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Class of meromorphic univalent functions with
positive coefficients associated with a generalized
q-Sălăgean operator
M. K. Aouf, A. O. Mostafa & F. Y. Al-Quhali
To cite this article: M. K. Aouf, A. O. Mostafa & F. Y. Al-Quhali (2020) Class of meromorphic
univalent functions with positive coefficients associated with a generalized q-
Sălăgean operator, Journal of Taibah University for Science, 14:1, 1544-1553, DOI:
10.1080/16583655.2020.1844994
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JOURNAL OF TAIBAH UNIVERSITY FOR SCIENCE
2020, VOL. 14, NO. 1, 1544–1553
https://doi.org/10.1080/16583655.2020.1844994
Class of meromorphic univalent functions with positive coefficients associated
with a generalized q-Sălăgean operator
M. K. Aouf, A. O. Mostafa and F. Y. Al-Quhali
Faculty of Science, Department of Mathematics, Mansoura University, Mansoura, Egypt
ABSTRACT
In this paper, using q−derivative operator, we define the class �∗
q,v(n,α,β , λ) of meromorphic
univalent functions and obtain coefficient estimates, distortion, closure theorems, modified
Hadamardproducts, some radii for functions in this class and family of integral operators. Further,
we determine partial sums results.
ARTICLE HISTORY
Received 19 June 2020
Revised 22 September 2020
Accepted 20 October 2020
KEYWORDS
Univalent function;
meromorphic function;
q-derivative; distortion;
modified Hadamard
products; partial sums
2010MATHEMATICS
SUBJECT CLASSIFICATION
30C45
1. Introduction
Let � be the class of analytic univalent meromorphic
functions of the form:
f (z) = 1
z
+
∞∑
k=1
akz
k ,
z ∈ U
∗ = {z : z ∈ C : 0 < |z| < 1} = U\ {0} , (1)
�∗(α) be the subclass of starlike functions of order α
satisfying
− Re
{
zf
′
f
}
> α (0 ≤ α < 1) (2)
and �c(α) be the subclass of convex functions of order
α satisfying (see [1–5]) (also see [6, 7]):
− Re
{
1 + zf
′′
f′
}
> α (0 ≤ α < 1) . (3)
Let �v be the subclass of � of the form
f (z) = 1
z
+
∞∑
k=1
akz
k (ak ≥ 0). (4)
For f ∈ �, Tang et al. [8] defined the q−derivative
∂q(f (z)) by (for analytic functions see [9–11]):
∂qf (z) = f (z) − (qz)
(1 − q)z
= − 1
qz2
+
∞∑
k=1
[
k
]
q akz
k−1 (0 < q < 1), (5)
[i]q = 1 − qi
1 − q
. (6)
If q → 1−, [i]q → i and
lim
q→1−
∂qf = f
′
. (7)
For f (z) ∈ �, n ∈ N0 = N ∪ {0},N = {1, 2, . . .}, λ ≥
0, 0 < q < 1, let :
∂0λ,qf = f , (8)
∂1λ,qf = (1 − λ)f + λ
z
∂q(z
2f ) (9)
...
∂nλ,qf = (1 − λ)(∂n−1
λ,q f ) + λ
z
∂q(z
2∂n−1
λ,q f ) (n ∈ N ), (10)
that is,
∂nλ,qf (z) = 1
z
+
∞∑
k=1
σ n
q (k, λ)akz
k (n ∈ N0), (11)
where
σ n
q (k, λ) = [
1 + λ
(
[k + 2]q − 1
)]n . (12)
CONTACT M. K. Aouf mkaouf127@yahoo.com Faculty of Science, Department of Mathematics, Mansoura University, Mansoura 35516, Egypt
© 2020 The Author(s). Published by Informa UK Limited, trading as Taylor & Francis Group.
This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted
use, distribution, and reproduction in any medium, provided the original work is properly cited.
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JOURNAL OF TAIBAH UNIVERSITY FOR SCIENCE 1545
We note that
(i) limq−→1− ∂nλ,qf (z) = D∗n
λ f (z) = 1
z +∑∞
k=1[1 + λ
(k + 1)]nakzk ;
(ii) limq−→1− ∂n1,qf (z) = D∗nf (z) = 1
z +∑∞
k=1(k + 2)n
akzk (see [12, 13] with p = 1).
From (11), we can obtain
λq2z∂q(∂
n
λ,qf (z)) = ∂n+1
λ,q f (z) − (λq + 1)∂nλ,qf (z)(λ > 0).
(13)
Definition 1.1: A function f ∈ � is in the class�∗
q(n,α,
β , λ) if ∣∣∣∣∣∣∣
qz∂q(∂nλ,qf (z))
∂nλ,qf (z)
+ 1
qz∂q(∂nλ,qf (z))
∂nλ,qf (z)
− 1 + 2α
∣∣∣∣∣∣∣ < β , (14)
for some 0 < β ≤ 1, 0 ≤ α < 1 and 0 < q < 1.
Let
�∗
q,v(n,α,β , λ) = �∗
q(n,α,β , λ) ∩ �v .
We note that
(i) limq→1− �∗
q,v(0,α,β , λ) = �∗
v (α,β) (see Mogra
et al. [14] and Aouf [[2], with A = −1 and B =
v = 1]);
(ii)
�∗
q,v(0,α,β , 1) = �∗
q,v(α,β)
= f ∈ � :
∣∣∣∣∣∣
qz∂qf (z)
f (z) + 1
qz∂qf (z)
f (z) − 1 + 2α
∣∣∣∣∣∣ < β .
2. Coefficient estimates
Unless indicated, we assume that 0 ≤ α < 1, 0 < β ≤
1, λ ≥ 0, n ∈ N0, σ n
q (k, λ) given by (12) and 0 < q < 1.
Theorem 2.1: Let f ∈ �. If
∞∑
k=1
[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
σ n
q (k, λ) |ak|
≤ 2β(1 − α), (15)
then f ∈ �∗
q(n,α,β , λ).
Proof: If (15) holds, then we may show that
∣∣∣qz∂q(∂nλ,qf (z)) + ∂nλ,qf (z)
∣∣∣
− β
∣∣∣qz∂q(∂nλ,qf (z) + ∂nλ,qf (z)(2α − 1)
∣∣∣ ≤ 0. (16)
We have∣∣∣qz∂q(∂nλ,qf (z)) + ∂nλ,qf (z)
∣∣∣
− β
∣∣∣qz∂q(∂nλ,qf (z) + ∂nλ,qf (z)(2α − 1)
∣∣∣
=
∣∣∣∣∣
∞∑
k=1
(
q
[
k
]
q + 1
)
σ n
q (k, λ)akz
k
∣∣∣∣∣
− β
∣∣∣∣∣2(α − 1)
1
z
+
∞∑
k=1
(
q
[
k
]
q + (2α − 1)
)
× σ n
q (k, λ)akz
k
∣∣∣∣
≤
∞∑
k=1
[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
× σ n
q (k, λ) |ak| |z|k+1 − 2β(1 − α). (17)
Since (17) holds for all r = |z|, 0 < r < 1, we have
∞∑
k=1
[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
σ n
q (k, λ) |ak|
− 2β(1 − α) ≤ 0
by (15), which yields (16). �
Theorem 2.2: Let f∈ �v . Then f ∈ �∗
q,v(n,α,β , λ) ⇔
∞∑
k=1
[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
σ n
q (k, λ)ak
≤ 2β(1 − α). (18)
Proof: In view of Theorem 2.1, we need to prove that if
f ∈ �∗
q,v(n,α,β , λ), then
∣∣∣∣∣∣∣
qz∂q(∂nλ,qf (z))
∂nλ,qf (z)
+ 1
qz∂q(∂nλ,qf (z))
∂nλ,qf (z)
− 1 + 2α
∣∣∣∣∣∣∣
=
∣∣∣∣∣∣∣∣∣∣∣
∑∞
k=1
(
q
[
k
]
q + 1
)
σ n
q (k, λ)akzk
2(1 − α) 1z −∑∞
k=1
(
q
[
k
]
q + (2α − 1)
)
×σ n
q (k, λ)akzk
∣∣∣∣∣∣∣∣∣∣∣
< β .
Since Ref(z) ≤ |f(z)| for all z, then
Re
⎧⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎩
∑∞
k=1
(
q
[
k
]
q + 1
)
σ n
q (k, λ)akzk+1
2(1 − α) −∑∞
k=1
(
q
[
k
]
q + (2α − 1)
)
×σ n
q (k, λ)akzk+1
⎫⎪⎪⎪⎪⎪⎬
⎪⎪⎪⎪⎪⎭
< β .
(19)
Taking values of z on the real axis so that
qz∂q(∂nλ,qf (z))
∂nλ,qf (z)
is
real and letting z → 1−, we obtain (2.4). �
1546 M. K. AOUF ET AL.
Corollary 2.1: For f (z) ∈ �∗
q,v(n,α,β , λ),
ak ≤ 2β(1 − α)[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
σ n
q (k, λ)
,
k ≥ 1. (20)
The result is sharp for
f (z) = 1
z
+ 2β(1 − α)[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
×σ n
q (k, λ)
zk . (21)
3. Growth and distortion theorems
Theorem 3.1: For f ∈ �∗
q,v(n,α,β , λ) and 0 < |z| = r <
1,
|f (z)| ≥ 1
r
− 2β(1 − α)[
q(1 + β) + β(2α − 1) + 1
]
σ n
q (1, λ)
r,
(22)
and
|f (z)| ≤ 1
r
+ 2β(1 − α)[
q(1 + β) + β(2α − 1) + 1
]
σ n
q (1, λ)
r.
(23)
The sharpness is attained for
f (z) = 1
z
+ 2β(1 − α)[
q(1 + β) + β(2α − 1) + 1
]
σ n
q (1, λ)
z
at z = ir, r. (24)
Proof: Since for k ≥ 1,
[
q(1 + β) + β(2α − 1) + 1
]
σ n
q (1, λ)
∞∑
k=1
ak
≤
∞∑
k=1
[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
σ n
q (k,λ)ak
≤ 2β(1 − α),
then
∞∑
k=1
ak ≤ 2β(1 − α)[
q(1 + β) + β(2α − 1) + 1
]
σ n
q (1, λ)
. (25)
From (4) and (25), we have
|f (z)| ≥ 1
r
− r
∞∑
k=1
ak ≥ 1
r
− 2β(1 − α)[
q(1 + β) + β(2α − 1) + 1
]
σ n
q (1, λ)
r,
and
|f (z)| ≤ 1
r
+ r
∞∑
k=1
ak ≤ 1
r
+ 2β(1 − α)[
q(1 + β) + β(2α − 1) + 1
]
σ n
q (1, λ)
r.
�
Theorem3.2: Let f ∈ �∗
q,v(n,α,β , λ). Then for 0 < |z| =
r < 1,
∣∣∂qf (z)∣∣ ≥ 1
qr2
− 2β(1 − α)[
q(1 + β) + β(2α − 1) + 1
]
σ n
q (1, λ)
(26)
and∣∣∂qf (z)∣∣
≤ 1
qr2
+ 2β(1 − α)[
q(1 + β) + β(2α − 1) + 1
]
σ n
q (1, λ)
.
(27)
Equalities in (26) and (27) hold for
∂qf (z) = − 1
qz2
+ 2β(1 − α)[
q(1 + β) + β(2α − 1) + 1
]
σ n
q (1, λ)
at z = ir, r. (28)
Proof: Note that f ∈ �∗
q,v(n,α,β , λ) ⇔ ∂qf ∈ �∗
q,v(n,α,
β , λ). Using Theorem 2.1, we have
[
q(1 + β) + β(2α − 1) + 1
]
σ n
q (1, λ)
∞∑
k=1
[
k
]
q ak
≤
∞∑
k=1
[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
σ n
q (k, λ)ak
≤ 2β(1 − α),
that is,
∞∑
k=1
[
k
]
q ak ≤ 2β(1 − α)[
q(1 + β) + β(2α − 1) + 1
]
σ n
q (1, λ)
.
(29)
It follows from (5) and (29) that
∣∣∂qf (z)∣∣ ≥ 1
qr2
−
∞∑
k=1
[
k
]
q ak ≥ 1
qr2
− 2β(1 − α)[
q(1 + β) + β(2α − 1) + 1
]
σ n
q (1, λ)
,
(30)
and
∣∣∂qf (z)∣∣ ≤ 1
qr2
+
∞∑
k=1
[
k
]
q ak ≤ 1
qr2
+ 2β(1 − α)[
q(1 + β) + β(2α − 1) + 1
]
σ n
q (1, λ)
.
(31)
This completes the proof. �
4. Closure theorems
Let fj be defined, for j = 1, 2, . . . ,m, by
fj(z) = 1
z
+
∞∑
k=1
ak,jz
k (ak,j ≥ 0). (32)
JOURNAL OF TAIBAH UNIVERSITY FOR SCIENCE 1547
Theorem 4.1: Let fj ∈ �∗
q,v(n,α,β , λ), j = 1, 2, . . . ,m.
Then
h(z) =
m∑
j=1
cjfj(z)
⎛
⎝cj ≥ 0,
m∑
j=1
cj ≤ 1
⎞
⎠ , (33)
also h(z) ∈ �∗
q,v(n,α,β , λ).
Proof: From (33),
h(z) = 1
z
+
∞∑
k=1
⎛
⎝ m∑
j=1
cjak,j
⎞
⎠ zk . (34)
Further, since fj ∈ �∗
q,v(n,α,β , λ), then
∞∑
k=1
[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
σ n
q (k, λ)ak,j
≤ 2β(1 − α). (35)
Hence
∞∑
k=1
[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
× σ n
q (k, λ)
⎛
⎝ m∑
j=1
cjak,j
⎞
⎠
=
m∑
j=1
cj
[ ∞∑
k=1
[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
× σ n
q (k, λ)ak,j
]
≤
⎛
⎝ m∑
j=1
cj
⎞
⎠ 2β (1 − α) = 2β (1 − α) , (36)
that is, h ∈ �∗
q,v(n,α,β , λ). �
Corollary 4.1: The class �∗
q,v(n,α,β , λ) is closed under
convex linear combination.
Proof: Let fj ∈ �∗
q,v(n,α,β , λ). Then we may show that
h(z) ∈ �∗
q,v(n,α,β , λ),
h(z) = μf1(z) + (1 − μ)f2(z) (0 ≤ μ ≤ 1). (37)
This accrues by taking m = 2, c1 = μ and c2 = 1 − μ
(0 ≤ μ ≤ 1) in Theorem 4.1. �
Theorem 4.2: Let f0(z) = 1
z and
fk(z) = 1
z
+ 2β(1 − α)[
q[k]q(1 + β) + β(2α − 1) + 1
]
σ n
q (k, λ)
× zk (k ≥ 1). (38)
Then f ∈ �∗
q,v(n,α,β , λ) ⇔
f (z) =
∞∑
k=0
μkfk(z), (39)
where μk ≥ 0 (k ≥ 0),
∑∞
k=0 μk = 1 and fk(z), k ≥ 1 are
the extreme points.
Proof: Suppose that
f (z) =
∞∑
k=0
μkfk(z) = 1
z
+
∞∑
k=1
× 2β(1 − α)[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
σ n
q (k, λ)
× μkz
k . (40)
Then
∞∑
k=1
[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
σ n
q (k, λ)
2β(1 − α)
· 2β(1 − α)[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
σ n
q (k, λ)
μk
=
∞∑
k=1
μk = 1 − μ0 ≤ 1. (41)
So by Theorem 2.2, f ∈ �∗
q,v(n,α,β , λ).
Conversely, let f ∈ �∗
q,v(n,α,β , λ). Then
ak ≤ 2β(1 − α)[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
σ n
q (k, λ)
(k ≥ 1). (42)
By setting
μk =
[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
σ n
q (k, λ)
2β(1 − α)
ak
(k ≥ 1), (43)
and
μ0 = 1 −
∞∑
k=1
μk , (44)
it can be seen that f can be written as in (39). �
5. Some radii of the class�∗
q,v(n,α,β,λ)
In the next theorem, we obtain the radii of close-to-
convexity, starlikeness and convexity.
Theorem 5.1: Let f ∈ �∗
q,v(n,α,β , λ). Then for 0 ≤ δ <
1, f is
(i) close -to- convex of order δ in |z| < r1,
r1 = r1(q,α,β , δ)
:= inf
k
⎡
⎢⎢⎢⎣
(1 − δ)[q[k]q(1 + β)
+β(2α − 1) + 1]σ n
q (k, λ)
2kβ(1 − α)
⎤
⎥⎥⎥⎦
1
(k+1)
. (45)
1548 M. K. AOUF ET AL.
(ii) starlike of order δ in |z| < r2,
r2 = r2(q,α,β , δ)
:= inf
k
⎡
⎢⎢⎢⎣
(1 − δ)[q
[
k
]
q (1 + β)
+β(2α − 1) + 1]σ n
q (k, λ)
2β(1 − α)(k + 2 − δ)
⎤
⎥⎥⎥⎦
1
(k+1)
. (46)
(iii) convex of order δ in |z| < r3,
r3 = r3(q,α,β , δ)
:= inf
k
⎡
⎢⎢⎢⎣
(1 − δ)[q
[
k
]
q (1 + β)
+β(2α − 1) + 1]σ n
q (k, λ)
2β(1 − α)k(k + 2 − δ)
⎤
⎥⎥⎥⎦
1
(k+1)
, (47)
f (z) given by (21) gives the sharpness.
Proof: To prove (i) it may be shown that∣∣∣z2f ′
(z) + 1
∣∣∣ ≤ 1 − δ for |z| < r1(q,α,β , ρ).
From (4), we have
∣∣∣z2f ′
(z) + 1
∣∣∣ ≤
∞∑
k=1
kak |z|k+1 .
Thus ∣∣∣z2f ′
(z) + 1
∣∣∣ ≤ 1 − δ,
if
∞∑
k=1
(
k
1 − δ
)
ak |z|k+1 ≤ 1. (48)
But, by Theorem 2.2 and (48) will be true if(
k
1 − δ
)
|z|k+1
≤
[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
σ n
q (k, λ)
2β(1 − α)
,
that is, if
|z| ≤
⎡
⎢⎢⎢⎣
(1 − δ)[q
[
k
]
q (1 + β)
+β(2α − 1) + 1]σ n
q (k, λ)
2kβ(1 − α)
⎤
⎥⎥⎥⎦
1
(k+1)
(k ≥ 1),
(49)
which gives (45).
To prove (ii) and (iii) it may be to show∣∣∣∣∣zf
′
(z)
f (z)
+ 1
∣∣∣∣∣ ≤ 1 − δ for |z| < r2, (50)
∣∣∣∣∣zf
′′
(z)
f ′
(z)
+ 2
∣∣∣∣∣ ≤ 1 − δ for |z| < r3, (51)
respectively, by using arguments as in proving (i), we
have the results. �
6. Modified Hadamard products
To prove this section, we will use the techniques of
Schild and Silverman [15].
Let fj bedefinedby (32). Then form = 2, themodified
Hadamard product is defined by
(f1 ∗ f2) = 1
z
+
∞∑
k=1
ak,1ak,2z
k = (f2 ∗ f1) . (52)
Theorem 6.1: If fj ∈ �∗
q,v(n,α,β , λ), then
(f1 ∗ f2) ∈ �∗
q,v(n, γ ,β , λ),
where
γ = 1 − 2 (q + 1) (1 + β)β (1 − α)2[
q(1 + β) + β(2α − 1) + 1
]2
σ n
q (1, λ)
+4β2 (1 − α)2
. (53)
The result is sharp.
Proof: We need to find the largest γ such that
∞∑
k=1
[
q
[
k
]
q (1 + β) + β(2γ − 1) + 1
]
σ n
q (k, λ)
2β(1 − γ )
ak,1ak,2
≤ 1. (54)
Since
∞∑
k=1
[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
σ n
q (k, λ)
2β (1 − α)
ak,j
≤ 1 (j = 1, 2), (55)
then Cauchy–Schwarz inequality yields
∞∑
k=1
[q[k]q(1 + β) + β(2α − 1) + 1]σ n
q (k, λ)
2β(1 − α)
√
ak,1ak,2
≤ 1. (56)
Thus it is sufficient to show that[
q
[
k
]
q (1 + β) + β(2γ − 1) + 1
]
σ n
q (k, λ)
2β(1 − γ )
ak,1ak,2
≤
[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
σ n
q (k, λ)
2β(1 − α)
× √
ak,1ak,2, (57)
that is, that
√
ak,1ak,2 ≤
[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
(1 − γ )[
q
[
k
]
q (1 + β) + β(2γ − 1) + 1
]
(1 − α)
(k ≥ 1). (58)
JOURNAL OF TAIBAH UNIVERSITY FOR SCIENCE 1549
Note that
√
ak,1ak,2 ≤ 2β (1 − α)[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
σ n
q (k, λ)
(k ≥ 1).
(59)
From (56) and (58), we need to prove that
2β (1 − α)[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
σ n
q (k, λ)
≤
[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
(1 − γ )[
q
[
k
]
q (1 + β) + β(2γ − 1) + 1
]
(1 − α)
(k ≥ 1), (60)
or, equivalently, that
γ = 1 −
2
(
q
[
k
]
q + 1
)
(1 + β)β (1 − α)2[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]2
×σ n
q (k, λ) + 4β2 (1 − α)2
(k ≥ 1).
(61)
Since
�q(k) = 1 −
2
(
q
[
k
]
q + 1
)
(1 + β)β (1 − α)2[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]2
×σ n
q (k, λ) + 4β2 (1 − α)2
, (62)
is an increasing function of k (k ≥ 1), letting k = 1
in (62), we obtain
γ ≤ �q(1) = 1 − 2 (q + 1) (1 + β)β (1 − α)2[
q(1 + β) + β(2α − 1) + 1
]2
×σ n
q (1, λ) + 4β2 (1 − α)2
, (63)
which is (53).
Finally, if
fj(z) = 1
z
+ 2β (1 − α)[
q(1 + β) + β(2α − 1) + 1
]
σ n
q (1, λ)
z,
(64)
we can see that the result is sharp. �
Theorem 6.2: Let f1 ∈ �∗
q,v(n,α,β , λ) and f2 ∈ �∗
q,v
(n, ρ,β , λ). Then
(f1 ∗ f2) ∈ �∗
q,v(n, ξ ,β , λ),
where
ξ = 1 − 2 (q + 1) (1 + β)β (1 − α) (1 − ρ)
[q(1 + β) + β(2α − 1) + 1][q(1 + β) + β(2ρ
−1) + 1]σ n
q (1, λ) + 4β2 (1 − α) (1 − ρ)
.
(65)
The result is the best possible for
f1(z) = 1
z
+ 2β (1 − α)[
q(1 + β) + β(2α − 1) + 1
]
σ n
q (1, λ)
z,
f2(z) = 1
z
+ 2β (1 − ρ)[
q(1 + β) + β(2ρ − 1) + 1
]
σ n
q (1, λ)
z.
(66)
Proof: Proceeding as in the proof of Theorem 6.1, we
get
ξ ≤ Bq(k)
= 1 − 2
(
q[k]q + 1
)
(1 + β)β (1 − α) (1 − ρ)[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
×
[
q
[
k
]
q (1 + β) + β(2ρ − 1) + 1
]
×σ n
q (1, λ) + 4β2 (1 − α) (1 − ρ)
(k ≥ 1).
(67)
Since Bq(k) is an increasing function of k (k ≥ 1), setting
k = 1 in (67), we get
ξ ≤ Bq(1)
= 1 − 2 (q + 1) (1 + β)β (1 − α) (1 − ρ)
[q(1 + β) + β(2α − 1) + 1][q(1 + β)
+β(2ρ − 1)+ 1]σ n
q (1, λ)+ 4β2(1− α)(1− ρ)
.
(68)�
Corollary 6.1: Let fj ∈ �∗
q,v(n,α,β , λ). Then
(f1 ∗ f2 ∗ f3) ∈ �∗
q,v(n, δ,β , λ),
where
δ = 1 − 4 (q + 1) (1 + β)β2 (1 − α)3[
q(1 + β) + β(2α − 1) + 1
]3
σ 2n
q (1, λ)
+8β3 (1 − α)3
, (69)
fj given by (64) gives the sharpness.
Proof: From Theorem 6.1, we have (f1 ∗ f2) ∈ �∗
q,v(n,
γ ,β , λ), where γ is given by (53). By using Theorem 6.2,
we get (f1 ∗ f2 ∗ f3) ∈ �∗
q,v(n, δ,β , λ), where
δ = 1 −
2
(
q
[
k
]
q + 1
)
(1 + β)β (1 − α) (1 − γ )[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
×
[
q
[
k
]
q (1 + β) + β(2γ − 1) + 1
]
×σ n
q (k, λ) + 4β2 (1 − α) (1 − γ )
= 1 − 4 (q + 1) (1 + β)β2 (1 − α)3[
q(1 + β) + β(2α − 1) + 1
]3
σ 2n
q (1, λ)
+8β3 (1 − α)3
.
�
1550 M. K. AOUF ET AL.
Theorem 6.3: Let fj ∈ �∗
q,v(n,α,β , λ)(j = 1, 2). Then
h(z) ∈ �∗
q,v(n, τ ,β , λ),
h(z) = 1
z
+
∞∑
k=1
(a2k,1 + a2k,2)z
k , (70)
and
τ = 1 − 4 (q + 1) (1 + β)β (1 − α)2[
q(1 + β) + β(2α − 1) + 1
]2
σ n
q (1, λ)
+8β2 (1 − α)2
, (71)
fj (j = 1, 2) defined by (64) gives the sharpness.
Proof: By (18), we have
∞∑
k=1
⎡
⎣
[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
σ n
q (k, λ)
2β (1 − α)
⎤
⎦
2
a2k,j
≤
⎡
⎣ ∞∑
k=1
[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
σ n
q (k, λ)
2β (1 − α)
ak,j
⎤
⎦
2
≤ 1 (j = 1, 2). (72)
It follows that
∞∑
k=1
1
2
⎡
⎣
[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
σ n
q (k, λ)
2β (1 − α)
⎤
⎦
2
× (
a2k,1 + a2k,2
) ≤ 1. (73)
Therefore, we need to find the largest τ such that[
q
[
k
]
q (1 + β) + β(2τ − 1) + 1
]
σ n
q (k, λ)
2β (1 − τ)
≤ 1
2
[
[q[k]q(1 + β) + β(2α − 1) + 1]σ n
q (k, λ)
2β (1 − α)
]2
(k ≥ 1), (74)
that is,
τ ≤ 1 −
4
(
q
[
k
]
q + 1
)
(1 + β)β (1 − α)2[(
q
[
k
]
q (1 + β) + β(2α − 1) + 1
)]2
×σ n
q (k, λ) + 8β2 (1 − α)2
(k ≥ 1). (75)
Since
Qq(k) = 1 −
4
(
q
[
k
]
q + 1
)
(1 + β)β (1 − α)2[(
q
[
k
]
q (1 + β) + β(2α − 1) + 1
)]2
×σ n
q (k, λ) + 8β2 (1 − α)2
,
(76)
is an increasing function of k (k ≥ 1), then
τ ≤ Qq(1) = 1 − 4 (q + 1) (1 + β)β (1 − α)2[
q(1 + β) + β(2α − 1) + 1
]2
×σ n
q (1, λ) + 8β2 (1 − α)2
, (77)
and the theorem follows. �
Theorem 6.4: Let f1(z) = 1
z +∑∞
k=1 ak,1z
k ∈ �∗
q,v(n,α,
β , λ) and f2(z) = 1
z +∑∞
k=1 |ak,2|zk with |ak,2| ≤ 1, k =
1, 2, . . .. Then (f1 ∗ f2) ∈ �∗
q,v(n,α,β , λ).
Proof: Since
∞∑
k=1
[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
σ n
q (k, λ)
∣∣ak,1ak,2∣∣
=
∞∑
k=1
[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
× σ n
q (k, λ)
∣∣ak,2∣∣ ak,1
≤
∞∑
k=1
[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
σ n
q (k, λ)ak,1
≤ 2β(1 − α),
by Theorem 2.2, it follows that (f1 ∗ f2)(z) ∈ �∗
q,v(n,α,
β , λ). �
7. A family of integral operators
In this section, a family of integral operators for func-
tions belonging to the class �∗
q,v(n,α,β , λ).
Theorem 7.1: Let f (z) ∈ �∗
q,v(n,α,β , λ). Then Fc(z)
defined by
Fc(z) = c
∫ 1
0
ucf (uz)du (c > 0), (78)
is in�∗
q,v(δ),with
δ = δ(q, n,α,β , λ, c) = 1
− 4β(1 − α)c
2β(1 − α)c + (c + 2)[
q(1 + β) + β(2α − 1) + 1
]
σ n
q (1, λ)
.
The result is best possible for f (z) = 1
z
+ 2β(1−α)
[q(1+β)+β(2α−1)+1]σ n
q (1,λ)
z.
Proof: Suppose f (z)= 1
z +∑∞
k=1 akz
k ∈ �∗
q,v(n,α,β , λ),
we have
Fc(z) = c
∫ 1
0
ucf (uz)du = 1
z
+
∞∑
k=1
c
k + c + 1
akz
k .
(79)
It is sufficient to show that
∞∑
k=1
k + δ
1 − δ
c
k + c + 1
akz
k ≤ 1. (80)
Since f (z) ∈ �∗
q,v(n,α,β , λ), we have
∞∑
k=1
[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
σ n
q (k, λ)
2β(1 − α)
ak ≤ 1.
(81)
JOURNAL OF TAIBAH UNIVERSITY FOR SCIENCE 1551
Thus (80) will be satisfied if
(k + δ) c
(1 − δ) (k + c + 1)
≤
[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
σ n
q (k, λ)
2β(1 − α)
for each k,
or
δ ≤
(k + c + 1)
[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
σ n
q (k, λ) − 2β(1 − α)ck
2β(1 − α)c + (k + c + 1)[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
σ n
q (k, λ)
.
(82)
Since the right side of (82) is an increasing function of k,
by putting k = 1 in (82),
δ ≤ 1 − 4β(1 − α)c
2β(1 − α)c + (c + 2)[
q(1 + β) + β(2α − 1) + 1
]
σ n
q (1, λ)
.
Hence the theorem. �
Remark 7.1: Putting q → 1− and n = 0 in the above
results, we obtain the results of Mogra et al. [14].
8. Partial sums
Let f of the form (1),
fs(z) = 1
z
+
s∑
k=1
akz
k .
and
�q,k(α,β)
=
[
q
[
k
]
q (1 + β) + β(2α − 1) + 1
]
σ n
q (k, λ). (83)
In the next theorem, we determine bounds for f (z)
fs(z)
.
Theorem 8.1: If f ∈ �, satisfies (15), then
Re
(
f(z)
fs(z)
)
≥ �q,s+1 − 2β(1 − α)
�q,s+1
, (84)
where
�q,k ≥
{
2β(1 − α), if k = 1, 2, . . . , s
�q,s+1, if k ≥ s + 1.
(85)
The result (84) is sharp for
f (z) = 1
z
+ 2β(1 − α)
�q,s+1
zs+1. (86)
Proof: Definew(z) by
1 + w(z)
1 − w(z)
= �q,s+1
2β (1 − α)
[
f (z)
fs(z)
− �q,s+1 − 2β (1 − α)
�q,s+1
]
=
1 +∑s
k=1 akz
k+1 +
(
�q,s+1
2β(1−α)
)
∑∞
k=s+1 akz
k+1
1 +∑s
k=1 akz
k+1
. (87)
It must be shown that |w| ≤ 1. Now (87) yields
w(z)=
(
�q,s+1
2β(1−α)
)∑∞
k=s+1 akz
k+1
2 + 2
∑s
k=1 akz
k+1+
(
�q,s+2
2β(1−α)
)∑∞
k=s+1 akz
k+1
.
Hence
|w(z)| ≤
(
�q,s+1
2β(1−α)
)∑∞
k=s+1 |ak|
2 − 2
∑s
k=1 |ak| −
(
�q,s+1
2β(1−α)
)∑∞
k=s+1 |ak|
.
Now |w| ≤ 1 if and only if
(
�q,s+1
β (1 − α)
) ∞∑
k=s+1
|ak| ≤ 2 − 2
s∑
k=1
|ak| ,
or, equivalently,
s∑
k=1
|ak| +
∞∑
k=m+1
�q,s+1
2β (1 − α)
|ak| ≤ 1.
From 15, it is sufficient to show that
s∑
k=1
|ak| +
∞∑
k=s+1
�q,s+1
2β (1 − α)
|ak| ≤
∞∑
k=1
�q,k
2β (1 − α)
|ak| ,
that is
s∑
k=1
(
�q,k − 2β (1 − α)
2β (1 − α)
)
|ak|+
∞∑
k=s+1
(
�q,k − �q,s+1
2β (1 − α)
)
× |ak| ≥ 0. (88)
For z = reiπ�s+2 we have
f (z)
fs(z)
= 1 + 2β (1 − α)
�q,s+1
zs+2 → 1 − 2β (1 − α)
�q,s+1
= �q,s+1 − 2β (1 − α)
�q,s+1
where r → 1−,
which shows that f (z) given by (86) gives the sharpness.
�
In the next theorem, we determine bounds for fs(z)
f (z) .
Theorem 8.2: If f ∈ �, satisfies (15), then
Re
(
fs(z)
f(z)
)
≥ �q,s+1
�q,s+1 + 2β (1 − α)
, (89)
where�q,s+1 defined by (83) and satisfies (85) and f given
by (86) gives the sharpness.
1552 M. K. AOUF ET AL.
Proof: The proof follows by defining
1 + w
1 − w
= �q,s+1 + 2β (1 − α)
2β (1 − α)
×
[
fs(z)
f (z)
− �q,s+1
�q,s+1 + 2β (1 − α)
]
.
The reminder part is as in Theorem 8.1. So, we omit it.
�
In Theorem 8.3, we determine bounds for ∂qf (z)
∂qfs(z)
and
∂qfs(z)
∂qf (z)
.
Theorem 8.3: If f ∈ �, satisfies (15), then
Re
(
∂qf(z)
∂qfs(z)
)
≥ �q,s+1 − 2 [s + 1]q β (1 − α)
�q,s+1
(90)
and
Re
(
∂qfs(z)
∂qf(z)
)
≥ �q,s+1
�q,s+1 + 2 [s + 1]q β (1 − α)
, (91)
where�q,s+1 ≥ 2[s + 1]qβ(1 − α) and
�q,k ≥
⎧⎪⎨
⎪⎩
2
[
k
]
q β (1 − α) , if k = 1, 2, . . . , s[
k
]
q
(
�q,s+1
[s + 1]q
)
, if k ≥ s.
(92)
f given by (86) gives the sharpness.
Proof: We write
1 + w
1 − w
= �q,s+1
2 [s + 1]q β (1 − α)
×
⎡
⎢⎢⎢⎣ ∂qf (z)
∂qfs(z)
−
⎛
⎜⎜⎜⎝
�q,s+1 − 2 [s + 1]q
β (1 − α)
�q,s+1
⎞
⎟⎟⎟⎠
⎤
⎥⎥⎥⎦ ,
where
w =
(
�q,s+1
2[s+1]qβ(1−α)
)∑∞
k=s+1
[
k
]
q akz
k+1
2 + 2
∑s+1
k=1
[
k
]
q akz
k+1 +
(
�q,s+1
2 [s + 1]q β (1 − α)
)
∑∞
k=s+1
[
k
]
q akz
k+1
.
Now |w| ≤ 1 if and only if
s∑
k=1
[
k
]
q |ak|+
(
�q,s+1
2 [s + 1]q β (1 − α)
) ∞∑
k=s+1
[
k
]
q |ak|≤1.
From (15), it is sufficient to show that
s∑
k=1
[
k
]
q |ak| +
(
�q,s+1
2 (s + 1) β (1 − α)
) ∞∑
k=s+1
[
k
]
q |ak|
≤
∞∑
k=1
�q,k
2β (1 − α)
|ak| ,
that is
s∑
k=1
(
�q,k − 2
[
k
]
q β (1 − α)
2β (1 − α)
)
|ak|
+
∞∑
k=s+1
(
[s + 1]q �q,k − [
k
]
q �q,s+1
2 [s + 1]q β (1 − α)
)
|ak| ≥ 0.
To prove (91), define the functionw by
1 + w
1 − w
= 2 [s + 1]q β (1 − α) + �q,s+1
2 [s + 1]q β (1 − α)
×
[
∂qfs(z)
∂qf (z)
− �q,s+1
2 [s + 1]q β (1 − α) + �n
q,s+1
]
,
and by similar arguments in the first part, we get the
desired result. �
Remark 8.1: Putting q → 1− and n = 0 in Section 8,
we get new results for the class �∗
v (α,β).
Remark 8.2: Putting n = 0 in our results, we have cor-
responding ones for the class �∗
q,v(α,β).
Acknowledgments
The authors express their sincere thanks to the referees for
their valuable comments and suggestions.
Disclosure statement
No potential conflict of interest was reported by the
authors.
References
[1] Aouf MK. On the coefficients of some meromorphic
classes of order α and type β . Rend. Math., Ser. (7).
1988;8:211–221.Roma.
[2] Aouf MK. A certain subclass of meromorphic starlike
functions with positive coefficients. Rend Mat Appl.
1989;9(7):225–235.
[3] Clunie J. On meromorphic Schlicht functions. J London
Math Soc. 1959;34:215–216.
[4] Miller JE. Convex meromorphic mapping and related
functions. Proc Amer Math Soc. 1970;25:220–228.
[5] Pommerenke CH. On meromorphic starlike functions.
Pacific J Math. 1963;13:221–235.
[6] Došenović T, Kopellaar H, Radenović S. On some known
fixed point results in the complex domain, survey.
Vojnotehnički glasnik/Military Technical Courier. 2018;
66(3):563–579.
[7] Hariri P, Klén R, Vuorinen M. Conformally invariant
metrics and quasiconformal mappings. Springer; 2020.
(Springer Monographs in Mathematics).
[8] Tang H, Zayed HM, Mostafa AO, et al. Fekete-Szeg
ö problems for certain classes of meromorphic func-
tions using q-derivative operator. J Math Res Appl.
2018;38(3):236–246.
[9] Gasper G, Rahman M. Basic hypergeometric series. Cam-
bridge: Cambridge Univ. Press; 1990.
JOURNAL OF TAIBAH UNIVERSITY FOR SCIENCE 1553
[10] Seoudy TM, Aouf MK. Convolution properties for cer-
tain classes of analytic functions defined by q−derivative
operator. Abstract and Appl Anal. 2014;2014:1–7.
[11] SeoudyTM,AoufMK.Coefficient estimatesof newclasses
of q-convex functions of complex order. J Math Ineq.
2016;10(1):135–145.
[12] Aouf MK, Hossen HM. New criteria for meromorphic p-
valent starlike functions. Tsukuba J Math. 1993;17(2):
481–486.
[13] Aouf MK, Seoudy TM. Subordination and superor-
dination of certain linear operator on meromorphic
function. Ann Univ Mariae Curie-Sklodowska Sec. A.
2010;64(1):1–16.
[14] Mogra ML, Reddy TR, Juneja OP. Meromorphic univa-
lent functionswithpositive coefficients. Bull AustralMath
Soc. 1985;32:161–176.
[15] Schild A, Silverman H. Convolution of univalent func-
tions with negative coefficients. Ann Univ Mariae Curie-
Sklodowska Sect A. 1975;29:99–106.
	1. Introduction
	2. Coefficient estimates
	3. Growth and distortion theorems
	4. Closure theorems
	5. Some radii of the class q,v(n,,,)
	6. Modified Hadamard products
	7. A family of integral operators
	8. Partial sums
	Acknowledgments
	Disclosure statement
	References
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