1+\lambda)+m{b}}. \end{equation}$

(3.27)

Proof  Suppose that $f\in R_{m}^{s}(b, k, \lambda)$. Also, let $\omega_{0}$ be any complex number such that $f(z)\neq\omega_{0}$ for $z\in\mathbb{U}$, then $\omega_{0}\neq0$ and

$\begin{equation}\label{59} \frac{\omega_{0}f(z)}{\omega_{0}-f(z)}=z+\left(a_{2}+\frac{1}{\omega_{0}}\right)z^{2}+\cdots \end{equation}$

(3.28)

for every univalent function $f$. This leads to

$\begin{equation}\label{60} |{a_{2}+\frac{1}{\omega_{0}}}|\leq2 \end{equation}$

(3.29)

and hence

$|{\omega _0}| \ge \frac{1}{{|{a_2}| + 2}}.$

(3.30)

Using (3.30) and Theorem 3.4, we obtain the required result.

Let $L_{r}f(z)$ denote the length of the image of the circle $|z|=r$ under $f(z)$. we finally show some basic properties of functions in the class $K_{m}^{s}(\alpha, b, k, \lambda, \delta)$ including arc-length and coefficient problems.

Theorem 3.6 Suppose that $f\in K_{m}^{s}(\alpha, b, k, \lambda, \delta)$ with $0<\lambda\leq1$ and $0<b\leq1$. Then

$\begin{equation}\label{62} L_{r}f(z)\leq \begin{cases} C(\alpha, b, \delta, m)M(r)^{1-\alpha}\left(\frac{1}{1-r}\right)^{\frac{4\alpha b+1}{2}}&\quad(0<\alpha\leq1), \\ C(\alpha, b, \delta, m)N(r)^{1-\alpha}\left(\frac{1}{1-r}\right)^{\frac{4\alpha b+1}{2}}&\quad(\alpha>1), \end{cases} \end{equation}$

(3.31)

where $N(r)=\min\limits_{{|z|}=r}{|f(z)|}$, $M(r)=\max\limits_{{|z|}=r}{|f(z)|}$, and $C(\alpha, b, \delta, m)$ is a constant which is determined by the parameters $\alpha, b, \delta$ and $m$.

Proof  Suppose that $f\in K_{m}^{s}(\alpha, b, k, \lambda, \delta)$. From definition (1.4), we know that

$\frac{{z{f^\prime }(z)}}{{f(z)}}{\left( {\frac{{f(z)}}{{{g_k}(z)}}} \right)^\alpha } = p(z)$

(3.32)

for some $p\in P_{m}(\delta)$. It follows that

$\begin{equation}\label{64} zf^{\prime}(z)=(f(z))^{1-\alpha}\left(g_{k}(z)\right)^{\alpha}p(z). \end{equation}$

(3.33)

For $0<\alpha\leq1$, we find from (3.33) that

$\begin{equation}\label{65}\begin{split} L_{r}f(z)=\int_{0}^{2\pi}{|zf^{\prime}(z)|}d\theta\leq\int_{0}^{2\pi}{|f(z)|}^{1-\alpha}{|g_{k}(z)|}^{\alpha}{|p(z)|}d\theta\leq M^{1-\alpha}(r)\int_{0}^{2\pi}{|g_{k}(z)|}^{\alpha}{|p(z)|}d\theta, \end{split} \end{equation}$

(3.34)

where $M(r)=\max\limits_{{|z|}=r}{|f(z)|}$. Since $g\in R_{2}^{s}(b, k, \lambda)$, from Theorem 3.1, we have

$\begin{equation}\label{66} (1-\lambda)g_{k}(z)+\lambda zg_{k}^{\prime}(z)=G_{k}(z)\in R_{2}(1-b)\equiv S^{*}(1-b). \end{equation}$

(3.35)

Let $q(z)=\frac{zg_{k}^{\prime}(z)}{g_{k}(z)}$. It follows from (3.35) that

$\begin{equation}\label{660} \frac{G_{k}(z)}{g_{k}(z)}=1-\lambda+\lambda q(z). \end{equation}$

(3.36)

Differentiate both sides of (3.36) logarithmically, we obtain

$\begin{equation}\label{67} q(z)+\frac{zq^{\prime}(z)}{q(z)+\frac{1-\lambda}{\lambda}}=\frac{zG_{k}^{\prime}(z)}{G_{k}(z)}\prec\frac{1+(2b-1)z}{1-z}. \end{equation}$

(3.37)

By noting that

${\rm{Re}}\left[\frac{1+(2b-1)z}{1-z}+\frac{1-\lambda}{\lambda}\right]>0\qquad(0<b\leq1;\ 0<\lambda\leq1), $

an application of Lemma 2.1 to (3.37) yields

$\begin{equation}\label{68} q(z)\prec\frac{1+(2b-1)z}{1-z}, \end{equation}$

(3.38)

which implies that $g_{k}(z)\in S^{*}(1-b)$. By Lemma 2.2, we have

$\begin{equation}\label{69} \frac{r}{(1+r)^{2b}}\leq{g_{k}(z)}\leq\frac{r}{(1-r)^{2b}}. \end{equation}$

(3.39)

Using (3.39) and Lemma 2.3, we deduce from (3.34) that

$\begin{equation}\label{70}\begin{split} L_{r}f(z)&\leq M(r)^{1-\alpha}\frac{r^{\alpha}}{(1-r)^{2\alpha b}}\int_{0}^{2\pi}{|p(z)|}d\theta\\ &\leq 2\pi M(r)^{1-\alpha}\frac{r^{\alpha}}{(1-r)^{2\alpha b}}\left(\frac{1}{2\pi}\int_{0}^{2\pi}{|p(z)|}^{2}d\theta\right)^{\frac{1}{2}}\\ &\leq 2\pi M(r)^{1-\alpha}\frac{r^{\alpha}}{(1-r)^{2\alpha b}}\left(\frac{1+[m^{2}(1-\delta)^{2}-1]r^{2}}{1-r^{2}}\right)^{\frac{1}{2}}\\ &=C(\alpha, b, \delta, m)M(r)^{1-\alpha}\left(\frac{1}{1-r}\right)^{\frac{4\alpha b+1}{2}}.\end{split} \end{equation}$

(3.40)

Similarly, for $\alpha>1$, we have

$L_{r}f(z)\leqq C(\alpha, b, \delta, m)N(r)^{1-\alpha}\left(\frac{1}{1-r}\right)^{\frac{4\alpha b+1}{2}}.$

Theorem 3.7 Let $f\in K_{m}^{s}(\alpha, b, k, \lambda, \delta)$ with $0<\lambda\leq1$ and $0<b\leq1$. Then

$\begin{equation}\label{72} \left| {{{a}_{n}}} \right|\leq \begin{cases} C_{1}(\alpha, b, \delta, m)M(n)^{{1-\alpha}}n^{\frac{4\alpha b-1}{2}}&\quad(0<\alpha\leq1), \\ C_{1}(\alpha, b, \delta, m)N(n)^{{1-\alpha}}n^{\frac{4\alpha b-1}{2}}&\quad(\alpha>1).\\ \end{cases} \end{equation}$

(3.41)

Proof  Suppose that $f\in K_{m}^{s}(\alpha, b, k, \lambda, \delta)$. For $n\geq1$ and $z=re^{i\theta}$, Cauchy's Theorem gives that

$\begin{equation}\label{73} na_{n}=\frac{1}{2\pi r^{n}}\int_{0}^{2\pi}zf^{\prime}(z)e^{-in\theta}d\theta. \end{equation}$

(3.42)

Using Theorem 3.6 for $0<\alpha\leq1$, we get

$n\left| {{a_n}} \right| \le \frac{1}{{2\pi {r^n}}}C(\alpha ,b,\delta ,m)M{(r)^{1 - \alpha }}{\left( {\frac{1}{{1 - r}}} \right)^{\frac{{4\alpha b + 1}}{2}}}.$

(3.43)

Taking $r=1-\frac{1}{n}$ in (3.43), we obtain

$\begin{equation}\label{75} \left| {{{a}_{n}}} \right|\leq C_{1}(\alpha, b, \delta, m)M(n)^{1-\alpha}n^{\frac{4\alpha b-1}{2}}. \end{equation}$

(3.44)

Using the similar techniques, we can prove the corresponding result for $\alpha>1$.

Theorem 3.8 Let $f\in K_{m}^{s}(\alpha, b, k, \lambda, \delta)$ with $0<\lambda\leq1$ and $0<b\leq1$. Then

$\begin{equation}\label{76} \left\| {{{a}_{{n}{\rm{ + 1}}}}\left| {\rm{ - }} \right|{{a}_{n}}} \right\|\leq \begin{cases} C_{2}(\alpha, b, \delta, m)M(r)^{1-\alpha}\left(\frac{1}{1-r}\right)^{\frac{1}{2}}&\quad(0<\alpha\leq1), \\ C_{2}(\alpha, b, \delta, m)N(r)^{1-\alpha}\left(\frac{1}{1-r}\right)^{\frac{4(\alpha-1)b+1}{2}}&\quad(\alpha>1).\\ \end{cases} \end{equation}$

(3.45)

Proof  It is known that for $\xi\in\mathbb{U}, z=re^{i\theta}$ and $n\geq1$, one has

$\begin{equation}\label{77} {|(n+1)\xi a_{n+1}-na_{n}|}\leq\int_{0}^{2\pi}{|z-\xi}{zf^{\prime}(z)|}d\theta. \end{equation}$

(3.46)

Since $f\in K_{m}^{s}(\alpha, b, k, \lambda, \delta)$, we obtain

$\begin{equation}\label{640} zf^{\prime}(z)=(f(z))^{1-\alpha}\left(g_{k}(z)\right)^{\alpha}p(z) \end{equation}$

(3.47)

and

$\begin{equation}\label{690} \frac{r}{(1+r)^{2b}}\leq{g_{k}(z)}\leq\frac{r}{(1-r)^{2b}}. \end{equation}$

(3.48)

For $0<\alpha\leq1$, combining (3.47), (3.48) and (3.46), we get

$\begin{equation}\label{80} {|(n+1)\xi a_{n+1}-na_{n}|}\leq M(r)^{1-\alpha}\frac{r^{\alpha-1}}{(1+r)^{2(\alpha-1) b}}\int_{0}^{2\pi}|{z-\xi}||{g_{k}(z)}||{p(z)|}d\theta. \end{equation}$

(3.49)

By Lemmas 2.3 and 2.4, we deduce that

$\begin{equation*}\label{81}\begin{split} {|(n+1)\xi a_{n+1}-na_{n}|}&\leq M(r)^{1-\alpha}\frac{r^{\alpha-1}}{(1+r)^{2(\alpha-1) b}}\frac{2r^{2}}{1-r^{2}}\int_{0}^{2\pi}{|p(z)|}d\theta\\ &\leq 2\pi M(r)^{1-\alpha}\frac{r^{\alpha-1}}{(1+r)^{2(\alpha-1) b}}\frac{2r^{2}}{1-r^{2}}\left(\frac{1}{2\pi}\int_{0}^{2\pi}{|p(z)|}^{2}d\theta\right)^{\frac{1}{2}}\\ &\leq 2\pi M(r)^{1-\alpha}\frac{r^{\alpha-1}}{(1+r)^{2(\alpha-1) b}}\frac{2r^{2}}{1-r^{2}}\left(\frac{1+[m^{2}(1-\delta)^{2}-1]r^{2}}{1-r^{2}}\right)^{\frac{1}{2}}.\end{split} \end{equation*}$

Putting ${\xi}=r=\frac{n}{n+1}$, it follows that

$\begin{equation*}\label{83} {{||a_{n+1}|}-{|a_{n}||}}\leq C_{2}(\alpha, b, \delta, m)M(n)^{1-\alpha}n^{\frac{1}{2}}. \end{equation*}$

Similarly, we can get the required result for $\alpha>1$.