MATEMATIQKI VESNIK

59 (2007), 65–73

UDK 517.524

originalni nauqni rad research paper

SOME SUBCLASSES OF CLOSE-TO-CONVEX AND QUASI-CONVEX FUNCTIONS

Zhi-Gang Wang

Abstract. In the present paper, the author introduce two new subclasses S_{sc}^{(k)}(α, β, γ) of close-to-convex functions and C_{sc}^{(k)}(α, β, γ) of quasi-convex functions with respect to 2k-symmetric conjugate points. The coefficient inequalities and integral representations for functions belonging to these classes are provided, the inclusion relationships and convolution conditions for these classes 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, [2,4,6,7,8].

Al-Amiri, Coman and Mocanu [1] once introduced and investigated a class S_{sc}^{(k)} of functions starlike with respect to 2k-symmetric conjugate points, which satisfy the following inequality

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 and investigate the following two more generalized subclasses S_{sc}^{(k)}(α, β, γ) and C_{sc}^{(k)}(α, β, γ) of A with respect to 2k-symmetric conjugate points, and obtain some interesting results.

AMS Subject Classification: 30C45.

Keywords and phrases: Close-to-convex functions, quasi-convex functions, differential subordination, Hadamard product, 2k-symmetric conjugate points..

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

|\frac{\frac{zf′(z)}{f_{2k}(z)} − 1}{β \frac{zf′(z)}{f_{2k}(z)} + (1 − γ)}| < 1 − α,  (1.3)

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

Note that S_{sc}^{(k)}(0, 1, 0) = S_{sc}^{(k)}, so the class S_{sc}^{(k)}(α, β, γ) is a generalization of the class S_{sc}^{(k)}.

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

Lemma 1. [3] Let H(z) = 1 + ∑_{n=1}^{∞} h_{n}z^{n} be analytic in U, 0 ≤ α < 1, 0 ≤ β ≤ 1 and 0 ≤ γ < 1, then the inequality

|\frac{H(z) − 1}{βH(z) + (1 − γ)}| < 1 − α (z ∈ U)

can be written as

H(z) ≺ \frac{1 + (1 − α)(1 − γ)z}{1 − (1 − α)βz} (z ∈ U),

where “≺” stands for the usual subordination.

Lemma 2. Let 0 ≤ α < 1, 0 ≤ β ≤ 1 and 0 ≤ γ < 1, then we have

S_{sc}^{(k)}(α, β, γ) ⊂ C ⊂ S.

Proof. Suppose that f(z) ∈ S_{sc}^{(k)}(α, β, γ), by Lemma 1, we know that the condition (1.3) can be written as

\frac{zf′(z)}{f_{2k}(z)} ≺ \frac{1 + (1 − α)(1 − γ)z}{1 − (1 − α)βz} (z ∈ U).  (1.4)

Thus we have

ℜ \{\frac{zf′(z)}{f_{2k}(z)}\} > 0 (z ∈ U)  (1.5)

since

ℜ \{\frac{1 + (1 − α)(1 − γ)z}{1 − (1 − α)βz} \} > 0 (z ∈ U).

Now it suffices to show that f_{2k}(z) ∈ S^{∗} ⊂ S. Substituting z by ε^{μ}z (μ = 0, 1, 2, …, k − 1) in (1.5), then (1.5) is also true, that is,

ℜ \{\frac{ε^{μ}zf′(ε^{μ}z)}{f_{2k}(ε^{μ}z)}\} > 0 (z ∈ U).  (1.6)

From inequality (1.6), we have

ℜ \{\frac{\overline{ε^{μ}\overline{z}} \overline{f′(ε^{μ}\overline{z})}}{\overline{f_{2k}(ε^{μ}\overline{z})}}\} > 0 (z ∈ U).  (1.7)

Note that f_{2k}(ε^{μ}z) = ε^{μ}f_{2k}(z) and \overline{f_{2k}(ε^{μ}\overline{z})} = ε^{−μ}f_{2k}(z), then inequalities (1.6) and (1.7) can be written as

ℜ \{\frac{zf′(ε^{μ}z)}{f_{2k}(z)}\} > 0 (z ∈ U),  (1.8)

and

ℜ \{\frac{z\overline{f′(ε^{μ}\overline{z})}}{f_{2k}(z)}\} > 0 (z ∈ U).  (1.9)

Summing inequalities (1.8) and (1.9), we can get

ℜ \{\frac{z (f′(ε^{μ}z) + \overline{f′(ε^{μ}\overline{z})})}{f_{2k}(z)}\} > 0 (z ∈ U).  (1.10)

Letting μ = 0, 1, 2, …, k − 1 in (1.10), 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})})]}{f_{2k}(z)}\} > 0 (z ∈ U),

or equivalently,

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

that is f_{2k}(z) ∈ S^{∗} ⊂ S. This means that S_{sc}^{(k)}(α, β, γ) ⊂ C ⊂ S, hence the proof of Lemma 2 is complete.

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

Lemma 3. Let 0 ≤ α < 1, 0 ≤ β ≤ 1 and 0 ≤ γ < 1, then we have

C_{sc}^{(k)}(α, β, γ) ⊂ C^{∗} ⊂ C.

Lemma 4. [5] Let −1 ≤ B_{2} ≤ B_{1} < A_{1} ≤ A_{2} ≤ 1, then we have

\frac{1 + A_{1}z}{1 + B_{1}z} ≺ \frac{1 + A_{2}z}{1 + B_{2}z} .

In the present paper, we shall provide the coefficient inequalities and integral representations for functions belonging to the classes S_{sc}^{(k)}(α, β, γ) and C_{sc}^{(k)}(α, β, γ), we shall also provide the inclusion relationships and convolution conditions for these classes.

68 Zhi-Gang Wang

2. Inclusion relationships We first give some inclusion relationships for the classes S_{sc}^{(k)}(α, β, γ) and

C_{sc}^{(k)}(α, β, γ).

Theorem 1. Let 0 ≤ β_{2} ≤ β_{1} ≤ 1, 0 ≤ α_{1} ≤ α_{2} < 1 and 0 ≤ γ_{1} ≤ γ_{2} < 1, then we have

S_{sc}^{(k)}(α_{2}, β_{2}, γ_{2}) ⊂ S_{sc}^{(k)}(α_{1}, β_{1}, γ_{1}).

Proof. Suppose that f(z) ∈ S_{sc}^{(k)}(α_{2}, β_{2}, γ_{2}), by (1.4), we have

\frac{zf′(z)}{f_{2k}(z)} ≺ \frac{1 + (1 − α_{2})(1 − γ_{2})z}{1 − (1 − α_{2})β_{2}z} .

Since 0 ≤ α_{1} ≤ α_{2} < 1, 0 ≤ β_{2} ≤ β_{1} ≤ 1 and 0 ≤ γ_{1} ≤ γ_{2} < 1, then we have

−1 ≤ −(1 − α_{1})β_{1} ≤ −(1 − α_{2})β_{2} < (1 − α_{2})(1 − γ_{2}) ≤ (1 − α_{1})(1 − γ_{1}) ≤ 1. Thus, by Lemma 4, we have

\frac{zf′(z)}{f_{2k}(z)} ≺ \frac{1 + (1 − α_{2})(1 − γ_{2})z}{1 − (1 − α_{2})β_{2}z} ≺ \frac{1 + (1 − α_{1})(1 − γ_{1})z}{1 − (1 − α_{1})β_{1}z} ,

that is f(z) ∈ S_{sc}^{(k)}(α_{1}, β_{1}, γ_{1}). This means that S_{sc}^{(k)}(α_{2}, β_{2}, γ_{2}) ⊂ S_{sc}^{(k)}(α_{1}, β_{1}, γ_{1}). Similarly, for the class C_{sc}^{(k)}(α, β, γ), we have Corollary 1. Let 0 ≤ β_{2} ≤ β_{1} ≤ 1, 0 ≤ α_{1} ≤ α_{2} < 1 and 0 ≤ γ_{1} ≤ γ_{2} < 1,

then we have C_{sc}^{(k)}(α_{2}, β_{2}, γ_{2}) ⊂ C_{sc}^{(k)}(α_{1}, β_{1}, γ_{1}).

3. Coefficient inequalities In this section, we give some coefficient inequalities for functions belonging to

the classes S_{sc}^{(k)}(α, β, γ) and C_{sc}^{(k)}(α, β, γ).

Theorem 2. Let f(z) = z + ∑_{n=2}^{∞} a_{n}z^{n} be analytic in U, if for 0 ≤ α < 1, 0 ≤ β ≤ 1 and 0 ≤ γ < 1, we have

∑_{n=2}^{∞} n[1+(1−α)β] |a_{n}|+ ∑_{l=1}^{∞}[(1−α)(1−γ)+1] |ℜ(a_{lk+1})| ≤ (1−α)(1+β −γ), (3.1)

then f(z) ∈ S_{sc}^{(k)}(α, β, γ).

Proof. Suppose that f(z) = z + ∑_{n=2}^{∞} a_{n}z^{n}, and f_{2k}(z) is defined by equality (1.2). We now let M be denoted by

M := |zf′(z) − f_{2k}(z)| − (1 − α) |βzf′(z) + (1 − γ)f_{2k}(z)|

= | ∑_{n=2}^{∞} na_{n}z^{n} − ∑_{n=2}^{∞} ℜ(a_{n})c_{n}z^{n}|

− (1 − α) |β (z + ∑_{n=2}^{∞} na_{n}z^{n}) + (1 − γ) (z + ∑_{n=2}^{∞} ℜ(a_{n})c_{n}z^{n})| ,

Some subclasses of close-to-convex and quasi-convex functions 69

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} (ε = exp(2πi/k); l ∈ N = \{1, 2, … \}).

(3.2)

Thus, for |z| = r < 1, we have

M ≤ ∑_{n=2}^{∞}(n |a_{n}| + |ℜ(a_{n})| c_{n})r^{n}

− (1 − α) [(1 + β − γ)r − ∑_{n=2}^{∞}[nβ |a_{n}| + (1 − γ) |ℜ(a_{n})| c_{n}]r^{n}]

< ( ∑_{n=2}^{∞}{n[1 + (1 − α)β] |a_{n}| + [(1 − α)(1 − γ) + 1] |ℜ(a_{n})| c_{n}}

− (1 − α)(1 + β − γ))r

< ∑_{n=2}^{∞}{n[1 + (1 − α)β] |a_{n}| + [(1 − α)(1 − γ) + 1] |ℜ(a_{n})| c_{n}} − (1 − α)(1 + β − γ)

= ∑_{n=2}^{∞} n[1 + (1 − α)β] |a_{n}| + ∑_{l=1}^{∞}[(1 − α)(1 − γ) + 1] |ℜ(a_{lk+1})| − (1 − α)(1 + β − γ).

From inequality (3.1), we know that M < 0, thus we can get inequality (1.3), that is f(z) ∈ S_{sc}^{(k)}(α, β, γ). This completes the proof of Theorem 2.

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

Corollary 2. Let f(z) = z + ∑_{n=2}^{∞} a_{n}z^{n} be analytic in U, if for 0 ≤ α < 1, 0 ≤ β ≤ 1 and 0 ≤ γ < 1, we have

∑_{n=2}^{∞} n^{2}[1+(1−α)β] |a_{n}|+ ∑_{l=1}^{∞}[(1−α)(1−γ)+1](lk+1) |ℜ(a_{lk+1})| ≤ (1−α)(1+β−γ),

then f(z) ∈ C_{sc}^{(k)}(α, β, γ).

4. Integral representations In this section, we provide the integral representations for functions belonging

to the classes S_{sc}^{(k)}(α, β, γ) and C_{sc}^{(k)}(α, β, γ).

Theorem 3. Let f(z) ∈ S_{sc}^{(k)}(α, β, γ), then we have

f_{2k}(z) = z · exp \{\frac{1}{2k} ∑_{μ=0}^{k−1} ∫_{0}^{z} \frac{(1 − α)(1 + β − γ)}{ζ}×

× [\frac{ω(ε^{μ}ζ)}{1 − (1 − α)βω(ε^{μ}ζ)} + \frac{\overline{ω(ε^{μ}\overline{ζ})}}{1 − (1 − α)β\overline{ω(ε^{μ}\overline{ζ})}} ] dζ\}, (4.1)

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)}(α, β, γ), by (1.4), we have
\frac{zf′(z)}{f_{2k}(z)} = \frac{1 + (1 − α)(1 − γ)ω(z)}{1 − (1 − α)βω(z)},  (4.2)

where ω(z) is analytic in U and ω(0) = 0, |ω(z)| < 1. Substituting z by ε^{μ}z (μ = 0, 1, 2, …, k − 1) in (4.2), we have

\frac{ε^{μ}zf′(ε^{μ}z)}{f_{2k}(ε^{μ}z)} = \frac{1 + (1 − α)(1 − γ)ω(ε^{μ}z)}{1 − (1 − α)βω(ε^{μ}z)}.  (4.3)

From equality (4.3), we have

\frac{\overline{ε^{μ}\overline{z}} \overline{f′(ε^{μ}\overline{z})}}{\overline{f_{2k}(ε^{μ}\overline{z})}} = \frac{1 + (1 − α)(1 − γ)\overline{ω(ε^{μ}\overline{z})}}{1 − (1 − α)β\overline{ω(ε^{μ}\overline{z})}}.  (4.4)

Summing equalities (4.3) and (4.4), and making use of the same method as in Lemma 2, we have

\frac{zf′_{2k}(z)}{f_{2k}(z)} = \frac{1}{2k} ∑_{μ=0}^{k−1} [\frac{1 + (1 − α)(1 − γ)ω(ε^{μ}z)}{1 − (1 − α)βω(ε^{μ}z)} + \frac{1 + (1 − α)(1 − γ)\overline{ω(ε^{μ}\overline{z})}}{1 − (1 − α)β\overline{ω(ε^{μ}\overline{z})}}] ,

(4.5)

from equality (4.5), we can get

\frac{f′_{2k}(z)}{f_{2k}(z)} − \frac{1}{z} = \frac{1}{2k} ∑_{μ=0}^{k−1} \frac{1}{z} \{[\frac{1 + (1 − α)(1 − γ)ω(ε^{μ}z)}{1 − (1 − α)βω(ε^{μ}z)}+

+ \frac{1 + (1 − α)(1 − γ)\overline{ω(ε^{μ}\overline{z})}}{1 − (1 − α)β\overline{ω(ε^{μ}\overline{z})}}] − 2\} . (4.6)

Integrating equality (4.6), we have

log \{\frac{f_{2k}(z)}{z}\} = \frac{1}{2k} ∑_{μ=0}^{k−1} ∫_{0}^{z} \frac{(1 − α)(1 + β − γ)}{ζ}×

× [\frac{ω(ε^{μ}ζ)}{1 − (1 − α)βω(ε^{μ}ζ)} + \frac{\overline{ω(ε^{μ}\overline{ζ})}}{1 − (1 − α)β\overline{ω(ε^{μ}\overline{ζ})}}] dζ. (4.7)

From equality (4.7), we can get equality (4.1) easily. This completes the proof of Theorem 3.

Theorem 4. Let f(z) ∈ S_{sc}^{(k)}(α, β, γ), then we have
f(z) = ∫_{0}^{z} exp \{\frac{1}{2k} ∑_{μ=0}^{k−1} ∫_{0}^{ξ} \frac{(1 − α)(1 + β − γ)}{ζ}[\frac{ω(ε^{μ}ζ)}{1 − (1 − α)βω(ε^{μ}ζ)}

+ \frac{\overline{ω(ε^{μ}\overline{ζ})}}{1 − (1 − α)β\overline{ω(ε^{μ}\overline{ζ})}}]dζ\} · \frac{1 + (1 − α)(1 − γ)ω(ξ)}{1 − (1 − α)βω(ξ)}dξ,  (4.8)

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

Proof. Suppose that f(z) ∈ S_{sc}^{(k)}(α, β, γ), from equalities (4.1) and (4.2), we can get

f′(z) = \frac{f_{2k}(z)}{z} · \frac{1 + (1 − α)(1 − γ)ω(z)}{1 − (1 − α)βω(z)}

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

+ \frac{\overline{ω(ε^{μ}\overline{ζ})}}{1 − (1 − α)β\overline{ω(ε^{μ}\overline{ζ})}}]dζ\} · \frac{1 + (1 − α)(1 − γ)ω(z)}{1 − (1 − α)βω(z)}.

Integrating the above equality, we can get equality (4.8) easily. Hence the proof of Theorem 4 is complete.

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

Corollary 3. Let f(z) ∈ C_{sc}^{(k)}(α, β, γ), then we have

f_{2k}(z) = ∫_{0}^{z} exp \{\frac{1}{2k} ∑_{μ=0}^{k−1} ∫_{0}^{ξ} \frac{(1 − α)(1 + β − γ)}{ζ}×

× [\frac{ω(ε^{μ}ζ)}{1 − (1 − α)βω(ε^{μ}ζ)} + \frac{\overline{ω(ε^{μ}\overline{ζ})}}{1 − (1 − α)β\overline{ω(ε^{μ}\overline{ζ})}}] dζ\} dξ,

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

Corollary 4. Let f(z) ∈ C_{sc}^{(k)}(α, β, γ), then we have
f(z) = ∫_{0}^{z} \frac{1}{t} ∫_{0}^{t} exp \{\frac{1}{2k} ∑_{μ=0}^{k−1} ∫_{0}^{ξ} \frac{(1 − α)(1 + β − γ)}{ζ}[\frac{ω(ε^{μ}ζ)}{1 − (1 − α)βω(ε^{μ}ζ)}

+ \frac{\overline{ω(ε^{μ}\overline{ζ})}}{1 − (1 − α)β\overline{ω(ε^{μ}\overline{ζ})}}]dζ\} · \frac{1 + (1 − α)(1 − γ)ω(ξ)}{1 − (1 − α)βω(ξ)} dξ dt,

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

5. Convolution conditions

Finally, we provide the convolution conditions for the classes S_{sc}^{(k)}(α, β, γ) and C_{sc}^{(k)}(α, β, γ). 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 5. A function f(z) ∈ S_{sc}^{(k)}(α, β, γ) if and only if

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

−\frac{1 + (1 − α)(1 − γ)e^{iθ}}{2} · \overline{(f ∗ h)(\overline{z})}\} ≠ 0  (5.1)

for all z ∈ U and 0 ≤ θ < 2π, where h(z) is given by (5.6).

Proof. Suppose that f(z) ∈ S_{sc}^{(k)}(α, β, γ), we know that the condition (1.3) can be written as (1.4), since (1.4) is equivalent to

\frac{zf′(z)}{f_{2k}(z)} ≠ \frac{1 + (1 − α)(1 − γ)e^{iθ}}{1 − (1 − α)βe^{iθ}}  (5.2)

for all z ∈ U and 0 ≤ θ < 2π. It is easy to know that the condition (5.2) can be written as

\frac{1}{z} \{[1 − (1 − α)βe^{iθ}] zf′(z) − [1 + (1 − α)(1 − γ)e^{iθ}] f_{2k}(z)\} ≠ 0.  (5.3)

On the other hand, it is well known that

zf′(z) = f(z) ∗ \frac{z}{(1 − z)^{2}} .  (5.4)

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

f_{2k}(z) = \frac{1}{2} [(f ∗ h)(z) + \overline{(f ∗ h)(\overline{z})}] ,  (5.5)

where

h(z) = \frac{1}{k} ∑_{υ=0}^{k−1} \frac{z}{1 − ε^{υ}z} .  (5.6)

Substituting (5.4) and (5.5) into (5.3), we can get (5.1) easily. This completes the proof of Theorem 5.

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

Corollary 5. A function f(z) ∈ C_{sc}^{(k)}(α, β, γ) if and only if

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

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

for all z ∈ U and 0 ≤ θ < 2π, where h(z) is given by (5.6).

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. The author would like to thank Prof. Chun-Yi Gao and Prof. Yue-Ping Jiang for their support and encouragement.

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] C.-Y. Gao, A subclass of close-to-convex functions, J. Changsha Commun. Univ. 10 (1994), 1–7.

[4] C.-Y. Gao, S.-M. Yuan and H.M. Srivastava, Some functional inequalities and inclusion relationships associated with certain families of integral operators, Comput. Math. Appl. 49 (2005), 1787–1795.

[5] M.-S. Liu, On a subclass of p-valent close-to-convex functions of order β and type α, J. Math. Study 30 (1997), 102–104.

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

[7] 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.

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

(received 06.12.2006)

School of Mathematics and Computing Science, Changsha University of Science and Technology, Changsha, 410076 Hunan, People’s Republic of China.

E-mail: zhigwang@163.com