International Journal of Analysis and Applications

Volume 18, Number 2 (2020), 254-261

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

DOI: 10.28924/2291-8639-18-2020-254

CERTAIN SUBFAMILY OF HARMONIC FUNCTIONS RELATED TO SĂLĂGEAN

q-DIFFERENTIAL OPERATOR

SH. NAJAFZADEH1, DEBORAH OLUFUNMILAYO MAKINDE2,∗

1Department of Mathematics, Payame Noor University, P. O. Box: 19395–3697, Tehran, Iran

2Department of Mathematics, Obafemi Awolowo University, 220005, Ile-Ife, Osun State, Nigeria

∗Corresponding author: funmideb@yahoo.com

Abstract. The theory of q–calculus operators are applied in many areas of sciences such as complex

analysis. In this paper we apply Sălăgean q–differential operator to harmonic functions and introduce

sharp coefficient bounds, extreme points, distortion inequalities and convexity results.

1. Introduction

We state some notations regarding to q–calculus used in this article, see [1, 4] and [6]. For 0 < q < 1

and positive integer n, the q–integer number is denoted by [n]q and introduced by:

[n]q =
1 −qn

1 −q
= 1 + q + q2 + . . . + qn−1. (1.1)

We can easily conclude that:

lim
q→1−

[n]q = n.

If f(z) be analytic in this open unit disk U = {z ∈ C : |z| < 1} and normalized by f(0) = f′(0) − 1 = 0,

then the q–difference operator of q–calculus operated on f given by:

∂qf(z) =
f(z) −f(qz)
z(1 −q)

, (1.2)

where lim
q→1−

∂qf(z) = f
′(z), for example see [2, 5] and [8].

Received 2019-04-24; accepted 2019-05-23; published 2020-03-02.

2010 Mathematics Subject Classification. 30C45, 30C50.

Key words and phrases. q–calculus; harmonic; univalent; Sălăgean operator; convex set.

c©2020 Authors retain the copyrights of

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

254

https://doi.org/10.28924/2291-8639
https://doi.org/10.28924/2291-8639-18-2020-254


Int. J. Anal. Appl. 18 (2) (2020) 255

For f(z) = z +
∑∞
k=2akz

k, the Sălăgean q–differential operator is defined by:

S0qf(z) = f(z)

S1qf(z) = z∂qf(z) =
f(z) −f(qz)

(1 −q)
...

Smq f(z) = z∂q
(
Sm−1q f(z)

)
= f(z) ∗

(
z +

∑∞
k=2[k]

m
q z

k
)

= z +

∞∑
k=2

[k]mq akz
k,

(1.3)

where m is a positive integer and “∗” is the familiar Hadamard product or convolution of two analytic

functions.

Since

lim
q→1−

Smq (z) = z +
∞∑
k=2

kmakz
k,

is the famous Sălăgean operator [9], so the operator Smq is called Sălăgean q–differential operator.

Let Sh denote the class of functions:

f = h + g (1.4)

which are harmonic, univalent and sense-preserving in U and normalized by f(0) = f′(0) − 1 = 0, where

h and g are analytic in U take the form:

h(z) = z +

∞∑
k=2

akz
k and g(z) =

∞∑
k=1

bkz
k, (0 6 b1 < 1). (1.5)

Also, we call h and g analytic part and co-analytic part of f respectively, see [3].

Hence f ∈Sh is of the type:

f(z) = z +

∞∑
k=2

akz
k +

∞∑
k=1

bkzk (1.6)

Now, we consider the Sălăgean q–differential operator of harmonic functions f = h + g, by:

Smq f(z) = S
m
q h(z) + (−1)

mSmq g(z), (1.7)

where Smq is defined by (1.3) and h and g are of the type (1.5). For more details see [7].

We denote by S∗h the family of functions of the type (1.4) where:

h(z) = z −
∞∑
k=2

|ak|zk , g(z) =
∞∑
k=1

|bk|zk, |b1| < 1. (1.8)

For A > 0, 0 6 B,C 6 1, 0 6 D < 1 and γ ∈ R let S∗
h(γ)

(A,B,C,D) denote the class of functions in S∗h
of the type (1.5) such that:

Re

{
(1 −A)(1 −B)

S0qf(z)
z

+ (A + B)

(
Smq f(z)

)′
z′

−Ceiγ
(
Smq f(z)

)′′
z′′

+
(
Ceiγ −AB

)}
> D, (1.9)



Int. J. Anal. Appl. 18 (2) (2020) 256

where

z′ =
∂

∂θ
(z) = iz,

(
Smq f(z)

)′
=

∂

∂θ

(
Smq f(re

iθ)
)

= iz
(
Smq h

)′ − iz(Smq g)′ (1.10)
z′′ =

∂2

∂θ2
(z) = −z,

(
Smq f(z)

)′′
=

∂2

∂θ2
(
Smq f(re

iθ)
)

= −z
(
Smq h

)′ −z2(Smq h)′′ −z(Smq g)′ −z2(Smq g)′′.
(1.11)

We further denote by S∗
h(γ)

(A,B,C,D) the subclass of S
h(γ)

(A,B,C,D) consisting of harmonic functions

f = h + g so that h and g are of the form (1.8) and satisfying (1.9).

2. Main results

In our first theorem, we introduce a sufficient coefficient condition for functions in S
h(γ)

(A,B,C,D)

and then we show that this condition is also necessary for f(z) ∈S∗
h(γ)

(A,B,C,D).

Theorem 2.1. Suppose f = h + g, where h and g be given by (1.5) and:

∞∑
k=2

∣∣(A + B)k − (1 −A−B + AB) −Ck2∣∣ [k]mq |ak|+
∞∑
k=1

∣∣(A + B)k − (1 −A−B + AB) −Ck2∣∣ [k]mq |bk|6 1 −D.
(2.1)

Then f(z) ∈S
h(γ)

(A,B,C,D).

Proof. In view of the fact that:

“ Re{W}> 0 ⇐⇒|W + 1 −D|> |w − 1 −D|, ”

and letting:

W = (1 −A)(1 −B)
Smq f(z)

z
+ (A + B)

(
Smq f(z)

)′
z′

−Ceiγ
(
Smq f(z)

)′′
z′′

+
(
Ceiθ −AB

)
,

it is enough to show that:

|W + 1 −D|− |W − 1 −D|> 0.

But by using (1.10) and (1.11) we have:

|W + 1 −D| =

∣∣∣∣∣(1 −A)(1 −B)
(

1 +

∞∑
k=2

ak[k]
m
q z

k−1 +

∞∑
k=1

bk[k]
m
q (z)

k−1
)

+ (A + B)
(

1 +

∞∑
k=2

kak[k]
m
q z

k−1 −
∞∑
k=1

kbk[k]
m
q (z)

k−1
)

−Ceiγ
(

1 +

∞∑
k=2

kak[k]
m
q z

k−1 +

∞∑
k=2

k(k − 1)ak[k]mq z
k−1

+

∞∑
k=1

kbk[k]
m
q (z)

k−1 +

∞∑
k=1

k(k − 1)bk[k]mq (z)
k−1

)



Int. J. Anal. Appl. 18 (2) (2020) 257

+ Ceiγ −AB + 1 −D

∣∣∣∣∣
6 2 −D −

∞∑
k=1

∣∣1 + (A + B)(k − 1) + AB −Ck2[k]mq ∣∣ |ak|
∣∣∣∣zkz
∣∣∣∣

−
∞∑
k=1

∣∣1 − (A + B)(k − 1) + AB −Ck2∣∣ [k]mq |bk|
∣∣∣∣zkz
∣∣∣∣ ,

and

|W − 1 −D|6 D +
∞∑
k=2

∣∣(A + B)(k − 1) + AB −Ck2∣∣ [k]mq |ak|
∣∣∣∣zkz
∣∣∣∣

+

∞∑
k=1

∣∣1 − (A + B)(k − 1) + AB −Ck2∣∣ [k]mq |bk|
∣∣∣∣zkz
∣∣∣∣ .

So by using (2.1), we get:

|W + 1 −D|− |W − 1 −D|>

2

[
1 −D −

∞∑
k=2

∣∣(A + B)k − (1 −A−B + AB) −Ck2∣∣ [k]mq |ak|−
∞∑
k=1

∣∣(A + B)k − (1 −A−B + AB) −Ck2∣∣ [k]mq |bk|
]
> 0.

�

Remark 2.1. The coefficient bound (2.1) is sharpt for the function:

F(z) = z +

∞∑
k=2

xk
|(A + B)k − (1 −A−B + AB) −Ck2| [k]mq

zk

+

∞∑
k=1

yk
|(A + B)k − (1 −A−B + AB) −Ck2| [k]mq

(z)k,

where

1

1 −D

(
∞∑
k=2

|xk| +
∞∑
k=1

|yk|

)
= 1.

Theorem 2.2. Let f = h + g ∈S∗h. Then f(z) ∈S
∗
h(γ)

(A,B,C,D) if and only if:

∞∑
k=2

∣∣(A + B)k − (1 −A−B + AB) −Ck2∣∣ [k]mq |ak|
+

∞∑
k=1

∣∣(A + B)k − (1 −A−B + AB) −Ck2∣∣ [k]mq |bk|6 1 −D.
(2.2)

Proof. From Theorem 2.1, and since S∗
h(γ)

(A,B,C,D) ⊂ S
h(γ)

(A,B,C,D), we conclude the “if” part.

For the “only if” part, suppose that f(z) ∈S∗
h(γ)

(A,B,C,D). Thus for z = reiθ ∈ U, we have:

Re

{
(1 −A)(1 −B)

Smq f(z)
z

+ (A + B)

(
Smq f(z)

)′
z′

−Ceiγ
(
Smq f(z)

)′′
z′′

+ Ceiγ + Ceiγ −AB

}

= Re

{
(1 −A)(1 −B)

(
1 +

∞∑
k=2

ak[k]
m
q z

k−1 +

∞∑
k=1

bk[k]
m
q (z)

k−1
)



Int. J. Anal. Appl. 18 (2) (2020) 258

+ (A + B)
(

1 +

∞∑
k=2

kak[k]
m
q z

k−1 −
∞∑
k=1

kbk[k]
m
q (z)

k−1
)

−Ceiγ
(

1 +

∞∑
k=2

kak[k]
m
q z

k−1 +

∞∑
k=2

k(k − 1)ak[k]mq z
k−1

+

∞∑
k=1

kbk[k]
m
q (z)

k−1 +

∞∑
k=1

k(k − 1)bk[k]mq (z)
k−1
)

+ Ceiγ −AB

}

> 1 −
∞∑
k=2

∣∣(A + B)k − (1 −A−B + AB) −Ck2∣∣ [k]mq |ak|rk−1
−
∞∑
k=1

∣∣(A + B)k − (1 −A−B + AB) −Ck2∣∣ [k]mq |bk|rk−1 > D.
The above inequality holds for all z = reiθ ∈ U. So if z = r → 1, we obtain the required result (2.2).

Now the proof is complete. �

3. Geometric properties of S∗
h(γ)

(A,B,C,D)

In this section, we first introduce extreme points of S∗
h(γ)

(A,B,C,D) and then we obtain the distortion

bounds for f ∈S∗
h(γ)

(A,B,C,D). Finally we show that the class S∗
h(γ)

(A,B,C,D) is a convex set.

Theorem 3.1. f = h + g ∈S∗
h(γ)

(A,B,C,D) if and only if it can be expressed:

f(z) = X1z +

∞∑
k=2

Xkhk(z) +

∞∑
k=1

Ykgk(z), (z ∈ U), (3.1)

where

hk(z) = z −
1 −D

|(A + B)k − (1 −A−B + AB) −Ck2| [k]mq
zk, (k = 2, 3, . . .), (3.2)

gk(z) =
1 −D

|(A + B)k − (1 −A−B + AB) −Ck2| [k]mq
(z)k, (k = 1, 2, . . .), (3.3)

X1 > 0, Y1 > 0, X1 +
∑∞
k=2Xk +

∑∞
k=1Yk = 1, Xk > 0 and Yk > 0 for k = 2, 3, . . ..

Proof. If f is given by (3.1), then:

f(z) = z −
∞∑
k=2

1 −D
|(A + B)k − (1 −A−B + AB) −Ck2| [k]mq

Xkz
k

+

∞∑
k=1

1 −D
|(A + B)k − (1 −A−B + AB) −Ck2| [k]mq

Yk(z)
k.

Since by (2.2), we have:

∞∑
k=2

∣∣(A + B)k − (1 −A−B + AB) −Ck2∣∣ [k]mq
×
( 1 −D
|(A + B)k − (1 −A−B + AB) −Ck2| [k]mq

)
|Xk|

+

∞∑
k=1

∣∣(A + B)k − (1 −A−B + AB) −Ck2∣∣ [k]mq ×
×
( 1 −D
|(A + B)k − (1 −A−B + AB) −Ck2| [k]mq

)
|Yk|



Int. J. Anal. Appl. 18 (2) (2020) 259

= (1 −D)
( ∞∑
k=2

|Xk| +
∞∑
k=1

|Yk|
)

= (1 −D)(1 −X1) 6 1 −D.

So f(z) ∈S∗
h(γ)

(A,B,C,D).

Conversely, suppose f(z) ∈S∗
h(γ)

(A,B,C,D). By putting:

X1 = 1 −
( ∞∑
k=2

Xk +

∞∑
k=1

Yk

)
,

where

Xk =

∣∣(A + B)k − (1 −A−B + AB) −Ck2∣∣ [k]mq
1 −D

|ak|,

Yk =

∣∣(A + B)k − (1 −A−B + AB) −Ck2∣∣ [k]mq
1 −D

|bk|,

we conclude the required representation (3.1), so the proof is complete. �

Theorem 3.2. If f(z) ∈S∗
h(γ)

(A,B,C,D), |z| = r < 1, then:

|f(z)|> (1 −|b1|)r

−
1

[2]mq

( 1 −D
(A + B) + (1 + AB) − 4C

−
2(A + B) − (1 + AB) −C
(A + B) + (1 + AB) − 4C

|b1|
)
r2,

(3.4)

and

|f(z)|6 (1 −|b1|)r

+
1

[2]mq

( 1 −D
(A + B) + (1 + AB) − 4C

−
2(A + B) − (1 + AB) −C
(A + B) + (1 + AB) − 4C

|b1|
)
r2.

(3.5)

Proof. Suppose f(z) ∈S∗
h(γ)

(A,B,C,D), then by (2.2), we have:

|f(z)| =

∣∣∣∣∣z −
∞∑
k=2

|ak|zk +
∞∑
k=1

|bk|(z)k
∣∣∣∣∣

=

∣∣∣∣∣z + |b1|(z) −
∞∑
k=2

(
|ak|zk −|bk|(z)k

)∣∣∣∣∣
> r −|b1|r −

1 −D
(A + B) + (1 + AB) − 4C

[
∞∑
k=2

((A + B) + (1 + AB) − 4C
1 −D

|ak|

+
(A + B) + (1 + AB) − 4C

1 −D
|bk|
)
rk
]
> (1 −|b1|)r

−
1 −D

(A + B) + (1 + AB) − 4C

[
∞∑
k=2

((A + B)(k − 1) + (1 + AB) − 4C
1 −D

|ak|

+
(A + B)(k − 1) − (1 + AB) − 4C

1 −D
|bk|
)
rk
]
> (1 −|b1|)r

−
1 −D

(A + B) + (1 + AB) − 4C

(
1 −

2(A + B) − (1 + AB) −C
1 −D

|b1|
)
r2

= (1 −|b1|)r −
1

[2]mq

( 1 −D
(A + B) + (1 + AB) − 4C

−
2(A + B) − (1 + AB) −C
(A + B) + (1 + AB) − 4C

|b1|
)
r2.

Relation (3.5) can be proved by using the similar statements. So the proof is complete. �



Int. J. Anal. Appl. 18 (2) (2020) 260

Theorem 3.3. If fj(z), j = 1, 2, . . ., belongs to S∗h(γ)(A,B,C,D), then the function F(z) =
∑∞
j=1λjfj(z)

is also in S∗
h(γ)

(A,B,C,D), where fj(z) defined by:

fj(z) = z −
∞∑
k=2

ak,jz
k +

∞∑
k=1

bk,j(z)
k,

(
j = 1, 2, . . . ,

∑∞
j=1λj = 1

)
.

In the other worlds, S∗
h(γ)

(A,B,C,D), is a convex set.

Proof. Since fj(z) ∈S∗h(γ)(A,B,C,D), so by (2.2), we get:

∞∑
k=2

∣∣(A + B)k − (1 −A−B + AB) −Ck2∣∣ [k]mq |ak|
∞∑
k=1

∣∣(A + B)k − (1 −A−B + AB) −Ck2∣∣ [k]mq |bk,j|6 1 −D, (j = 1, 2, . . .).
Also

F(z) =

∞∑
j=1

λjfj(z) = z −
∞∑
k=k

( ∞∑
j=1

λjak,j

)
zk +

∞∑
k=1

( ∞∑
j=1

λjbk,j

)
(z)k.

Now, according to Theorem 2.2, we have:

∞∑
k=2

∣∣(A + B)k − (1 −A−B + AB) −Ck2∣∣ [k]mq
∣∣∣∣∣∣
∞∑
j=1

λjak,j

∣∣∣∣∣∣
+

∞∑
k=1

∣∣(A + B)k − (1 −A−B + AB) −Ck2∣∣ [k]mq
∣∣∣∣∣∣
∞∑
j=1

λjbk,j

∣∣∣∣∣∣
=

∞∑
j=1

(
∞∑
k=2

∣∣(A + B)k − (1 −A−B + AB) −Ck2∣∣ [k]mq |ak,j|
+

∞∑
k=1

∣∣(A + B)k − (1 −A−B + AB) −Ck2∣∣ [k]mq |bk,j|
)
λj

6 (1 −D)
∞∑
j=1

λj = 1 −D.

Thus, F(z) ∈S∗
h(γ)

(A,B,C,D). �

Conflicts of Interest: The author(s) declare that there are no conflicts of interest regarding the publi-

cation of this paper.

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	1. Introduction
	2. Main results
	3. Geometric properties of 
	References