International Journal of Analysis and Applications

Volume 17, Number 6 (2019), 958-973

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

DOI: 10.28924/2291-8639-17-2019-958

ON GENERALIZED K-UNIFORMLY CLOSE-TO-CONVEX FUNCTIONS OF

JANOWSKI TYPE

AFIS SALIU1,2,∗

1Department of Mathematics, COMSATS University, Chak Shahzad, Islamabad, 44000, Pakistan

2Department of Mathematics, Gombe State University, Gombe State, Nigeria

∗Corresponding author: afissaliu@gsu.edu.ng

Abstract. This work is concerned with the class of analytic functions that maps open unit disk onto conic

domains. Necessary condition, arc length, growth rate of coefficients, radius problems and property of some

integral transformation under the class are examined.

1. Introduction

Let A be the class of functions f(z) whose series representation is given by

f(z) = z +

∞∑
n=2

anz
n, (1.1)

and regular in an open unit disk ∆ = {z : |z| < 1}. Let S denotes the class of univalent functions in ∆

and C(ρ),S∗(ρ) and K(ρ), 0 ≤ ρ < 1 be the well known subclasses of S which consist of convex, starlike

and close-to-convex functions of order ρ respectively. C(0) ≡ C,S∗(0) ≡ S∗ and K(0) ≡ K are the classes

of convex, starlike and close-to-convex functions in ∆ respectively. A function f ∈ A is subordinate to

g ∈ A (written as f(z) ≺ g(z)) if there exists a function w(z) with |w(z)| < 1 and w(0) = 0 such that

f(z) = g(w(z)). In addition, if g(z) is univalent in ∆, then f(0) = g(0) and f(∆) ⊂ g(∆) [4].

2010 Mathematics Subject Classification. 30C45, 30C50, 30C55.

Key words and phrases. Analytic functions; Janowski functions; Conic domains.

c©2019 Authors retain the copyrights

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

958

https://doi.org/10.28924/2291-8639
https://doi.org/10.28924/2291-8639-17-2019-958


Int. J. Anal. Appl. 17 (6) (2019) 959

Let H be the class of functions p(z) = 1 +
∞∑
n=1

cnz
n that are regular in ∆ with p(0) = 1. Then p ∈

P [A,B], −1 ≤ B < A ≤ 1 if and only if p (z) ≺ 1+Az
1+Bz

, or equivalently

p (z) =
(A + 1)h(z) − (A− 1)
(B + 1)h(z) − (B − 1)

,

where h ∈ P[1,−1] = P, the class of functions with positive real part. This class of functions was first

considered and study extensively by Janowski [6].

Kanas and Wisniowska [7,8] introduced the class of k-uniformly convex functions and k-uniformly starlike

function denoted by k − UCV and k − UST respectively, which were defined subject to the conic domain

Ωk,k ≥ 0, given by

Ωk = {u + iv : u > k
√

(u− 1)2 + v2 } (1.2)

This domain represents the right half plane for k = 0, the right branch of hyperbola for 0 < k < 1 and an

ellipse when k > 1. The function pk(z) plays an extremal role for all functions that maps ∆ onto Ωk and it

is given by

pk(z) =




1+z
1−z , k = 0,

1 + 2
π2

(
log

1+
√
z

1−
√
z

)2
, k = 1,

1 + 2
1−k2 sinh

2
[
( 2
π

arccos k) arctan
√
z
]
, 0 < k < 1,

1 + 1
1−k2 sin

[
π

2R(t)

u(z)√
t∫

0

1√
1−x2
√

1−(tx)2
dx

]
+ 1

k2−1, k > 1,

(1.3)

where u(z) = z−
√
t

1−
√
tz
, t ∈ (0, 1),z ∈ ∆ and t is chosen such that k = cosh(πR

′(t)
4R(t)

), R(t) is Legendre’s complete

elliptic integral of the first kind and R′(t) is the complementary integral of R(t) [7].

We denote by P(pk), the class of functions that are subordinate to pk(z). Ronning [20] proved that for

p ∈ P(pk), there exists a function h ∈P such that p(z) = hγ(z) and γ is given as :

γ =
2

π
arctan

(
1

k

)
(1.4)

It was also shown in [9] that for pk(z) = 1 + δkz + · · · ∈ P(pk),

δk =




8(arccos k)2

π2(1−k2) , 0 ≤ k < 1,
8
π2
, k = 1,

π2

4(k2−1)
√
t(1+t)R2(t)

, k > 1.

(1.5)

Very recently, K.I. Noor [18] extended the domain Ωk to that of Janowski type, Ωk[A,B],

− 1 ≤ B < A ≤ 1 and defined it as

Ωk[A,B] =

{
w: Re

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

)
> k

∣∣∣∣(B − 1)h(z) − (A− 1)(B + 1)h(z) − (A + 1) − 1
∣∣∣∣
}

(1.6)



Int. J. Anal. Appl. 17 (6) (2019) 960

or alternatively,

Ωk[A,B] =
{
u + iv : [(B2 − 1)(u2 + v2) − 2(AB − 1)u + (A2 − 1)]2

>k2[(−2(B + 1)(u2 + v2) + 2(A + B + 2)u− 2(A + 1))2 + 4(A−B)2v2]
}
. (1.7)

Geometrically, the effect of Ωk[A,B] on Ωk was described in [18].

We denote by k−P [A,B], −1 ≤ B < A ≤ 1, the class of functions that map ∆ onto Ωk[A,B]. Equivalently,

we say p ∈ k −P [A,B] if and only if

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

, k ≥ 0, −1 ≤ B < A ≤ 1, (1.8)

where pk is given by (1.3). Also, it is worthy to note that p ∈ k − P [A,B] ⊂ P(β1) which implies that

p(z) = (1 −β1)h1(z) + β1, (see [18]) where h1 ∈P and β1 is given by

β1 =
2k + 1 −A
2k + 1 −B

(1.9)

We extend the class k −P [A,B] as follows :

Definition 1.1. Let p(z) ∈H. Then p ∈ k −Pµ[A,B] if and only if for µ ≥ 2,k ≥ 0,−1 ≤ B < A ≤ 1, we

have

p(z) =
µ + 2

4
p1(z) −

µ− 2
4

p2(z), p1,p2 ∈ k −P [A,B]. (1.10)

Definition 1.2. Let p(z) ∈H. Then p ∈ k −Pµ[A,B,α] if and only if for µ ≥ 2,k ≥ 0,

− 1 ≤ B < A ≤ 1,α ∈ [0, 1), we have

p(z) = (1 −α)h(z) + α, h(z) ∈ k −Pµ[A,B]. (1.11)

For k = 0,A = 1,B = −1,α = 0, we have the class Pµ introduced and studied in [22]. Also, when

µ = 2,α = 0, we get the class k − P [A,B], which was first considered by K.I. Noor in [18]. The class

k −Pµ[1 − 2γ∗,−1, 0], γ∗ ∈ [0, 1) is the same as the class k −Pµ(γ∗) studied in [17].

We now define the following classes of functions.

Definition 1.3. Let f ∈ A. Then f ∈ k −URµ[A,B,α], k ≥ 0, µ ≥ 2, α ∈ [0, 1),−1 ≤ B < A ≤ 1, if and

only if
zf′(z)

f(z)
∈ k −Pµ[A,B,α].

Definition 1.4. Let f ∈ A. Then f ∈ k −UVµ[A,B,α], k ≥ 0, µ ≥ 2, α ∈ [0, 1),−1 ≤ B < A ≤ 1, if and

only if
(zf′(z))′

f′(z)
∈ k −Pµ[A,B,α].

It is obvious to note that f ∈ k −UVµ[A,B,α] ⇔ zf′(z) ∈ k −URµ[A,B,α]. Also,

0 −UVµ[1,−1, 0] = Vµ, is the class of functions of bounded boundary rotation(see [1], [22])



Int. J. Anal. Appl. 17 (6) (2019) 961

Definition 1.5. Let f ∈A. Then f ∈ k−Tµ[A,B,C,D,α], k ≥ 0, µ ≥ 2, α ∈ [0, 1),−1 ≤ B < A ≤ 1,−1 ≤

D < C ≤ 1 if and only if there exists g ∈ k −URµ[C,D,α] such that

zf′(z)

g(z)
∈ k −P [A,B]

or equivalently as

f′(z)

g′(z)
∈ k −P [A,B]

where g ∈ k −UVµ[C,D,α].

We note the following special cases.

i 0 −T2[A,B,C,D, 0] = K[A,B,C,D], is the class of functions studied by Silvia [23].

ii 0 −T2[1,−1, 1,−1, 0] = K, is the class of close to convex functions examined by Kaplan [10].

iii k −T2[A,B,C,D, 0] = k −UK[A,B,C,D], the class considered in [12]

iv 1−T2[1,−1, 1,−1, 0] = UCC, is the class of uniformly close to convex functions explored by Kumar

and Ramesha [11].

v 1−T2[1−2ρ,−1, 1,−1, 0] = UCC(ρ) is the class of uniformly close to convex of order ρ, −1 ≤ ρ < 1

that was taken into account in [24].

vi 0 −Tµ[1 − 2ρ,−1, 1 − 2ρ,−1, 0] = Tµ(ρ),ρ ∈ [0, 1), is the class of functions that was studied by K.I.

Noor in [16]

vii 0 − Tµ[1,−1, 1,−1, 0] = Tµ, is the class of generalized close-to-convex functions introduced and

studied in [15].

Let

G(a,b,c; z) =
Γ(c)

Γ(a)Γ(c−a)

∫ 1
0

ua−1(1 −u)c−a−1(1 −zu)−bdu, (1.12)

where Rea > 0,Re(c−a) > 0, Γ denotes the Gamma function and G(a,b,c; z) is hypergeometric function.

Unless if otherwise stated, we lay down once and for all that k ≥ 0, µ ≥ 2, α ∈ [0, 1),−1 ≤ B < A ≤

1,−1 ≤ D < C ≤ 1.

2. Some Preliminary Lemmas

We need the following lemmas.

Lemma 2.1. [1] Every function f ∈Vµ is a close to convex function of order µ2 − 1.

Lemma 2.2. [13] Let p ∈ P [A,B]. Then

|arg p(z)| ≤ sin−1
(A−B)r
1 −ABr2

. (2.1)



Int. J. Anal. Appl. 17 (6) (2019) 962

Lemma 2.3. [5] Let p(z) ∈H with Rep (z) > 0,z = reiθ(0 < r < 1). Then

2π∫
0

|p(reiθ)|λdθ < C(λ)
1

(1 −r)λ−1
, (2.2)

where C(λ) is a constant and λ > 1.

Lemma 2.4. [1] Let g ∈Vµ. Then for µ > 3 there exists s1 ∈S∗(0) and p ∈P such that

zg′(z) = s1(z)(p(z))
µ
2
−1

Lemma 2.5. [19] Let q(z) be analytic in ∆ with q(0) = 1. If % ≥ 1, Rec ≥ 0, 0 ≤ θ1 < θ2 ≤ 2π, z = reiθ,

then
θ2∫
θ1

Re

{
q(z) +

%zq′(z)

c% + q(z)

}
dθ > −σπ, (0 < σ ≤ 1),

implies

θ2∫
θ1

Req(z)dθ > −σπ.

3. Main Result

In this section, we present our main work.

Theorem 3.1. Let f ∈ k −UVµ[C,D,α]. Then f′(z) = (φ′(z))1−β, where φ ∈ Vµ, β = α + (1 −α)β1 and

β1 is given by (1.9).

Proof. Let f ∈ k −UVµ[C,D,α]. Then

(zf′(z))′

f′(z)
= (1 −α)

[
µ + 2

4
p1(z) −

µ− 2
4

p2(z)

]
+ α, p1,p2 ∈ k −P [C,D] ⊂P(β2).

where β2 =
2k+1−C
2k+1−D . Therefore,

(zf′(z))′

f′(z)
=
µ + 2

4
h1(z) −

µ− 2
4

h2(z), h1,h2 ∈P(β),

where

β = (1 −α)β2 + α =
2k + 1 −C + α(C −D)

2k + 1 −D
. (3.1)

Thus, there exist f1,f2 ∈ C(β) such that

(zf′(z))′

f′(z)
=
µ + 2

4

(zf′1(z))
′

f′1(z)
−
µ− 2

4

(zf′2(z))
′

f′2(z)



Int. J. Anal. Appl. 17 (6) (2019) 963

Integrating and using the result due to Brannan [1], we have

f′(z) =
(f′1(z))

(
µ+2
4

)

(f′2(z))
(
µ−2
4

)

=ϕ′(z), ϕ ∈Vµ(β)

=(φ′(z))1−β, φ ∈Vµ.

�

Theorem 3.2. Let f ∈ k −Tµ[A,B,C,D,α]. Then for µ > 2, 0 ≤ θ1 < θ2 ≤ 2π,z = reiθ,

θ2∫
θ1

Re
(zf′(z))′

f′(z)
dθ > −π

[
(1 −α)

(
C −D

2k + 1 −D

)
(
µ

2
− 1) + γ

]
,

where γ is given by (1.4).

Proof. Let

f′(z) = g′(z)h(z), (3.2)

where g ∈ k −UVµ[C,D,α], h ∈ k −P [A,B]. From Theorem 3.1, we write

zg′(z) = zβ(zφ′(z))1−β.

By logarithmic differentiation and using Lemma 2.1, it follows that

θ2∫
θ1

(zg′(z))′

g′(z)
dθ =β(θ2 −θ1) + (1 −β)

θ2∫
θ1

(zφ′(z))′

φ′(z)
dθ

>−π(1 −β)(
µ

2
− 1). (3.3)

Since h ∈ k − P [A,B], there exists h1 ∈ P [A,B] such that h(z) = (h1(z))γ, where γ is given by (1.4).

Therefore,

f′(z) = g′(z)(h1(z))
γ

and thus,
θ2∫
θ1

Re
(reiθf′(reiθ))′

f′(reiθ)
dθ =

θ2∫
θ1

Re
(reiθg′(reiθ))′

g′(reiθ)
dθ + γ

θ2∫
θ1

Re
reiθh′1(re

iθ)

h1(reiθ)
dθ. (3.4)

But
θ2∫
θ1

Re
reiθh′1(re

iθ)

h1(reiθ)
dθ = arg h1(re

iθ2 ) − arg h1(reiθ1 ),

which implies

max
h1∈P[A,B]

∣∣∣∣∣
θ2∫
θ1

Re
reiθh′1(re

iθ)

h1(reiθ)
dθ

∣∣∣∣∣ ≤ maxh1∈P[A,B] |arg h1(reiθ2 )| + maxh1∈P[A,B] |arg h1(reiθ1 )| . (3.5)



Int. J. Anal. Appl. 17 (6) (2019) 964

Using Lemma 2.2 in (3.5), we get

max
h1∈P[A,B]

∣∣∣∣∣
θ2∫
θ1

Re
reiθh′1(re

iθ)

h1(reiθ)
dθ

∣∣∣∣∣ ≤ π − 2 cos−1 (A−B)r1 −ABr2 .
This means that

θ2∫
θ1

Re
reiθh′1(re

iθ)

h1(reiθ)
dθ ≥−π. (3.6)

Using (3.3) and (3.6) in (3.4), we have the result and this completes the proof. �

Corollary 3.1. [14] Let f ∈ k −Tµ[1,−1, 1,−1, 0]. Then for µ > 2, 0 ≤ θ1 < θ2 ≤ 2π,z = reiθ,

θ2∫
θ1

Re
(zf′(z))′

f′(z)
dθ > −π

(
µ− 2

2(k + 1)
+ γ

)
.

Corollary 3.2. Let f ∈ 1 −Tµ[1,−1, 1,−1, 0]. Then for µ > 2, 0 ≤ θ1 < θ2 ≤ 2π,z = reiθ,

θ2∫
θ1

Re
(zf′(z))′

f′(z)
dθ > −π

µ

4
.

Corollary 3.3. [15] Let f ∈ 0 −Tµ[1,−1, 1,−1, 0]. Then for 0 ≤ θ1 < θ2 ≤ 2π,z = reiθ,

θ2∫
θ1

Re
(zf′(z))′

f′(z)
dθ > −π

µ

2
.

Using Goodman result in [3], we have the following.

Corollary 3.4. The function f ∈ k −Tµ[A,B,C,D,α] is univalent in ∆ for

µ < 2
(

1−γ
1−α

)(
2k+1−D
C−D

)
+ 2.

Remark 3.1. Let f ∈ k−Tµ[A,B,C,D,α]. Then setting
zf′(z)
f(z)

= p(z) in Theorem 3.2 and applying Lemma

2.5, we have
θ2∫
θ1

Re
zf′(z)

f(z)
dθ > −π

[
(1 −α)

(
C −D

2k + 1 −D

)
(
µ

2
− 1) + γ

]
.

Theorem 3.3. Let f ∈ k −Tµ[A,B,C,D,α]. Then

|arg f′(z)| ≤ (1 −β)µ sin−1 r + γ sin−1
(A−B)r
1 −ABr2

,

where γ is given by (1.4).

Proof. Let

f′(z) = (φ′(z))1−β(h(z))γ, φ ∈Vµ,h ∈ P [A,B],



Int. J. Anal. Appl. 17 (6) (2019) 965

where β is given by (3.1). Therefore,

|arg f′(z)| ≤ (1 −β)|arg(φ′(z))||arg(h(z))γ|.

Since it is well known in [21], that for φ ∈ Vµ, |arg(φ′(z))| ≤ µ sin−1 r. Using this results and Lemma 2.2,

the proof is complete. �

Theorem 3.4. Let p(z) = 1 +
∞∑
n=1

cnz
n ∈ k −Pµ[A,B,α]. Then for n ≥ 1,

|cn| ≤
µ(1 −α)(A−B)|δk|

4
,

where δk is given by (1.5).

Proof. Since p ∈ k −Pµ[A,B,α], then we write

p(z) =
µ + 2

4
[(1 −α)h1(z) + α] −

µ− 2
4

[(1 −α)h2(z) + α], h1,h2 ∈ k −P [A,B]. (3.7)

Let h1(z) = 1 +
∞∑
n=1

anz
n and h2(z) = 1 +

∞∑
n=1

bnz
n. Then from (3.7), we have

1 +

∞∑
n=1

cnz
n = 1 +

∞∑
n=1

[
µ + 2

4
an −

µ− 2
4

bn

]
zn.

Comparing the coefficient of zn, we obtain

|cn| ≤ (1 −α)
[
µ + 2

4
|an| +

µ− 2
4
|bn|

]
.

It has been established in [18] that for h(z) = 1 +
∞∑
n=1

dnz
n ∈ k −P [A,B],

|dn| ≤
(A−B)|δk|

2
, n ≥ 1

and δk is given by (1.5). Using this result, it follows that

|cn| ≤
µ(1 −α)(A−B)|δk|

4

and this completes the proof. �

Corollary 3.5. Let p(z) = 1 +
∞∑
n=1

cnz
n ∈ 1 −P2[1,−1, 0]. Then for n ≥ 1,

|cn| ≤
8

π2

Corollary 3.6. [13] Let p(z) = 1 +
∞∑
n=1

cnz
n ∈ 0 −P2[A,B, 0]. Then for n ≥ 1,

|cn| ≤ A−B

Corollary 3.7. [4] Let p(z) = 1 +
∞∑
n=1

cnz
n ∈ 0 −P2[1,−1, 0]. Then for n ≥ 1,

|cn| ≤ 2.



Int. J. Anal. Appl. 17 (6) (2019) 966

Theorem 3.5. Let p(z) = 1 +
∞∑
n=1

cnz
n ∈ k −Pµ[A,B,α]. Then for z = reiθ(0 < r < 1),

1

2π

2π∫
0

|p(z)|2dθ ≤
1 +

[(
µ(1−α)(A−B)|δk|

4

)2
− 1
]
r2

1 −r2
.

Proof. Using Perseval’s identity, we have

1

2π

2π∫
0

|p(z)|2dθ = 1 +
∞∑
n=1

|cn|2r2n.

Applying Theorem 3.4, we get

1

2π

2π∫
0

|p(z)|2dθ ≤1 +
[
µ(1 −α)(A−B)|δk|

4

]2 ∞∑
n=1

r2n

=

1 +

[(
µ(1−α)(A−B)|δk|

4

)2
− 1
]
r2

1 −r2
.

�

Corollary 3.8. For p ∈ 0 −P2[A,B, 0] = P [A,B],

1

2π

2π∫
0

|p(z)|2dθ ≤
1 + [(A−B)2 − 1]r2

1 −r2
.

Corollary 3.9. [16] For p ∈ 0 −P2[1 − 2γ,−1, 0] = P(γ),

1

2π

2π∫
0

|p(z)|2dθ ≤
1 + [µ2(1 −γ)2 − 1]r2

1 −r2
.

Theorem 3.6. [Arc Length Problem] Let f ∈ k −Tµ[A,B,C,D,α]. Then for µ > 2−γ1−β + 2,

Lr(f) ≤ C(γ,β,A,B)
(

1

1 −r

)(1−β)( µ
2

+1)+γ−1

(r → 1),

where C(γ,β,A,B)
(

1
1−r

)
is a constant that only depends on γ,β,A,B, where γ and β are respectively given

by (1.4) and (3.1).

Proof. For f ∈ k −Tµ[A,B,C,D,α], and application of the result due to Brannan [1], we have

zf′(z) = zβ
(
s1(z)(p1(z))

µ
2
−1
)1−β

(h1(z))
γ, (3.8)

where s1 ∈S∗,p1 ∈P and h1 ∈ P [A,B]. Therefore, for z = reiθ (r < 1),

Lr(f) =
2π∫
0

|zf′(z)|dθ



Int. J. Anal. Appl. 17 (6) (2019) 967

Using (3.8), distortion theorem for starlike function s1(z), Hölder’s inequality, Corollary 3.8 and Lemma 2.3,

it follows that

Lr(f) ≤
2π

(1 −r)2(1−β)

(
1

2π

∫ 2π
0

|p1(z)|
(1−β)(µ−2)

2−γ dθ

)2−γ
2
(

1

2π

∫ 2π
0

|h1(z)|2 dθ
)γ

2

≤
2π

(1 −r)2(1−β)

(
1 + [(A−B)2 − 1]r2

1 −r2

)γ
2 (

1

2π

∫ 2π
0

|p1(z)|
(1−β)(µ−2)

2−γ dθ

)2−γ
2

≤C(γ,β,A,B,α)
(

1

1 −r

)(1−β)( µ
2

+1)+γ−1

(r → 1),

where

C(γ,β,A,B,α) = [2π(A−B)2]
γ
2 (C(λ))

2−γ
2

and

λ =
(1 −β)(µ− 2)

2 −γ
> 1.

�

Theorem 3.7. Let f ∈ k −Tµ[A,B,C,D,α]. Then for µ > 2−γ1−β + 2,

an = O(1)n
(1−β)( µ

2
+1)+γ−2 (n →∞),

where O(1) is a constant depending on γ,β,A and B, and γ, β are respectively given by (1.4) and (3.1).

Proof. By Cauchy Theorem, we have for z = reiθ

n|an| ≤
1

2πrn

∫ 2π
0

|zf′(z)|dθ

=
1

2πrn
Lr(f).

Now, by applying Theorem 3.6 and setting r = 1 − 1
n

as n →∞, we have the result. �

Theorem 3.8. Let f ∈ k −Tµ[A,B,C,D,α]. Then for µ > 2−γ1−β + 2,∣∣|an+1|− |an| ∣∣ ≤ D(γ,β,µ,A,B)nη−1,
where η =

2(γ−β−1)+µ(1−β)−2
2

and D(γ,β,µ,A,B) is a constant that depends only on γ,β,µ,A,B, and γ, β

are given by (1.4) and (3.1) respectively.

Proof. For f ∈ k −Tµ[A,B,C,D,α], we write

zf′(z) = zβ
(
s1(z)(p1(z))

µ
2
−1
)1−β

(h1(z))
γ,

where s1 ∈S∗,p1 ∈P and h1 ∈ P [A,B]. Let z1 be complex number with |z1| = r. Then by Cauchy Theorem,∣∣∣∣z1(n + 1)|an+1|−n|an|
∣∣∣∣ ≤ 12πrn+1

∫ 2π
0

|z1 −z|
∣∣∣∣zβ (s1(z)(p1(z)) µ2 −1)1−β (h1(z))γ

∣∣∣∣.



Int. J. Anal. Appl. 17 (6) (2019) 968

Since s1 ∈ S∗, then by a result due to Golusin [2],

|(z1 −z)s1| ≤
2r2

1 −r2
,

we have

∣∣∣∣z1(n + 1)|an+1|−n|an|
∣∣∣∣ ≤ 2rβ+22πrn+1(1 −r2)

∫ 2π
0

∣∣∣∣(s1(z))−β(p1(z))( µ2 −1)(1−β)(h1(z))γ
∣∣∣∣.

Now, using Distortion Theorem for starlike function s1(z), Holder’s inequality, Corollary 3.8 and Lemma

2.3, we have

∣∣∣∣z1(n + 1)|an+1|−n|an|
∣∣∣∣ ≤ 2rn+1(1 −r)1−2β

(
1

2π

∫ 2π
0

|p1(z)|
(1−β)(µ−2)

2−γ dθ

)2−γ
2

×
(

1

2π

∫ 2π
0

|h1(z)|2 dθ
)γ

2

≤
2

rn+1(1 −r)1−2β

(
1 + [(A−B)2 − 1]r2

1 −r2

)γ
2

×
(∫ 2π

0

|p1(z)|
(1−β)(µ−2)

2−γ dθ

)2−γ
2

≤
D1(γ,β,µ,A,B)

rn+1

(
1

1 −r

)2γ−2β+µ(1−β)−2
2

,

where D1 is a constant. Setting r = 1 − 1n and z1 =
n
n+1

, the proof is completed. �

Corollary 3.10. For f ∈ k −T2[A,B,C,D, 0],

∣∣∣∣|an+1|− |an|
∣∣∣∣ ≤ D1(γ,β,A,B)n(γ−2β)−1,

where D1(γ,β,A,B) is a constant that depends on γ,β,A,B.

Corollary 3.11. [15] For f ∈ 0 −Tµ[1,−1, 1,−1, 0],

∣∣∣∣|an+1|− |an|
∣∣∣∣ ≤ D1(µ)nµ2 −1,

where D1(γ,β,A,B) is a constant that depends only on µ.

Theorem 3.9. f ∈ k −Tµ[A,B,C,D,α]. maps the disk |z| < r∗ onto a convex domain, where r∗ is the

smallest positive real number of the equation

1 −λ1r + λ2r2 + λ3r3 + λ4r4 = 0 (3.9)



Int. J. Anal. Appl. 17 (6) (2019) 969

where

λ1 =A + B + (1 −β)µ + γ(A−B),

λ2 =1 + AB + (1 −β)µ(A + B),

λ3 =(2β − 1)(A + B) − (1 −β)µAB + γ(A−B),

λ4 =AB(1 − 2β),

and γ, β are respectively given by (1.4) and (3.1).

Proof. We set

zf′(z) = zβ (zφ′(z))
1−β

(h(z))γ,

where φ ∈Vµ and h ∈ P[A,B]. By logarithmic differentiation, we have

Re
(zf′(z))′

f(z)
= β + (1 −β)Re

(zφ′(z))′

φ(z)
+ γRe

zh′(z)

h(z)
.

For h ∈ P [A,B] and φ ∈Vµ , it is known in [6] and [22] respectively that

Re
zh′(z)

h(z)
≥−

(A−B)r
(1 −Ar)(1 −B)r

and

Re
(zφ′(z))′

φ(z)
≥
r2 −µr + 1

1 −r2
.

Using these results, we have that

Re
(zf′(z))′

f(z)
≥
[
β(1 −r2)(1 −Ar)(1 −Br) + (1 −β)(r2 −µr + 1)(1 −Ar)(1 −Br)

−γ(A−B)r(1 −r2)
]/

(1 −r2)(1 −Ar)(1 −Br).

If we let

T(r) = β(1 −r2)(1 −Ar)(1 −Br) + (1 −β)(r2 −µr + 1)(1 −Ar)(1 −Br) −γ(A−B)r(1 −r2).

Then

T(1) = (1 −β)(2 −µ)(1 −A)(1 −B) ≤ 0, since µ ≥ 2

and

T(0) = 1 > 0.

Thus, r ∈ (0, 1). Hence, the theorem is proved. �

Corollary 3.12. Let f ∈ 0 −T2[1,−1, 1,−1, 0]. Then f ∈ C for

|z| < 2 −
√

3.



Int. J. Anal. Appl. 17 (6) (2019) 970

Corollary 3.13. Let f ∈ 0 −Tµ[1,−1, 1,−1, 0]. Then f(z) maps the disc

|z | <
1

2
[ µ + 2 −

√
µ2 + 4µ ]

onto a convex domain

Using the well-known distortion theorems for φ ∈Vµ and h1 ∈P, we prove the following.

Theorem 3.10. Let f ∈ k −Tµ[1,−1,C,D,α]. Then

A2(a,b,c,r1) ≤ |f(z) | ≤ A1(a,b,c,r1),

where

A1(a,b,c,r1) =
2b−1

a

[
ra1 G(a,b,c,−r1) −G(a,b,c,−1)

]
and

A2(a,b,c,r1) =
2b−1

a

[
G(a,b,c,−1) −r−a1 G(a,b,c,−r

−1
1 )
]
,

where

a =(
µ

2
− 1)(1 −β)γ + 1,

b =2β,

c =(
µ

2
− 1)(1 −β)γ + 2,

and, γ and β are given by (1.4) and (3.1) respectively.

Proof. Let

f′(z) = (φ′(z))1−β(h1(z))
γ, φ ∈Vµ, h1 ∈P.

Then

|f(z) | ≤
∫ |z |

0

|φ(t) |1−β|h1(t) |γ dt

≤
∫ |z |

0

(1 + t)(
µ
2
−1)(1−β)

(1 − t)(
µ
2

+1)(1−β)
(1 + t)γ

(1 − t)γ
dt

=

∫ |z |
0

(
1 + t

1 − t

)( µ
2

+1)(1−β)
(1 + t)γ

(1 − t)2(1−β)+γ
.dt

Let ξ = 1+t
1−t, then 1 − t =

2
1+ξ

,dt = 2
(1+ξ)2

dξ, (1 + t) = 2ξ
1+ξ

. Therefore,

|f(z) | ≤22β−1
∫ 1+|z |

1−|z |

1

ξ(
µ
2
−1)(1−β)+γ(1 + ξ)−2βdξ

=2b−1(I1 − I2), (3.10)



Int. J. Anal. Appl. 17 (6) (2019) 971

where

I2 =

∫ 1
0

ξa−1(1 − ξ)c−a−1(1 − (1ξ))−bdξ

=
1

a
G(a,b,c,−1). (3.11)

To calculate I1, let ξ = r1u and r1 =
1+r
1−r . Then

I1 =

∫ 1
0

(r1u)
(
µ
2
−1)(1−β)+γ(1 + r1u)

−2βr1du

=ra1

∫ 1
0

ua−1(1 + r1u)
−bdu

=
ra1
a
G(a,b,c,−r1). (3.12)

Using (3.10), (3.11) and (3.12), the upper bound is obvious.

Now, we proceed to calculate the lower bound. Let dr denotes the radius of the largest schlicht disk

centered at the origin contained in the image of |z | < r under the function f(z). Then there exists a point

z0 with |z0 | = r, such that |f(z) | = dr. The ray from 0 to f(z) lies entirely in the image of ∆ and the

inverse image of this ray is a curve Γ in |z | < r. Thus,

dr =|f(z0) | =
∫

Γ

|f′(z) ||dr |

=

∫ |z |
0

|φ′(ρ) |1−β|h1(ρ) |γdρ

≥
∫ |z |

0

(
1 −ρ
1 + ρ

)( µ
2

11)(1−β)
(1 −ρ)γ

(1 + ρ)2(1−β)+γ
dρ.

Let ν = 1−ρ
1+ρ

. Then dρ = −2
(1+ν)2

dν and 1 −ρ = 2ν
1+ν

. Going through the same process as it has been done

for the upper bound, we easily obtain the lower bound and this completes the proof. �

Theorem 3.11. Let f,g ∈ k−Tµ[A,B,C,D,α], η1,c1,δ and v be positively real, η1 < 1, c1 > η1, η1 = v +δ.

Then for the function F(z) defined by

[F(z)]η1 = c1z
η1−c1

∫ z
0

t(c1−δ−v)−1(f(t))δ(g(t))vdt, (3.13)

∫ θ2
θ1

Re
zF ′(z)

F(z)
> −π

[
(1 −α)

(
C −D

2k + 1 −D

)(
µ

2
− 1
)

+ γ

]
,

where 0 ≤ θ1 < θ2 ≤ 2π.



Int. J. Anal. Appl. 17 (6) (2019) 972

Proof. It has been shown in [17] that the integral transformation (3.13) is a well defined analytic function

in ∆. Let H(z) =
zF′(z)
F(z)

. Then, differentiating (3.13), we obtain

F(z)η1 [(c1 −η1) + η1H(z)] = c1f(z)δg(z)v.

Differentiating logarithmically and with some simple computations, we have

H(z) +

1
η
zH′(z)

(c1−η1)
η1

+ H(z)
=

δ

η1

zf′(z)

f(z)
+

v

η1

zg′(z)

g(z)
.

Now, since f,g ∈ k −Tµ[A,B,C,D,α], then application of Remark 3.1 gives∫ θ2
θ1

Re

{
H(z) +

1
η
zH′(z)

(c1−η1)
η1

+ H(z)

}
> −π

[
(1 −α)

(
C −D

2k + 1 −D

)(
µ

2
− 1
)

+ γ

]
.

On using Lemma 2.5 with % = 1
η
, c = c1 −η, we complete the proof. �

Theorem 3.12. Let f,g ∈ k −Tµ[A,B,C,D,α] and −1 < ρ ≤ 0, 0 < α1 ≤ 1. Then for the function J(z)

defined by

J(z) =

[
(ρ +

1

α1
)z

1− 1
α1

∫ z
0

t
1
α1
−2
f(t)ρg(t)dt

] 1
1+ρ

, (3.14)

∫ θ2
θ1

Re
zJ′(z)

J(z)
> −π

[
(1 −α)

(
C −D

2k + 1 −D

)(
µ

2
− 1
)

+ γ

]
,

where 0 ≤ θ1 < θ2 ≤ 2π.

Proof. Set
zJ′(z)
J(z)

= G(z). Then differentiating (3.14) logarithmically and with some simple calculations, we

get (
1

α1
− 1
)

+ (ρ + 1)G(z) =
z

1
α1
−1
f(z)ρg(z)∫ z

0
t

1
α1
−2
f(t)ρg(t)dt

.

This implies (
ρ +

1

α1

)
f(z)ρg(z) =

((
1

α1
− 1
)

+ (ρ + 1)G(z)

)
J(z)ρ+1.

Differentiating logarithmically once again, we have

H(z) +

1
1+ρ

zH′(z)

( 1
α1
−1)

1+ρ
+ H(z)

=
ρ

1 + ρ

zf′(z)

f(z)
+

1

1 + ρ

zg′(z)

g(z)
.

Applying Lemma 2.5 with % = 1
1+ρ

, c = 1
α1
− 1, it follows that

∫ θ2
θ1

Re
zJ′(z)

J(z)
> −π

[
(1 −α)

(
C −D

2k + 1 −D

)(
µ

2
− 1
)

+ γ

]
.

�



Int. J. Anal. Appl. 17 (6) (2019) 973

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	1. Introduction
	2. Some Preliminary Lemmas
	3. Main Result
	References