Lobachevskii Journal of Mathematics

http://ljm.ksu.ru
ISSN 1818-9962

Vol. 26, 2007, 125–135 \begin{matrix}c \\ ◯ \end{matrix} Zhi-Gang Wang and Da-Zhao Chen

Zhi-Gang Wang and Da-Zhao Chen

ON SUBCLASSES OF CLOSE-TO-CONVEX AND QUASI-CONVEX FUNCTIONS WITH RESPECT TO 2K-SYMMETRIC CONJUGATE POINTS

(submitted by F. G. Avkhadiev)

Abstract. In the present paper, the authors introduce two new subclasses S_{sc}^{(k)}(λ, α) of close-to-convex functions and C_{sc}^{(k)}(λ, α) of quasiconvex functions with respect to 2k-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.

1. Introduction

Let A denote the class of functions of the form

f(z) = z + ∑_{n=2}^{∞} a_{n}z^{n},  (1.1)

which are analytic in the open unit disk U = {z ∈ 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 2k-symmetric conjugate points, which satisfy the following inequality

2000 Mathematical Subject Classification. 30C45.

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

ℜ \{\frac{zf′(z)}{f_{2k}(z)}\} > 0 (z ∈ U),

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

f_{2k}(z) = \frac{1}{2k} ∑_{ν=0}^{k−1} [ε^{−ν}f(ε^{ν}z) + ε^{ν}\overline{f(ε^{ν}\overline{z})}] (ε = exp(2πi/k); z ∈ U).

(1.2)

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

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

ℜ \{\frac{zf′(z) + λz^{2}f′′(z)}{(1 − λ)f_{2k}(z) + λzf′_{2k}(z)}\} > α (z ∈ U),  (1.3)

where 0 ≤ λ ≤ 1, 0 ≤ α < 1 and f_{2k}(z) is defined by equality (1.2). And a function f(z) ∈ A is in the class C_{sc}^{(k)}(λ, α) if and only if zf′(z) ∈ S_{sc}^{(k)}(λ, α).

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

Lemma 1. Let γ ≥ 0 and f ∈ C, then

F (z) = \frac{1 + γ}{z^{γ}} ∫_{0}^{z} f(t)t^{γ−1}dt ∈ C.

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

Lemma 2 [3]. Let 0 < λ ≤ 1 and f ∈ C^{∗}, then

F (z) = \frac{1}{λ}z^{1− \frac{1}{λ}} ∫_{0}^{z} f(t)t^{\frac{1}{λ} −2}dt ∈ C^{∗} ⊂ C.

Lemma 3. Let 0 ≤ λ ≤ 1 and 0 ≤ α < 1, then we have S_{sc}^{(k)}(λ, α) ⊂ C ⊂ S.

Proof. Let F (z) = (1 − λ)f(z) + λzf′(z), F_{2k}(z) = (1 − λ)f_{2k}(z) + λzf′_{2k}(z) with f(z) ∈ S_{sc}^{(k)}(λ, α), substituting z by ε^{μ}z (μ = 0, 1, 2, …, k− 1) in (1.1), we get

ℜ \{\frac{ε^{μ}zf′(ε^{μ}z) + λ(ε^{μ}z)^{2}f′′(ε^{μ}z)}{(1 − λ)f_{2k}(ε^{μ}z) + λε^{μ}zf′_{2k}(ε^{μ}z)}\} > α.  (1.4)

CLOSE-TO-CONVEX AND QUASI-CONVEX FUNCTIONS 127

From inequality (1.4), we have

ℜ \{\frac{\overline{ε^{μ}\overline{z}} \overline{f′(ε^{μ}\overline{z})} + λ\overline{(ε^{μ}\overline{z})^{2}} \overline{f′′(ε^{μ}\overline{z})}}{(1 − λ)\overline{f_{2k}(ε^{μ}\overline{z})} + λ\overline{ε^{μ}\overline{z}} \overline{f′_{2k}(ε^{μ}\overline{z})}}\} > α.  (1.5)

Note that f_{2k}(ε^{μ}z) = ε^{μ}f_{2k}(z), f′_{2k}(ε^{μ}z) = f′_{2k}(z), \overline{f_{2k}(ε^{μ}\overline{z})} = ε^{−μ}f_{2k}(z) and \overline{f′_{2k}(ε^{μ}\overline{z})} = f′_{2k}(z), thus, inequalities (1.4) and (1.5) can be written as

ℜ \{\frac{zf′(ε^{μ}z) + λz^{2}ε^{μ}f′′(ε^{μ}z)}{(1 − λ)f_{2k}(z) + λzf′_{2k}(z)}\} > α,  (1.6)

and

ℜ \{\frac{z\overline{f′(ε^{μ}\overline{z})} + λz^{2}ε^{−μ}\overline{f′′(ε^{μ}\overline{z})}}{(1 − λ)f_{2k}(z) + λzf′_{2k}(z)}\} > α.  (1.7)

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

ℜ \{\frac{z [f′(ε^{μ}z) + \overline{f′(ε^{μ}\overline{z})}] + λz^{2} [ε^{μ}f′′(ε^{μ}z) + ε^{−μ}\overline{f′′(ε^{μ}\overline{z})}]}{(1 − λ)f_{2k}(z) + λzf′_{2k}(z)}\} > 2α.

(1.8)

Let μ = 0, 1, 2, …, k − 1 in (1.8), respectively, and summing them we can get

ℜ \{\frac{z \frac{1}{2k} ∑_{μ=0}^{k−1} [f′(ε^{μ}z) + \overline{f′(ε^{μ}\overline{z})}] + λz^{2} \frac{1}{2k} ∑_{μ=0}^{k−1} [ε^{μ}f′′(ε^{μ}z) + ε^{−μ}\overline{f′′(ε^{μ}\overline{z})}]}{(1 − λ)f_{2k}(z) + λzf′_{2k}(z)}\} > α,

or equivalently,

ℜ \{\frac{zf′_{2k}(z) + λz^{2}f′′_{2k}(z)}{(1 − λ)f_{2k}(z) + λzf′_{2k}(z)}\} = ℜ \{\frac{zF′_{2k}(z)}{F_{2k}(z)}\} > α,

that is F_{2k}(z) ∈ 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)+λzf′(z) ∈ C. We now split it into two cases to prove.

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

Case 2. When 0 < λ ≤ 1. From F (z) = (1 − λ)f(z) + λzf′(z) and 0 < λ ≤ 1, we have

f(z) = \frac{1}{λ}z^{1− \frac{1}{λ}} ∫_{0}^{z} F (t)t^{\frac{1}{λ} −2}dt.

Since γ = \frac{1}{λ} − 1 ≥ 0, by Lemma 1, we obtain that f(z) ∈ C. Hence S_{sc}^{(k)}(λ, α) ⊂ C ⊂ 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 ≤ λ ≤ 1 and 0 ≤ α < 1, then we have C_{sc}^{(k)}(λ, α) ⊂ C^{∗} ⊂ C.

In the present paper, we shall provide the integral representations and convolution conditions for the classes S_{sc}^{(k)}(λ, α) and C_{sc}^{(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_{sc}^{(k)}(λ, α) and C_{sc}^{(k)}(λ, α).

Theorem 1. Let f(z) ∈ S_{sc}^{(k)}(λ, α) with 0 < λ ≤ 1, then we have

f_{2k}(z) = \frac{1}{λ}z^{1− \frac{1}{λ}} ∫_{0}^{z}

exp \{\frac{1}{2k} ∑_{μ=0}^{k−1} ∫_{0}^{u} \frac{2(1 − α)}{ζ} [\frac{ω(ε^{μ}ζ)}{1 − ω(ε^{μ}ζ)} + \frac{\overline{ω(ε^{μ}\overline{ζ})}}{1 − \overline{ω(ε^{μ}\overline{ζ})}}] dζ\}
u^{\frac{1}{λ} −1}du, (2.1)

where f_{2k}(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) ∈ S_{sc}^{(k)}(λ, α), we know that the condition (1.3) can be written as

\frac{zf′(z) + λz^{2}f′′(z)}{(1 − λ)f_{2k}(z) + λzf′_{2k}(z)} ≺ \frac{1 + (1 − 2α)z}{1 − z} ,

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

\frac{zf′(z) + λz^{2}f′′(z)}{(1 − λ)f_{2k}(z) + λzf′_{2k}(z)} = \frac{1 + (1 − 2α)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

\frac{(1 − λ)zf′_{2k}(z) + λz(zf′_{2k}(z))′}{(1 − λ)f_{2k}(z) + λzf′_{2k}(z)}

= \frac{1}{2k} ∑_{μ=0}^{k−1} [\frac{1 + (1 − 2α)ω(ε^{μ}z)}{1 − ω(ε^{μ}z)} + \frac{1 + (1 − 2α)\overline{ω(ε^{μ}\overline{z})}}{1 − \overline{ω(ε^{μ}\overline{z})}}] . (2.3)

CLOSE-TO-CONVEX AND QUASI-CONVEX FUNCTIONS 129 From equality (2.3), we get

\frac{(1 − λ)f′_{2k}(z) + λ(zf′_{2k}(z))′}{(1 − λ)f_{2k}(z) + λzf′_{2k}(z)} − \frac{1}{z}

= \frac{1}{2k} ∑_{μ=0}^{k−1} [\frac{2(1 − α)ω(ε^{μ}z)}{z(1 − ω(ε^{μ}z))} + \frac{2(1 − α)\overline{ω(ε^{μ}\overline{z})}}{z (1 − \overline{ω(ε^{μ}\overline{z})})}] . (2.4)

Integrating equality (2.4), we have

log \{\frac{(1 − λ)f_{2k}(z) + λzf′_{2k}(z)}{z}\}

= \frac{1}{2k} ∑_{μ=0}^{k−1} ∫_{0}^{z} [\frac{2(1 − α)ω(ε^{μ}ζ)}{ζ(1 − ω(ε^{μ}ζ))} + \frac{2(1 − α)\overline{ω(ε^{μ}\overline{ζ})}}{ζ (1 − \overline{ω(ε^{μ}\overline{ζ})})}] dζ,

that is,

(1 − λ)f_{2k}(z) + λzf′_{2k}(z)

= z · exp \{\frac{1}{2k} ∑_{μ=0}^{k−1} ∫_{0}^{z} \frac{2(1 − α)}{ζ} [\frac{ω(ε^{μ}ζ)}{1 − ω(ε^{μ}ζ)} + \frac{\overline{ω(ε^{μ}\overline{ζ})}}{1 − \overline{ω(ε^{μ}\overline{ζ})}}] dζ\} . (2.5)

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

Theorem 2. Let f(z) ∈ S_{sc}^{(k)}(λ, α) with 0 < λ ≤ 1, then we have

f(z) = \frac{1}{λ}z^{1− \frac{1}{λ}} ∫_{0}^{z} ∫_{0}^{u}

exp \{\frac{1}{2k} ∑_{μ=0}^{k−1} ∫_{0}^{t} \frac{2(1 − α)}{ζ} [\frac{ω(ε^{μ}ζ)}{1 − ω(ε^{μ}ζ)} + \frac{\overline{ω(ε^{μ}\overline{ζ})}}{1 − \overline{ω(ε^{μ}\overline{ζ})}}] dζ\}

· \frac{1 + (1 − 2α)ω(t)}{1 − ω(t)}dtu^{\frac{1}{λ} −2}du, (2.6)

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

Proof. Suppose that f(z) ∈ S_{sc}^{(k)}(λ, α), from equalities (2.2) and (2.5), we can get

(1 − λ)f′(z) + λ(zf′(z))′

= \frac{(1 − λ)f_{k}(z) + λzf′_{k}(z)}{z} · \frac{1 + (1 − 2α)ω(z)}{1 − ω(z)}

= exp \{\frac{1}{2k} ∑_{μ=0}^{k−1} ∫_{0}^{z} \frac{2(1 − α)}{ζ} [\frac{ω(ε^{μ}ζ)}{1 − ω(ε^{μ}ζ)} + \frac{\overline{ω(ε^{μ}\overline{ζ})}}{1 − \overline{ω(ε^{μ}\overline{ζ})}}] dζ\}

· \frac{1 + (1 − 2α)ω(z)}{1 − ω(z)}.

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

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

Corollary 1. Let f(z) ∈ C_{sc}^{(k)}(λ, α) with 0 < λ ≤ 1, then we have

f_{2k}(z) = \frac{1}{λ}z^{1− \frac{1}{λ}} ∫_{0}^{z} ∫_{0}^{u}

exp \{\frac{1}{2k} ∑_{μ=0}^{k−1} ∫_{0}^{ξ} \frac{2(1 − α)}{ζ} [\frac{ω(ε^{μ}ζ)}{1 − ω(ε^{μ}ζ)} + \frac{\overline{ω(ε^{μ}\overline{ζ})}}{1 − \overline{ω(ε^{μ}\overline{ζ})}}] dζ\}
dξu^{\frac{1}{λ} −2}du,

where f_{2k}(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) ∈ C_{sc}^{(k)}(λ, α) with 0 < λ ≤ 1, then we have

f(z) = \frac{1}{λ}z^{1− \frac{1}{λ}} ∫_{0}^{z} ∫_{0}^{u}

\frac{1}{ξ} ∫_{0}^{ξ} exp \{\frac{1}{2k} ∑_{μ=0}^{k−1} ∫_{0}^{t} \frac{2(1 − α)}{ζ} [\frac{ω(ε^{μ}ζ)}{1 − ω(ε^{μ}ζ)} + \frac{\overline{ω(ε^{μ}\overline{ζ})}}{1 − \overline{ω(ε^{μ}\overline{ζ})}}] dζ\}

· \frac{1 + (1 − 2α)ω(t)}{1 − ω(t)}dtdξu^{\frac{1}{λ} −2}du,

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_{sc}^{(k)}(λ, α) and C_{sc}^{(k)}(λ, α). Let f, g ∈ A, where f(z) is given by (1.1) and g(z) is defined by

g(z) = z + ∑_{n=2}^{∞} b_{n}z^{n},

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

(f ∗ g)(z) = z + ∑_{n=2}^{∞} a_{n}b_{n}z^{n} = (g ∗ f)(z).

Theorem 3. A function f(z) ∈ S_{sc}^{(k)}(λ, α) if and only if

\frac{1}{z}\{f ∗ \{(1 − λ) [\frac{z}{(1 − z)^{2}} (1 − e^{iθ}) − \frac{1 + (1 − 2α)e^{iθ}}{2} h]

+ λz [\frac{z}{(1 − z)^{2}} (1 − e^{iθ}) − \frac{1 + (1 − 2α)e^{iθ}}{2} h]′ \}(z)

− [1 + (1 − 2α)e^{iθ}] · \overline{f ∗ (\frac{1 − λ}{2}h + \frac{λ}{2}zh′) (\overline{z})} \} ≠ 0 (3.1)

for all z ∈ U and 0 ≤ θ < 2π, where h(z) is given by (3.6).

Proof. Suppose that f(z) ∈ S_{sc}^{(k)}(λ, α), since (1.3) is equivalent to

\frac{zf′(z) + λz^{2}f′′(z)}{(1 − λ)f_{2k}(z) + λzf′_{2k}(z)} ≠ \frac{1 + (1 − 2α)e^{iθ}}{1 − e^{iθ}}  (3.2)

for all z ∈ U and 0 ≤ θ < 2π. And the condition (3.2) can be written as

\frac{1}{z} \{[(1 − λ)zf′(z) + λz(zf′(z))′ ](1 − e^{iθ})

−[(1 − λ)f_{2k}(z) + λzf′_{2k}(z)][1 + (1 − 2α)e^{iθ}]} ≠ 0. (3.3)

On the other hand, it is well known that

zf′(z) = f(z) ∗ \frac{z}{(1 − z)^{2}} .  (3.4)

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

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

where

h(z) = \frac{1}{k} ∑_{υ=0}^{k−1} \frac{z}{1 − ε^{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_{sc}^{(k)}(λ, α), we have

Corollary 3. A function f(z) ∈ C_{sc}^{(k)}(λ, α) if and only if

\frac{1}{z}\{f ∗ \{z\{(1 − λ) [\frac{z}{(1 − z)^{2}} (1 − e^{iθ}) − \frac{1 + (1 − 2α)e^{iθ}}{2} h]

+ λz [\frac{z}{(1 − z)^{2}} (1 − e^{iθ}) − \frac{1 + (1 − 2α)e^{iθ}}{2} h]′ \}′\}(z)

− [1 + (1 − 2α)e^{iθ}] · \overline{f ∗ [z (\frac{1 − λ}{2}h + \frac{λ}{2}zh′)′ ] (\overline{z})} \} ≠ 0

for all z ∈ U and 0 ≤ θ < 2π, 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_{sc}^{(k)}(λ, α) and C_{sc}^{(k)}(λ, α).

Theorem 4. Let 0 ≤ λ ≤ 1 and 0 ≤ α < 1. If

∑_{n=1}^{∞}[(1−λ)+λ(nk+1)][|(nk + 1)a_{nk+1} − ℜ(a_{nk+1})|+(1−α) |ℜ(a_{nk+1})|]

+ ∑_{n≠lk+1^{n=2}}^{∞} n[(1 − λ) + λn] |a_{n}| ≤ 1 − α, (4.1)

then f(z) ∈ S_{sc}^{(k)}(λ, α).

Proof. It suffices to show that

|\frac{zf′(z) + λz^{2}f′′(z)}{(1 − λ)f_{2k}(z) + λzf′_{2k}(z)} − 1| < 1 − α.

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

|\frac{zf′(z) + λz^{2}f′′(z)}{(1 − λ)f_{2k}(z) + λzf′_{2k}(z)} − 1|

= |\frac{∑_{n=2}^{∞}[(1 − λ) + λn](na_{n} − ℜ(a_{n})c_{n})z^{n−1}}{1 + ∑_{n=2}^{∞}[(1 − λ) + λn]ℜ(a_{n})c_{n}z^{n−1}} |

≤ \frac{∑_{n=2}^{∞}[(1 − λ) + λn] |na_{n} − ℜ(a_{n})c_{n}| |z|^{n−1}}{1 − ∑_{n=2}^{∞}[(1 − λ) + λn]c_{n} |ℜ(a_{n})| |z|^{n−1}}

≤ \frac{∑_{n=2}^{∞}[(1 − λ) + λn] |na_{n} − ℜ(a_{n})c_{n}|}{1 − ∑_{n=2}^{∞}[(1 − λ) + λn]c_{n} |ℜ(a_{n})|},

where

c_{n} = \frac{1}{k} ∑_{ν=0}^{k−1} ε^{(n−1)ν} = \{ \begin{matrix}1, & n = lk + 1, \\ 0, & n ≠ lk + 1. \end{matrix}  (4.2)

This last expression is bounded above by 1 − α if

∑_{n=2}^{∞}[(1 − λ) + λn][|na_{n} − ℜ(a_{n})c_{n}| + c_{n}(1 − α) |ℜ(a_{n})|] ≤ 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_{sc}^{(k)}(λ, α), we have

Corollary 4. Let 0 ≤ λ ≤ 1 and 0 ≤ α < 1. If

∑_{n=1}^{∞}(nk + 1)[(1 − λ) + λ(nk + 1)]

· [|(nk + 1)a_{nk+1} − ℜ(a_{nk+1})| + (1 − α) |ℜ(a_{nk+1})|]

+ ∑_{n≠lk+1^{n=2}}^{∞} n^{2}[(1 − λ) + λn] |a_{n}| ≤ 1 − α,

then f(z) ∈ C_{sc}^{(k)}(λ, α).

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

f(z) = z − ∑_{n=2}^{∞} a_{n}z^{n} (a_{n} ≥ 0).

For convenience, we write S_{sc}^{(k)}(λ, α) ∩ T as T S_{sc}^{(k)}(λ, α) and C_{sc}^{(k)}(λ, α) ∩ T simple as T C_{sc}^{(k)}(λ, α). We now provide the necessary and sufficient coefficient conditions for functions belonging to the classes T S_{sc}^{(k)}(λ, α) and T C_{sc}^{(k)}(λ, α).

Theorem 5. Let 0 ≤ λ ≤ 1, 0 ≤ α < 1 and f(z) ∈ T , then f(z) ∈ T S_{sc}^{(k)}(λ, α) if and only if

∑_{n=1}^{∞}[(1 − λ) + λ(nk + 1)][(nk + 1) − α]a_{nk+1} + ∑_{n≠lk+1^{n=2}}^{∞} n[(1 − λ) + λn]a_{n}

≤ 1 − α. (4.4)

Proof. In view of Theorem 4, we need only to prove the necessity. Suppose that f(z) ∈ T S_{sc}^{(k)}(λ, α), then from (1.3), we can get

ℜ \{\frac{1 − ∑_{n=2}^{∞} n[(1 − λ) + λn]a_{n}z^{n−1}}{1 − ∑_{n=2}^{∞}[(1 − λ) + λn]c_{n}a_{n}z^{n−1}} \} > α,  (4.5)

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

\frac{1 − ∑_{n=2}^{∞} n[(1 − λ) + λn]a_{n}}{1 − ∑_{n=2}^{∞}[(1 − λ) + λn]c_{n}a_{n}} ≥ α,

or equivalently,

∑_{n=2}^{∞}[(1 − λ) + λn](n − αc_{n})a_{n} ≤ 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_{sc}^{(k)}(λ, α), we have

Corollary 5. Let 0 ≤ λ ≤ 1, 0 ≤ α < 1 and f(z) ∈ T , then f(z) ∈ T C_{sc}^{(k)}(λ, α) if and only if

∑_{n=1}^{∞}(nk+1)[(1−λ)+λ(nk+1)][(nk+1)−α]a_{nk+1}+ ∑_{n≠lk+1^{n=2}}^{∞} n^{2}[(1−λ)+λn]a_{n}

≤ 1 − α.

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