International Journal of Analysis and Applications

Volume 16, Number 6 (2018), 783-792

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

DOI: 10.28924/2291-8639-16-2018-783

ANALYTIC FUNCTIONS RELATED WITH MOCANU CLASS

AKHTER RASHEED1,∗, SAQIB HUSSAIN2, MUHAMMAD ASAD ZAIGHUM1 AND

ZAHID SHAREEF 3

1Department of Mathematics and Statistics, Riphah International University, Islamabad, Pakistan

2Department of Mathematics, COMSATS University Islamabad, Abbottabad Campus, Pakistan

3Division of Engineering, Higher Colleges of Technology, P.O. Box 4114, Fujairah, UAE

∗Corresponding author: akhter@ciit.net.pk

Abstract. In this article, we define a new class of analytic functions. This class generalizes the mocanu

class. We obtain relationships of this class with other subclasses of analytic functions and derived many

interesting results.

1. Introduction

Let A denote the class of functions f analytic in ∆ = {z ∈ C : |z| < 1}, normalized by f (0) = 0 and

f
′
(0) = 1. So each f ∈A has series representation of form

f (z) = z +

∞∑
n=2

anz
n. (1.1)

A function f ∈A is in class S if and only if f (z1) = f (z2) implies z1 = z2, for all z1, z2 ∈ ∆. An analytic

function f is subordinate to an analytic function g (written as f ≺ g) if and only if there exists an analytic

function w with w (0) = 0 and |w (z)| < 1 for z ∈ ∆ such that f (z) = g (w (z)) .

Received 2018-01-24; accepted 2018-03-14; published 2018-11-02.

2010 Mathematics Subject Classification. 30C45, 30C50.

Key words and phrases. convex functions; strongly starlike functions; subordination.

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.

783

https://doi.org/10.28924/2291-8639
https://doi.org/10.28924/2291-8639-16-2018-783


Int. J. Anal. Appl. 16 (6) (2018) 784

For 0 ≤ α ≤ 1, Mocanu [15] introduced the class Mα of functions f ∈ A such that
f(z)f

′
(z)

z
6= 0 for all

z ∈ ∆ and

Re

{
(1 −α)

zf′ (z)

f (z)
+ α

(zf′ (z))
′

f′ (z)

}
> 0, z ∈ ∆. (1.2)

Mocanu defined the class Mα geometrically as a class of functions that maps the circle centered at the origin

on α−convex arcs and derived the condition (1.2).

The class Mα was extensively studied in literature by several authors, for instance, see [4–6, 18–21]. For

particular values of α, we obtain a number of interesting classes of analytic functions having nice geometry, for

instance M0 = S∗ and M1 = C are well known classes of starlike and convex univalent functions introduced

by Alexander [1]. By S∗ (δ) and C (δ) , 0 ≤ δ < 1, we mean the subclasses of starlike and convex function of

order δ given by (1.3) and (1.4) respectively.

Re

{
zf′ (z)

f (z)

}
> δ, z ∈ ∆, (1.3)

and

Re

{
1 +

zf′′ (z)

f′ (z)

}
> δ, z ∈ ∆. (1.4)

We denote the classes of uniformly starlike and uniformly convex functions by UST and UCV, see [7,13,14].

A function f ∈S is uniformly starlike if f maps every circular arc γ contained in ∆ with center ζ ∈ ∆ onto

a starlike arc with respect to f (ζ). A function f ∈ S is uniformly convex if f maps every circular arc γ

contained in ∆ with center ζ ∈ ∆ onto a convex arc.

In 1999, Kanas and Wisnoiska [9] introduced the class k −UCV, (k ≥ 0) of k-uniformly convex functions

as:

f ∈ k −UCV ⇐⇒ f ∈A and Re
{

1 +
zf′′ (z)

f′ (z)

}
> k

∣∣∣∣zf′′ (z)f′ (z)
∣∣∣∣ , z ∈ ∆, (1.5)

where its geometric definition and connections with the conic domains were considered. It is worth mentioning

that 1 −UCV = UCV. In recent years many authors investigated interesting properties of these classes. For

some details see [2, 8–12, 26, 30] and references cited there in. Let SS∗ (λ) denote the class of strongly starlike

functions of order λ as:

SS∗ (λ) =
{
f ∈A :

∣∣∣∣arg zf′ (z)f (z)
∣∣∣∣ < λπ2

}
, (1.6)

where λ ∈ (0, 1). This class of functions was introduced and discussed by [3, 27].

In our current investigation, we extend the work of J. sokol [25] and introduced a new class of analytic

function as:

Definition 1.1. Let f ∈A and k ≥ 0, 0 ≤ α ≤ 1. Then f ∈ k −UMα if and only if

Re

{
(1 −α)

zf′ (z)

f (z)
+ α

(zf′ (z))
′

f′ (z)

}
> k

∣∣∣∣∣(zf
′ (z))

′

f′ (z)
− 1

∣∣∣∣∣ , z ∈ ∆. (1.7)



Int. J. Anal. Appl. 16 (6) (2018) 785

For special values of parameters k and α, we obtain a number of known classes of analytic functions. Here

we present few of them.

(i) 0 −UMα = Mα, [15].

(ii) 0 −UM0 = S∗, [1].

(iii) 1 −UM1 = UCV, [7].

(iv) k −UM1 = k −UCV, [9].

Lemma 1.1. Let u = u1 + iu2 and v = v1 + iv2 and let Ψ (u,v) : C2 ×∆ → C be a complex-valued function

satisfying the conditions:

(i) Ψ (u,v) is continuous in a domain D ⊂ C2,

(ii) (1, 0) ∈ D and Ψ (1, 0) > 0.

(iii) Re Ψ (iu2,v1) ≤ 0 whenever (iu2,v1) ∈ D and v1 ≤−12
(
1 + u22

)
.

If h (z) = 1 + c1z + c2z
2 + · · · is a function that is analytic in ∆ such that (h(z),zh′(z)) ∈ D and

Re Ψ (h(z),zh′(z)) > 0 for z ∈ ∆, then Re h(z) > 0.

This result is due to Miller [16].

Lemma 1.2. Let β > 0 and 0 ≤ γ < 1. Let

p (z) +
βzp′ (z)

p (z)
≺

1 + (1 − 2γ)
1 −z

.

Then

p (z) ≺
1 + (1 − 2δ)

1 −z
,

where

δ =
(2γ −β) +

√
(2γ −β)2 + 8β
4

. (1.8)

Lemma 1.3. [17] If −1 ≤ B < A ≤ 1, λ > 0 and the complex number γ satisfies Re{γ}≥−λ (1 −A) / (1 −B) ,

then the differential equation

q (z) +
zq′ (z)

λq (z) + γ
=

1 + Az

1 + Bz
, z ∈ ∆,

has a univalent solution in E given by

q (z) =




zλ+γ(1+Bz)λ(A−B)/B

λ

z∫
0

tλ+γ−1(1+Bt)λ(A−B)/Bdt

− γ
λ
, B 6= 0,

zλ+γeλAz

λ

z∫
0

tλ+γ−1eλAtdt

− γ
λ
, B = 0.



Int. J. Anal. Appl. 16 (6) (2018) 786

If h (z) = 1 + c1z + c2z
2 + . . . is analytic in ∆ and satisfies

h (z) +
zh′ (z)

λh (z) + γ
≺

1 + Az

1 + Bz
, z ∈ ∆,

then

h (z) ≺ q (z) ≺
1 + Az

1 + Bz
,

and q (z) is the best dominant.

Lemma 1.4. [29] Let ε be a positive measure on [0, 1] . Let g be a complex-valued function defined on

∆×[0, 1] such that g (., t) is analytic in ∆ for each t ∈ [0, 1] and g (z, .) is ε-integrable on [0, 1] for all z ∈ ∆. In

addition, suppose that Re g (z,t) > 0, g (−r,t) is real and Re{1/g (z,t)} ≥ 1/g (−r,t) for |z| ≤ r < 1 and

t ∈ [0, 1] . If g (z) =
1∫
0

g (z,t) dε (t) , then Re{1/g (z)}≥ 1/g (−r) .

Lemma 1.5. [28, Chapter 14] Let a, b and c 6= 0,−1,−2 . . . be complex numbers. Then, for Re c > Re b > 0,

(i) 2F1 (a,b,c; z) =
Γ (c)

Γ (c− b) Γ (b)

1∫
0

tb−1 (1 − t)c−b−1 (1 − tz)−a dt.

(ii) 2F1 (a,b,c; z) = 2F1 (b,a,c; z) .

(iii) 2F1 (a,b,c; z) = (1 −z)
−a

2F1

(
a,c− b,c;

z

z − 1

)
.

Lemma 1.6. Let p be analytic in ∆ of the form

p (z) = 1 +

∞∑
n=m

cnz
n, cm 6= 0,

with p (z) 6= 0 in ∆. If there exists a point z0, |z0| < 1 such that |arg p (z)| < πϕ2 (|z| < |z0|) and |arg p (z0)| =
πϕ
2

for some ϕ > 0, then we have
z0p
′(z0)

p(z0)
= ilϕ, where

 l ≥
m
2

(
a + 1

a

)
, when arg p (z0) =

πϕ
2
,

l ≤−m
2

(
a + 1

a

)
, when arg p (z0) = −πϕ2 ,

where (p (z0))
1/ϕ

= ±ia (a > 0) .

This result is generalization of the Nunokawa’s lemma [23].

2. Results and Discussion

Theorem 2.1. Let f ∈ k −UMα. Then f ∈S∗ (δ) , where

δ =
(2γ −β) +

√
(2γ −β)2 + 8β
4

, (2.1)

with β = α+k
1+k

and γ = k
1+k

.



Int. J. Anal. Appl. 16 (6) (2018) 787

Proof. Let

zf′ (z)

f (z)
= p (z) ,

where p is analytic in · with p (0) = 1. We obtain

Re

{
(1 −α) p (z) + α

(
p (z) +

zp′ (z)

p (z)

)}
> k

∣∣∣∣p (z) + zp′ (z)p (z) − 1
∣∣∣∣

= k

∣∣∣∣1 −p (z) − zp′ (z)p (z)
∣∣∣∣

Re

{
(1 −α) p (z) + α

(
p (z) +

zp′ (z)

p (z)

)}
> Re k

{
(1 −p (z)) −

zp′ (z)

p (z)

}
,

hence

Re

{
p (z) +

zp′ (z)
1
β
p (z)

}
> γ, (2.2)

where β = α+k
1+k

and γ = k
1+k

. The above relation can be written in the following Briot-Bouquet differential

subordination

p (z) +
zp′ (z)
1
β
p (z)

≺
1 + (1 − 2γ) z

1 −z
.

Now using Lemma 2, we have p ∈S∗ (δ) , where δ is given by (2.1) . �

Special Cases

(i) For α = 0, k = 1,we have β = γ = 1
2
. Then for f ∈ 1 −UM0, we have f ∈S∗ (δ) , where δ ' 0.64.

(ii) For α = 0, we have β = k
1+k

and γ = k
1+k

. Then

δ1 =
k +
√

9k2 + 8k

4(k + 1)
.

In other words for f ∈ k −UM0, we have f ∈S∗ (δ1) .

(iii) If α = 1, β = 1, γ = k
1+k

, then

δ2 =
(k − 1) +

√
(k − 1)2 + 8 (k + 1)2

4(k + 1)
.

In other words f ∈ k −UCV implies f ∈S∗ (δ2) .

Theorem 2.2. Let f ∈ k −UMα and of the form

f (z) = z +

∞∑
n=m+1

anz
n, am+1 6= 0.

Then f is strongly starlike of order β0, where

β0 = min
β∈(0,1)




1 − 2aβ cos βπ
2

+
(
k2 − 1

)(
aβ cos βπ

2

)2
+(

βm(a2+1)
2a

+ aβ sin βπ
2

)2

 . (2.3)



Int. J. Anal. Appl. 16 (6) (2018) 788

Proof. Let

zf′ (z)

f (z)
= p (z) .

Then p is the form

p (z) = 1 +

∞∑
n=m

cnz
n, cm 6= 0.

Now using the definition of the class k −UMα, we have

Re

{
(1 −α) p (z) + α

(
p (z) +

zp′ (z)

p (z)

)}
> k

∣∣∣∣p (z) + zp′ (z)p (z) − 1
∣∣∣∣ . (2.4)

If there exists a point z0 ∈ ∆ such that

|arg p (z)| <
βπ

2
, |z| < |z0|

and

|arg p (z0)| =
βπ

2
,

then from Lemma 1.6, we have
z0p
′(z0)

p(z0)
= ilβ, where (p (z0))

1/β
= ±ia (a > 0) and

l :


 l ≥

m
2

(
a + 1

a

)
, when arg p (z0) =

βπ
2
,

l ≤−m
2

(
a + 1

a

)
, when arg p (z0) = −βπ2 .

(2.5)

For the case arg p (z0) =
βπ
2
, we obtain

Re

{
p (z0) +

αz0p
′ (z0)

p (z0)

}
= Re

{
(ia)

β
+ iαlβ

}
= aβ cos

βπ

2
. (2.6)

Also, we have

k

∣∣∣∣p (z0) − 1 + z0p′ (z0)p (z0)
∣∣∣∣ = k ∣∣∣(ia)β + ilβ − 1∣∣∣

= k

∣∣∣∣aβ cos βπ2 − 1 + i
(
aβ sin

βπ

2
+ lβ

)∣∣∣∣
= k

√(
aβ cos

βπ

2
− 1
)2

+

(
aβ sin

βπ

2
+ lβ

)2
. (2.7)

Then from l ≥ m
(
a2 + 1

)
/2a for β ≥ β0, we have

0 ≤ 1 − 2aβ cos β
π

2
+
(
k2 − 1

)(
aβ cos β

π

2

)2
+

(
βm

(
a2 + 1

)
2a

+ aβ sin β
π

2

)2
. (2.8)

Therefore

0 ≤ 1 − 2aβ0 cos β0
π

2
+
(
k2 − 1

)(
aβ0 cos β0

π

2

)2
+

(
β0m

(
a2 + 1

)
2a

+ aβ0 sin β0
π

2

)2
. (2.9)

By using (2.6) and (2.7) , we have

Re

{
(1 −α) p (zo) + α

{
p (zo) +

zp′ (zo)

p (zo)

}}
< k

∣∣∣∣p (z0) + zp′ (z0)p (z0) − 1
∣∣∣∣ .

which is contradiction, therefore |arg p (z)| < βπ
2

for |z| < 1.



Int. J. Anal. Appl. 16 (6) (2018) 789

Similarly we can prove the case arg p (z0) = −βπ2 by using the same method as the above we will get a

contradiction. This proves that f is strongly starlike of order β0. �

Corollary 2.1. Let f ∈UCV and of the form

f (z) = z +

∞∑
n=m+1

anz
n, am+1 6= 0.

Then f is strongly starlike of order β0, where

β0 = min
β∈(0,1)


1 − 2aβ cos βπ2 +

(
βm

(
a2 + 1

)
2a

+ aβ cos β
π

2

)2
 .

This result is due to Nunokawa and Sokol [24].

Theorem 2.3. If f ∈ k −UMα, then

zf′ (z)

f (z)
≺ q (z) =

1

g (z)
, (2.10)

where g (z) =
[
2F1

(
1
β

(1 −γ) , 1; 1
β

+ 1; z
z−1

)]
with β = k+α

1+k
and γ = k

1+k
.

Proof. Let
zf′ (z)

f (z)
= p (z) ,

where p is analytic in ∆ with p (0) = 1. Now using (1.7), we obtain

Re

{
(1 −α) p (z) + α

(
p (z) +

zp′ (z)

p (z)

)}
> k

∣∣∣∣p (z) + zp′ (z)p (z) − 1
∣∣∣∣

= k

∣∣∣∣1 −p (z) − zp′ (z)p (z)
∣∣∣∣

Re

{
(1 −α) p (z) + α

(
p (z) +

zp′ (z)

p (z)

)}
> Re k

{
(1 −p (z)) −

zp′ (z)

p (z)

}
.

This implies that

Re

{
p (z) +

zp′ (z)
1
β
p (z)

}
> γ,

where β = k+α
1+k

and γ = k
1+k

. The above relation can be written in the following Briot-Bouquet differential

subordination

p (z) +
zp′ (z)
1
β
p (z)

≺
1 + (1 − 2γ) z

1 −z
. (2.11)

Using Lemma 1.3 for λ = 1
β

and γ = 0, we have

p (z) ≺ q (z) =
1

g (z)
=

1

1/β

1∫
0

t1/β−1
(

1−tz
1−z

)−2(γ−1)/β
dt

=

{
2F1

(
2

β
(1 −γ) , 1;

1

β
+ 1;

z

z − 1

)}−1
.

which is the required result. �



Int. J. Anal. Appl. 16 (6) (2018) 790

Theorem 2.4. If f ∈ k −UMα. Then

Re
zf′ (z)

f (z)
>

1

g (−1)
= γ0 =

{
2F1

(
2

β
(1 −γ) , 1;

1

β
+ 1;

1

2

)}−1
.

In other words k −UMα ⊂S∗ (γ0) , where

γ0 =

{
2F1

(
2

β
(1 −γ) , 1;

1

β
+ 1;

1

2

)}−1
. (2.12)

Proof. To prove k −UMα ⊂ S∗ (γ0) , we show that inf
|z|<1

{Re q (z)} = q (−1) . Now for a = 2
β

(1 −γ) , b =
1
β
, c = 1

β
+ 1, we have

g (z) = (1 + Bz)
a

1∫
0

tb−1 (1 + Btz)
−a
dt.

=
Γ (b)

Γ (c)
2F1

(
1,a,c;

z

z − 1

)
. (2.13)

To prove that inf
|z|<1

{Re q (z)} = q (−1) , we need to show that

Re{1/g (z)}≥ 1/g (−1) .

Now by using Lemma 1.4, (2.13) it can easily follows that

g (z) =

1∫
0

g (z,t) dε (t) ,

where

g (z,t) =
1 −z

1 − (1 − t) z
, (0 ≤ t ≤ 1) ,

dε (t) =
Γ (b)

Γ (a) Γ (c−a)
ta−1 (1 − t)c−a−1 dt,

which is a positive measure on [0, 1] . It is clear that Re g (z,t) > 0 and g (−r,t) is real for 0 ≤ |z| ≤ r < 1

and t ∈ [0, 1] . Also

Re

{
1

g (z,t)

}
= Re

{
1 − (1 − t) z

1 −z

}
≥

1 + (1 − t) r
1 + r

=
1

g (−r,t)

for |z| ≤ r < 1. Therefore using Lemma 1.4, we have

Re{1/g (z)}≥ 1/g (−r) .

Now letting r → 1−, it follows

Re{1/g (z)}≥ 1/g (−1) .

Therefore k −UMα ⊂S∗ (γ0) . �



Int. J. Anal. Appl. 16 (6) (2018) 791

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	1. Introduction
	2. Results and Discussion
	References