International Journal of Analysis and Applications

Volume 16, Number 2 (2018), 239-253

URL: https://doi.org/10.28924/2291-8639

DOI: 10.28924/2291-8639-16-2018-239

NEW SUBCLASS OF ANALYTIC FUNCTIONS IN CONICAL DOMAIN ASSOCIATED

WITH RUSCHEWEYH q-DIFFERENTIAL OPERATOR

SHAHID KHAN1,∗, SAQIB HUSSAIN2, MUHAMMAD ASAD ZAIGHUM1, MUHAMMAD

MUMTAZ KHAN3

1Department of Mathematics, Riphah International University Islamabad, Pakistan

2Department of Mathematics, COMSATS Institute of Information Technology, Abbottabad, Pakistan

3Department of Public Health Riphah University of Haripur, Pakistan

∗Corresponding author: shahidmath761@gmail.com

Abstract. In this paper, we consider a new class of analytic functions which is defined by means of a

Ruscheweyh q-differential operator. We investigated some new results such as coefficients inequalities and

other interesting properties of this class. Comparison of new results with those that were obtained in earlier

investigation are given as Corollaries.

1. Introduction

Let A denote the class of functions f analytic in the open unit disk

E = {z : z ∈ C and |z| < 1}

and satisfying the normalization condition

f (0) = 0 and f′ (0) = 1.

Received 2017-09-25; accepted 2017-12-07; published 2018-03-07.

2010 Mathematics Subject Classification. Primary 05A30, 30C45; Secondary 11B65, 47B38.

Key words and phrases. analytic functions; Ruscheweyh q-differential operator; q-derivative operator; conic domains.

c©2018 Authors retain the copyrights

of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.

239

https://doi.org/10.28924/2291-8639
https://doi.org/10.28924/2291-8639-16-2018-239


Int. J. Anal. Appl. 16 (2) (2018) 240

Thus, the functions in A are represented by the Taylor-Maclaurin series expansion given by

f(z) = z +

∞∑
n=2

anz
n, (z ∈ E) . (1.1)

Let S be the subset of A consisting of the functions that are univalent in E. Given functions f, g ∈A, f is

said to be subordinate to g in E, denoted by

f ≺ g or f (z) ≺ g (z) (z ∈ E) ,

if there exists a function w ∈P0 where

P0 = {w ∈A : w (0) = 0, and |w (z)| < 1 (z ∈ E)} ,

such that

f (z) = g (w (z)) (z ∈ E) .

If g is univalent in E, then it follows that

f (z) ≺ g (z) (z ∈ E) , ⇒ f (0) = 0 and f (E) ⊂ g (E) .

Kanas and Wísniowska [5, 6] introduced the conic domain Ωk,k ≥ 0 as

Ωk =

{
u + iv : u > k

√
(u− 1)2 + v2

}
.

We note that Ωk is a region in the right half-plane, symmetric with respect to real axis, and contains the

point (1, 0). More precisely for k = 0, Ω0 is the right half-plane, for 0 < k < 1, Ωk is an unbounded region

having boundary ∂Ωk, a rectangular hyperbola; for k = 1, Ω1 is still an unbounded region where ∂Ω1 is a

parabola, and for k > 1, Ωk is a bounded region enclosed by an ellipse. The extremal functions for these

conic regions are

pk (z) =




1+z
1−z , k = 0,

1 + 2
π2

(
log

1+
√
z

1−
√
z

)2
, k = 1,

1
1−k2 cosh

{(
2
π

arccos k
)

log
1+
√
z

1−
√
z

}
− k

2

1−k2 , 0 < k < 1,

1
k2−1 sin

(
π

2K(κ)

∫ u(z)√
κ

0
dt√

1−t2
√
1−κ2t2

)
+ k

2

k2−1, k > 1,

(1.2)

where

u(z) =
z −
√
κ

1 −
√
κz

(z ∈ E)

and κ ∈ (0, 1) is chosen such that k = cosh (πK′(κ)/(4K(κ))). Here K(κ) is Legender’s complete elliptic

integral of first kind and K′(κ) = K(
√

1 −κ2) and K′ (t) is the complementary integral of K (t) for details



Int. J. Anal. Appl. 16 (2) (2018) 241

see [1, 5, 6] and more recently [9, 12, 14]. If pk (z) = 1 + L1 (k) z + L2 (k) z
2 + ...,z ∈ E. Then it was shown

in [6] that for (1.2) one can have

L1 (k) =




2A2

1−k2 , 0 ≤ k < 1
8
π2
, k = 1,

π2

4K2(t)2(1+t)
√
t
, k > 1,

(1.3)

L2 (k) = D (k) L1 (k) ,

where

D (k) =




A2+2
3

, 0 ≤ k < 1
8
π2
, k = 1,

(4K(t))2(t2+6t+1)−π2

24K(t)2(1+t)
√
t

, k > 1,

(1.4)

with A = 2
π

arccos k.

Furthermore a function p is said to be in the class k −P [A,B] , if and only if

p (z) ≺
(A + 1) pk (z) − (A− 1)
(B + 1) pk (z) − (B − 1)

, k ≥ 0,

where pk is defined in (1.2) and −1 ≤ B < A ≤ 1. Geometrically, the function p ∈ k − P [A,B] takes all

values from the domain Ωk[A,B], −1 ≤ B < A ≤ 1, k ≥ 0 which is defined as:

Ωk[A,B] =

{
w : <

(
(B − 1) w − (A− 1)
(B + 1) w − (A + 1)

)
> k

∣∣∣∣(B − 1) w − (A− 1)(B + 1) w − (A + 1) − 1
∣∣∣∣
}
,

or equivalently Ωk[A,B] is a set of numbers w = u + iv such that

[(
B2 − 1

)(
u2 + v2

)
− 2 (AB − 1) u +

(
A2 − 1

)]2
> k

[(
−2 (B + 1)

(
u2 + v2

)
+ 2 (A + B + 2) u− 2 (A + 1)

)2
+ 4 (A−B)2 v2

]
.

This domain represents the conic type regains for detail see [11]. It can be easily seen that 0 −P [A,B] =

P [A,B] introduced in [4] and k −P [1,−1] = P (pk) introduced in [5].

We now recall some basic concept details of the q-calculus which are used in this paper. Throughout this

paper we assume q to be a fixed number between 0 and 1.

For any non-negative integer n, the q-integer number n, [n,q] is defined by:

[n,q] =
1 −qn

1 −q
= 1 + q + ... + qn−1, [0,q] = 0. (1.5)

The q-number shifted factorial is defined by [0,q]! = 1 and [n,q]! = [1,q] [2,q] ... [n,q] . Clearly, lim
q→1

[n,q] = n

and lim
q→1

[n,q]! = n!. In general we will denote [t,q] = 1−q
t

1−q also for a non-integer number. Let f ∈A, and let

the q-derivative operator or q-difference operator be defined by

∂qf (z) =
f (qz) −f (z)

(q − 1) z
(z ∈ E) .



Int. J. Anal. Appl. 16 (2) (2018) 242

It is easy to observed that for n ∈ N := {1, 2, 3, ...} and z ∈ E

∂qz
n = [n,q] zn−1.

Let the q-generalized Pochhammer symbol be defined as

[t,q]n = [t,q] [t + 1,q] [t + 2,q] ... [t + n− 1,q] ,

and for t > 0 let the q-gamma function be defined as

Γq (t + 1) = [t] Γq (t) and Γq (1) = 1.

The study of opretors play in important role in Geomatric Functions Theory. Several diffierential and

integral operators were introduced and studied, see for example [2, 3, 13]. Kannas et al. define Ruscheweyh

q-differential operator as follow:

Definition 1.1. [7] For f ∈A, let the Ruscheweyh q-differential operator be defined as follows:

Rλqf(z) = f(z) ∗Fq,λ+1(z), (z ∈ E, λ > −1) (1.6)

where

Fq,λ+1(z) = z +

∞∑
n=2

Γq(n + λ)

[n− 1,q]!Γq(1 + λ)
zn,

= z +

∞∑
n=2

[λ + 1,q]n−1
[n− 1,q]!

zn,

= z +

∞∑
n=2

ϕn−1z
n. (1.7)

Where

ϕn−1 =
Γq(n + λ)

[n− 1,q]!Γq(1 + λ)
=

[λ + 1,q]n−1
[n− 1,q]!

.

From (1.6) we obtain that

R0qf(z) = f(z), R
1
qf(z) = z∂qf(z)

and

Rmq f(z) =
z∂mq (z

m−1f(z))

[m,q]!
, (m ∈ N).

Making use of (1.6) and (1.7), the power series of Rλqf(z) is given by

Rλqf(z) = z +

∞∑
n=2

Γq(n + λ)

[n− 1,q]!Γq(1 + λ)
anz

n = z +
∞∑
n=2

[λ + 1,q]n−1
[n− 1,q]!

anz
n. (1.8)

Note that

lim
q→1

Fq,λ+1(z) =
z

(1 −z)λ+1



Int. J. Anal. Appl. 16 (2) (2018) 243

and

lim
q→1

Rλqf(z) = f(z) ∗
z

(1 −z)λ+1
.

Thus, we can say that Ruscheweyh q-differential operator reduces to the differential operator defined by

Ruscheweyh [16] in the case when q → 1. It is easy to check that

z∂ (Fq,λ+1(z)) =

(
1 +

[λ,q]

qλ

)
Fq,λ+2(z) −

[λ,q]

qλ
Fq,λ+1(z). (1.9)

Making use of (1.6), (1.9) and the properties of Hadamard product, we obtain the following equality

z∂
(
Rλqf(z)

)
=

(
1 +

[λ,q]

qλ

)
Rλ+1q f(z) −

[λ,q]

qλ
Rλqf(z). (1.10)

If q → 1, the equality (1.10) implies

z
(
Rλf(z)

)′
= (1 + λ) Rλ+1f(z) −λRλf(z).

which is the well known recurrent formula for Ruscheweyh differential operator.

Using Ruscheweyh differential operator various new classes of convex and starlike functions have been defined.

Now by using Ruscheweyh q-differential operator we introduce the following class of functions.

Definition 1.2. A function f(z) ∈ A is said to be in the class k−USq(λ,A,B,β), k ≥ 0,−1 ≤ B < A ≤ 1,

if and only if

<
(

(B − 1)G (z) − (A− 1)
(B + 1)G (z) − (A + 1)

)
> k

∣∣∣∣(B − 1)G (z) − (A− 1)(B + 1)G (z) − (A + 1) − 1
∣∣∣∣ ,

where

G (z) =
z∂qR

λ
qf(z)

Rλqf(z)
+ β

z2∂2qR
λ
qf(z)

Rλqf(z)
,

or equivalently

z∂qR
λ
qf(z)

Rλqf(z)
+ β

z2∂2qR
λ
qf(z)

Rλqf(z)
∈ k −P[A,B]. (1.11)

Remark 1.1. It is easily see that

lim
q→1−

k −USq(0,A,B, 0) = k −ST (A,B)

where k −ST (A,B) is a functions class, intrioduced and studied by Noor and Sarfraz [11].

Each of the following lemmas will be needed in our present investigation.

Lemma 1.1. [15] Let h(z) = 1 +
∑∞
n=1 cnz

n be subordinate to H(z) = 1 +
∑∞
n=1 Cnz

n. If H(z) is univalent

in E and H(E) is convex, then

|cn| ≤ |C1| , n ≥ 1.



Int. J. Anal. Appl. 16 (2) (2018) 244

Lemma 1.2. ( [8], [10]) If q(z) = 1 + c1z + c2z
2+... is an analytic function with positive real part in E,

then

∣∣c2 −vc21∣∣ ≤ 2 max{1, |2v − 1|} .
The result is sharp for the functions

q(z) =
1 + z2

1 −z2
, or q(z) =

1 + z

1 −z
.

Lemma 1.3. [8] Let the function w ∈ E be given by

w(z) = c1z + c2z
2 + ... z ∈ E.

Then for every complex number v,

∣∣c2 −vc21∣∣ ≤ 1 + (|v|− 1) |c1|2 .
Lemma 1.4. [11] Let k ∈ [0,∞) be a fixed and

qk(z) =
(A + 1)pk(z) − (A− 1)
(B + 1)pk(z) − (B − 1)

,

then

qk(z) = 1 + H1(k)z + H2(k)z
2 + ...,z ∈ E.

and

H1 := H1(k) =
A−B

2
L1(k),

H2 := H2(k) =
A−B

4
{2D(k) − (B + 1)H1}L1(k)

where L1(k) and D(k) are defined in (1.3) and (1.4).

2. Main Results

Theorem 2.1. A function f ∈A and of the form (1.1) is in the class k−USq(λ,A,B,β), if it satisfies the

condition

∞∑
n=2




{2(k + 1){1 − [n,q] −β[n,q][n− 1,q]}

+ |{(B + 1) [n,q] + β[n,q][n− 1,q] − (A + 1)}|}


ϕn−1 |an| ≤ |B −A| . (2.1)

where −1 ≤ B < A ≤ 1, β ≥ 0 and k ≥ 0.



Int. J. Anal. Appl. 16 (2) (2018) 245

Proof. Assume (2.1) is hold, then it suffices to show that


k

∣∣∣∣∣∣
(B−1)

(
z∂qR

λ
q f(z)

Rλq f(z)
+β

z2∂2qR
λ
q f(z)

Rλq f(z)

)
−(A−1)

(B+1)

(
z∂qR

λ
q f(z)

Rλq f(z)
+β

z2∂2qR
λ
q f(z)

Rλq f(z)

)
−(A+1)

− 1

∣∣∣∣∣∣
−<


(B−1)

(
z∂qR

λ
q f(z)

Rλq f(z)
+β

z2∂2qR
λ
q f(z)

Rλq f(z)

)
−(A−1)

(B+1)

(
z∂qR

λ
q f(z)

Rλq f(z)
+β

z2∂2qR
λ
q f(z)

Rλq f(z)

)
−(A+1)

− 1






< 1,

We have 


k

∣∣∣∣∣∣
(B−1)

(
z∂qR

λ
q f(z)

Rλq f(z)
+β

z2∂2qR
λ
q f(z)

Rλq f(z)

)
−(A−1)

(B+1)

(
z∂qR

λ
q f(z)

Rλq f(z)
+β

z2∂2qR
λ
q f(z)

Rλq f(z)

)
−(A+1)

− 1

∣∣∣∣∣∣
−<


(B−1)

(
z∂qR

λ
q f(z)

Rλq f(z)
+β

z2∂2qR
λ
q f(z)

Rλq f(z)

)
−(A−1)

(B+1)

(
z∂qR

λ
q f(z)

Rλq f(z)
+β

z2∂2qR
λ
q f(z)

Rλq f(z)

)
−(A+1)

− 1







≤ (k + 1)

∣∣∣∣∣(B − 1)
(
z∂qR

λ
qf(z) + βz

2∂2qR
λ
qf(z)

)
− (A− 1)Rλqf(z)

(B + 1)
(
z∂qRλqf(z) + βz

2∂2qR
λ
qf(z)

)
− (A + 1)Rλqf(z)

− 1

∣∣∣∣∣
= 2(k + 1)

∣∣∣∣∣ R
λ
qf(z) −z∂qRλqf(z) −βz2∂2qRλqf(z)

(B + 1)
(
z∂qRλqf(z) + βz

2∂2qR
λ
qf(z)

)
− (A + 1)Rλqf(z)

∣∣∣∣∣
= 2(k + 1)

∑∞
n=2 (1 − [n,q] −β[n,q][n− 1,q]) ϕn−1anz

n

(B −A) z +
∑∞
n=2

{
(B + 1) [n,q]q + β[n,q][n− 1,q] − (A + 1)

}
ϕn−1anzn

≤ 2(k + 1)
∑∞
n=2 (1 − [n,q] −β[n,q][n− 1,q]) ϕn−1 |an|

|B −A|−
∑∞
n=2

{
(B + 1) [n,q]q + β[n,q][n− 1,q] − (A + 1)

}
ϕn−1 |an|

< 1 (by (2.1)) .

�

When A = 1 − 2α, B = −1,β = 0 with 0 ≤ α < 1, then we have the following known result, proved by

Kanas and Raducanu in [7].

Corollary 2.1. A function f ∈ A and of the form (1.1) is in the class k−USq(λ, 1−2α,−1) , if it satisfies

the condition
∞∑
n=2

{(k + 1) [n,q] −k −α}ϕn−1 |an| ≤ 1 −α.

When q → 1, β = 0,λ = 0, then we have the following known result, proved by Noor and Sarfraz [11].

Corollary 2.2. A function f ∈ A and of the form (1.1 is in the class k −ST (A,B), if it satisfies the

condition
∞∑
n=2

{2(k + 1)(n− 1) + |n(B + 1) − (A + 1)|} |an| ≤ |B −A| .



Int. J. Anal. Appl. 16 (2) (2018) 246

When q → 1, λ = 0,β = 0, A = 1−2α, B = −1 with 0 ≤ α < 1, then we have the following known result,

proved by Shams et-al. in [18].

Corollary 2.3. A function f ∈ A and of the form (1.1) is in the class k −UST (1 − 2α,−1), if it satisfies

the condition
∞∑
n=2

{n(k + 1) − (k + α)}|an| ≤ 1 −α,

where 0 ≤ α < 1 and k ≥ 0.

When λ = 0,β = 0, A = 1 − 2α, B = −1 with 0 ≤ α < 1 and k = 0, then we have the following known

result, proved by Selverman in [17].

Corollary 2.4. A function f ∈ A and of the form (1.1) is in the class 0 −UST (1 − 2α,−1), if it satisfies

the condition
∞∑
n=2

{n−α}|an| ≤ 1 −α, 0 ≤ α < 1.

Theorem 2.2. If f(z) ∈ k −USq(λ,A,B,β) and is of the form (1.1). Then

|an| ≤
n−2∏
j=0

(
|L1(k)(A−B) − 2[j,q]B|

2 [j + 1,q]{q + β[j + 2,q]}ϕj+1

)
, n ≥ 2, (2.2)

where L1(k) is defined by (1.3).

Proof. Let

z∂qR
λ
qf(z)

Rλqf(z)
+ β

z2∂2qR
λ
qf(z)

Rλqf(z)
= p(z). (2.3)

Then

p(z) ≺
(A + 1)pk(z) − (A− 1)
(B + 1)pk(z) − (B − 1)

= [(A + 1)pk(z) − (A− 1)] [(B + 1)pk(z) − (B − 1)]
−1

=
(A− 1)
(B − 1)

[
1 −

(A + 1)

(A− 1)
pk(z)

][
1 +

∑((B + 1)
(B − 1)

pk(z)

)n]
=

(A− 1)
(B − 1)

+

(
(A− 1)(B + 1)

(B − 1)2
−

(A + 1)

(B − 1)

)
(pk(z))

+

(
(A− 1)(B + 1)2

(B − 1)3
−

(A + 1)(B + 1)

(B − 1)2

)
(pk(z))

2
+ ....

By taking

pk(z) = 1 + L1(k)z + L2(k)z
2 + ...,

after some simplification, we obtain

p(z) ≺
∞∑
n=1

−2(B + 1)n−1

(B − 1)n
+

{
∞∑
n=1

−2n(A−B)(B + 1)n−1

(B − 1)n+1

}
L1(k) + ....



Int. J. Anal. Appl. 16 (2) (2018) 247

Now we see that the series
∑∞
n=1

−2(B+1)n−1
(B−1)n and

∑∞
n=1

−2n(A−B)(B+1)n−1
(B−1)n+1 are convergent and converge to 1

and A−B
2

respectively. Therefore,

p(z) ≺ 1 +
A−B

2
L1(k)z + ....

Now if p(z) = 1 +
∑∞
n=1 cnz

n, then by Lemma 1, we have

|cn| ≤
A−B

2
L1(k), n ≥ 1. (2.4)

Now from (2.3), we have

z∂qR
λ
qf(z) + βz

2∂2qR
λ
qf(z) = R

λ
qf(z)p(z),

which implies that

z +

∞∑
n=2

{[n,q] + β[n,q][n− 1,q]}ϕn−1anzn =

(
1 +

∞∑
n=1

cnz
n

)(
z +

∞∑
n=2

ϕn−1anz
n

)
.

Equating coefficients of zn on both sides, we have

[n− 1,q]{q + β[n,q]}ϕn−1an =
n−1∑
j=1

ϕj−1ajcn−j, a1 = 1.

This implies that

|an| ≤
1

[n− 1,q]{q + β[n,q]}ϕn−1

n−1∑
j=1

ϕj−1 |aj| |cn−j| , a1 = 1.

Using (2.4), we have

|an| ≤
(A−B) |L1(k)|

2[n− 1,q]{q + β[n,q]}ϕn−1

n−1∑
j=1

ϕj−1 |aj| , a1 = 1. (2.5)

Now we prove that

(A−B) |L1(k)|
2[n− 1,q]{q + β[n,q]}ϕn−1

n−1∑
j=1

ϕj−1 |aj| ≤
n−2∏
j=0

(
|L1(k)(A−B) − 2[j,q]B|
2 [j + 1,q]{q + β[j + 2,q]}

)
. (2.6)

For this we use the induction method

For n = 2, from (2.5), we have

|a2| ≤
(A−B) |L1(k)|
2{q + β[2,q]}ϕ1

.

From (2.2), we have

|a2| ≤
(A−B) |L1(k)|
2{q + β[2,q]}ϕ1

.

For n = 3 from (2.5), we have

|a3| ≤
(A−B) |L1(k)|

2[2,q]{q + β[3,q]}ϕ2
{1 + ϕ1a2}

≤
(A−B) |L1(k)|

2[2,q]{q + β[3,q]}ϕ2

{
1 +

(A−B) |L1(k)|
2{q + β[2,q]}

}
.



Int. J. Anal. Appl. 16 (2) (2018) 248

From (2.2), we have

|a3| ≤
(A−B) |L1(k)|
2{q + β[2,q]}ϕ1

{(
|(A−B)L1(k) − 2B|
2 [2,q]{q + β[3,q]}ϕ2

)}
≤

(A−B) |L1(k)|
2{q + β[2,q]}ϕ1

{(
(A−B) |L1(k)| + 2 |B|
2 [2,q]{q + β[3,q]}ϕ2

)}
≤

(A−B) |L1(k)|
2 [2,q]{q + β[3,q]}ϕ2

{
(A−B) |L1(k)|
2{q + β[2,q]}ϕ1

+
1

{q + β[2,q]}ϕ1

}
.

Let the hypothesis be true for n = m. From (2.4), we have

|am| ≤
(A−B) |L1(k)|

2[m− 1,q]{q + β[m,q]}ϕm−1

n−1∑
j=1

|aj| , a1 = 1

From (2.2), we have

|am| ≤
m−2∏
j=0

(
|L1(k)(A−B) − 2[j,q]B|

2 [j + 1,q]{q + β[j + 2,q]}ϕj+1

)
, n ≥ 2

≤
m−2∏
j=0

(
|L1(k)|(A−B) + 2[j,q]

2 [j + 1,q]{q + β[j + 2,q]}ϕj+1

)
, n ≥ 2.

By the induction hypothesis, we have

(A−B) |L1(k)|
2[m− 1,q]{q + β[m,q]}ϕm−1

m−1∑
j=1

ϕj−1 |aj| ≤
m−2∏
j=0

(
|L1(k)|(A−B) + 2[j,q]

2 [j + 1,q]{q + β[j + 2,q]}ϕj+1

)
. (2.7)

Multiplying both sides by (2.7)

(A−B) |L1(k)| + 2[m− 1,q]{q + β[m,q]}
2[m− 1,q]{q + β[m,q]}ϕm−1

,

we have

m−2∏
j=0

(
|L1(k)|(A−B) + 2[j,q]

2 [j + 1,q]{q + β[j + 1,q]}ϕj+1

)

≥
{

(A−B) |L1(k)| + 2[m− 1,q]{q + β[m,q]}
2[m− 1,q]{q + β[m,q]}ϕm−1

}
(A−B) |L1(k)|

2[m− 1,q]{q + β[m,q]}ϕm−1

m−1∑
j=1

ϕj−1 |aj|

=
(A−B) |L1(k)|

2[m− 1,q]{q + β[m,q]}ϕm−1



{

(A−B)|L1(k)|+2[m−1,q]{q+β[m,q]}
2[m−1,q]{q+β[m,q]}ϕm−1

∑m−1
j=1 ϕj−1 |aj|

}

+
∑m−1
j=1 ϕj−1 |aj|




≥
(A−B) |L1(k)|

2[m− 1,q]{q + β[m,q]}ϕm−1


|am| +

m−1∑
j=1

ϕj−1 |aj|




=
(A−B) |L1(k)|

2[m− 1,q]{q + β[m,q]}ϕm−1

m∑
j=1

ϕj−1 |aj| .



Int. J. Anal. Appl. 16 (2) (2018) 249

That is,

(A−B) |L1(k)|
2[m− 1,q]{q + β[m,q]}ϕm−1

m∑
j=1

ϕj−1 |aj| ≤
m−2∏
j=0

(
|L1(k)|(A−B) + 2[j,q]

2 [j + 1,q]{q + β[j + 1,q]}ϕj+1

)
.

which shows that inequality (2.7) is true for n = m + 1. Hence the required result. �

When q → 1, λ = 0 and β = 0, then we have the following known result, proved by Noor and Sarfraz

in [11].

Corollary 2.5. A function f ∈ A and of the form (1.1) is in the class k −ST [A,B] , if it satisfies the

condition

|an| ≤
n−2∏
j=0

(
|L1(k)(A−B) − 2jB|

2 (j + 1)

)
.

When λ = 0,A = 1,B = −1 and β = 0 then we have the following known result, proved by Kanas and

Wisniowska in [6].

Corollary 2.6. A function f ∈ A and of the form (1.1) is in the class k −UST [A,B] , if it satisfies the

condition

|an| ≤
n−2∏
j=0

(
|L1(k) + j|

(j + 1)

)
.

When λ = 0,A = 1−2α, β = 0,B = −1 with 0 ≤ α < 1, then we have the following known result, proved

by Shams et al. in [18].

Corollary 2.7. A function f ∈ A and of the form (1.1) is in the class SD(k,α), if it satisfies the condition

|an| ≤
n−2∏
j=0

(
|L1(k)(1 −α) + j|

(j + 1)

)
.

where 0 ≤ α < 1 and k ≥ 0.

When λ = 0, β = 0,k = 0, then T1(k) = 2 and we get the following known result, proved in [4]

Corollary 2.8. A function f ∈ A and of the form (1.1) is in the class S∗[A,B], if it satisfies the condition

|an| ≤
n−2∏
j=0

(
|(A−B) − jB|

(j + 1)

)
, − 1 ≤ B < A ≤ 1.

When λ = 0, β = 0, A = 1 − 2α, B = −1 with 0 ≤ α < 1 and k = 0, then we have the following known

result, proved by Selverman in [17].

Corollary 2.9. A function f ∈ A and of the form (1.1) is in the class S∗(α), if it satisfies the condition

|an| ≤

n−2∏
j=0

(j − 2α)

(n− 1)!
, 0 ≤ α < 1.



Int. J. Anal. Appl. 16 (2) (2018) 250

Theorem 2.3. Let −1 ≤ B < A ≤ 1and 0 ≤ k < ∞ be fixed and let f(z) ∈ k −USq(λ,A,B,β) and is of

the form (1.1) Then for a complex number µ.

∣∣a3 −µa22∣∣ ≤




(A−B)L1(k)
2[2,q]{q+[3,q]β}ϕ2

∣∣∣{2 + 2D(k)−(1+B)L1(k)2 [2 + 2D(k)−(1+B)L1(k)2
− (A−B)

2{q+β[2,q]}L1(k)
(

1 −µ ϕ2
(ϕ1)

2

)]}∣∣∣ , (µ > δ1) ,
(A−B)L1(k)

2[2,q]{q+[3,q]β}ϕ2
. (δ1 ≤ µ ≤ δ2) ,

(A−B)L1(k)
2[2,q]{q+[3,q]β}ϕ2

[
2D(k)−(1+B)L1(k)

2

(A−B)
2{q+β[2,q]}L1(k)

(
1 −µ ϕ2

(ϕ1)
2

)] (µ < δ2) .

(2.8)

Where

δ1 =
(ϕ1)

2

ϕ2(A−B)L1(k)




(q + β[2,q]){2 + 2D(k) − (1 + B)L1(k)}

+(A−B)L1(k)


 , (2.9)

δ2 =
(ϕ1)

2

ϕ2(A−B)L1(k)




(q + β[2,q]){2D(k) − (1 + B)L1(k) − 2}

+(A−B)L1(k)


 . (2.10)

and L1(k), D(k) are defined in (1.3) and (1.4).

Proof. If f(z) ∈ k −USq(λ,A,B,β) then it follows that

z∂qR
λ
qf(z)

Rλqf(z)
+ β

z2∂2qR
λ
qf(z)

Rλqf(z)
≺ qk(z) = 1 +

A−B
2

L1(k)z +
[2D(k) − (1 + B)L1(k)] (A−B)

4
L1(k)z

2 + ....

(2.11)

Now by the definition of subordination there exists a function w analytic in E with w(0) = 0 and |w(z)| < 1

such that

z∂qR
λ
qf(z)

Rλqf(z)
+ β

z2∂2qR
λ
qf(z)

Rλqf(z)
= 1 +

A−B
2

L1(k)w(z) +
[2D(k) − (1 + B)L1(k)] (A−B)

4
L1(k)w

2(z) + ....

(2.12)

Now from Lemma 3, equation (2.11) and equation (2.12), we have

a2 =
(A−B)L1(k)

2{q + β[2,q]}ϕ1
c1,

and

a3 =
(A−B)L1(k)

2 [2,q]{q + β[3,q]}ϕ2

{
c2 +

{
2D(k) − (1 + B)L1(k)

2
+

(A−B)
2{q + β[2,q]}

L1(k)

}
c21

}
.



Int. J. Anal. Appl. 16 (2) (2018) 251

Therefore

∣∣a3 −µa22∣∣ = (A−B)L1(k)2 [2,q]{q + β[3,q]}ϕ2
∣∣∣∣c2 +

{
2D(k) − (1 + B)L1(k)

2

+
(A−B)

2{q + β[2,q]}
L1(k)

(
1 −µ

ϕ2

(ϕ1)
2

)}
c21

∣∣∣∣∣ . (2.13)
This gives

∣∣a3 −µa22∣∣ = (A−B)L1(k)2 [2,q]{q + β[3,q]}ϕ2
∣∣∣∣c2 − c21 +

{
1 +

2D(k) − (1 + B)L1(k)
2

+
(A−B)

2{q + β[2,q]}
L1(k)

(
1 −µ

ϕ2

(ϕ1)
2

)}
c21

∣∣∣∣∣ . (2.14)
Suppose that µ > δ1, then using the estimate

∣∣c2 − c21∣∣ ≤ 1 from Lemma 3 and the well known estimate
|c1| ≤ 1 of the Schwarz lemma, we obtain∣∣a3 −µa22∣∣ ≤ (A−B)L1(k)2 [2,q]{q + β[3,q]}ϕ2

∣∣∣∣
{

2 +
2D(k) − (1 + B)L1(k)

2

−
(A−B)

2{q + β[2,q]}
L1(k)

(
1 −µ

ϕ2

(ϕ1)
2

)∣∣∣∣∣ . (2.15)
The inequality (2.15) is our required assertion (2.8) for µ > δ1. On the other hand if µ < δ2, then (2.13)

gives

∣∣a3 −µa22∣∣ ≤ (A−B)L1(k)2 [2,q]{q + β[3,q]}ϕ2
[
|c2| +

{
2D(k) − (1 + B)T1(k)

2

+
(A−B)

2{q + β[2,q]}
L1(k)

(
1 −µ

ϕ2

(ϕ1)
2

)}
|c1|

2

]
.

Applying the estimates |c2| ≤ 1 −|c1|
2

of Lemma 3 and |c1| ≤ 1, we have∣∣a3 −µa22∣∣ ≤ (A−B)L1(k)2 [2,q]{q + β[3,q]}ϕ2
[{

2D(k) − (1 + B)T1(k)
2

+
(A−B)

2{q + β[2,q]}
L1(k)

(
1 −µ

ϕ2

(ϕ1)
2

)}]
.

This is the last inequality in (2.8). Finally if δ1 < µ < δ2, then∣∣∣∣∣2D(k) − (1 + B)L1(k)2 + (A−B)2{q + β[2,q]}L1(k)
(

1 −µ
ϕ2

(ϕ1)
2

)∣∣∣∣∣ ≤ 1.
Therefore (2.13), yields

∣∣a3 −µa22∣∣ ≤ (A−B)L1(k)2 [2,q]{q + β[3,q]}ϕ2
{
|c2| + |c1|

2
}
,

≤
(A−B)L1(k)

2 [2,q]{q + β[3,q]}ϕ2

{
1 −|c1|

2
+ |c1|

2
}
,

≤
(A−B)L1(k)

2 [2,q]{q + β[3,q]}ϕ2
.

We get the middle inequality in (2.8). This completes the proof. �



Int. J. Anal. Appl. 16 (2) (2018) 252

Theorem 2.4. Let 0 ≤ k < ∞, −1 ≤ B < A ≤ 1, be fixed and let f(z) ∈ k −USq(λ,A,B,β) and is of the

form (1.1) Then for a complex number µ.

∣∣a3 −µa22∣∣ ≤ (A−B)L1(k)2[2,q]{q + [3,q]β}ϕ2 max{1, |2v − 1|} ,
where v is given by (2.17).

Proof. From (2.13) we have

∣∣a3 −µa22∣∣ = (A−B)L1(k)2[2,q]{q + [3,q]β}ϕ2
∣∣∣∣c2 −

{
(1 + B)L1(k) − 2D(k)

2

−
(A−B)

2{q + β[2,q]}
L1(k)

(
1 −µ

ϕ2

(ϕ1)
2

)}
c21

∣∣∣∣∣ ,
=

(A−B)L1(k)
2[2,q]{q + [3,q]β}ϕ2

∣∣c2 −vc21∣∣ (2.16)
where

v =
(1 + B)L1(k) − 2D(k)

2
−

(A−B)
2{q + β[2,q]}

L1(k)

(
1 −µ

ϕ2

(ϕ1)
2

)
. (2.17)

Applying the Lemma 2 on equation (2.16), we obtain the required result. �

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	1. Introduction
	2. Main Results
	References