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Some Investigations on a Class of Analytic and Univalent Functions Involving q-Differentiation

Ayotunde Olajide Lasode∗ , Timothy Oloyede Opoola
Department of Mathematics, Faculty of Physical Sciences, University of Ilorin, Ilorin, Nigeria

lasode_ayo@yahoo.com, opoola.to@unilorin.edu.ng
∗Correspondence: lasode_ayo@yahoo.com

Abstract. We use the concept of q-differentiation to define a class Eq(β,δ) of analytic and univalentfunctions. The investigations thereafter includes coefficient estimates, inclusion property and someconditions for membership of some analytic functions to be in the class Eq(β,δ). Our results generalizesome known and new ones.

1. Introduction and Definitions
We let UD = {z : z ∈ C, |z| < 1} represent the unit disk and A represent the class ofnormalized analytic functions of the form

f (z)= z +

∞∑
m=2

amz
m, z ∈UD (1)

where f (0) = 0 = f ′(0)−1. Also, let S represent a subset of A containing functions univalentin UD. A function f in S is a member of class BT (δ) of bounded turning functions of order δ if itsatisfies the geometric condition
Ref ′(z) > δ ∈ [0,1), z ∈UD.

Let BT (0)=BT represent the class of bounded turning functions. It is known (see [1]) that f ∈BTare univalent functions. Also, a function f in S is a member of class CV(δ) of convex functions oforder δ if it satisfies the geometric condition
Re
(
z
f ′′(z)

f ′(z)
+1

)
> δ ∈ [0,1), z ∈UD.

Let CV(0)= CV represent the class of convex functions.The importance of operators in geometric function theory cannot be underrated. For instancesee [2, 13, 15] for some known ones.In 1908, Jackson [7] (see also [3, 4, 8–11]) initiated the concept of q-calculus as follows.
Received: 21 Jan 2022.
Key words and phrases. Analytic functions; Carathéodory functions; univalent functions; bounded turning function;coefficient bound; inclusion property and q-calculus. 1

https://adac.ee
https://doi.org/10.28924/ada/ma.2.12
https://orcid.org/0000-0002-2657-7698


Eur. J. Math. Anal. 10.28924/ada/ma.2.12 2
Definition 1.1. For q ∈ (0,1), the q-differentiation of function f ∈A is defined by

Dqf (0)= f ′(0), Dqf (z)=
f (z)− f (qz)
z(1−q)

(z 6=0) and D2qf (z)=Dq(Dqf (z)). (2)
Obviously, applying (2) in (1) gives us

Dqf (z)=1+
∞∑
m=2

[m]qamz
m−1 and zD2qf (z)= ∞∑

m=2

[m−1]q[m]qamzm−1 (3)
where [m]q = 1−qm1−q and lim

q↑1
[m]q = m.For example if f (z)= zm, then by using (2),

Dqf (z)=Dq(zm)=
1−qm

1−q
zm−1 = [m]qz

m−1

and observe that
lim
q↑1
Dqf (z)= lim

q↑1

(
[m]qz

m−1) = mzm−1 = f ′(z)
where f ′(z) is the classical differentiation.In this work, the q-differential operator was used to define a class of analytic functions andgeneralize some results.

2. Relevant Lemmas
We represent by P the well-known class of analytic functions of the form

p(z)=1+

∞∑
m=1

cmz
m, Re p(z) > 0, z ∈UD (4)

and by P(δ)⊆P(0)=P the class whose members are of the form
pδ(z)=1+

∞∑
m=1

(1−δ)cmzm, Re p(z) > δ ∈ [0,1), z ∈UD. (5)
The following lemmas shall be required to proof our results.

Lemma 2.1 ( [14]). Let g(z) = ∞∑
m=1

amz
m ≺ G(z) =

∞∑
m=1

bmz
m, z ∈ UD where G(z) is univalent

in UD and G(UD) is a convex domain, then |am| ≤ |b1|, m ∈ N. Equality holds for the function
g(z)= G(τzm), |τ|=1.

The lemmas that follow are the q-analogous versions of the original ones as referenced.
Lemma 2.2 ( [6]). Let p(z) be analytic in UD such that p(0)=1. If

Re
(
zDq(p(z))
p(z)

+1

)
>
3δ −1
2δ

, z ∈UD,

then for α =(δ −1)/δ (δ ∈ [1/2,1)), Re p(z) > 2α. The constant 2α is the best possible.

Lemma 2.3 ( [5]). Let u = u1+u2i and v = v1+v2i such that γ(u,v) :C2 −→C is a complex-valued
function such that

https://doi.org/10.28924/ada/ma.2.12


Eur. J. Math. Anal. 10.28924/ada/ma.2.12 3
(1) γ(u,v) is continuous in Π ⊂C2,(2) (1,0)∈ Π and Re(γ(1,0)) > 0 and(3) Re(γ(ξ+(1−ξ)u2i,v1))≤ ξ (0≤ ξ < 1)) if (ξ+(1−ξ)u2i,v1)∈ Π and

v1 ≤−12(1−ξ)(1+u
2
2) and Re(γ(ξ+(1−ξ)u2i,v1))≥ ξ (ξ > 1) if (ξ+(1−ξ)u2i,v1)∈ Π

and v1 ≥ 12(1−ξ)(1+u
2
2).

If p(z)∈P for (p(z),zDqp(z))∈ Π and Re(γ(p(z), zDqp(z))) > ξ, z ∈UD, then Rep(z) > ξ
in UD.

3. Main Results
The definition of the investigated class is as follows.A function f (z)∈A is a member of the class Eq(β,δ) if the condition

Re
(
Dqf (z)+

1+eiβ

2
zD2qf (z)

)
> δ, δ ∈ [0,1), β ∈ (−π,π], z ∈UD (6)

holds.When parameters in (6) are varied, the class Eq(β,δ) reduces to some well-known classes ofanalytic functions that have been studied by some authors. These are cited in our corollaries andremarks.The following are the proved results.
Theorem 3.1. Let β ∈ (−π,π] and δ ∈ [0,1), if condition (6) holds, then

Eq(β,δ)⊂BT q(δ).

BT q(δ) is the class of q-bounded turning function of order δ.

Proof. Let p(z) = Dqf (z) so that Dqp(z) = D2qf (z) and for κ = (1+ eiβ)/2, then (6) can beexpressed as
Re(p(z)+κzDqp(z)) > δ. (7)

In view of the conditions in Lemma 2.3 and for p(z) in (7), we define the function
γ(u,ν)= u +κν

on the domain Π of C2, then
(i) clearly, γ(u,ν) satisfies the condition (1) in Lemma 2.3,(ii) for (1,0)∈ Π, γ(1,0)=1 =⇒ Re(γ(1,0)) > 0 and(iii) γ(δ +(1−δ)u2i,ν1)= δ + 1+cosδ2 ν1 +((1−δ)u2 + sinδ2 ν1) i, thus,

Re(γ(δ +(1−δ)u2i,ν1))= δ +
1+cosβ

2
ν1 ≤ δ

for ν1 ≤−12(1−δ)(1+u22).

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Eur. J. Math. Anal. 10.28924/ada/ma.2.12 4
Now since γ(u,ν) satisfies all the conditions (1−3) in Lemma 2.3, then it implies that

Rep(z)=Re(Dqf (z)) > δ, z ∈UD

hence the proof is complete. �
Corollary 3.2 ( [1]). Since class BT q(δ) is well-known to consist of univalent functions, then
Eq(β,δ)⊂BT q(δ) consists of univalent functions.

Corollary 3.3. lim
q↑1
Eq(β,δ)⊂BT (δ), z ∈UD.

Theorem 3.4. If f ∈A is such that

Re

(
zDq(Dqf (z)+κzD2qf (z))
Dqf (z)+κzD2qf (z)

)
>
δ −1
2δ

, (8)
then

Re(Dqf (z)+κzD2qf (z)) > 2
(δ−1)/δ, δ ∈ [1/2,1), z ∈UD

and κ =(1+eiβ)/2.

Proof. From (6), let p(z)=Dqf (z)+κzD2qf (z), then by logarithmic q-differentiation we obtain
zDqp(z)
p(z)

+1=
zDq(Dqf (z)+κzD2qf (z))
Dqf (z)+κzD2qf (z)

+1.

Now applying Lemma 2.2 gives
Re
(
zDqp(z)
p(z)

+1

)
=Re

(
zDq(Dqf (z)+κzD2qf (z))
Dqf (z)+κzD2qf (z)

+1

)
>
3δ −1
2δ

implies that
Re

(
zDq(Dqf (z)+κzD2qf (z))
Dqf (z)+κzD2qf (z)

)
>
δ −1
2δ

and by the same Lemma 2.2 the proof in complete. �
Corollary 3.5. If f ∈A satisfies condition (8), then f ∈Eq(β,2(δ−1)/δ).
Corollary 3.6. If f ∈ lim

q↑1
Eq(β,1/2) is such that

Re
(
z(1+κ)f ′′(z)+κz2f ′′′(z)

f ′(z)+κzf ′′(z)

)
> −
1

2
,

then

Re(f ′(z)+κzf ′′(z)) > 1/2, z ∈UD.

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Eur. J. Math. Anal. 10.28924/ada/ma.2.12 5
Corollary 3.7. If f ∈Eq(π,1/2) is such that

Re
(
zDq(Dqf (z))
Dqf (z)

)
> −
1

2
, (9)

then

Re(Dqf (z)) >
1

2
.

This means that if condition (9) holds, then f is a q-bounded turning function of order 1/2. Now
if q ↑ 1, then

Re
(
zf ′′(z)

f ′(z)

)
> −
1

2
, (10)

implies

Re(f ′(z)) >
1

2
z ∈UD.

This means that if condition (10) holds, then f is a bounded turning function of order 1/2.
Corollary 3.8. If f ∈Eq(0,1/2) is such that

Re

(
zDq(Dqf (z)+zD2qf (z))
Dqf (z)+zD2qf (z)

)
> −
1

2
, (11)

then

Re(Dqf (z)+zD2qf (z)) >
1

2

and if q ↑ 1,

Re
(
2zf ′′(z)+z2f ′′′(z)

f ′(z)+zf ′′(z)

)
> −
1

2

implies that

Re(f ′(z)+zf ′′(z)) > 1/2, z ∈UD.

Theorem 3.9. Let β ∈ (−π,π] and δ ∈ [0,1), then the function

f (z)= z +amz
m ∈Eq(β,δ), m = {2,3, . . .} (12)

if

|am| ≤
2

[m]q
{
|Xm|− ((2+[m−1]q)cosθ+[m−1]q cos(β +θ0))

} (13)
where

Xm =2+[m−1]q(1+eiβ)

|Xm|=
√
2
{
2+[m−1]q(2+[m−1]q)(1+cosβ)

}
≥ 2

 (14)
and θ0 attains minimum at

θ0 = π +arctan

(
−[m−1]q sinβ

2+[m−1]q(1+cosβ)

)
. (15)

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Eur. J. Math. Anal. 10.28924/ada/ma.2.12 6
Proof. Firstly, applying (2) in (12) gives

Dqf (z)=1+[m]qamzm−1

zD2qf (z)= [m−1]q[m]qamzm−1

}
. (16)

Note that it suffices to study the condition that for |z|=1,∣∣∣∣Dqf (z)+ 1+eiβ2 zD2qf (z)−1
∣∣∣∣ < Re {Dqf (z)+ 1+eiβ2 zD2qf (z)

} (17)
so that by putting (16) into (17) we obtain∣∣∣∣[m]qamzm−1 + 12[m−1]q[m]q(1+eiβ)amzm−1

∣∣∣∣
< Re

{
1+[m]qamz

m−1 +
1

2
[m−1]q[m]q(1+eiβ)amzm−1

}
.

Now letting |am|= r , amzm−1 = reiθ and using (14) we obtain∣∣∣∣12[m]qreiθXm
∣∣∣∣ ≤Re {1+[m]qreiθ + 12[m−1]q[m]q(1+eiβ)reiθ

} (18)
so that

1

2
[m]qr|Xm| ≤ReF (19)

where
F =1+[m]qreiθ +

1

2
[m−1]q[m]q(1+eiβ)reiθ

in (18). Further simplification gives
F =1+[m]qr cosθ+

1

2
[m−1]q[m]qr cosθ+

1

2
[m−1]q[m]qr cos(β +θ)+Im(F)

so that
ReF =1+

1

2
[m]qr{2cosθ+[m−1]q cosθ+[m−1]q cos(β +θ)}= ψ. (20)

Now (19) becomes
1

2
[m]qr|Xm| ≤ 1+

1

2
[m]qr{(2+[m−1]q)cosθ+[m−1]q cos(β +θ)}

and by simplification we obtain (13).To know the values of θ where (20) attains minimum implies that
∂ψ

∂θ
=−

r[m]q
2

{
(2+[m−1]q)sinθ+[m−1]q sin(β +θ)

}
implies that

(2+[m−1]q)sinθ+[m−1]q sin(β +θ)=0

so that
tanθ =

−[m−1]q sinβ
2+[m−1]q(1+cosβ)which simplifies to (15). �

https://doi.org/10.28924/ada/ma.2.12


Eur. J. Math. Anal. 10.28924/ada/ma.2.12 7
Corollary 3.10. Let f (z)= z +amzm ∈Eq(0,δ) and m = {2,3, . . .}, then

|am| ≤
1

[m]q

{√
1+2[m−1]q +[m−1]2q +1+[m−1]q

}
and if q ↑ 1, then

|am| ≤
1

2m2
.

Corollary 3.11. Let f (z)= z +amzm ∈Eq(π,δ) and m = {2,3, . . .}, then

|am|5
1

2[m]q

and if q ↑ 1, then

|am| ≤
1

2m
.

Remark 3.12. Let q ↑ 1, then Theorem 3.9 becomes the result in [18].
Theorem 3.13 (Coefficient Estimates). Let β ∈ (−π,π], δ ∈ [0,1) and let G(z) = 1+ b1z +
b2z

2 + · · · ∈ CV(δ). If f ∈A belongs to Eq(β,δ), then

|am| ≤
2(1−δ)|b1|
[m]q|Xm|

, m = {2,3, . . .} (21)
where |Xm| is defined in (14).
Proof. Let f (z)∈Eq(β,δ), therefore from (6) and using (5),

Dqf (z)+
1+eiβ

2
zD2qf (z)= δ +(1−δ)p(z), z ∈UD. (22)

Now putting (3) and (4) into (22) and simplifying gives
1+

∞∑
m=2

{
1+[m−1]q

(
1+eiβ

2

)}
[m]qamz

m−1 =1+

∞∑
m=2

(1−δ)cm−1zm−1

which implies that
{2+[m−1]q(1+eiβ)}

[m]q
2
am =(1−δ)cm−1, m = {2,3, . . .}

where by applying (14) we obtain
Xm

[m]q
2(1−δ)

am = cm−1, m = {2,3, . . .}. (23)
Since G(UD) is a convex domain, then from Lemma 2.1, (23) becomes∣∣∣∣Xm [m]q2(1−δ)am

∣∣∣∣ = |cm−1| ≤ |b1|
and simplifying further we obtain (21). �

https://doi.org/10.28924/ada/ma.2.12


Eur. J. Math. Anal. 10.28924/ada/ma.2.12 8
Corollary 3.14. Let f (z)∈Eq(0,δ), then

|am| ≤
(1−δ)|b1|√

1+2[m−1]q +[m−1]2q

and if q ↑ 1, then

|am| ≤
(1−δ)|b1|

m
, m = {2,3, . . .}.

Corollary 3.15. Let f ∈Eq(π,δ), then

|am| ≤
(1−δ)|b1|
[m]q

and if q ↑ 1, then

|am| ≤
(1−δ)|b1|

m
, m = {2,3, . . .}

Remark 3.16. Let p(z)∈P and φ(z)=1+ 2
π2

(
ln
1+
√
z

1−
√
z

)2. If q ↑ 1,
(1) β = π and G(z)= p(z), then Theorem 3.13 becomes the result in [12].(2) and G(z)= p(z), then Theorem 3.13 becomes the result in [16].(3) and G(z)= φ(z), then Theorem 3.13 becomes the result in [18].(4) and β =0, then Theorem 3.13 becomes the result in [17].

Acknowledgment. The authors would like to thank the referees for their careful reading of thismanuscript and their valuable suggestions.
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	1. Introduction and Definitions
	2. Relevant Lemmas
	3. Main Results
	References

