Research Article Sufficiency Criteria for a Class of Complex Order
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Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 517296, 4 pages
http://dx.doi.org/10.1155/2013/517296
Research Article
Sufficiency Criteria for a Class of π-Valent Analytic Functions of
Complex Order
Muhammad Arif
Department of Mathematics, Abdul Wali Khan University, Mardan 23200, Pakistan
Correspondence should be addressed to Muhammad Arif; marifmaths@yahoo.com
Received 21 December 2012; Accepted 26 February 2013
Academic Editor: Fuding Xie
Copyright © 2013 Muhammad Arif. This is an open access article distributed under the Creative Commons Attribution License,
which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
In the present paper, we consider a subclass of π-valent analytic functions and obtain certain simple sufficiency criteria by using
three different methods for the functions belonging to this class. Many known results appear as special consequences of our work.
Now we define the subclass Mπ (π, π; π(π§)) of π΄ π (π) by
1. Introduction
Let π΄ π (π) be the class of functions π(π§) analytic and π-valent
in the open unit disk U = {π§ : |π§| < 1} and of the form
∞
π (π§) = π§π + ∑ ππ π§π (π ∈ N) .
In particular, π΄ π (1) = π΄ π , π΄ 1 (π) = π΄(π), and π΄ 1 (1) = π΄. By
S∗π (π, π) and Cπ (π, π), π, π ∈ N and π ∈ C \ {0}, we mean the
subclasses of π΄ π (π) which are defined, respectively, by
Re {1 +
π§πσΈ (π§)
1
− π)} > 0,
(
π + π − 1 π (π§)
π§πσΈ σΈ (π§)
1
− π + 1)} > 0,
( σΈ
π + π − 1 π (π§)
(π§ ∈ U) ,
(π§ ∈ U) .
(2)
For π = 1, π = 1, π = 1, the previous two classes defined in
(2) reduce to the well-known classes of starlike and convex,
respectively.
For functions π(π§), π(π§) ∈ π΄ π (π) of the form (1), we define the convolution (Hadamard product) of π(π§) and π(π§)
by
∞
(π β π) (π§) = π§π + ∑ ππ ππ π§π ,
π=π+π
π§(π (π§) β π (π§))
1
(
− π)} > 0,
π+π−1
π (π§) β π (π§)
(1)
π=π+π
Re {1 +
σΈ
Re {1 +
(π§ ∈ U) .
(3)
(4)
(π§ ∈ U) .
Sufficient conditions were studied by various authors for
different subclasses of analytic and multivalent functions, for
some of the related work see [1–8]. The object of the present
paper is to obtain sufficient conditions for the subclass
Mπ (π, π; π(π§)) of π΄ π (π). We also consider some special cases
of our results which lead to various interesting corollaries and
relevances of some of these with other known results also
being mentioned.
We will assume throughout our discussion, unless otherwise stated, that π ∈ N, π ∈ N, π ∈ C \ {0}.
2. Preliminary Results
To obtain our main results, we need the following Lemma’s.
Lemma 1 (see [9]). If π(π§) ∈ π΄(π) with π ≥ 1 and satisfies the
condition
σ΅¨
σ΅¨σ΅¨ σΈ
σ΅¨σ΅¨π (π§) − 1σ΅¨σ΅¨σ΅¨ <
σ΅¨
σ΅¨
π+1
√(π + 1)2 + 1
(π§ ∈ U) ,
(5)
2
Abstract and Applied Analysis
Thus using (10), we have
then
∗
π (π§) ∈ S (π, 1) .
(6)
Lemma 2 (see [10]). If π(π§) ∈ π΄(π) satisfing the condition
σ΅¨ π
σ΅¨σ΅¨
σ΅¨σ΅¨arg πσΈ (π§)σ΅¨σ΅¨σ΅¨ < πΏπ (π§ ∈ U) ,
σ΅¨ 2
σ΅¨
(7)
2 tan [π (1 − πΏπ )] + π (1 − 2πΏπ ) = 0,
(8)
π (π§) ∈ S∗ (π, 1) .
(9)
Lemma 3 (see [11]). Let Ω be a set in the complex plane C, and
suppose that Ψ is a mapping from C2 × U to C which satisfies
Ψ(ππ₯, π¦, π§) ∉ Ω for π§ ∈ U and for all real π₯, π¦ such that π¦ ≤
(−π/2)(1 + π₯2 ). If π(π§) = 1 + ππ π§π + ⋅ ⋅ ⋅ is analytic in U and
Ψ (π(π§), π§πσΈ (π§), π§) ∈ Ω for all π§ ∈ U, then Re π(π§) > 0.
3. Main Results
Theorem 4. If π(π§) ∈ π΄ π (π) satisfies
σ΅¨σ΅¨
σΈ
σ΅¨σ΅¨ π (π§) ∗ π (π§) 1/(π+π−1) π§(π (π§) ∗ π (π§))
σ΅¨σ΅¨(
+ π − 1}
)
{
σ΅¨σ΅¨
π§π
π (π§) ∗ π (π§)
σ΅¨σ΅¨
σ΅¨σ΅¨
σ΅¨
− (π + π − 1) σ΅¨σ΅¨σ΅¨
(10)
σ΅¨σ΅¨
π+1
√(π + 1)2 + 1
Proof. Let us set a function π(π§) by
π (π§) ∗ π (π§) 1/(π+π−1)
)
π§π
ππ+π ππ+π
(π + π − 1)
(π§ ∈ U) .
(11)
π§π + ⋅ ⋅ ⋅
for π(π§) ∈ π΄ π (π). Then clearly (11) shows that π(π§) ∈ π΄(π).
Differentiating (11) logarithmically, we have
π (π§) ∗ π(π§)σΈ π
πσΈ (π§)
1
1
=
[
− ]+
π (π§)
π + π − 1 π (π§) ∗ π (π§) π§
π§
which gives
σ΅¨
σ΅¨σ΅¨ σΈ
σ΅¨σ΅¨π (π§) − 1σ΅¨σ΅¨σ΅¨
σ΅¨
σ΅¨
σ΅¨σ΅¨
σ΅¨σ΅¨ π (π§) ∗ π (π§) 1/(π+π−1)
1
= σ΅¨σ΅¨σ΅¨(
)
π
σ΅¨σ΅¨
π§
π
+
π−1
σ΅¨
σ΅¨σ΅¨
σΈ
π§(π (π§) ∗ π (π§))
σ΅¨σ΅¨
×{
+ π − 1} − 1σ΅¨σ΅¨σ΅¨ .
σ΅¨σ΅¨
π (π§) ∗ π (π§)
σ΅¨
(14)
Hence, using Lemma 1, we have π(π§) ∈ S∗ (π, 1).
From (12), we can write
(15)
Since π(π§) ∈ S∗ (π, 1), it implies that Re(π§πσΈ (π§)/π(π§)) > 0.
Therefore, we get
σΈ
Re {1 +
π§(π (π§) ∗ π (π§))
1
(
− π)}
π+π−1
π (π§) ∗ π (π§)
π§πσΈ (π§)
> 0,
= Re
π (π§)
(16)
and this implies that π(π§) ∈ Mπ (π, π; π(π§)).
Setting π = π = 1 and π(π§) = π§/(1 − π§) in Theorem 4, we
get the following.
Corollary 5. If π(π§) ∈ π΄ satisfies
σ΅¨σ΅¨
σ΅¨σ΅¨
σ΅¨σ΅¨ π (π§) 1/π π§πσΈ (π§)
σ΅¨σ΅¨ 2 |π|
σ΅¨σ΅¨(
σ΅¨σ΅¨ <
+
π
−
1}
−
π
{
)
σ΅¨σ΅¨ √5 (π§ ∈ U) ,
σ΅¨σ΅¨σ΅¨ π§
π
(π§)
σ΅¨σ΅¨
σ΅¨
(17)
∗
then π(π§) ∈ S (π), the class of starlike functions of complex
order π.
Corollary 6. If π(π§) ∈ π΄ satisfies
σ΅¨σ΅¨ σΈ
σ΅¨
σ΅¨σ΅¨(π (π§))(1−π)/π {π§πσΈ σΈ (π§) + ππσΈ (π§)} − πσ΅¨σ΅¨σ΅¨ < 2 |π|
σ΅¨σ΅¨
σ΅¨σ΅¨
σ΅¨
σ΅¨ √5
then π(π§) ∈ Mπ (π, π; π(π§)).
=π§+
,
Putting π = π = 1 and π(π§) = π§/(1 − π§)2 in Theorem 4,
we have the following.
σ΅¨
σ΅¨σ΅¨
σ΅¨σ΅¨π + π − 1σ΅¨σ΅¨σ΅¨ (π§ ∈ U) ,
π (π§) = π§(
√(π + 1)2 + 1
σΈ
then
<
π+1
π§(π (π§) ∗ π (π§))
π§πσΈ (π§)
1
=
− π] + 1.
[
π (π§)
π+π−1
π (π§) ∗ π (π§)
where πΏπ is the unique root of the equation
−1
σ΅¨σ΅¨ σΈ
σ΅¨
σ΅¨σ΅¨π (π§) − 1σ΅¨σ΅¨σ΅¨ ≤
σ΅¨
σ΅¨
(12)
(13)
(π§ ∈ U) ,
(18)
then π(π§) ∈ C(π), the class of convex functions of complex
order π.
Remark 7. If we put π = 1−πΌ in Corollaries 5 and 6, we get the
results proved by UyanΔ±k et al. [1]. Furthermore, for π = 1, we
obtain the results studied by Mocanu [2] and Nunokawa et al.
[3], respectively. Also if we set π = 1−πΌ with π(π§) = π§π /(1−π§π )
2π
and π(π§) = π§π /(1 − π§π ) in Theorem 4, we obtain the results
due to Goyal et al. [4].
Theorem 8. If π(π§) ∈ π΄ π (π) satisfies
σ΅¨σ΅¨
1/(π+π−1)
σ΅¨σ΅¨
σ΅¨σ΅¨arg ( π (π§) ∗ π (π§) )
σ΅¨σ΅¨σ΅¨
π§π
σ΅¨
σ΅¨σ΅¨
σΈ
σ΅¨σ΅¨
π§(π (π§) ∗ π (π§))
1
+ arg {
(
+ π − 1)}σ΅¨σ΅¨σ΅¨ (19)
σ΅¨σ΅¨
π+π−1
π (π§) ∗ π (π§)
σ΅¨
π
< πΏπ (π§ ∈ U) ,
2
where πΏπ is the unique root of (8), then π(π§) ∈ Mπ (π, π; π(π§)).
Abstract and Applied Analysis
3
Proof. Let π(π§) be given by (11), which clearly belongs to the
class π΄(π).
Now differentiating (11), we have
πσΈ (π§) = (
π (π§) ∗ π (π§) 1/(π+π−1)
)
π§π
Theorem 12. If π(π§) ∈ π΄ π (π) satisfies
σΈ
π§(π (π§) ∗ π (π§))
1
×
{
+ π − 1}
π+π−1
π (π§) ∗ π (π§)
(20)
σΈ
Re [
σΈ σΈ
π§(π (π§)∗π (π§))
π§(π (π§)∗π (π§))
1
(π
+ 1)}
{
σΈ
π+π−1
π (π§)∗π (π§)
(π (π§)∗π (π§))
+π − 1] >
which gives
σ΅¨
σ΅¨σ΅¨
σ΅¨σ΅¨arg πσΈ (π§)σ΅¨σ΅¨σ΅¨
σ΅¨
σ΅¨
σ΅¨σ΅¨
σ΅¨σ΅¨
π (π§) ∗ π (π§) 1/(π+π−1)
= σ΅¨σ΅¨σ΅¨arg (
)
σ΅¨σ΅¨
π§π
σ΅¨
π2
+ π,
4πΏ
(26)
where 0 ≤ π ≤ 1 and
(21)
σ΅¨σ΅¨
σ΅¨σ΅¨
π§(π (π§) ∗ π (π§))
1
+π−1)}σ΅¨σ΅¨σ΅¨ .
+ arg {
(
σ΅¨σ΅¨
π+π−1
π (π§) ∗ π (π§)
σ΅¨
σΈ
π
πΏ = π (π + Re π − 1 + ) ,
2
π = 2π Im π,
π = π (( ((Re π)2 − (Im π)2 − Re π)
Thus using (19), we have
σ΅¨σ΅¨
σ΅¨ π
σ΅¨σ΅¨arg πσΈ (π§)σ΅¨σ΅¨σ΅¨ ≤ πΏπ
σ΅¨
σ΅¨ 2
Remark 11. For putting π = 1, πΌ = 0 in Corollary 10 and π =
1 in Corollary 9, we obtain the results proved by Mocanu [10]
and UyanΔ±k et al. [1], respectively.
(π§ ∈ U) ,
(22)
× (π + Re π − 1) + (Im π)2 (2 Re π − 1)
where πΏπ is the root of (8). Hence, using Lemma 2, we have
π(π§) ∈ S∗ (π, 1).
From (20), we can write
(23)
Since π(π§) ∈ S∗ (π, 1), it implies that Re(π§πσΈ (π§)/π(π§)) > 0.
Therefore, we get (16), and hence π(π§) ∈ Mπ (π, π; π(π§)).
Making π = 1, π = 1 − πΌ with 0 ≤ πΌ < π and π(π§) =
π§π /(1 − π§), we have the following.
Corollary 9. If π(π§) ∈ π΄ π satisfies
σ΅¨σ΅¨
σ΅¨σ΅¨
π (π§)
π§πσΈ (π§)
σ΅¨σ΅¨
σ΅¨σ΅¨
σ΅¨σ΅¨arg ( π ) + (π − πΌ) arg {
σ΅¨σ΅¨
−
πΌ}
σ΅¨σ΅¨
σ΅¨σ΅¨
π§
π
(π§)
σ΅¨
σ΅¨
π
< πΏ1 (π − πΌ) (π§ ∈ U) ,
2
−1
π
2
×((π + Re π − 1) + (Im π)2 ) ) − ) ,
2
then π(π§) ∈ Mπ (π, π; π(π§)).
σΈ
π§(π (π§) ∗ π (π§))
π§πσΈ (π§)
1
=
− π] + 1.
[
π (π§)
π+π−1
π (π§) ∗ π (π§)
Proof. Let us set
σΈ
π§(π (π§) ∗ π (π§))
= (π + π − 1) π (π§) − π + 1.
π (π§) ∗ π (π§)
(28)
Then π(π§) is analytic in U with π(0) = 1.
Taking logarithmic differentiation of (28) and then by
simple computation, we obtain
σΈ
(24)
π§(π (π§) ∗ π (π§))
1
{
π+π−1
π (π§) ∗ π (π§)
σΈ σΈ
× (π
where πΏ1 is the unique root of (8) with π = 1, then π(π§) ∈
S∗π (πΌ), the class of π-valent starlike functions of order πΌ.
π§(π (π§) ∗ π(π§))
(π (π§) ∗ π (π§))
σΈ
+ 1) + π − 1}
(29)
= π΄π§πσΈ (π§) + π΅π2 (π§) + πΆπ (π§) + π·
Also if we take π = 1, π = 1 − πΌ with 0 ≤ πΌ < π and
π(π§) = π§π /(1 − π§)2π in Theorem 8, we obtain the following
result.
= Ψ (π (π§) , π§πσΈ (π§) , π§)
with
Corollary 10. If π(π§) ∈ π΄ π satisfies
σ΅¨σ΅¨
σ΅¨σ΅¨
πσΈ (π§)
π§πσΈ σΈ (π§)
σ΅¨σ΅¨
σ΅¨σ΅¨
σ΅¨σ΅¨arg ( π−1 ) + (π − πΌ) arg { σΈ
σ΅¨σ΅¨
+
1
−
πΌ}
σ΅¨σ΅¨
σ΅¨σ΅¨
ππ§
π
(π§)
σ΅¨
σ΅¨
π
< πΏ1 (π − πΌ) (π§ ∈ U) ,
2
(27)
π΄ = π,
(25)
where πΏ1 is the unique root of (8) with π = 1, then π(π§) ∈
Cπ (πΌ), the class of π-valent convex functions of order πΌ.
π΅ = π (π + π − 1) ,
πΆ = −2ππ + π + 1,
π·=
ππ2 − ππ
.
π+π−1
(30)
Now for all real π₯ and π¦ satisfying π¦ ≤ −(π/2)(1+π₯2 ), we have
Ψ (ππ₯, π¦, π§) = π΄π¦ − π΅π₯2 + πΆ (ππ₯) + π·.
(31)
4
Abstract and Applied Analysis
Reputing the values of π΄, π΅, πΆ, π· and then taking real part,
we obtain
Re Ψ (ππ₯, π¦, π§) ≤ −πΏπ₯2 + ππ₯ + π
π 2 π2
) +
= −(√πΏπ₯ −
+π
4πΏ
2√πΏ
(32)
π2
+ π,
4πΏ
where πΏ, π, π are given in (27).
Let Ω = {π€ : Re π€ > (π2 /4πΏ)+π}. Then Ψ (β(π§), π§βσΈ (π§),
π§) ∈ Ω and Ψ(ππ₯, π¦, π§) ∉ Ω, for all real π₯ and π¦ satisfying π¦ ≤
−(π/2)(1 + π₯2 ), π§ ∈ U. Using Lemma 3, we have Re π(π§) > 0.
This implies that
<
σΈ
Re {1 +
π§(π (π§) ∗ π (π§))
1
(
− π)} > 0,
π+π−1
π (π§) ∗ π (π§)
(33)
and hence π(π§) ∈ Mπ (π, π; π(π§)).
If we put π = π = 1, π = 1−πΌ and π(π§) = π§/(1 − π§)2(1−πΌ) in
Theorem 12, we obtain the following result proved in [12].
Corollary 13. If π(π§) ∈ π΄ satisfies
Re {
π§πσΈ σΈ (π§)
π§πσΈ (π§)
(π σΈ
+ 1)}
π (π§)
π (π§)
π
1
> πΌπ (πΌ − ) + (πΌ − ) ,
2
2
(34)
then π(π§) ∈ S∗ (πΌ).
Furthermore, for πΌ = 0 in Corollary 13, we have the
following result proved in [13].
Corollary 14. If π(π§) ∈ π΄ satisfies
Re {
π§πσΈ σΈ (π§)
π§πσΈ (π§)
π
(π σΈ
+ 1)} > − ,
π (π§)
π (π§)
2
(35)
then π(π§) ∈ S∗ .
Acknowledgment
The author is thankful to the Prof. Dr. Ihsan Ali, Vice chancellor of Abdul Wali Khan University Mardan, for providing
research facilities in AWKUM.
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