# Some Applications of Ordinary Differential Operator to Certain Multivalent Functions

Authors: Müfit Şan; Hüseyin Irmak; Ayhan Şerbetçi
Journal: Journal of Mathematics
 (2015)

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## Abstract

The aim of this paper is to apply the well-known ordinary differential operator to certain multivalent functions which are analytic in the certain domains of the complex plane and then to determine some criteria concerning analytic and geometric properties of the related complex functions.


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## Body

## 1. Introduction, Notations, and Definitions

LetA
(
κ
;
p
) denote the class of functions f
(
z
) of the following form:(1)
f
z
=
z
κ
+
a
p
+
1
z
p
+
1
+
a
p
+
2
z
p
+
2
+
⋯
a
k
∈
C
;
κ
∈
Z
=
±
1
,
±
2
,
±
3
,
…
;
p
∈
N
=
1,2
,
3
,
…
,
Z
-
=
Z
-
N
,which areanalytic andmultivalent in the domain (2)
Δ
=
U
,
when
κ
∈
N
D
,
when
κ
∈
Z
-
,where C is the set of complex numbers. As is known, the domains U and D are known as unit open disk and punctured open unit disk, respectively. Also let M
(
p
)
≔
A
(
-
p
;
p
) and T
(
p
)
≔
A
(
p
;
p
) when p
∈
N.By differentiating both sides of the functionf
(
z
) in the form (1), q-times with respect to complex variable z, one can easily derive the following (ordinary) differential operator: (3)
f
q
z
=
p
!
p
-
q
!
z
p
-
q
+
∑
k
=
p
+
1
∞
k
!
k
-
q
!
a
k
z
k
-
q
,
if
f
∈
T
p
p
+
q
-
1
!
p
-
1
!
-
1
q
z
-
p
-
q
+
∑
k
=
p
+
1
∞
k
!
k
-
q
!
a
k
z
k
-
q
,
if
f
∈
M
p
,where p
≥
q, p
∈
N, and q
∈
N
0
≔
N
∪
{
0
}
.In this investigation, by applying the differential operator, defined by (3), to certain analytic functions which are multivalent in U or meromorphic multivalent in D, several criteria, which also include both analytic and geometric properties of univalent functions (see [1, 2]), for functions f
(
z
) in the classes M
(
p
) and T
(
p
), are then determined. In the literature, by using certain operators, several researchers obtained some results concerning functions belonging to the general class A
(
κ
;
p
). In this paper, we also determined many results which includestarlikeness,convexity, close-to-convexity, andclose-to-starlikeness of analytic functions in the second section of this paper. One may refer to some results determined by ordinary differential operator in [3, 4], some properties of certain linear operators in [5–7], and also certain results appertaining to multivalent functions and some of their geometric and analytic properties in [8, 9] in the references.For the proofs of the main results, we then need to recall the well-known method which was obtained by Jack [10] (see also [11]) and given by the following lemma.Lemma 1.
Let the functionw
(
z
) given by (4)
w
z
=
c
n
z
n
+
c
n
+
1
z
n
+
1
+
c
c
+
2
z
c
+
2
+
⋯
n
∈
Nbe analytic in U with w
(
z
)
≢
0
(
z
∈
U
)
.
If(5)
z
0
=
r
e
i
θ
0
,
w
z
0
=
max
z
≤
z
0
⁡
w
z
,then (6)
z
0
w
′
z
0
=
c
w
z
0
,where c is real number and c
≥
n
≥
1.


## 2. The Main Results and Their Applications

Theorem 2.
Letf
(
z
)
∈
M
(
p
) and z
∈
D, and also let the following inequality:(7)
q
+
R
e
z
f
1
+
q
z
f
q
z
<
1
2be true. Then (8)
R
e
z
p
+
q
f
q
z
>
0
,
when
q
is even
<
0
,
when
q
is odd
.Proof.
Letf
(
z
)
∈
M
(
p
)
. By applying the differential operator, defined in (3), to the function f
(
z
), one easily get that (9)
z
p
+
q
f
q
z
γ
p
,
q
=
1
+
∑
k
=
p
+
1
∞
Γ
k
,
q
γ
p
,
q
a
k
z
k
+
p
=
1
+
∑
k
=
p
+
1
∞
d
k
z
k
+
p
=
1
+
d
p
+
1
z
2
p
+
1
+
d
p
+
2
z
2
p
+
2
+
⋯
=
1
+
w
z
,where γ
(
p
,
q
) and Γ
(
k
,
q
) are defined by (10)
γ
p
,
q
≔
p
+
q
-
1
!
p
-
1
!
-
1
q
p
≥
q
;
p
∈
N
;
q
∈
N
0
,
(11)
Γ
k
,
q
≔
k
!
k
-
q
!
k
=
p
+
1
,
p
+
2
,
…
;
p
≥
q
;
p
∈
N
;
q
∈
N
0
,respectively. It is clear that the defined function w
(
z
) has the form in Lemma 1; that is, it is analytic in U with w
(
z
)
≢
0. Upon differentiating of the identity (9), one easily obtains that (12)
p
+
z
f
1
+
q
z
f
q
z
+
q
=
z
w
′
z
1
+
w
z
,
z
∈
U
;
w
z
≠
-
1
.
Now suppose that there exists a pointz
0
∈
U such that (13)
max
⁡
w
z
:
z
≤
z
0
=
w
z
0
=
1
,by applying Lemma 1; we then have z
0
w
′
(
z
0
)
=
c
w
(
z
0
)
(
c
≥
n
≥
1
)
. Thus, in view of the above equality, it can be calculated that (14)
R
e
p
+
z
f
1
+
q
z
f
q
z
+
q
z
=
z
0
=
R
e
z
w
′
z
1
+
w
z
z
=
z
0
=
c
2
≥
2
p
+
1
2
,which contradicts (7). Hence, we conclude that |
w
(
z
)
|
<
1 for all z
∈
U, and Definition (9) yields the inequality (15)
z
p
+
q
f
q
z
γ
p
,
q
-
1
=
w
z
<
1
,which is equivalent to (8). Therefore, this completes the desired proof.Theorem 3.
Letf
(
z
)
∈
M
(
n
), g
(
z
)
∈
M
(
m
), q
∈
N
0, and, also, the following inequality (16)
R
e
g
q
z
z
g
1
+
q
z
>
α
,
α
>
0
;
z
∈
U
;
q
∈
N
0be satisfied. If (17)
f
1
+
q
z
g
1
+
q
z
<
1
+
α
m
-
n
·
γ
n
,
q
γ
m
,
q
,then (18)
f
1
+
q
z
g
1
+
q
z
<
γ
n
,
q
γ
m
,
q
,
m
>
n
;
n
,
m
∈
N
.Proof.
Let the functionsf
(
z
)
∈
M
(
n
) and g
(
z
)
∈
M
(
m
) be in the form (19)
f
z
=
z
-
n
+
∑
k
=
p
+
1
∞
a
k
z
k
,
g
z
=
z
-
m
+
∑
k
=
p
+
1
∞
b
k
z
k
,respectively. Then, from related differential operator and definitions of the functions f
(
z
) and g
(
z
), determine (20)
γ
m
,
q
γ
n
,
q
·
f
q
z
g
q
z
=
z
-
n
-
q
+
∑
k
=
p
+
1
∞
Γ
k
,
q
/
γ
p
,
q
a
k
z
k
+
p
z
-
m
-
q
+
∑
k
=
p
+
1
∞
Γ
k
,
q
/
γ
p
,
q
b
k
z
k
-
q
=
z
m
-
n
+
c
m
-
n
+
1
z
m
-
n
+
1
+
⋯
∈
T
m
-
n
,where γ
(
k
,
q
) and Γ
(
p
,
q
) are defined by (10) and (11), respectively. Define v
(
z
) by (21)
γ
m
,
q
γ
n
,
q
·
f
q
z
g
q
z
=
v
z
.Clearly, it is easily seen that v
(
z
) satisfies the conditions of Lemma 1. The definition in (21) clearly gives us (22)
γ
m
,
q
γ
n
,
q
·
f
1
+
q
z
g
1
+
q
z
=
v
z
1
+
z
v
′
z
v
z
·
g
q
z
z
g
1
+
q
z
.Assume now that there exists a point z
0
∈
U such that (23)
max
z
≤
z
0
⁡
w
z
=
w
z
0
=
1
,
z
∈
U
.Then, applying Lemma 1, it follows from (16) and (22) that (24)
γ
m
,
q
γ
n
,
q
·
f
1
+
q
z
g
1
+
q
z
z
=
z
0
=
v
z
1
+
z
v
′
z
v
z
·
g
q
z
z
g
1
+
q
z
z
=
z
0
≥
1
+
R
e
z
v
′
z
v
z
·
g
q
z
z
g
1
+
q
z
z
=
z
0
=
1
+
c
·
R
e
g
q
z
z
g
1
+
q
z
z
=
z
0
≥
1
+
m
-
n
α
since
c
≥
m
-
n
,which is contradicting to the assertion (17). Hence, |
v
(
z
)
|
<
1 for all z in the disk U. Thus, the modulus of the identity (21), that is, (25)
γ
m
,
q
γ
n
,
q
·
f
q
z
g
q
z
=
v
z
,requires inequality (18). This completes the proof of Theorem 3.Theorem 4.
Letf
(
z
)
∈
M
(
n
), g
(
z
)
∈
M
(
m
), q
<
min
⁡
{
n
,
m
}, n
∈
N, m
∈
N, and q
∈
N
0
.
If(26)
R
e
z
f
1
+
q
z
f
q
z
-
g
1
+
q
z
g
q
z
<
m
-
n
+
1
2
,
z
∈
D
,then (27)
R
e
z
n
-
m
f
q
z
g
q
z
>
0
,
z
∈
D
.Proof.
Let the functionsf
(
z
)
∈
M
(
n
) and g
(
z
)
∈
M
(
m
) be given by (19). Then, in view of Definition (20), define again an analytic function w
(
z
) by (28)
γ
m
,
q
γ
n
,
q
·
f
q
z
g
q
z
=
z
m
-
n
1
+
c
1
z
+
c
2
z
2
+
c
3
z
3
+
⋯
=
z
m
-
n
1
+
w
z
.It is obvious that w
(
z
) is analytic in U with w
(
0
)
=
0. By differentiating w
(
z
) in (28) logarithmically, we find that (29)
z
f
1
+
q
z
f
q
z
-
g
1
+
q
z
g
q
z
=
m
-
n
+
z
w
′
z
1
+
w
z
,
z
∈
D
,and also suppose that there exists a point z
0
∈
U such that (30)
max
z
≤
z
0
⁡
w
z
=
w
z
0
=
1
,
w
z
0
=
e
i
θ
≠
-
1
.Then, applying Lemma 1, it follows from (29) that (31)
R
e
z
f
1
+
q
z
f
q
z
-
g
1
+
q
z
g
q
z
z
=
z
0
=
m
-
n
+
c
·
R
e
i
θ
1
+
e
i
θ
≥
m
-
n
+
1
2
since
c
≥
1
,which contradicts the assumption given by (26). Thus, for all z
∈
U, |
v
(
z
)
|
<
1 holds. So, (28) immediately yields inequality (27). Therefore, the desired proof is completed.Theorem 5.
Letf
(
z
)
∈
M
(
p
), p
≥
q, and z
∈
D, and also let the following inequality: (32)
R
e
p
+
z
f
1
+
q
z
f
q
z
+
q
<
2
δ
-
1
,
when
δ
>
0
>
2
δ
-
1
,
when
δ
<
0hold. Then (33)
R
e
z
p
+
q
f
q
z
γ
p
,
q
δ
>
0
,
δ
≠
0
,where the value of the above complex power is taken to be as its principal value and γ
(
p
,
q
) is given by (10).Proof.
Letf
(
z
)
∈
M
(
p
). In view of (9), we again define a function w
(
z
) by (34)
z
p
+
q
f
q
z
γ
p
,
q
δ
=
1
+
w
z
;we can easily see that w
(
0
)
=
0 and also w
(
z
) is analytic in U. By differentiating logarithmically (34), we receive (35)
p
+
z
f
1
+
q
z
f
q
z
+
q
=
1
δ
·
z
w
′
z
1
+
w
z
,
z
∈
U
;
w
z
≢
-
1
.Now assuming that there exists a point z
0
∈
U such that max
⁡
{
w
z
:
|
z
|
≤
|
z
0
|
}
=
|
w
(
z
0
)
|
=
1, by using Lemma 1, we then have z
0
w
′
(
z
0
)
=
c
w
(
z
0
)
(
c
≥
n
≥
1
) and also take w
(
z
0
)
=
e
i
θ
≠
-
1. Thus, with the help of (35), it can be determined that (36)
R
e
p
+
z
f
1
+
q
z
f
q
z
+
q
z
=
z
0
=
1
δ
·
R
e
z
w
′
z
1
+
w
z
z
=
z
0
≥
2
δ
-
1
,
when
δ
>
0
,
≤
2
δ
-
1
,
when
δ
<
0
,which contradict the cases of (32), respectively. Hence, we conclude that |
w
(
z
)
|
<
1 for all z
∈
U, and Definition (34) immediately yields inequality (33). This completes the related proof.This paper includes several useful results which will be important and/or interesting for analytic and geometric function theory [1, 2]. We want to point only two special results of the main results out. The others are here omitted.Remark 6.
If we takeδ
=
1 in Theorem 5, then Theorem 5 requires Theorem 2.If we takeq
=
0 in Theorem 5, then we receive the following corollary.Corollary 7.
Letf
(
z
)
∈
M
(
p
), δ
≠
0, and z
∈
D. Then (37)
p
+
R
e
z
f
′
z
f
z
<
1
2
δ
,
if
δ
>
0
>
1
2
δ
,
if
δ
<
0
⟹
R
e
z
p
f
z
δ
>
0
,or, equivalently, (38)
R
e
z
p
f
z
δ
≤
0
⟹
p
+
R
e
z
f
′
z
f
z
≥
1
2
δ
,
if
δ
>
0
≤
1
2
δ
,
if
δ
<
0
,where the value of the above complex power is considered to be its principal value.



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*Source: 825903-2015-08-18.xml*