on certain classes of meromorphic harmonic concave ... · ekstrim, batas konvolusi, konvolusi...

Journal of Quality Measurement and Analysis JQMA 12(1-2) 2016, 53-65

Jurnal Pengukuran Kualiti dan Analisis

ON CERTAIN CLASSES OF MEROMORPHIC HARMONIC CONCAVE

FUNCTIONS DEFINED BY AL-OBOUDI OPERATOR (Berkenaan Kelas Fungsi Cekung Meromorfi Harmonik Tertentu

yang Ditakrif oleh Pengoperasi Al-Oboudi)

KHALID CHALLAB & MASLINA DARUS

ABSTRACT

In this work, a class of meromorphic harmonic concave functions defined by Oboudi operator in

the punctured unit disc is introduced. Coefficient conditions, distortion inequalities, extreme

points, convolution bounds, geometric convolution, integral convolution and convex

combinations for functions f belonging to this class are obtained.

Keywords: Meromorphic function; harmonic function; concave function; Al-Oboudi operator

ABSTRAK

Dalam kajian ini, kelas fungsi cekung meromorfi harmonik yang ditakrif oleh pengoperasi Al-

Oboudi diperkenalkan dalam cakera unit terpancit. Syarat pekali, ketaksamaan erotan, titik

ekstrim, batas konvolusi, konvolusi geometri, konvolusi kamiran dan gabungan cembung untuk

fungsi f dalam kelas tersebut diperoleh.

Kata kunci: fungsi meromorfi; fungsi harmonik; fungsi cekung; pengoperasi Al-Oboudi

1. Introduction

Conformal maps of the unit disc onto convex domain are very classic. We note that Avkhadiev and Wirths (2005) discovered the conformal mapping of a unit disc onto concave domains (the complements of convex closed sets). It is quite interesting to see other related results in this direction as not many problems are discussed thoroughly towards this approach. Let 𝑈 = {𝑧 ∈ ℂ: |𝑧| < 1} denote the open unit disc, where f has the form given by

2

n

n

n

f z z a z

(1)

that maps U conformally onto a domain whose complement with respect to C is convex and

that satisfies the normalisation (1) f . In addtion, they imposed on these functions the

condition that the opening angle of ( )f U at infinity is less than or equal to , (1,2]. These families of functions are denoted by 0 ( )C . The class 0 ( )C is referred to as the class

of concave univalent functions.

Chuaqui et al. (2012) defined the concept of meromorphic concave mappings. A

conformal mapping of meromorphic function in U , where 𝑈∗ = 𝑈\{0} is said to be a concave mapping if its image is the complement of a compact convex set.

If f has the form

20 1 21

,f z a a z a zz

then a necessary and sufficient condition for f to be a concave mapping is

Khalid Challab & Maslina Darus

54

''

'

( )1 0, 1.

( )

f zRe z z

f z

A continuous function f u iv is a complex valued harmonic function in a domain if both u and v are real harmonic in In any simply connected domain, we write where h and g are analytic in A necessary and sufficient condition for f to be locally univalent and orientation preserving in U is that h g in U (see Clunie and Sheil-Small 1984). Hengartner and Schober (1987) investigated functions harmonic in the exterior

of the unit disc . They showed that complex valued, harmonic, sense preserving, univalent mapping f must admit the representation.

,f z h z g z Alog z

where h z and ( )g z are defined by

1 1

, n nn nn n

h z z a z g z z b z

for 0 and , **z U ,where

We call h the analytic part and g the co-analytic part of f .

For*z U , let HM be the class of functions:

1 1

1( ) k kk k

k k

f z h z g z a z b zz

(2)

which are analytic in the punctured unit disc*U , where ( )h z and ( )g z are analytic in U and

*U , respectively, and ( )h z has a simple pole at the origin with residue 1 (see Al-Shaqsi and Darus 2008).

A function Hf M is said to be in the subclass HMS

of meromorphically harmonic

starlike functions in *U if it satisfies the condition

'

*'( ) 0, .( )

zh z zg zRe z U

h z g z

Also, a function Hf M is said to be in the subclass HMC of meromorphically harmonic

convex functions in *U if it satisfies the condition

'' ''*' '( ) 0, .

' '( )

zh z h z zg z g zRe z U

h z g z

Note that the classes of harmonic meromorphic starlike functions, harmonic meromorphic

convex functions and harmonic meromorphic concave functions ( 0MHC ) have been studied by Jahangiri and Silverman (1999), Jahangiri (2000), Jahangiri (1998), Aldawish and Darus

(2015) and Challab and Darus (2016). Other works related to concave can be read in

Aldawish and Darus (2015), Darus et al. (2015) and Aldawish et al. (2014.)

AL-Oboudi (2004) introduced the operator nD for f A which is the class of function of

the form (1) analytic in the unit disc , and defined the following differential operator

0 , D f z f z

On certain classes of meromorphic harmonic concave functions defined by Al-Oboudi operator

55

1 '1 , 0,D f z zf z D f z (3)

1 .n nD f z D D f z (4)

If f is given by (1), then from (3) and (4) we see that

(5)

Al-Oboudi (2004) introduced the operator nO for 0f MHC which is the class of

functions f h g that are harmonic univalent and sense-preserving in the unit disc

for which { (0) (0) (0) 1 0zf h f }.

Now, we define nO for f h g given by (2) as

*( ) ( ) ( ) , 0,1,2, , ,...n n nO f n O h n O g n n z U where

1

( 1)1 1 ,

nnn k

k

k

O h z k a zz

1

g(z) 1 1 .nn k

k

k

O k b z

This work is an attempt to give a connection between harmonic function and

meromorphic concave functions defined by Oboudi operator by introducing a class 0nO MHC

of meromorphic harmonic concave functions.

Definition 1.1. For 0 {1,2,...}n N , 1,2 , 1k , let 0nO MHC denote the class of

meromorphic harmonic concave functions ( )nO f z defined by Oboudi differential

operator of the form,

1 1

( ) ( 1) / [1 ( 1) ] [1 ( 1) ] ( )n n n k n kk kk k

O f z z k a z k b z

(6)

such that

''

'1 0.

( )

n

n

z O f zRe

O f z

2. Coefficient Conditions

In this section, sufficient condition for a function 0( )n nO f z O MHC is derived.


56

Theorem 2.1. Let ( )nO f z h g be of the form

1 1

(1 1 .

1)1 1

nn nn k k

k k

k k

O f z k a z k b zz

If 21

1 1 1,n

k k

k

k k a b

then nO f z is harmonic univalent, sense preserving in

Proof: First for 1 20 1z z , we have

1 2 1 1 1

1 1

2 2 2

1 1

| ( ) ( ) | | ( 1) / [1 ( 1) ] [1 ( 1) ] ( )

( 1) / [1 ( 1) ] [1 ( 1) ] ( ) |

n n n n k n kk k

k k

n n k n kk k

k k

O f z O f z z k a z k b z

z k a z k b z

1 2

1 2 1 2

1 1

1/ | | 1/ | |

[1 ( 1) ] | || | [1 ( 1) ] | || |n k k n k kk kk k

z z

k a z z k b z z

n1 2

1 2 k k

1 2 k 1

z zz z k 1 k 1 γ a b

z z

2 21 2

1

1 2 2| | / | || | [1 | | [1 ( 1) ] (| | | |)]n

k

k kz z z z z k k a b

21 2 1 2

1

| | / | || | [1 [1 ( 1) ] (| | | |)].n k kk

z z z z k k a b

The last expression is non negative by 2

1

[1 ( 1) ] (| | | |) 1n k kk

k k a b

and ( )nO f z is

univalent in *U .

Now we want to show f is sense preserving in**U , we need to show that ' '( ) ( )h z g z

in *U .

1'

21

11 1

n k

k

k

h z k k a zz

121

1 1 1

n kk

k

k k a rr


57

1

1 1 1n

k

k

k k a

2

1

1 1 1n

k

k

k k a

2

1

1 1n

k

k

k k b

11 '

1 1

1 1 1 1 .n n kk

k k

k k

k k b r k k b z g z

Thus this completes the proof of the Theorem 2.1.

Theorem 2.2. Let gnO f z h be of the form,

1 1

(1 1 .

1)1 1

nn nn k k

k k

k k

O f z k a z k b zz

Then ( )n nO f z O MHC if the inequality

2

1

[1 ( 1) ] (| | | |) 1n k kk

k k a b

(7)

holds for coefficient ( ) .nO f z h g

Proof: Using the fact that 1

( ) 0 11

wRe w

w

, it suffices to show

11

1

w

w

.

Let

''

'1

( )

n

n

z O f zw Re

O f z

, such that

' ( )

( )

zg zw

g z , where ( ) '.ng z z O f z

Now,


58

2 2

1 11

2 2

1 1

( ) 1 1 ( ) 1 11

1 2 1( ) 1 1 ( ) 1 1

n nk kk kk k

nn nk k

k kk k

k k k a z k k k b zw

wk k k a z k k k b z

z

2 2

1 1

2 2

1 1

( ) 1 1 ( ) 1 11.

1 2 ( ) 1 1 ( ) 1 1

n n

k kk k

n n

k kk k

k k k a k k k bw

w k k k a k k k b

(8)

The last expression is bounded above by 1 if

2 2

1 1

( ) 1 1 ( ) 1 1n n

k k

k k

k k k a k k k b

2 2

1 1

2 ( )[1 ( 1) ] | | ( )[1 ( 1) ] | |n nk kk k

k k k a k k k b

which is equivalent to our condition by

2

1

[1 ( 1) ] (| | | |) 1.n k kk

k k a b

Theorem 2.3. Let ( )nO f z h g be of the form

1 1

(1 1 .

1)1 1

nn nn k k

k k

k k

O f z k a z k b zz

A necessary and sufficient condition for nO f z to be in nO MHC is that

21

1 1 1.n

k k

k

k k a b

Proof: In view of Theorem 2.2, we assume that 2

1

[1 ( 1) ] (| | | |) 1.n k kk

k k a b

Since 0( )n nO f z O MHC , then

1 {( ( ( )) ) / ( ( )) }n nRe z O f z O f z equivalent to

2

2 1 2 1

2 1 1'

1

1 1

11 1 1 1

( )

( ) ( 1)1 1 1 1

nn nk k

k kk k

nn nk k

k kk k

z k k a z k k b zzzg z

Re Reg z

k k a z k k b zz


59

12 1 2 1

1 1

1

1 1

( 1)1 1 1 1

0( 1)

1 1 1 1

nn nk k

k kk k

nn nk k

k kk k

k k a z k k b zzRe

k k a z k k b zz

for 1z r , the above expression reduce to

1 2

1

1

1

1 1 1 ( ) ( )0.

( )1 1 1 ( )

nn kk kk

nn kk kk

k k a b r A rRe

B rk k a b r

As we assumed2

1

[1 ( 1) ] (| | | |) 1,n k kk

k k a b

then A(r) and B(r) are positive for r

sufficiently close to 1. Thus there exists a 1z r for which the quotient is positive. This

contradicts the required condition that ( )

0( )

A r

B r , so the proof is complete.

3. Distortion bounds and extreme points

Bounds and extreme points for functions belonging to the class nO MHC are estimated in

this section.

Theorem 3.1. If n n n nk k kO f O h O g O MHC and 0 1z r then

21

n kr

O f zr

and

21( ) .n k

rO f z

r

Proof: Let 0 .n n n n

k k kO f O h O g O MHC Taking the absolute value of ,n

kO f we obtain

1 1

11 1 1 1

nn nn k k

k k k

k k

O f z k a z k b zz

1

11 1

nn kk k k

k

O f z k a b rr

21

11 1

n k

k k

k

k k a b rr

21

11 1

n

k k

k

k k a b rr

by applying

2

1

[1 ( 1) ] (| | | |) 1,n k kk

k k a b

then


60

21 1

.n kr

O f z rr r

Now,

1 1

11 1 1 1

nn nn k k

k k k

k k

O f z k a z k b zz

1

11 1

n kk k

k

k a b rr

21

11 1 .

n

k k

k

k k a b rr

By applying2

1

[1 ( 1) ] (| | | |) 1,n k kk

k k a b

then

21 1

.n kr

O f z rr r

Theorem 3.2. Let gn n nn n nO f O h O and ( ) ( ) g ( )n n n

n n nO f z O h z O z

That is

1 1

11 1 1 1 .

nn nn k k

n k k

k k

O f z k a z k b zz

Set

,0 ,0

1g

n

n nn nO h O

z

and

, 2

1 1( ) ,

n

n k

n kO h z zz k

For 1,2,3,...k and let

, 2

1 1g ,

n

n kn kO z z

z k

for 1,2,3,...k then n nnO f O MHC if and only if n

nO f can be expressed as

, , ,0

g ,n n nn k k n k k n kk

O f O h z O z

where 0, 0k k and 0

( ) 1k kk

. In particular the extreme points of nO MHC are

,n n kO h and ,{ .}n n kO g


61

Proof: For functions gn n nn n nO f O h O , where n

nO h and gn

nO are given by

1 1

11 1 1 1 ,

nn nn k k

n k k

k k

O f z k a z k b zz

we have

, , ,0

( ) gn n nn k k n k k n kk

O f z O h z O z

0 ,0 0 ,0 , ,1

g gn n n nn n k n k k n kk

O h O O h z O z

0 ,0 0 ,0 2 2

1 1

1 11 1g

n n

n n k kn n k k

k k

O h O z zz zk k

20 1

1 1( ) .

n

k kk k k k

k k

z zz k

Now by Theorem 2.2,

2 20 02 2

1 1

1 11 1,k k k k

k k

k kk k

We have ( , )( ) 0.n n

n k zO f O MHC

Conversely, suppose that , ( )n n

n kO f z O MHC . Setting

2 1 1 , 1n

k kk k a k

and

2 | [1 ( 1) ] |, 1.nk kk k b k

We define

0 0

1 1

1 .k kk k

Therefore ( )n nO f z can be written as

1 1

1 1 1 1 1

nn nn k k

n k k

k k

O f z k a z k b zz


62

2 21

1 1 1

n

k kk k

k

z zz k k

, ,

1 1

1 1 1g

n n n

n nn k k n k k

k k

O h z O zz z z

, ,1 1 1 1

11 g

n

n nk k n k k n k k

k k k k

O h z O zz

0 0 , ,1 1

1g

n

n nn k k n k k

k k

O h z O zz

and finally

, ,0

g .n n nn k n k k n kk

O f z O h z O z

The proof is completed, therefore ( , )( ),n

n k zO h and ( , )( )n

n k zO g are extreme points. .

4. Convolution Properties

In this section, we define and study the convolution, geometric convolution and integral

convolution of the class nO MHC .

For harmonic function ,n nn nO f O F is defined as follows:

1 1

11 1 1 1

nn nn k k

n k k

k k

O f z k a z k b zz

(9)

and

1 1

( ) ( 1) / [1 ( 1) ] | | [1 ( 1) ] | | .n n n k n kn k kk k

O F z z k A z k B z

(10)

The convolution of n nO f and n

nO F is given by

( * )( ) ( )* ( ),n n n nn n n nO f O F z O f z O F z

1 1

( 1) / [1 ( 1) ] | || | [1 ( 1) ] | || .|n n k n kk k k kk k

z k a A z k b B z

The geometric convolution n nO f and n

nO F is given by

( )( ) ,n n n nn n n nO f O F z O f z O F z

1 1

( 1) / [1 ( 1) ] | | [1 ( 1) ] | | .n n k n kk k k kk k

z k a A z k b B z


63

The integral convolution of n nO f and n

nO F is given by

1 1

( 1) / [1 ( 1) ] | | / [1 ( 1) ] | | / .n n k n kk k k kk k

z k a A kz k b B k z

Theorem 4.1. Let 0n n

nO f O MHC and 0.n n

nO F O MHC Then the convolution

0* .n n n

n nO f O F O MHC

Proof: For n nO f and n

nO F given by (9) and (10), then the convolution is given by (10). We

show that the coefficients of *n nn nO f O F satisfy the required condition given in Theorem

2.2.

For 0n n

nO F O MHC , we note that 1kA and 1kB . Now for convolution function

* ,n nn nO f O F we obtain

2 21 1

1 1 1 1n n

k k k k

k k

k k a A k k b B

2 2

1 1

1 1 1 1 1.n n

k k

k k

k k a k k b

Therefore * n n nn nO f O F O MHC , this proves the required result.

Theorem 4.2. If n nO f and n

nO F of the form (9) and (10) belong to the class 0nO MHC , then

the geometric convolution n nn nO f O F also belong to 0.nO MHC

Proof: Since 0, ,n n n

n nO f O F O MHC it follows that

2

1

1 1 ( ) 1n

k k

k

k k a b

and

2

1

1 1 ( ) 1.n

k k

k

k k A B

Hence by Cauchy-Schwartz's inequality it is noted that

2

1

1 1 ( ) 1.n

k k k k

k

k k a A b B

The proof is complete.

Theorem 4.3. If n nO f and n

nO F of the form (8) and (9) belong to the class nO MHC then

the integral convolution also belong to the

nO MHC .


64

Proof: Since 0, ,n n n

n nO f O F O MHC it follows that 1kA and 1.kB Then

because

2 2

1 1

1 1 1 1n nk k k k

k k

a A b Bk k k k

k k

2 2

1 1

1 1 1 1n nk k

k k

a bk k k k

k k

2 2

1 1

1 1 1 1 1.n n

k k

k k

k k a k k b

This proves the required result.

5. Convex Combinations

In this section, we show that the class nO MHC is invariant under convex combinations of its

members.

Theorem 5.1. The class 0nO MHC is closed under convex combinations.

Proof: For 1,2,3,...i suppose that 0( ) ,n n

iO f z O MHC where n

iO f is given by

1 1

11 1 1 1 ,

nn nn k k

i ik ik

k k

O f z k a z k b zz

0, 0,ik ika b then by Theorem 2.2,

2

1

[1 ( 1) ] (| | | |) 1.n i ik

k k a k b k

For 1

1,0 1,i ik

t t

the convex combinations of n iO f may be written as

1 1 1 1

( ) ( 1) / [1 ( 1) ] ( ) [1 ( 1) ] ( ) .i

n n n k n ki i i ik i ik

i k k

t O f z z k t a z k t b z

Then by

21

1 1 1.n

ik ik

k

k k a b

2

1 1 1

1 1n

i ik i ik

k i i

k k t a t b

21 1

1 1n

i ik ik

i k

t k k a b

1

1.ii

t

Thus 01

( ) .n ni ii

t O f z O MHC


65

Acknowledgements The work here is supported by MOHE grant FRGS/1/2016/STG06/UKM/01/1.

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School of Mathematical Sciences

Faculty of Science and Technology

Universiti Kebangsaan Malaysia

43600 UKM Bangi

Selangor DE, MALAYSIA

E-mail: [email protected], [email protected]*

*Corresponding author

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