International Journal of Analysis and Applications

Volume 19, Number 6 (2021), 890-903

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

DOI: 10.28924/2291-8639-19-2021-890

GENERALIZED CLOSE-TO-CONVEXITY RELATED WITH BOUNDED BOUNDARY

ROTATION

KHALIDA INAYAT NOOR1, MUHAMMAD ASLAM NOOR1,∗ AND MUHAMMAD

UZAIR AWAN2

1COMSATS University Islamabad, Islamabad, Pakistan

2Government College University Faisalabad, Pakistan

∗Corresponding author: noormaslam@gmail.com

Abstract. The class Pα,m[A,B] consists of functions p, analytic in the open unit disc E with p(0) = 1 and

satisfy

p(z) =

(
m

4
+

1

2

)
p1(z) −

(
m

4
−

1

2

)
p2(z), m ≥ 2,

and p1, p2 are subordinate to strongly Janowski function
(

1+Az
1+Bz

)α
, α ∈ (0, 1] and −1 ≤ B < A ≤ 1. The

class Pα,m[A,B] is used to define Vα,m[A,B] and Tα,m[A,B; 0; B1], B1 ∈ [−1, 0). These classes generalize

the concept of bounded boundary rotation and strongly close-to-convexity, respectively. In this paper, we

study coefficient bounds, radius problem and several other interesting properties of these functions. Special

cases and consequences of main results are also deduced.

1. Introduction

Let A denote the class of analytic functions defined in the open unit disc

E = {z : |z| < 1} and be given by

(1.1) f(z) = z +

∞∑
n=2

anz
n, z ∈ E.

Received August 6th, 2021; accepted September 23rd, 2021; published October 28th, 2021.

2010 Mathematics Subject Classification. 30C45.

Key words and phrases. Janowski function; Subordination bounded boundary rotation; univalent; starlike; close-to-convex;

integral operator; coefficient problem.

©2021 Authors retain the copyrights

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

890

https://doi.org/10.28924/2291-8639
https://doi.org/10.28924/2291-8639-19-2021-890


Int. J. Anal. Appl. 19 (6) (2021) 891

Let S ⊂ A be the class of univalent functions in E and let C, S? and K be the subclasses of S consisting of

convex, starlike and close-to-convex functions, respectively. For details, see [3].

For f,g ∈ A, we say f is subordinate to g in E, written as f(z) ≺ g(z), if there exists a Schwartz function

w(z) such that

f(z) = g(w(z)), g(z) = z +

∞∑
n=2

bnz
n.

Furthermore, if the function g is univalent in E, then we have the following

equivalence

f(z) ≺ g(z) ⇔ f(0) = g(0) and f(E) ⊂ g(E).

Convolution of f and g is defined as

(g ∗f)(z) = z +
∞∑
n=2

anbnz
n.

The class Pα[A,B] of strongly Janowski functions is defined as follows.

Definition 1.1. Let p be analytic in E with p(0) = 1. Then p ∈ Pα[A,B], if p(z) ≺
(

1+Az
1+Bz

)α
, where

α ∈ (0, 1] and −1 ≤ B < A ≤ 1.

We denote Pα[0,B1] as Pα[B1], −1 ≤ B1 < 0.

The class Pα[A,B] is generalized as:

Definition 1.2. An analytic function p : p(z) = 1 +
∞∑
n=1

cnz
n is in the class Pα,m[A,B], if and only if, there

exist p1,p2 ∈ Pα[A,B] such that

(1.2) p(z) =

(
m

4
+

1

2

)
p1(z) −

(
m

4
−

1

2

)
p2(z), m ≥ 2.

It is obvious Pα,2[A,B] = Pα[A,B]. For the class P1[A,B] = P [A,B], we refer to [6].

About the class Pα[A,B], we observe the following.

(i) p(z) ≺
(

1+Az
1+Bz

)α
implies p ∈ Pα[A,B] and it can easily be shown that φα(A,B; z) =

(
1+Az
1+Bz

)α
is

convex univalent in E. In fact simple calculation yield that

Reφ′α(A,B; z) ≥ α|A−B|
(1 −|A|)α−1

(1 −|B|)α+1
> 0, z ∈ E.

This shows φα(A,B; z) is univalent in E.

Also

Re

{
(zφ′α(A,B; z))

′

φ′α(A,B; z)

}
≥

T(r)

(1 + Ar)(1 + Br)
,

where

T(r) = 1 −α(A−B)r −ABr2



Int. J. Anal. Appl. 19 (6) (2021) 892

is decreasing on (0, 1) and T(0) = 1.

This implies Re
[
(zφ′α(A,B;z))

′

φ′α(A,B;z)

]
≥ 0 in E.

(ii) For A = 1, B = −1, p ∈ φα(1,−1; z) implies

∣∣ arg p(z)∣∣ ≤ απ
2
, z ∈ E.

Definition 1.3. Let f,g ∈ A, (g∗f)
′(z)
z

6= 0, z ∈ E. Then f ∈ Vα,m[A,B; g], if and only if,

(z(g ∗f)′)′

(g ∗f)′
∈ Pα,m[A,B], z ∈ E,

with F = zf′, F ∈ Rα,m[A,B; g], if and only if, f ∈ Vα,m[A,B; g] in E.

Special Cases.

(i) V1,m[A,B;
z

1−z ] = Vm[A,B] ⊂ Vm[1,−1] = Vm, where Vm is the well known class of functions of

bounded boundary rotation. See, for example, [2, 10, 12].

(ii) R1,m[A,B;
z

1−z ] = Rm[A,B] ⊂ Rm and Rm is the class of functions with bounded radius rotation,

see [9].

(iii) Vα,m[A,B;
z

(1−z)2 ] = Rα,m[A,B;
z

1−z ] = Rα,m[A,B].

Definition 1.4. Let f,g ∈ A with (g ∗ f)(z) 6= 0. Then f ∈ Tα,m[A,B; 0; B1; g], if there exists ψ ∈

Vα,m[A,B; g] such that, for B1 ∈ [−1, 0),

(g ∗f)′

(g ∗ψ)′
∈ Pα[B1], z ∈ E.

We note that T1[A,B; 0;−1; z1−z ] = Tm[A,B]. For certain special cases, see [8, 11, 12].

2. The class Vα,m[A,B; g]

Theorem 2.1. Let f ∈ Vα,m[A,B; g] and let g(z) = z +
∞∑
n=2

bnz
n. Then, with f given by (1.1), An = anbn,

An = O(1)n
σ, σ =

{(m
2

+ 1
)

(1 −ρ) − (ρ + 2)
}
,

where ρ =
(

1−A
1−B

)α
, m ≥ 2(1+ρ)

1−ρ and O(1) denotes a constant.

Proof. Let F = f ∗g. Then F ∈ Vα,m[A,B]. Since p ∈ Pα[A,B] implies Rep(z) > ρ, ρ =
(

1−A
1−B

)α
, it follows

that Vα,m[A,B] ⊂ Vm(ρ).

Now, F ∈ Vm(ρ), we can write

(2.1) F ′1(z) = (F
′
1(z))

1−ρ
, F1 ∈ Vm,



Int. J. Anal. Appl. 19 (6) (2021) 893

see [13].

Using a result due to Brannan [2], we can write

(2.2) zF ′1(z) =
(s1(z))

( m4 +
1
2 )(1−ρ)

(s2(z))
( m4 −

1
2 )(1−ρ)

, s1,s2 ∈ S?.

Therefore, from (2.1), (2.2) and Cauchy Theorem with z = reιθ, we have

n2|An| ≤
1

2πrn

∫ 2π
0

∣∣F ′1(z)h(z)∣∣1−ρdθ, h ∈ Pα,m[A,B] ⊂ Pm(ρ)
=

1

2πrn+1

∫ 2π
0

|s1(z)|(
m
4
+ 1

2 )(1−ρ)

|s2(z)|(
m
4
−1

2 )(1−ρ)
· |h(z)|1−ρdθ.(2.3)

Applying distortion result for s2 ∈ S? and Holder’s inequality in (2.3), we get

n2|An| ≤
1

rn+1

(
4

r

)( m4 −12 )(1−ρ) ( 1
2π

∫ 2π
0

∣∣s1(z)∣∣{( m4 + 12 )(1−ρ)} 21+ρ dθ)
1+ρ
2

·
(

1

2π

∫ 2π
0

∣∣h(z)∣∣2dθ)1−ρ2(2.4)
Now, for h(z) = 1 +

∞∑
n=1

cnz
n, we use Parsval identity to have

1

2π

∫ 2π
0

∣∣h(z)∣∣2dθ = ∞∑
n=0

|cn|2r2n

≤ 1 + m2(1 −ρ)2
∞∑
n=1

r2n

=
1 + [m2(1 −ρ)2 − 1]r2

1 −r2
,(2.5)

where we have used coefficient bounds |cn| ≤ m(1 −ρ), for h ∈ Pm(ρ).

From (2.5) together with subordination for starlike functions, and a result due to Hayman [5] for m ≥ 2(1+ρ)
1−ρ ,

we have

(2.6) n2|An| ≤ c1(m,ρ)
(

1

1 −r

){( m2 +1)(1−ρ)}−ρ
,

where c1(m,ρ) denotes a constant.

Taking r = 1 − 1
n

in (2.6), we obtain the required result. �

Special Cases.

(i) Let g(z) = z
1−z , then An = an. Take A = 0, and in this case f ∈ Vm. This leads us to a known

coefficient result that an = O(1)n
( m2 −1).

(ii) Let f ∈ V1,m
[
0,−1, z

(1−z)2

]
= Rm

(
1
2

)
. Then an = O(1)n

m
4
−2, m ≥ 6.



Int. J. Anal. Appl. 19 (6) (2021) 894

Theorem 2.2. Let f ∈ Vα,m[A,B; g]. Then, for F = f ∗g, z = reιθ,

0 ≤ θ1 < θ2 ≤ 2π, we have

(2.7)

∫ θ2
θ1

Re

{
(zF ′(z))′

F ′(z)

}
dθ > −

(m
2
− 1
)

(1 −ρ)π, ρ =
(

1 −A
1 −B

)α
.

Proof. Proof is straight forward, since Vα,m[A,B] ⊂ Vm(ρ) and F ∈ Vm(ρ) implies there exist F1 ∈ Vm with

F ′(z) = (F ′1(z))
(1−ρ). Now, using essentially the same method given in [2], the required result follows. �

Remark 2.1. Let β
(
m
2
− 1
)

(1 − ρ). Then, from a result of Goodman [4] and from (2.7), it follows that

F = f∗g ∈ Vα,m[A,B] is univalent for β =
(
m
2
− 1
)

(1−ρ) ≤ 1. That is F ∈ S for m ≤ 2(2−ρ)
1−ρ . As a special

case, with g(z) = z
1−z , A = 0, B = −1 and α = 1, we have F = f, ρ =

1
2

. Then f ∈ V1,m[0,−1] implies

∫ θ2
θ1

Re

{
(zF ′(z))′

F ′(z)

}
dθ > −

(
m

4
−

1

2

)
π

For this, we can conclude that

V1,m[0,−1] ⊂ S for 2 ≤ m ≤ 6.

Also, with g(z) = z
1−z , A = 1, B = −1, we have a well known result that f ∈ Vm is univalent for 2 ≤ m ≤ 4.

Theorem 2.3. Let f ∈ Vα,m[A,B; g], m ≤
2(2−ρ)
1−ρ and ρ =

(
1−A
1−B

)α
. Then F(E) with F = f ∗g, contains

the disc d:

d =

{
w : |w| <

4

8 + αm|A−B|

}
Proof. From Theorem 2.2, F is univalent in E. Let w0 (w0 6= 0) be any complex number such that F(z) 6= w0

for z ∈ E. Then the function

F1(z) =
w0F(z)

w0 −F(z)
= z +

(
A2 +

1

w0

)
z2 + . . .

is analytic and univalent in E. Using the well known Bieberbach Theorem for the best bound for second

coefficient of univalent functions, see [3], we have

1

|w0|
− |A2| ≤

∣∣A2 + 1
w0

∣∣ ≤ 2.
This gives us

1

|w0|
≤ 2 + |A2|

≤ 2 +
αm|A−B|

4
=

8 + αm|A−B|
4

.

This completes the proof. �



Int. J. Anal. Appl. 19 (6) (2021) 895

Special Cases.

(i) Let A = 1, B = −1, α = 1; (ρ = 0) and so F(E) contains the disc |w| < 2
4+m

, m ≤ 4.

(ii) With A = 0, B = −1, α = 1
2
, we have ρ = 1

4
, and F(E) contains the disc |w| < 8

16+m
, m ≤ 14

3
.

The following properties of the class Vα,m[A,B; g] can easily be proved with simple computations and well

known results and therefore we omit the proof.

Theorem 2.4. (i) The class Vα,m[A,B; g] is preserved under the integral operator L : A → A defined

as

L(z) =

∫ z
0

(L′1(ξ))
β

(L′2(ξ))
γ
dξ,

where Li ∈ Vα,m[A,B; g], i = 1, 2 and β,γ are positively real with β + γ = 1.

(ii) Let f ∈ Vα,m
[
A,B; z

1−z

]
. Then, with ρ =

(
1−A
1−B

)α
, z ∈ E and z = reιθ, we have

(1 −Br)(1−ρ)(
m
4
+ 1

2 )

(1 + Br)(1−ρ)(
m
4
−1

2 )
≤ |f′(z)| ≤

(1 + Br)(1−ρ)(
m
4
+ 1

2 )

(1 −Br)(1−ρ)(
m
4
−1

2 )
.

For α = 1, f ∈ Vm[A,B] and A = 1, B = −1,the result reduces to f ∈ Vm studied in [2].

(iii) Let f ∈ Vα,2
[
A,B; z

1−z

]
and define F ∈ A as

F(z) =
β + 1

zβ

∫ z
0

tβ−1f(t)dt,β > 0.

Then F is convex of order γ(ρ), ρ =
(

1−A
1−B

)α
, where

γ = γ(ρ) =

{
(β + 1)

2F1
(
2(1 −ρ), 1; (β + 2); 1

2

) −β
}
,

2F1 represents Gauss hypergeometric function.

(iv) The set of all points log f′(z) for a fixed z ∈ E and f ranging over the class Vα,m[A,B; g] is convex.

(v) Let f ∈ Vα,m
[
A,B; z

1−z

]
, B 6= 0. Then f is close-to-convex for |z| < r1, where

r1 =

{
sin

(
π

B(γ − 2)

)
, B 6= 0, m >

2

γ
, γ = 1 −

(
1 −A
1 −B

)α
(vi) Let f ∈ Vα,m [A,B; g], and let F = f ∗ g. Then F is convex of order ρ =

(
1−A
1−B

)α
for |z| < rm,

where

r(m) =
m−

√
m2 − 4
2

, m ≥ 2.

Theorem 2.5. Let f1,f2 ∈ Vα,m [A,B; g], β, δ, c and ν be positively real,

c ≥ β ≥ 1, (ν + δ) = β. Let F = F1 ∗g, Gi = fi ∗g, i = 1, 2 and define

(2.8) [F(z)]
β

= cz(β−c)
∫ z
0

tc−1 (G′1(t))
δ

(G′2(t))
ν
dt.



Int. J. Anal. Appl. 19 (6) (2021) 896

Then, for z = reιθ, 0 ≤ θ1 < θ2 ≤ 2π, zF
′

F
= p, we have

∫ θ2
θ1

Re

{
p(z) +

1
β
zp′(z)

p(z) + 1
β

(c−β)

}
dθ > −(1 −ρ)

(m
2
− 1
)
π, ρ =

(
1 −A
1 −B

)α
.

Proof. First we show that there exists a function F ∈ A satisfying (2.8). We assume F1 ∗g 6= 0, fi ∗g 6= 0,

z ∈ E. Let

Q(z) = (G′1(z))
δ

(G′2(z))
ν

= 1 + d1z + d2z
2 + . . .

and choose the branches which equal 1, when z = 0.

For K(z) = zc−1 (G′1(z))
δ

(G′2(z))
ν

= zc−1Q(z), we have

N(z) =
c

zc

∫ z
0

K(t)dt = 1 +
c

c + 1
d1z + . . .

Hence N is well defined and analytic.

Now let

F(z) =
[
zβN(z)

] 1
β = z [N(z)]

1
β ,

where we choose the branch of [N(z)]
1
β which equal 1 when z = 0. Thus F ∈ A and satisfies (2.8). We write

(2.9)
zF ′(z)

F(z)
= p(z), F = F1 ∗g.

From (2.8) and (2.9) with some calculations

βp(z) +
βzp′(z)

(c−β) + βp(z)
= δ

[
(zG′1(z))

′

G′1(z)

]
+ ν

[
(zG′1(z))

′

G′2(z)

]
.

That is

p(z) +

1
β
zp′(z)

p(z) + 1
β

(c−β)
=

]

δ
β

[
(zG′1(z))

′

G′1(z)

]
+
ν

β

[
(zG′1(z))

′

G′2(z)

]
.

We now apply Theorem 2.2 and obtain the required result.

For m ≤ 2(2−ρ)
1−ρ and applying a result proved in [14], it can easily be deduced that∫ θ2

θ1

Re{p(z)}dθ > −π, p(z) =
z(F1 ∗g)′

F1 ∗g
.

Taking g(z) = z
(1−z)2 , it follows that F1 ∈ S in E, see [4]. �

3. The class Tα,m[A,B; 0; B1; g]

Theorem 3.1. Let f ∈ Tα,m
[
A,B; 0; B1;

z
1−z

]
= Tα,m[A,B; 0; B1]. Then, for z = re

ιθ, 0 ≤ θ1 < θ2 ≤ 2π,

ρ1 =
(
1
2

)α
, ρ =

(
1−A
1−B

)α
,

∫ θ2
θ1

Re

{
(zf′(z))

′

f′(z)

}
dθ > −βπ, β =

[
(1 −ρ1) +

(m
2
− 1
)

(1 −ρ)
]
.



Int. J. Anal. Appl. 19 (6) (2021) 897

Proof. For f ∈ Tα,m [A,B; 0; B1], we can write

f′(z)

ψ′(z)
= h(z), ψ ∈ Vα,m[A,B], h ∈ Pα[0,B1].

To prove this result, we shall essentially follow the method due to Kaplan [4].

For ψ ∈ Vα,m[A,B], it implies that ψ ∈ Vm(ρ), where ρ =
(

1−A
1−B

)α
.

Also h ∈ Pα[0,B1], B1 ∈ [−1, 0) is equivalent to h ≺
(

1
1+B1z

)α
. That is,

h ∈ P(α1) ⊂ P, α1 =
(
1
2

)α
.

Now, with z = reιθ, write p(z) = arg f′(z) and q(z) = arg ψ′(z). Then

(3.1) |p(z) −q(z)| <
(

1 −
(

1

2

)α)
π

2

Let P(r,θ) = p(reιθ) + θ, Q(r,θ) = q(reιθ) + θ be defined for 0 ≤ r < 1 and

for all θ. This gives us

(3.2) |P(r,θ) −Q(r,θ)| <
(

1 −
(

1

2

)α)
π

2
.

From Theorem 2.2, for ψ ∈ Vα,m[A,B] ⊂ Vm(ρ), we have∫ θ2
θ1

Re

{
(zψ′(z))

′

ψ′(z)

}
dθ > −

(m
2
− 1
)(

1 −
(

1 −A
1 −B

)α)
π, (z = reιθ).

Thus

(3.3) |Q(r,θ1) −Q(r,θ2)| <
(

1 −
(

1 −A
1 −B

)α)(m
2
− 1
)
π.

From (3.2) and (3.3), it follows that

|P(r,θ1) −P(r,θ2)|

= |{P(r,θ1) −Q(r,θ1)}−{P(r,θ2) −Q(r,θ2)} + {Q(r,θ1) −Q(r,θ2)}|

<

(
1 −

(
1

2

)α)
π

2
+

(
1 −

(
1

2

)α)
π

2
+

(
1 −

(
1 −A
1 −B

)α)(m
2
− 1
)
π

=

[(
1 −

(
1

2

)α)
+

(
1 −

(
1 −A
1 −B

)α)(m
2
− 1
)]
π

=
[
(1 −ρ1) +

(m
2
− 1
)

(1 −ρ)
]
π = βπ,

and this proves our result. �

Special Cases.

(i) Let α = 1, A = 1 and B = −1. Then β = m−1
2

= 1 for m = 3. This implies

f ∈ T1,m[1,−1; 0;−1] is univalent for 2 ≤ m ≤ 3.

(ii) For A = 0, B = −1, α = 1 we have β = m
4

and, in this case, f is univalent for 2 ≤ m ≤ 4.



Int. J. Anal. Appl. 19 (6) (2021) 898

Remark 3.1. For F ∈ A, Goodman [4] introduced a class K(β) as∫ θ2
θ1

Re

{
(zF ′(z))

′

F ′(z)

}
dθ > −βπ, z = reιθ, 0 ≤ θ1 < θ2 ≤ 2π,

and β ≥ 0. When 0 ≤ β ≤ 1, K(β) consists of univalent functions

(close-to-convex), whilest for β > 1, F need not even be finitely-valent, see [4].

We note that, for ρ1 =
(
1
2

)α
, ρ =

(
1−A
1−B

)α
.

Tα,m[A,B, 0,B1] ⊂ K
(m

2
(1 −ρ) + (ρ−ρ1)

)
.

This implies F ∈ Tα,m[A,B; 0;−1] is univalent for m ≤ 2
[
1 + ρ1

1−ρ

]
.

Theorem 3.2. For g(z) = z
1−z , let f ∈ Tα,2[A,B, 0,B1] and for γ,β > 0, let

F1 be defined by

(3.4) F1(z) =

[
(1 + β)z−β

∫ z
0

tβ−1fγ(t)dt

]1
γ

.

Then F1 ∈ T1,2[A,B; 0; B1] in E.

Proof. We can write (3.4) as

(3.5) F1(z) =

[(
f(z)

z

)γ
∗
(
φγ,β(z)

z

)]1
γ

,

where

(3.6) φγ,β(z) =

∞∑
n=1

(
zn

n + γ + β

)
is convex in E.

Since f ∈ Tα,2[A,B; 0; B1], there exists ψ1 = zψ′ ∈ Rα,2[A,B]

such that f
′

ψ′
∈ Pα[0,B1], ψ = Vα,2[A,B] in E. Let

(3.7) G1(z) =

[
(β + 1)z−β

∫ z
0

tβ−1ψ
γ
1 (t)dt

]1
γ

, G1 = zG
′.

We first show that G ∈ Vα,2[A,B].

From (3.7), it follows that

(3.8)
{
zβG

γ
1 (z)

}′
= zβ−1 (ψ

γ
1 (z))

That is

(3.9) (G
γ
1 (z)) [β + γH1(z)] = ψ

γ
1 (z), H1(z) =

zG′1(z)

G1(z)

Logarithmic differentiation of (3.9) and simple computations give us

(3.10) H1(z) +
zH′1(z)

γH1(z) + β
=

zψ′1
ψ1(z)

≺
(

1 + Az

1 + Bz

)α
≺
(

1 + Az

1 + Bz

)
.



Int. J. Anal. Appl. 19 (6) (2021) 899

Now, using Theorem 3.3 of [7, p: 109], It follows from (3.10) that H1 ∈ P [A,B] and G1 = zG′ belongs to

R1,2[A,B] = S
?[A,B]. Therefore G ∈ V1,2[A,B] = C[A,B].

From (3.4), we have

zF ′1(z)F
γ−1
1 (z)

G
γ
1 (z)

=
φγ,β(z) ∗z

(
ψ1(z)
z

)γ (
zf′(z) · f

γ−1(z)
ψ
γ
1 (z)

)
φγ,β(z) ∗z

(
ψ1(z)
z

)γ
=
φγ,β(z) ∗z

(
ψ1(z)
z

)γ
h(z)

φγ,β(z) ∗z
(
ψ1(z)
z

)γ , h ∈ Pα(B1).
Since h(z) is analytic in E, h(0) = 1, and φγ,β(z) is convex, ψ1 ∈ S?, we use a result due to Ruscheweyh

and Sheil-Small [17] to conclude that
(
zF′1F

γ−1

G
γ
1

)
(E) ⊂ C̄oh(E), where C̄oh(E) denotes convex hull of h(E).

This implies F1 ∈ T1,2[A,B; 0; B1] in E. or γ = 1 in (3.4), we obtain the well known Bernardi integral

operator, see [7]. �

Theorem 3.3. Let F = f ∗g, f ∈ Tα,m[A,B; 0; B], B 6= 0. Then with

ρ =
(

1−A
1−B

)α
and γ = A−B

3B
,

(i)

(3.11)

(
1

1 + Br

)α
(1 −Br)γ(1−ρ)(

m
4
+ 1

2
)

(1 + Br)γ(1−ρ)(
m
4
−1

2
)
≤ |F ′(z)| ≤

(1 + Br)γ(1−ρ)(
m
4
+ 1

2
)

(1 −Br)γ(1−ρ)(
m
4
−1

2
)
·
(

1

1 −Br

)α
(ii)

2γ(1−ρ)

a|B|
[
G12(a,b; c;−1) −r−a1 G12(a,b; c;−r1)

]
≤ |F(z)|

≤
2γ(1−ρ)

a|B|
·
[
G12(a,b; c;−1) −r−a2 G12(a,b; c;−r2)

]
,

where r1 = −r−12 =
1+Br
1−Br , m ≤

[
4(1−α)
γ(1−ρ) + 2

]
and a is given in (3.16).

Proof. We can write for F ∈ Tα,m[A,B; 0; B],

F ′(z) = G′(z)h(z), h ∈ Pα[B1], G = ψ ∗g ∈ Vα,m[A,B].

Since h ∈ Pα[B], it easily follows that

(3.12)

(
1

1 + Br

)α
≤ |h(z)| ≤

(
1

1 −Br

)α
From Theorem 2.4 (ii) and (3.12), the proof of (i) is established.

We know proceed to prove (ii).

Let dr denote the radius of the largest schlicht disc centered at the origin contained in the image of |z| < r

under F(z). Then there is a point z0, |z0| = r, such that |F(z0)| = dr. The ray from 0 to F(z0) lies entirely



Int. J. Anal. Appl. 19 (6) (2021) 900

in the image and the inverse image of this ray is a curve in |z| < r.

Using (3.11), we have

dr = |F(z0)| =
∫
C

|F ′(z)||dz|, r =
A−B

2B

≥
∫ |z|
0

[
(1 −Bs)γ{(1−ρ)(

m
4
+ 1

2
)}

(1 + Bs){γ(1−ρ)(
m
4
−1

2
)+α}

]
ds

=

∫ |z|
0

[(
1 −Bs
1 + Bs

){γ(1−ρ)(m
4
−1

2
)+α}

· (1 −Bs)γ(1−ρ−α)
]
ds,(3.13)

Let 1+Bs
1−Bs = t. Then

2B
(1−Bs)2 = dt, and 1 − Bs =

2
1+t

. This implies ds = 2
B

(
1

1+t

)2
dt. Therefore, from

(3.13), we have

|F(z0)| ≥
∫ 1+Br

1−Br

1

t−{(1−ρ)(
m
4
−1

2
)−α} ·

(
2

1 + t

)1−ρ−α
·

2

B

(
1

1 + t

)2
dt

=
−2(1−ρ)

|B|

[∫ 1+Br
1−Br

0

tγ(1−ρ)(
m
4
−1

2
)−α · (1 + t)γ(1−ρ−α)dt

−
∫ 1
0

tr(1−ρ)(
m
4
−1

2
)+α · (1 + t)r(1−ρ−α)dt

]
=

2γ(1−ρ)

|B|
[I1 + I2].(3.14)

Now put t = r1u with r1 =
1+Br
1−Br . Then dt = r1du and

I1 =

∫ 1
0

(r1u)
−[γ(1−ρ)(m4 −

1
2
)−α] · (1 + r1u)1−ρ−αdu

= r
−{γ(1−ρ)(m

4
−1

2
)+α−1}

1

∫ 1
0

u−γ(1−ρ)(
m
4
−1

2
)−α · (1 + r1u)−{γ(1−ρ)+α}du

= r
−{γ(1−ρ)(m

4
−1

2
)+α−1}

1 ·
Γ(a)Γ(c−a)

Γ(c)
G12(a,b; c;−r1),(3.15)

where Γ and G12, respectively denote gamma and Gauss hypergeometric functions. Also, here, b,c are

positively real for m ≤ 2
{

1 +
2(1−α)
1−ρ

}
and are given as

a = −γ(1 −ρ)
(
m

4
−

1

2

)
−α + 1, γ =

A−B
2B

, B 6= 0

b = −γ(1 −ρ) + α,

c = −γ(1 −ρ)
(
m

4
−

1

2

)
−α + 2, (c−a) > 0.(3.16)

Similarly, we calculate I2 and have

I2 =
Γ(a)Γ(c−a)

Γ(c)
G12(a,b; c;−1).(3.17)



Int. J. Anal. Appl. 19 (6) (2021) 901

Using (3.15), (3.16) and (3.17) in (3.14), we obtain the lower bound of |F(z)|. For the upper bound, we

proceed in similar way and have

|F(z)| ≤
∫ |z|
0

(1 + Bs)γ(1−ρ)(
m
4
+ 1

2
)

(1 −Bs)γ(1−ρ)(
m
4
−1

2
)
·
(

1

1 −Bs

)α
ds

=

∫ |z|
0

(
1 + Bs

1 −Bs

)γ(1−ρ)(m
4
−1

2
)+α

· (1 + Bs)(1−ρ−α)ds.

Now similar computations yield the required bound and the proof is complete. �

By choosing suitable and permissible values of involved parameters, we obtain several new and also known

results.

Remark 3.2. (i) We use a result of Pommerenke [16] together with

Theorem 3.1 and easily deduce that the class Tα,m[A,B; 0;−1],

m ≤ 2
{

1 + ρ1
1−ρ

}
, ρ1 =

(
1
2

)α
, ρ =

(
1−A
1−B

)α
, is a linearly invariant family of order B2 ={

m
2

(1 −ρ) + (ρ−ρ1) + 1
}

. With similar argument given in [16], we have the covering result for

Tα,m[A,B; 0;−1] as:

The image of E under F = f ∗g ∈ Tα,m[A,B; 0;−1] contains the Schlicht disc |z| = 12B2 , where

B2 =
{
m
2

(1 −ρ) + 1 + ρ−ρ1
}

, and F(z) = z +
∞∑
n=2

Anz
n.

(ii) Let F∗ be defined as

F∗(z) =
1

B1

[(
1 + z

1 −z

)B2
− 1

]
= z +

∞∑
n=2

A∗nz
n,

where

B1 =
{m

2
(1 −ρ) + (ρ−ρ1) + 2

}
,

B2 =
{m

2
(1 −ρ) + (ρ−ρ1) + 1

}
.

It can be shown, with some computations, that F∗ belongs to the linearly invariant family of Tα,m[A,B; 0;−1].

Using this concept, together with the same argument of Pommerenke [16], we have |An| ≤ |A∗n|, n ≥ 1

and Lr(F) ≤ Lr(F∗), F ∈ Tα,m[A,B; 0;−1] when Lr(F) is the length of the image of the circle |z| = r

under F , 0 ≤ r < 1.

Theorem 3.4. Let f ∈ Tα,m[A,B; 0;−1; g] and let F = f ∗g 6= 0 in E with

F(z) = z +

∞∑
n=2

Anz
n.

Then, for m > 2,

An = O(1) ·nγ1, γ1 =
{m

2
(1 −ρ) + [ρ1 − (1 + ρ)]

}
,

where O(1) is a constant depending on m, α, A and B only.



Int. J. Anal. Appl. 19 (6) (2021) 902

Proof. For F ∈ Tα,m[A,B; 0;−1], we can write

F ′ = G′h, G ∈ Vα,m[A,B], G = ψ ∗g, ψ ∈ Vα,m[A,B; g].

Since Vα,m[A,B] ⊂ Vm(ρ), ρ =
(

1−A
1−B

)α
, and it is well known that there exists Gi ∈ Vm such that

G′(z) = (G′(z))
1−ρ

for z ∈ E.

Also h ≺
(

1
1+z

)α
, which implies |arg h(z)| < ρ1π

2
, ρ1 =

(
1
2

)α
.

Therefore we have

F ′ = (G′1)
1−ρ

(h1)
ρ1, Reh1 > 0

in E.

For G1 ∈ Vm, there exists s ∈ S? such that G′1 = sh
(m

2
−1)

2 , m > 2 and

Reh2 > 0 in E, see [1].

Thus, for F ∈ Tα,m[A,B; 0;−1], it follows that

(3.18) F ′ = (s)1−ρ(h2)
(1−ρ)(m

2
−1)(h1)

ρ1, hi ∈ P, i = 1, 2

So, by Cauchy Theorem and (3.18), we have for z = reιθ.

n|An| ≤
1

2πrn

∫ 2π
0

|s|1−ρ|h1|ρ1|h2|(1−ρ)(
m
2
−1)dθ

≤
1

rn

(
r

(1 −r)2

)(1−ρ) ( 1
2π

∫ 2π
0

|h1|2dθ
)ρ1

2

·
(

1

2π

∫ 2π
0

|h2|
2δ

2−ρ1 dθ

)2−ρ1
2


 ,

where δ = (1 −ρ)
(
m
2
− 1
)

and we have used distortion result for s ∈ S? and Holder inequality.

Now, for m >
{

2 + 2−ρ1
1−ρ

}
, we apply a result due to Hayman [5] for hi ∈ P and obtain

(3.19) n|An| ≤ c(ρ,ρ1,m) ·
(

1

1 −r

)1+δ+ρ1−2ρ
where c(ρ,ρ1,δ) is a constant.

Setting r = 1 − 1
n

, n →∞ in (3.19), the required result follows as

An = O(1) ·n{
m
2
(1−ρ)+[ρ1−(ρ+1)]}, ρ1 =

(
1

2

)α
, ρ =

(
1 −A
1 −B

)α
,

and m >
{

2 + 2−ρ1
1−ρ

}
, n ≥ 2. �

Special Cases.

(i) A = 1 implies that ρ = 0 and for α = 1, ρ1 =
1
2
. Then

An = O(1) ·n
m
2
−1

2 , m >
7

2

Taking m = 4, we have An = O(1) ·n
3
2 .



Int. J. Anal. Appl. 19 (6) (2021) 903

(ii) A = 1
2
, B = −1, α = 1 ⇒ ρ = 1

4
. Also ρ1 =

1
2
. Then m = 5 > 4 implies An = O(1) ·n

9
8 .

Acknowledgement: This research was supported by HEC Pakistan under project No: 8081/Punjab/NRPU

/R&D/HEC/2017.

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

of this paper.

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ment. Math. Helv. 48(1973), 119-135.


	1. Introduction
	2. The class V, m [A, B; g]
	3. The class T, m [A, B; 0; B1; g]
	References