PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 75(91) (2005), 93–98

ON SOME RESULTS FOR λ-SPIRALLIKE AND λ-ROBERTSON FUNCTIONS OF COMPLEX ORDER

M. K. Aouf, F. M. Al-Oboudi, and M. M. Haidan

Communicated by Miroljub Jevtić

Abstract. We give some results of various kinds concerning λ-spirallike functions of complex order and λ-Robertson functions of complex order in the unit disc U = {z : |z| < 1}. They represent extensions and generalizations of many previous results. We mainly used the subordination method.

1. Introduction

Let A denote the class of functions of the form:

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

which are analytic in the unit disc U = {z : |z| < 1}.

For a function f(z) belonging to the class A we say that f(z) is λ-spirallike of complex order in U if and only if

(1.2)  Re \{\frac{1}{b cos λ}[e^{iλ} \frac{zf′(z)}{f(z)} − (1 − b) cos λ − i sin λ]\} > 0,

for some real λ, |λ| < π/2, b ≠ 0, complex. We denote this class by S^{λ}(b). It was introduced and studied by Al-Oboudi and Haidan [1].

Also for a function f(z) belonging to the class A we say that f(z) is λ-Robertson function of complex order in U if and only if

(1.3)  Re \{\frac{1}{b cos λ}[e^{iλ}(1 + \frac{zf′′(z)}{f′(z)}) − (1 − b) cos λ − i sin λ]\} > 0,

for some real λ, |λ| < π/2, b ≠ 0, complex. We denote this class by C^{λ}(b). If follows from (1.2) and (1.3) that

(1.4)  f(z) ∈ C^{λ}(b) if and only if zf′(z) ∈ S^{λ}(b).

2000 Mathematics Subject Classification: Primary 30C45.

Key words and phrases: λ-Spirallike, λ-Robertson functions, subordination.

We note that:

(1) S^{λ}(1) = S^{λ}, is the class of λ-spirallike univalent functions defined by Spacek [16], S^{0}(b) = S(b), is the class of starlike functions of complex order studied by Nasr and Aouf [7], S^{λ}(1 − α) = S^{λ}(α), 0 ⩽ α < 1, is the class of λ-spirallike functions of order α studied by Libera [4] and S^{0}(1 − α) = S^{∗}(α), 0 ⩽ α < 1, is the class of starlike functions of order α, studied by Robertson [12].

(2) C^{λ}(1) = C^{λ}, is the class of λ-Robertson functions studied by Robertson [13], C^{λ}(1 − α) = C^{λ}(α), 0 ⩽ α < 1, is the class of λ-Robertson functions of order α studied by Chichra [3] and C^{0}(b) = C(b), is the class of convex functions of complex order studied by Waitrowski [18], Nasr and Aouf [8] and Aouf [2] and C^{0}(1 − α) = C(α), 0 ⩽ α < 1, is the class of convex functions of order α studied by Robertson [12].

The object of this paper is to obtain some results for the classes S^{λ}(b) and C^{λ}(b) using mainly the method of subordination. In that sense, we give some definitions, notations and lemmas we need in the next part.

Let f and F be analytic in the unit disc U. The function f is subordinate to F, written f ≺ F or f(z) ≺ F (z), if F is univalent, f(0) = F (0) and f(U) ⊂ F (U).

The general theory of differential subordinations was introduced by Miller and Mocanu [5]. Some classes of the first-order differential subordinations were considered by the same authors in [6]. Namely let ψ : C^{2} → C be analytic in a domain D, let h be univalent in U, and let p(z) be analytic in U with (p(z), zp′(z)) ∈ D when z ∈ U, then p(z) is said to satisfy the first-order differential subordination if

(1.5)  ψ(p(z), zp′(z)) ≺ h(z).

The univalent function q is said to be a dominant of the differential subordination (1.5) if p ≺ q for all p satisfying (1.5). If q̃ is a dominant of (1.5) and q̃ ≺ q for all dominants q of (1.5), then q̃ is said to be the best dominant of (1.5).

First we cite the following lemma on differential subordinations due to Miller and Mocanu [6].

Lemma 1. Let q be univalent in U and let θ and ϕ be analytic in a domain D containing q(U), with ϕ(w) ≠ 0 when w ∈ q(U). Set

Q(z) = zq′(z)ϕ(q(z)), h(z) = θ(q(z)) + Q(z)

and suppose that

(i) Q is starlike (univalent) in U with Q(0) = 0 and Q′(0) ≠ 0,

(ii) Re \{z \frac{h′(z)}{Q(z)}\} = Re \{\frac{θ′(q(z))}{ϕ(q(z))} + z \frac{Q′(z)}{Q(z)}\} > 0, z ∈ U.

If p is analytic in U, with p(0) = q(0), p(U) ⊂ D and

(1.6)  θ(p(z)) + zp′(z)ϕ(p(z)) ≺ θ(q(z)) + zq′(z)ϕ(q(z)) = h(z),

then p ≺ q, and q is the best dominant of (1.6).

For the proof of Theorem 2, we need the following result of Robertson [14].

ON SOME RESULTS FOR λ-SPIRALLIKE AND λ-ROBERTSON 95

Lemma 2. Let f(z) ∈ A be univalent in U. For 0 ⩽ t ⩽ 1, let F (z, t) be analytic in U, let F (z, 0) ≡ f(z) and F (0, t) ≡ 0. Let r be positive real number for which

F (z) = lim_{t→0^{+}} \frac{F (z, t) − F (z, 0)}{zt^{r}}

exists. Let F (z, t) be subordinate to f(z) in U for 0 ⩽ t ⩽ 1. Then Re \{\frac{F (z)}{f′(z)}\} ⩽ 0,

z ∈ U. If, in addition, F (z) is also analytic in U and {F (0)} ≠ 0, then

(1.7)  Re \{\frac{f′(z)}{F (z)}\} < 0, z ∈ U.

2. Results and consequences

First we use the differential subordinations to obtain:

Theorem 1. Let f ∈ S^{λ}(b) (|λ| < π/2, b ≠ 0, complex), then

(2.1)  (\frac{f(z)}{z})^{a} ≺ \frac{1}{(1 − z)^{2abe^{−iλ} cos λ}} ,

where a ≠ 0 is complex and either |2abe^{−iλ} cos λ + 1| ⩽ 1 or |2abe^{−iλ} cos λ − 1| ⩽ 1, and this is the best dominant.

Proof. If we put q(z) = (1 − z)^{−2abe^{−iλ} cos λ}, ϕ(w) = (abe^{−iλ} cos λ)^{−1}w^{−1} and θ(w) = 1 in Lemma 1, then it is easy to check that the conditions (i) and (ii) in that lemma are satisfied. Namely, q(z) is univalent in U [15], while

h(z) = θ(q(z)) + zq′(z)ϕ(q(z)) = \frac{1 + z}{1 − z} ·

Consequently, for p(z) = 1 + p_{1}z + ⋯ analytic in U with p(z) ≠ 0 for 0 < |z| < 1, from (1.6) we get

(2.2)  1 + \frac{e^{iλ}}{ab cos λ} \frac{zp′(z)}{p(z)} ≺ \frac{1 + z}{1 − z} ⇒ p(z) ≺ q(z)·

Now, if in (2.2) we choose p(z) = (\frac{f(z)}{z})^{a}, then we have

(\frac{f(z)}{z})^{a} ≺ \frac{1}{(1 − z)^{2abe^{−i^{λ} cos λ}}} ,

which evidently completes the proof of Theorem 1. □

Remark 1. (1) Putting (i) λ = 0, (ii) b = 1, and (iii) b = 1 − α, 0 ⩽ α < 1, in Theorem 1, we get the results obtained by Obradović, Aouf and Owa [10] for the classes S(b), S^{λ} and S^{λ}(α), respectively.

(2) Putting (i) λ = 0 and b = 1, (ii) λ = 0 and b = 1−α, 0 ⩽ α < 1, in Theorem 1, we get the corresponding results for the classes S^{∗} and S^{∗}(α), especially, the wellknown results for the classes S^{∗} and S^{∗}(α) when a = 1.

From Theorem 1 and using (1.4), we directly get:

Corollary 1. Let f(z) ∈ C^{λ}(b), (|λ| < π/2, b ≠ 0, complex), and let a ≠ 0 be a complex number and either |2abe^{−iλ} cos λ + 1| ⩽ 1 or |2abe^{−iλ} cos λ − 1| ⩽ 1. Then

(f′(z))^{a} ≺ (1 − z)^{−2abe^{−iλ} cos λ}

and this is the best dominant.

Putting b = 1 − α, 0 ⩽ α < 1, in Corollary 1, we get the following result for the class C^{λ}(α):

Corollary 2. Let f(z) ∈ C^{λ}(α) (|λ| < π/2, 0 ⩽ α < 1), and let a be a complex number such that

|2a(1 − α) cos λe^{−iλ} − 1| ⩽ 1 or |2a(1 − α) cos λe^{−iλ} + 1| ⩽ 1.

Then

(2.3)  (f′(z))^{a} ≺ (1 − z)^{−2a(1−α)e^{−iλ} cos λ}

and this is the best dominant.

Putting λ = 0 in Corollary 2 we get the result obtained by Obradović, Aouf and Owa [10].

If we put a = −\frac{e^{iλ}}{2b cos λ} in Theorem 1, we get:
Corollary 3. Let f(z) ∈ S^{λ}(b) (|λ| < π/2, b ≠ 0), then
(2.4)  (\frac{z}{f(z)})^{\frac{e^{iλ}}{2b cos λ}} ≺ (1 − z),

and this is the best dominant.

From (2.4), we have the following inequality for f(z) ∈ S^{λ}(b)

(2.5)  |(\frac{z}{f(z)})^{\frac{e^{iλ}}{2b cos λ}} − 1| ⩽ |z|, z ∈ U.

Remark 2. (i) Putting λ = 0 in (2.5), we get the result obtained by Obradović, Aouf and Owa [10], (ii) Putting b = 1 − α, 0 ⩽ α < 1 and λ = 0 in (2.5), we get the result obtained by Obradović, Aouf and Owa [10] and Todorov [17].

By using Lemma 2 we give a criterion for a function f(z) ∈ A to be in the class S^{λ}(b).

Theorem 2. Let f(z) ∈ A with f(z)/z ≠ 0 in U, and let the function

(2.6)  g(z) = \frac{e^{iλ}}{b cos λ}[f(z) − (1 − be^{−iλ} cos λ) ∫_{0}^{z} \frac{f(s)}{s}ds] = z + ⋯,

be univalent in U, where |λ| < π/2 and b ≠ 0, is a complex number. If the function

(2.7) G(z, t) = \frac{e^{iλ}}{b cos λ}[(1−tbe^{−iλ} cos λ)f(z)−(1−be^{−iλ} cos λ)(1−t^{2}) ∫_{0}^{z} \frac{f(s)}{s}ds]

is subordinate to g(z) for a fixed b, |λ| < π/2, and for each 0 ⩽ t ⩽ 1, then f(z) ∈ S^{λ}(b).

Proof. It is evident that G(z, 0) ≡ g(z) and G(0, t) ≡ 0. In Lemma 2, we choose r = 1 and F (z, t) to be the function G(z, t) defined by (2.7). Then we have

G(z) = lim_{t→0^{+}} \frac{G(z, t) − G(z, 0)}{zt} = \frac{1}{z} lim_{t→0} \frac{∂G(z, t)}{∂t} = −\frac{f(z)}{z}

and G(z) is analytic in U with Re{G(0)} = −1 ≠ 0. Since from (2.6)

g′(z) = \frac{e^{iλ}}{b cos λ}[f′(z) − (1 − be^{−iλ} cos λ)\frac{f(z)}{z}],

then from (1.7) we have Re \{\frac{g′(z)}{G(z)}\} < 0, z ∈ U, which is equivalent to

Re \{1 + \frac{e^{iλ}}{b cos λ}(\frac{zf′(z)}{f(z)} − 1)\} > 0 z ∈ U,

i.e., f(z) ∈ S^{λ}(b). □

Remark 3. (1) Putting (i) λ = 0 (ii) λ = 0 and b = 1 − α, 0 ⩽ α < 1, in Theorem 2, we get the results for the classes S(b) and S^{∗}(α) obtained by Obradović, Aouf and Owa [10] and Obradović [9], respectively.

(2) Putting b = 1 − α, 0 ⩽ α < 1, Theorem 2 we get the following result for the class S^{λ}(α) (|λ| < π/2, 0 ⩽ α < 1).

Corollary 4. Let f(z) ∈ A, and let the function g(z) defined by

g(z) = \frac{e^{iλ}}{(1 − α) cos λ}[f(z) − (1 − (1 − α)e^{−iλ} cos λ) ∫_{0}^{z} \frac{f(s)}{s}ds] = z + ⋯

be univalent in U, where |λ| < π/2 and 0 ⩽ α < 1. If the function

G(z, t) = \frac{e^{iλ}}{(1 − α) cos λ}[(1 − t(1 − α)e^{−iλ} cos λ)f(z)

− (1 − (1 − α)e^{−iλ} cos λ)(1 − t^{2})∫_{0}^{z} \frac{f(s)}{s}ds],

is subordinate to g(z) in the unit disc U for fixed λ (|λ| < π/2) and α (0 ⩽ α < 1), and for each t (0 ⩽ t ⩽ 1), then f(z) is in the class S^{λ}(α).

Remark 4. The result obtain in Corollary 4 corrected the result obtained by Obradović and Owa [11, Theorem 3] for the class S^{λ}.

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Mathematics Department (Received 21 10 2003) Girls College Of Education Science Sections Jeddah, Saudi Arabia

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