Lobachevskii Journal of Mathematics http://ljm.ksu.ru Vol. 26, 2007, 125–135

Zhi-Gang Wang and Da-Zhao Chen
ON SUBCLASSES OF CLOSE-TO-CONVEX AND QUASI-CONVEX FUNCTIONS WITH RESPECT TO $2 K$ -SYMMETRIC CONJUGATE POINTS
(submitted by F. G. Avkhadiev)

Abstract. In the present paper, the authors introduce two new subclasses $S_{s c}^{(k)} (λ, α)$ of close-to-convex functions and $C_{s c}^{(k)} (λ, α)$ of quasi-convex functions with respect to $2 k$ -symmetric conjugate points. The integral representations and convolution conditions for these classes are provided. Some coefficient inequalities for functions belonging to these classes and their subclasses with negative coefficients are also provided.

________________

2000 Mathematical Subject Classiﬁcation. 30C45.

Key words and phrases. Close-to-convex functions, quasi-convex functions, Hadamard product, $2 k$ -symmetric conjugate points.

1. Introduction

Let $A$ denote the class of functions of the form

f (z) = z + \sum_{n = 2}^{\infty} a_{n} z^{n}, (1.1)

which are analytic in the open unit disk $U = {z \in C : |z| < 1}$ . Let $S$ , $S^{*}$ , $K$ , $C$ and $C^{*}$ denote the familiar subclasses of $A$ consisting of functions which are univalent, starlike, convex, close-to-convex and quasi-convex in $U$ , respectively (see, for details, [2, 3, 4, 5]).

Al-Amiri, Coman and Mocanu [1] once introduced and investigated a class of functions starlike with respect to $2 k$ -symmetric conjugate points, which satisfy the following inequality

ℜ \{\frac{z f^{'} (z)}{f_{2 k} (z)}\} > 0 (z \in U),

where $k \geq 2$ is a ﬁxed positive integer and $f_{2 k} (z)$ is deﬁned by the following equality

f_{2 k} (z) = \frac{1}{2 k} \sum_{ν = 0}^{k - 1} [ɛ^{- ν} f (ɛ^{ν} z) + ɛ^{ν} \bar{f (ɛ^{ν} \bar{z})}] (ɛ = e x p (2 π i ∕ k); z \in U) . (1.2)

In the present paper, we introduce the following two classes of analytic functions with respect to $2 k$ -symmetric conjugate points, and obtain some interesting results.

Deﬁnition 1. Let $S_{s c}^{(k)} (λ, α)$ denote the class of functions in $A$ satisfying the following inequality

ℜ \{\frac{z f^{'} (z) + λ z^{2} f^{''} (z)}{(1 - λ) f_{2 k} (z) + λ z f_{2 k}^{'} (z)}\} > α (z \in U), (1.3)

where $0 \leq λ \leq 1, 0 \leq α < 1$ and $f_{2 k} (z)$ is deﬁned by equality (1.2). And a function $f (z) \in A$ is in the class $C_{s c}^{(k)} (λ, α)$ if and only if $z f^{'} (z) \in S_{s c}^{(k)} (λ, α)$ .

In our proposed investigation of the classes $S_{s c}^{(k)} (λ, α)$ and $C_{s c}^{(k)} (λ, α)$ , we shall also make use of the following lemmas.

Lemma 1. Let $γ \geq 0$ and $f \in C$ , then

F (z) = \frac{1 + γ}{z^{γ}} \int_{0}^{z} f (t) t^{γ - 1} d t \in C .

This lemma is a special case of Theorem 4 in [6].

Lemma 2 [3]. Let $0 < λ \leq 1$ and $f \in C^{*}$ , then

F (z) = \frac{1}{λ} z^{1 - \frac{1}{λ}} \int_{0}^{z} f (t) t^{\frac{1}{λ} - 2} d t \in C^{*} \subset C .

Lemma 3. Let $0 \leq λ \leq 1$ and $0 \leq α < 1$ , then we have $S_{s c}^{(k)} (λ, α) \subset C \subset S$ .

Proof. Let $F (z) = (1 - λ) f (z) + λ z f^{'} (z), F_{2 k} (z) = (1 - λ) f_{2 k} (z) + λ z f_{2 k}^{'} (z)$ with $f (z) \in S_{s c}^{(k)} (λ, α)$ , substituting $z$ by $ɛ^{μ} z (μ = 0, 1, 2, \dots, k - 1)$ in (1.1), we get

ℜ \{\frac{ɛ^{μ} z f^{'} (ɛ^{μ} z) + λ {(ɛ^{μ} z)}^{2} f^{''} (ɛ^{μ} z)}{(1 - λ) f_{2 k} (ɛ^{μ} z) + λ ɛ^{μ} z f_{2 k}^{'} (ɛ^{μ} z)}\} > α . (1.4)

From inequality (1.4), we have

ℜ \{\frac{\bar{ɛ^{μ} \bar{z}} \bar{f^{'} (ɛ^{μ} \bar{z})} + λ \bar{{(ɛ^{μ} \bar{z})}^{2}} \bar{f^{''} (ɛ^{μ} \bar{z})}}{(1 - λ) \bar{f_{2 k} (ɛ^{μ} \bar{z})} + λ \bar{ɛ^{μ} \bar{z}} \bar{f_{2 k}^{'} (ɛ^{μ} \bar{z})}}\} > α . (1.5)

Note that $f_{2 k} (ɛ^{μ} z) = ɛ^{μ} f_{2 k} (z)$ , $f_{2 k}^{'} (ɛ^{μ} z) = f_{2 k}^{'} (z)$ , $\bar{f_{2 k} (ɛ^{μ} \bar{z})} = ɛ^{- μ} f_{2 k} (z)$ and $\bar{f_{2 k}^{'} (ɛ^{μ} \bar{z})} = f_{2 k}^{'} (z)$ , thus, inequalities (1.4) and (1.5) can be written as

ℜ \{\frac{z f^{'} (ɛ^{μ} z) + λ z^{2} ɛ^{μ} f^{''} (ɛ^{μ} z)}{(1 - λ) f_{2 k} (z) + λ z f_{2 k}^{'} (z)}\} > α, (1.6)

and

ℜ \{\frac{z \bar{f^{'} (ɛ^{μ} \bar{z})} + λ z^{2} ɛ^{- μ} \bar{f^{''} (ɛ^{μ} \bar{z})}}{(1 - λ) f_{2 k} (z) + λ z f_{2 k}^{'} (z)}\} > α . (1.7)

Summing inequalities (1.6) and (1.7), we can obtain

ℜ \{\frac{z [f^{'} (ɛ^{μ} z) + \bar{f^{'} (ɛ^{μ} \bar{z})}] + λ z^{2} [ɛ^{μ} f^{''} (ɛ^{μ} z) + ɛ^{- μ} \bar{f^{''} (ɛ^{μ} \bar{z})}]}{(1 - λ) f_{2 k} (z) + λ z f_{2 k}^{'} (z)}\} > 2 α . (1.8)

Let $μ = 0, 1, 2, \dots, k - 1$ in (1.8), respectively, and summing them we can get

ℜ \{\frac{z \frac{1}{2 k} \sum_{μ = 0}^{k - 1} [f^{'} (ɛ^{μ} z) + \bar{f^{'} (ɛ^{μ} \bar{z})}] + λ z^{2} \frac{1}{2 k} \sum_{μ = 0}^{k - 1} [ɛ^{μ} f^{''} (ɛ^{μ} z) + ɛ^{- μ} \bar{f^{''} (ɛ^{μ} \bar{z})}]}{(1 - λ) f_{2 k} (z) + λ z f_{2 k}^{'} (z)}\} > α,

or equivalently,

ℜ \{\frac{z f_{2 k}^{'} (z) + λ z^{2} f_{2 k}^{''} (z)}{(1 - λ) f_{2 k} (z) + λ z f_{2 k}^{'} (z)}\} = ℜ \{\frac{z F_{2 k}^{'} (z)}{F_{2 k} (z)}\} > α,

that is $F_{2 k} (z) \in S^{*} (α)$ , which is the class of starlike functions of order $α$ in $U$ . Note that $S^{*} (0) = S^{*}$ , this implies that $F (z) = (1 - λ) f (z) + λ z f^{'} (z) \in C$ . We now split it into two cases to prove.

Case 1. When $λ = 0$ . It is obvious that $f (z) = F (z) \in C$ .

Case 2. When $0 < λ \leq 1$ . From $F (z) = (1 - λ) f (z) + λ z f^{'} (z)$ and $0 < λ \leq 1$ , we have

f (z) = \frac{1}{λ} z^{1 - \frac{1}{λ}} \int_{0}^{z} F (t) t^{\frac{1}{λ} - 2} d t .

Since $γ = \frac{1}{λ} - 1 \geq 0$ , by Lemma 1, we obtain that $f (z) \in C$ . Hence $S_{s c}^{(k)} (λ, α) \subset C \subset S$ , and the proof is complete.

By means of Lemma 2, using the similar method as in Lemma 3, we may prove the following Lemma.

Lemma 4. Let $0 \leq λ \leq 1$ and $0 \leq α < 1$ , then we have $C_{s c}^{(k)} (λ, α) \subset C^{*} \subset C$ .

In the present paper, we shall provide the integral representations and convolution conditions for the classes $S_{s c}^{(k)} (λ, α)$ and $C_{s c}^{(k)} (λ, α)$ , we shall also provide some coefficient inequalities for functions belonging to these classes and their subclasses with negative coefficients.

2. Integral Representations

We ﬁrst give the integral representations of functions in the classes $S_{s c}^{(k)} (λ, α)$ and $C_{s c}^{(k)} (λ, α)$ .

Theorem 1. Let $f (z) \in S_{s c}^{(k)} (λ, α)$ with $0 < λ \leq 1$ , then we have

\begin{matrix} f_{2 k} (z) = \frac{1}{λ} z^{1 - \frac{1}{λ}} \int_{0}^{z} \\ exp \{\frac{1}{2 k} \sum_{μ = 0}^{k - 1} \int_{0}^{u} \frac{2 (1 - α)}{ζ} [\frac{ω (ɛ^{μ} ζ)}{1 - ω (ɛ^{μ} ζ)} + \frac{\bar{ω (ɛ^{μ} \bar{ζ})}}{1 - \bar{ω (ɛ^{μ} \bar{ζ})}}] d ζ\} \\ u^{\frac{1}{λ} - 1} d u, \end{matrix}

where $f_{2 k} (z)$ is deﬁned by equality (1.2), $ω (z)$ is analytic in $U$ and $ω (0) = 0$ , $|ω (z)| < 1$ .

Proof. Suppose that $f (z) \in S_{s c}^{(k)} (λ, α)$ , we know that the condition (1.3) can be written as

\frac{z f^{'} (z) + λ z^{2} f^{''} (z)}{(1 - λ) f_{2 k} (z) + λ z f_{2 k}^{'} (z)} ≺ \frac{1 + (1 - 2 α) z}{1 - z},

where “ $≺$ ” stands for the usual subordination, it follows that

\frac{z f^{'} (z) + λ z^{2} f^{''} (z)}{(1 - λ) f_{2 k} (z) + λ z f_{2 k}^{'} (z)} = \frac{1 + (1 - 2 α) ω (z)}{1 - ω (z)}, (2.2)

where $ω (z)$ is analytic in $U$ and $ω (0) = 0$ , $|ω (z)| < 1$ . By applying the similar method as in Lemma 3 to equality (2.2), we can obtain

\begin{matrix} \frac{(1 - λ) z f_{2 k}^{'} (z) + λ z {(z f_{2 k}^{'} (z))}^{'}}{(1 - λ) f_{2 k} (z) + λ z f_{2 k}^{'} (z)} \\ = \frac{1}{2 k} \sum_{μ = 0}^{k - 1} [\frac{1 + (1 - 2 α) ω (ɛ^{μ} z)}{1 - ω (ɛ^{μ} z)} + \frac{1 + (1 - 2 α) \bar{ω (ɛ^{μ} \bar{z})}}{1 - \bar{ω (ɛ^{μ} \bar{z})}}] . \end{matrix}

From equality (2.3), we get

\begin{matrix} \frac{(1 - λ) f_{2 k}^{'} (z) + λ {(z f_{2 k}^{'} (z))}^{'}}{(1 - λ) f_{2 k} (z) + λ z f_{2 k}^{'} (z)} - \frac{1}{z} \\ = \frac{1}{2 k} \sum_{μ = 0}^{k - 1} [\frac{2 (1 - α) ω (ɛ^{μ} z)}{z (1 - ω (ɛ^{μ} z))} + \frac{2 (1 - α) \bar{ω (ɛ^{μ} \bar{z})}}{z (1 - \bar{ω (ɛ^{μ} \bar{z})})}] . \end{matrix}

Integrating equality (2.4), we have

\begin{matrix} log \{\frac{(1 - λ) f_{2 k} (z) + λ z f_{2 k}^{'} (z)}{z}\} \\ = \frac{1}{2 k} \sum_{μ = 0}^{k - 1} \int_{0}^{z} [\frac{2 (1 - α) ω (ɛ^{μ} ζ)}{ζ (1 - ω (ɛ^{μ} ζ))} + \frac{2 (1 - α) \bar{ω (ɛ^{μ} \bar{ζ})}}{ζ (1 - \bar{ω (ɛ^{μ} \bar{ζ})})}] d ζ, \end{matrix}

that is,

\begin{matrix} (1 - λ) f_{2 k} (z) + λ z f_{2 k}^{'} (z) \\ = z \cdot exp \{\frac{1}{2 k} \sum_{μ = 0}^{k - 1} \int_{0}^{z} \frac{2 (1 - α)}{ζ} [\frac{ω (ɛ^{μ} ζ)}{1 - ω (ɛ^{μ} ζ)} + \frac{\bar{ω (ɛ^{μ} \bar{ζ})}}{1 - \bar{ω (ɛ^{μ} \bar{ζ})}}] d ζ\} . \end{matrix}

From equality (2.5), we can get equality (2.1) easily. Hence the proof is complete.

Theorem 2. Let $f (z) \in S_{s c}^{(k)} (λ, α)$ with $0 < λ \leq 1$ , then we have

\begin{matrix} f (z) = \frac{1}{λ} z^{1 - \frac{1}{λ}} \int_{0}^{z} \int_{0}^{u} \\ exp \{\frac{1}{2 k} \sum_{μ = 0}^{k - 1} \int_{0}^{t} \frac{2 (1 - α)}{ζ} [\frac{ω (ɛ^{μ} ζ)}{1 - ω (ɛ^{μ} ζ)} + \frac{\bar{ω (ɛ^{μ} \bar{ζ})}}{1 - \bar{ω (ɛ^{μ} \bar{ζ})}}] d ζ\} \\ \cdot \frac{1 + (1 - 2 α) ω (t)}{1 - ω (t)} d t u^{\frac{1}{λ} - 2} d u, \end{matrix}

where $ω (z)$ is analytic in $U$ and $ω (0) = 0$ , $|ω (z)| < 1$ .

Proof. Suppose that $f (z) \in S_{s c}^{(k)} (λ, α)$ , from equalities (2.2) and (2.5), we can get

\begin{matrix} (1 - λ) f^{'} (z) + λ {(z f^{'} (z))}^{'} \\ = \frac{(1 - λ) f_{k} (z) + λ z f_{k}^{'} (z)}{z} \cdot \frac{1 + (1 - 2 α) ω (z)}{1 - ω (z)} \\ = exp \{\frac{1}{2 k} \sum_{μ = 0}^{k - 1} \int_{0}^{z} \frac{2 (1 - α)}{ζ} [\frac{ω (ɛ^{μ} ζ)}{1 - ω (ɛ^{μ} ζ)} + \frac{\bar{ω (ɛ^{μ} \bar{ζ})}}{1 - \bar{ω (ɛ^{μ} \bar{ζ})}}] d ζ\} \\ \cdot \frac{1 + (1 - 2 α) ω (z)}{1 - ω (z)} . \end{matrix}

Integrating this equality, we can get equality (2.6) easily. Hence the proof is complete.

Similarly, for the class $C_{s c}^{(k)} (λ, α)$ , we have

Corollary 1. Let $f (z) \in C_{s c}^{(k)} (λ, α)$ with $0 < λ \leq 1$ , then we have

\begin{matrix} f_{2 k} (z) = \frac{1}{λ} z^{1 - \frac{1}{λ}} \int_{0}^{z} \int_{0}^{u} \\ exp \{\frac{1}{2 k} \sum_{μ = 0}^{k - 1} \int_{0}^{ξ} \frac{2 (1 - α)}{ζ} [\frac{ω (ɛ^{μ} ζ)}{1 - ω (ɛ^{μ} ζ)} + \frac{\bar{ω (ɛ^{μ} \bar{ζ})}}{1 - \bar{ω (ɛ^{μ} \bar{ζ})}}] d ζ\} \\ d ξ u^{\frac{1}{λ} - 2} d u, \end{matrix}

where $f_{2 k} (z)$ is deﬁned by equality (1.2), $ω (z)$ is analytic in $U$ and $ω (0) = 0$ , $|ω (z)| < 1$ .

Corollary 2. Let $f (z) \in C_{s c}^{(k)} (λ, α)$ with $0 < λ \leq 1$ , then we have

\begin{matrix} f (z) = \frac{1}{λ} z^{1 - \frac{1}{λ}} \int_{0}^{z} \int_{0}^{u} \\ \frac{1}{ξ} \int_{0}^{ξ} exp \{\frac{1}{2 k} \sum_{μ = 0}^{k - 1} \int_{0}^{t} \frac{2 (1 - α)}{ζ} [\frac{ω (ɛ^{μ} ζ)}{1 - ω (ɛ^{μ} ζ)} + \frac{\bar{ω (ɛ^{μ} \bar{ζ})}}{1 - \bar{ω (ɛ^{μ} \bar{ζ})}}] d ζ\} \\ \cdot \frac{1 + (1 - 2 α) ω (t)}{1 - ω (t)} d t d ξ u^{\frac{1}{λ} - 2} d u, \end{matrix}

where $ω (z)$ is analytic in $U$ and $ω (0) = 0$ , $|ω (z)| < 1$ .

3. Convolution Conditions

In this section, we give the convolution conditions for the classes $S_{s c}^{(k)} (λ, α)$ and $C_{s c}^{(k)} (λ, α)$ . Let $f, g \in A$ , where $f (z)$ is given by (1.1) and $g (z)$ is deﬁned by

g (z) = z + \sum_{n = 2}^{\infty} b_{n} z^{n},

then the Hadamard product (or convolution) $f * g$ is deﬁned (as usual) by

(f * g) (z) = z + \sum_{n = 2}^{\infty} a_{n} b_{n} z^{n} = (g * f) (z) .

Theorem 3. A function $f (z) \in S_{s c}^{(k)} (λ, α)$ if and only if

\begin{matrix} \frac{1}{z} \{f * \{(1 - λ) [\frac{z}{{(1 - z)}^{2}} (1 - e^{i θ}) - \frac{1 + (1 - 2 α) e^{i θ}}{2} h] \\ + λ z {[\frac{z}{{(1 - z)}^{2}} (1 - e^{i θ}) - \frac{1 + (1 - 2 α) e^{i θ}}{2} h]}^{'} \} (z) \\ - [1 + (1 - 2 α) e^{i θ}] \cdot \bar{f * (\frac{1 - λ}{2} h + \frac{λ}{2} z h^{'}) (\bar{z})} \} \neq 0 \end{matrix}

for all $z \in U$ and $0 \leq θ < 2 π$ , where $h (z)$ is given by (3.6).

Proof. Suppose that $f (z) \in S_{s c}^{(k)} (λ, α)$ , since (1.3) is equivalent to

\frac{z f^{'} (z) + λ z^{2} f^{''} (z)}{(1 - λ) f_{2 k} (z) + λ z f_{2 k}^{'} (z)} \neq \frac{1 + (1 - 2 α) e^{i θ}}{1 - e^{i θ}} (3.2)

for all $z \in U$ and $0 \leq θ < 2 π$ . And the condition (3.2) can be written as

\begin{matrix} \frac{1}{z} \{[(1 - λ) z f^{'} (z) + λ z {(z f^{'} (z))}^{'}] (1 - e^{i θ}) \\ - [(1 - λ) f_{2 k} (z) + λ z f_{2 k}^{'} (z)] [1 + (1 - 2 α) e^{i θ}]\} \neq 0 . \end{matrix}

On the other hand, it is well known that

z f^{'} (z) = f (z) * \frac{z}{{(1 - z)}^{2}} . (3.4)

And from the deﬁnition of $f_{2 k} (z)$ , we know

f_{2 k} (z) = \frac{1}{2} [(f * h) (z) + \bar{(f * h) (\bar{z})}], (3.5)

where

h (z) = \frac{1}{k} \sum_{υ = 0}^{k - 1} \frac{z}{1 - ɛ^{υ} z} . (3.6)

Substituting (3.4) and (3.5) into (3.3), we can get (3.1) easily. This completes the proof of Theorem 3.

Similarly, for the class $C_{s c}^{(k)} (λ, α)$ , we have

Corollary 3. A function $f (z) \in C_{s c}^{(k)} (λ, α)$ if and only if

\begin{matrix} \frac{1}{z} \{f * \{z \{(1 - λ) [\frac{z}{{(1 - z)}^{2}} (1 - e^{i θ}) - \frac{1 + (1 - 2 α) e^{i θ}}{2} h] \\ + λ z {[\frac{z}{{(1 - z)}^{2}} (1 - e^{i θ}) - \frac{1 + (1 - 2 α) e^{i θ}}{2} h]}^{'} {\}}^{'} \} (z) \\ - [1 + (1 - 2 α) e^{i θ}] \cdot \bar{f * [z {(\frac{1 - λ}{2} h + \frac{λ}{2} z h^{'})}^{'}] (\bar{z})} \} \neq 0 \end{matrix}

for all $z \in U$ and $0 \leq θ < 2 π$ , where $h (z)$ is given by (3.6).

4. Coefficient Inequalities

In this section, we ﬁrst provide the sufficient conditions for functions belonging to the classes $S_{s c}^{(k)} (λ, α)$ and $C_{s c}^{(k)} (λ, α)$ .

Theorem 4. Let $0 \leq λ \leq 1$ and $0 \leq α < 1$ . If

\begin{matrix} \sum_{n = 1}^{\infty} [(1 - λ) + λ (n k + 1)] [|(n k + 1) a_{n k + 1} - ℜ (a_{n k + 1})| + (1 - α) |ℜ (a_{n k + 1})|] \\ + \sum_{\overset{n = 2}{n \neq l k + 1}}^{\infty} n [(1 - λ) + λ n] |a_{n}| \leq 1 - α, \end{matrix}

then $f (z) \in S_{s c}^{(k)} (λ, α)$ .

Proof. It suffices to show that

|\frac{z f^{'} (z) + λ z^{2} f^{''} (z)}{(1 - λ) f_{2 k} (z) + λ z f_{2 k}^{'} (z)} - 1| < 1 - α .

Note that for $|z| = r < 1$ , we have

\begin{matrix} |\frac{z f^{'} (z) + λ z^{2} f^{''} (z)}{(1 - λ) f_{2 k} (z) + λ z f_{2 k}^{'} (z)} - 1| \\ = |\frac{\sum_{n = 2}^{\infty} [(1 - λ) + λ n] (n a_{n} - ℜ (a_{n}) c_{n}) z^{n - 1}}{1 + \sum_{n = 2}^{\infty} [(1 - λ) + λ n] ℜ (a_{n}) c_{n} z^{n - 1}}| \\ \leq \frac{\sum_{n = 2}^{\infty} [(1 - λ) + λ n] |n a_{n} - ℜ (a_{n}) c_{n}| {|z|}^{n - 1}}{1 - \sum_{n = 2}^{\infty} [(1 - λ) + λ n] c_{n} |ℜ (a_{n})| {|z|}^{n - 1}} \\ \leq \frac{\sum_{n = 2}^{\infty} [(1 - λ) + λ n] |n a_{n} - ℜ (a_{n}) c_{n}|}{1 - \sum_{n = 2}^{\infty} [(1 - λ) + λ n] c_{n} |ℜ (a_{n})|}, \end{matrix}

where

c_{n} = \frac{1}{k} \sum_{ν = 0}^{k - 1} ɛ^{(n - 1) ν} = \{\begin{matrix} 1, & n = l k + 1, \\ 0, & n \neq l k + 1 . \end{matrix} (4.2)

This last expression is bounded above by $1 - α$ if

\sum_{n = 2}^{\infty} [(1 - λ) + λ n] [|n a_{n} - ℜ (a_{n}) c_{n}| + c_{n} (1 - α) |ℜ (a_{n})|] \leq 1 - α . (4.3)

Since inequality (4.3) can be written as inequality (4.1), hence $f (z)$ satisﬁes the condition (1.3). This completes the proof of Theorem 4.

Similarly, for the class $C_{s c}^{(k)} (λ, α)$ , we have

Corollary 4. Let $0 \leq λ \leq 1$ and $0 \leq α < 1$ . If

\begin{matrix} \sum_{n = 1}^{\infty} (n k + 1) [(1 - λ) + λ (n k + 1)] \\ \cdot [|(n k + 1) a_{n k + 1} - ℜ (a_{n k + 1})| + (1 - α) |ℜ (a_{n k + 1})|] \\ + \sum_{\overset{n = 2}{n \neq l k + 1}}^{\infty} n^{2} [(1 - λ) + λ n] |a_{n}| \leq 1 - α, \end{matrix}

then $f (z) \in C_{s c}^{(k)} (λ, α)$ .

Let $T$ be the subclass of $A$ consisting of all functions which are of the form

f (z) = z - \sum_{n = 2}^{\infty} a_{n} z^{n} (a_{n} \geq 0) .

For convenience, we write $S_{s c}^{(k)} (λ, α) \cap T$ as ${T S}_{s c}^{(k)} (λ, α)$ and $C_{s c}^{(k)} (λ, α) \cap T$ simple as ${T C}_{s c}^{(k)} (λ, α)$ . We now provide the necessary and sufficient coefficient conditions for functions belonging to the classes ${T S}_{s c}^{(k)} (λ, α)$ and ${T C}_{s c}^{(k)} (λ, α)$ .

Theorem 5. Let $0 \leq λ \leq 1, 0 \leq α < 1$ and $f (z) \in T$ , then $f (z) \in {T S}_{s c}^{(k)} (λ, α)$ if and only if

\begin{matrix} \sum_{n = 1}^{\infty} [(1 - λ) + λ (n k + 1)] [(n k + 1) - α] a_{n k + 1} + \sum_{\overset{n = 2}{n \neq l k + 1}}^{\infty} n [(1 - λ) + λ n] a_{n} \\ \leq 1 - α . \end{matrix}

Proof. In view of Theorem 4, we need only to prove the necessity. Suppose that $f (z) \in {T S}_{s c}^{(k)} (λ, α)$ , then from (1.3), we can get

ℜ \{\frac{1 - \sum_{n = 2}^{\infty} n [(1 - λ) + λ n] a_{n} z^{n - 1}}{1 - \sum_{n = 2}^{\infty} [(1 - λ) + λ n] c_{n} a_{n} z^{n - 1}}\} > α, (4.5)

where $c_{n}$ is given by (4.2). By letting $z \to 1^{-}$ through real values in (4.5), we can get

\frac{1 - \sum_{n = 2}^{\infty} n [(1 - λ) + λ n] a_{n}}{1 - \sum_{n = 2}^{\infty} [(1 - λ) + λ n] c_{n} a_{n}} \geq α,

or equivalently,

\sum_{n = 2}^{\infty} [(1 - λ) + λ n] (n - α c_{n}) a_{n} \leq 1 - α . (4.6)

Substituting (4.2) into inequality (4.6), we can get inequality (4.4) easily. This completes the proof of Theorem 5.

Similarly, for the class ${T C}_{s c}^{(k)} (λ, α)$ , we have

Corollary 5. Let $0 \leq λ \leq 1, 0 \leq α < 1$ and $f (z) \in T$ , then $f (z) \in {T C}_{s c}^{(k)} (λ, α)$ if and only if

\begin{matrix} \sum_{n = 1}^{\infty} (n k + 1) [(1 - λ) + λ (n k + 1)] [(n k + 1) - α] a_{n k + 1} + \sum_{\overset{n = 2}{n \neq l k + 1}}^{\infty} n^{2} [(1 - λ) + λ n] a_{n} \\ \leq 1 - α . \end{matrix}

Acknowledgements

This work was supported by the Scientiﬁc Research Fund of Hunan Provincial Education Department and the Hunan Provincial Natural Science Foundation (No. 05JJ30013) of People’s Republic of China.

References

[1] H. Al-Amiri, D. Coman and P.T. Mocanu, Some properties of starlike functions with respect to symmetric conjugate points, Internat. J. Math. Math. Sci. 18 (1995), 469-474.

[2] P. L. Duren, Univalent Functions, Springer-Verlag, New York, 1983.

[3] K.I. Noor, On quasi-convex functions and related topics, Internat. J. Math. Math. Sci. 10 (1987), 241-258.

[4] S. Owa, M.Nunokawa, H. Saitoh and H.M. Srivastava, Close-to-convexity, starlikeness, and convexity of certain analytic functions, Appl. Math. Lett. 15 (2002), 63-69.

[5] H.M. Srivastava and S. Owa (Eds.), Current Topics in Analytic Function Theory, World Scientiﬁc, Singapore, 1992.

[6] Z.-R. Wu, The integral operator of starlikeness and the family of Bazilevič functions, Acta Math. Sinica 27 (1984), 394-409.

Changsha University of Science and Technology, Changsha, 410076 Hunan, People’s Republic of China

Department of Mathematics, Shaoyang University, Shaoyang, 422000 Hunan, People’s Republic of China

E-mail address: zhigwang@163.com

Received May 12, 2007