On Certain Classes of Meromorphically P-Valent Functions with Positive Coefficients Defined by Linear Operator

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International Journal of Basic & Applied Sciences IJBAS-IJENS Vol:13 No:05 32

ON CERTAIN CLASSES OF MEROMORPHICALLY P-VALENT FUNCTIONS WITH POSITIVE COEFFICIENTS DEFINED BY

LINEAR OPERATOR

ABDUL RAHMAN S. JUMA1_{, HAZHA ZIRAR HUSSAIN}2

DEPARTMENT OF MATHEMATICS, ALANBAR UNIVERSITY, RAMADI-IRAQ1_,

DEPARTMENT OF MATHEMATICS, SALAHADDIN UNIVERSITY,ERBIL-IRAQ2_,

Abstract. A purpose of this paper is to introduce the classNp,q,s(α1;α, β, γ)

of meromorphically p-valent functions by using linear operator. We study

var-ious properties such as coefficient inequality, growth and distortion theorems,

closure theorems, convolution properties, radii of meromorphically p-valent starlikeness and convexity, weighted mean and arithmetic mean.

Keywords :p- Valent, Hadamard product, Meromorphic, Positive coefficients.

AMS Subject Classifications : 30C45

1. _{Introduction and definitions}

LetP

p denote the class of functions of the form

f(z) = 1 zp +

∞

X

k=0

ap+kzp+k,(ap+k ≥0;p∈N={1,2, ...}), (1)

which are analytic and p-valent in the punctured unit disk∗ ={z∈C: 0<|z|< 1}=−{0}.

The Hadamard product (or convolution) of two functions,f given by (1) and

g(z) = 1 zp +

∞

X

k=0

bp+kzp+k,(bp+k ≥0),

is defined by

(f∗g)(z) = 1 zp +

∞

X

k=0

ap+kbp+kzp+k.

For real or complex numbersα1, α2, ..., αq andβ1, β2, ..., βs(βj 6= 0,−1,−2, ...;j =

1,2, ..., s), we define the generalized hypergeometric functionqFs(α1, α2, ..., αq;β1, β2, ..., βs;z) by (see for example, [13, p. 19]),

qFs(α1, α2, ..., αq;β1, β2, ..., βs;z) = ∞

X

k=0

(α1)k,(α2)k, ...,(αq)k (β1)k,(β2)k, ...,(βs)k(1)k

zk

1_E-mail:

dr−[email protected]

2_E-mail:

[email protected]

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International Journal of Basic & Applied Sciences IJBAS-IJENS Vol:13 No:05 33

(q≤s+ 1;s, q∈_N0=N∪ {0};z∈),

where (x)n is the Pochhammer symbol, defined in terms of the gamma function Γ by,

(x)n=

Γ(x+n) Γ(x) =

(

1 (n= 0;x∈_C− {0}, x(x+ 1)(x+ 2)...(x+n−1) (n∈N;x∈C, corresponding to the function

hp(α1, α2, ..., αq;β1, β2, ..., βs;z) =z−p qFs(α1, α2, ..., αq;β1, β2, ..., βs;z), Liu and Srivastava [8] (see for details [5] and [6]) introduced a linear operator:

Hp,q,s(α1, α2, ..., αq;β1, β2, ..., βs) :

X

p

→X

p

,

which is defined by the Hadamard product:

Hp,q,s(α1, α2, ..., αq;β1, β2, ..., βs)f(z) =hp(α1, α2, ..., αq;β1, β2, ..., βs;z)∗f(z),

(βj 6= 0,−1,−2, ...;j= 1,2, ..., s, q≤s+ 1;s, q∈N0=N∪ {0};z∈), For notational simplicity, we use

Hp,q,s(α1) =Hp,q,s(α1, α2, ..., αq;β1, β2, ..., βs).

For a function of the form (1), we have

Hp,q,s(α1)f(z) =

1 zp +

∞

X

k=0

Γkap+kzp+k,

where

Γk =

(α1)p+k,(α2)p+k, ...,(αq)p+k (β1)p+k,(β2)p+k, ...,(βs)p+k(1)p+k Then we can easily verify that:

z(Hp,q,s(α1)f(z))0=α1Hp,q,s(α1)f(z)−(α1+p)Hp,q,s(α1)f(z).

The linear operator Hp,q,s(α1) was investigated recently by Liu and Srivastava [9],

Aouf [2], and Aouf and Yassen [4]. Now we define the class Np,q,s(α1;α, β, γ) as

follows:

Definition 1.1 : A function f(z) of the form (1) is said to be in the class Np,q,s(α1;α, β, γ) if it satisfies the following inequality:

| z

p+1₍_H

p,q,s(α1)f(z))0+p

(2γ−1)zp+1₍_H

p,q,s(α1)f(z))0+ (2γα−p)

|< β, (2)

where 0 ≤α < p,0 < β ≤ 1,1₂ ≤ γ ≤1, α1 > 0, p ∈ N, q ≤ s+ 1;s, q ∈ N0 =

N∪ {0};z∈∗.

The following are special classes of the classNp,q,s(α1;α, β, γ):

(1) Np,q,s(α1;α,1,1) = {f ∈ Pp : <{−z p+1₍_H

p,q,s(α1)f(z))0} > α,0 ≤ α <

p, α1>0, p∈N, q≤s+ 1;s, q∈N0=N∪ {0};z∈∗}.

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International Journal of Basic & Applied Sciences IJBAS-IJENS Vol:13 No:05 34

(2) Np,2,1(a,1;c;α,1,1) = {f ∈ P_p : <{−zp+1(`p(a, c)f(z))0 > 0,0 ≤ α <

p, a, c >0, p∈N, z∈∗}, see [8].

(3) Np,2,1(x+p, p;p;α,1,1) ={f ∈P_p:<{−zp+1(Dx+p−1f(z))0 >0,0≤α <

p, x >−p, p∈N, x∈∗}, see [1], [3].

(4) Np,2,1(x,1;x+ 1;α,1,1) = {f ∈ Pp : <{−z p+1₍_F

x,pf(z))0 > 0,0 ≤ α <

p, x >0, p∈_N, x∈∗_}_{, see [1], [14], [15].}

Meromorphic multivalent functions have been studied (for example) by many outhers such as Rain and Srivastava [11], Yang [15], EL- Ashwah [7], Saif and Kilicman [12], Mostafa [10] and others.

In this paper, we derive several interesting properties for the classNp,q,s(α1;α, β, γ)

such as coefficient inequality, growth and distortion theorems, closure theorems, Hadamard properties, radii of meromorphically p-valent starlikeness, convexity and weighted mean and arithmetic mean for these functions.

2. _{Coefficient Estimates}

Theorem 2.1 : A function f(z) defined by (1) is said to be in the class Np,q,s(α1;α, β, γ) if and only if

∞

X

k=0

(p+k)(1 + 2βγ−β)Γkap+k ≤2βγ(p−α), (3)

where

Γk =

(α1)p+k,(α2)p+k, ...,(αq)p+k (β1)p+k,(β2)p+k, ...,(βs)p+k(1)p+k

,

and

0≤α < p,0< β≤1,₂1 ≤γ≤1, α1>0, p∈N, q≤s+ 1;s, q∈N0=N∪ {0};z∈∗.

proof: Assume that (3) holds. Then

|zp+1(Hp,q,s(α1)f(z))0+p| −β|(2γ−1)zp+1(Hp,q,s(α1)f(z))0+ (2γα−p)|<0

|zp+1(−pz−p−1+ ∞

X

k=0

(p+k)Γkap+kzp+k−1) +p|

−β|(2γ−1)zp+1(−pz−p−1+ ∞

X

k=0

(p+k)Γkap+kzp+k−1) + (2γα+p)|<0

|

∞

X

k=0

(p+k)Γkap+kz2p+k| −β|2γ(p−α) + ∞

X

k=0

(p+k)(2γ−1)Γkap+kz2p+k|<0.

For|z|=r <1 ∞

X

k=0

(p+k)Γkap+kr2p+k−2βγ(p−α) + ∞

X

k=0

β(p+k)(2γ−1)Γkap+kr2p+k

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International Journal of Basic & Applied Sciences IJBAS-IJENS Vol:13 No:05 35

<

∞

X

k=0

(p+k)(1 + 2βγ−β)Γkap+k−2βγ(p−α)≤0.

Thusf ∈Np,q,s(α1;α, β, γ).

Conversely, assume thatf ∈Np,q,s(α1;α, β, γ).

Then by (2), we have

| z

p+1₍_H

p,q,s(α1)f(z))0+p

(2γ−1)zp+1₍_H

p,q,s(α1)f(z))0+ (2γα−p) |, z∈∗

|

P∞

k=0(p+k)Γkap+kz2p+k 2γ(p−γ) +P∞

k=0(p+k)(2γ−1)Γkap+kz2p+k

|< β, z∈∗.

Since|<(z)| ≤ |z|for allz, then

<{

P∞

k=0(p+k)Γkap+kz2p+k 2γ(p−γ) +P∞

k=0(p+k)(2γ−1)Γkap+kz2p+k

}< β, z∈∗_. ₍₄₎

Now choosing the values ofzon the real axis so that the functionzp+1(Hp,q,s(α1)f(z))0

is real. By clearing the denominator in (4) and letting z → 1− through positive values, we get:

∞

X

k=0

(p+k)(1 + 2βγ−β)Γkap+k ≤2βγ(p−α)

Hence the proof is complete.

Corollary 2.1Let the functionf(z) defined by (1) be in the classNp,q,s(α1;α, β, γ).

Then

ap+k≤

2βγ(p−α) (p+k)(1 + 2βγ−β)Γk

,(k≥0, p∈N)

The result is sharp for the function:

f(z) = 1 zp +

zp+k,(k≥0, p∈_N) (5)

3. Growth and Distortion Theorem

A growth and distortion property for the functionfto be in the classNp,q,s(α1;α, β, γ)

is given as follows:

Theorem 3.1: Let the functionf(z) defined by (1) be in the classNp,q,s(α1;α, β, γ).

Then for 0<|z|=r <1, we have 1

rp −

2βγ(p−α) p(1 + 2βγ−β)Γ0

rp≤ |f(z)| ≤ 1

rp +

2βγ(p−α) p(1 + 2βγ−β)Γ0

rp, (6)

and p rp+1 −

2βγ(p−α) (1 + 2βγ−β)Γ0

rp−1≤ |f0(z)| ≤ p

rp+1 +

2βγ(p−α) (1 + 2βγ−β)Γ0

rp−1, (7)

with equality for

f(z) = 1 zp +

2βγ(p−α) p(1 + 2βγ−β)Γ0

zp(p∈N) (8)

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International Journal of Basic & Applied Sciences IJBAS-IJENS Vol:13 No:05 36

proof: By Theorem 2.1, we have

p(1 + 2βγ−β)Γ0

∞

X

k=0

ap+k≤ ∞

X

k=0

(p+k)(1 + 2βγ−β)Γkap+k≤2βγ(p−α).

Then

∞

X

k=0

ap+k≤

2βγ(p−α) p(1 + 2βγ−β)Γ0

for 0<|z|=r <1,

|f(z)| ≤ 1

rp + ∞

X

k=0

ap+krp+k,

≤ 1

rp +r p

∞

X

k=0

ap+k

≤ 1

rp +

2βγ(p−α) p(1 + 2βγ−β)Γ0

rp

and

|f(z)| ≥ 1

rp − ∞

X

k=0

ap+krp+k,

≥ 1

rp −r p

∞

X

k=0

ap+k,

≥ 1

rp −

2βγ(p−α) p(1 + 2βγ−β)Γ0

rp

which, together, yield (6). Also from Theorem 2. 1, it follows that ∞

X

k=0

(p+k)ap+k≤

2βγ(p−α) (1 + 2βγ−β)Γ0

.

Thus

|f0(z)| ≤ | −p

zp+1|+

∞

X

k=0

(p+k)ap+kzp+k−1,

|f0(z)| ≤ p

rp+1 +r

p−1

∞

X

k=0

(p+k)ap+k

≤ p

rp+1 +

2βγ(p−α) (1 + 2βγ−β)Γ0

rp−1,

and

|f0(z)| ≥ | −p

zp+1| −

∞

X

k=0

(p+k)ap+kzp+k−1,

|f0(z)| ≥ p

rp+1 −r

p−1

∞

X

k=0

(p+k)ap+k

≥ p

rp+1 −

2βγ(p−α) (1 + 2βγ−β)Γ0

rp−1,

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which, together, yield (7).

It is clear that the function given by (8) is extremal function. Hence the proof is complete.

4. _{closure Theorems}

Theorem 4.1: Let

fp−1(z) =

1 zp, and

fp+k(z) = 1 zp +

zp+k,(k≥0, p∈_N).

Thenf(z) is in the class Np,q,s(α1;α, β, γ) if and only if it can be expressed in the

form

f(z) = ∞

X

k=−1

λp+kfp+k(z),

whereλp+k≥0 andPk∞=−1λp+k = 1.

proof: First suppose thatf(z) can be expressed of the form

f(z) = ∞

X

k=−1

λp+kfp+k(z)

= 1 zp +

∞

X

k=0

2βγ(p−α)λp+k (p+k)(1 + 2βγ−β)Γk

zp+k

Then

∞

X

k=0

(p+k)(1 + 2βγ−β)Γk 2βγ(p−α)

2βγ(p−α)λp+k (p+k)(1 + 2βγ−β)Γk

= ∞

X

k=0

λp+k= 1−λp−1≤1.

Hence f ∈Np,q,s(α1;α, β, γ).

Conversely, suppose thatf ∈Np,q,s(α1;α, β, γ), then

ap+k≤

,(k≥0, p∈N)

Setting

λp+k =

(p+k)(1 + 2βγ−β)Γk 2βγ(p−α) ap+k and

λp−1= 1−

∞

X

k=0

λp+k,

we get

f(z) = ∞

X

k=−1

λp+kfp+k(z)

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International Journal of Basic & Applied Sciences IJBAS-IJENS Vol:13 No:05 38

Theorem 4.2: Let the functions fi(z) = _z1p+ P∞

k=0ap+k,iz

p+k_{, i}_{= 1}_,₂_{, ..., n} be in the classNp,q,s(α1;α, β, γ). Then the function

F(z) =Pn

i=1λifi(z), where

Pn

i=1λi= 1,

is also in the classNp,q,s(α1;α, β, γ).

proof: From Theorem 2.1, we have ∞

X

k=0

(p+k)(1 + 2βγ−β)Γkap+k ≤2βγ(p−α).

Since

F(z) = 1 zp +

∞

X

k=0

( n

X

i=1

λiap+k,i)zp+k

Then

∞

X

k=0

(p+k)(1 + 2βγ−β)Γk( n

X

i=1

λiap+k,i)

= n

X

i=1

λi ∞

X

k=0

(p+k)(1 + 2βγ−β)Γkap+k,i

≤2βγ(p−α) n

X

i=1

λi= 2βγ(p−α).

This completes the proof of the theorem.

Theorem 4.3: The classNp,q,s(α1;α, β, γ) is convex.

proof : In order to proof the theorem it is enough to show that the function h(z) defined by

h(z) =λf(z) + (1−λ)g(z),(0≤λ≤1)

is in the classNp,q,s(α1;α, β, γ),

where

f(z) = 1 zp +

∞

X

k=0

ap+kzp+k, ap+k≥0

g(z) = 1 zp +

∞

X

k=0

bp+kzp+k, bp+k ≥0,

are in the classNp,q,s(α1;α, β, γ).

Then

h(z) = 1 zp +

∞

X

k=0

(λap+k+ (1−λ)bp+k)zp+k.

By using Theorem 2. 1, we get ∞

X

k=0

(p+k)(1 + 2βγ−β)Γk(λap+k+ (1−λ)bp+k)

≤λ2βγ(p−α) + (1−λ)2βγ(p−α)

= 2βγ(p−α).

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Thush(z)∈Np,q,s(α1;α, β, γ).

5. _{Convolution Properties}

Theorem 5.1: Let the functions

fi(z) = 1 zp +

∞

X

k=0

ap+k,izp+k,(ap+k,i≥0;i= 1,2), (9)

be in the classNp,q,s(α1;α, β, γ), then (f1∗f2)∈Np,q,s(α1;η, β, γ), where

η=p− 2βγ(p−α) 2

p(1 + 2βγ−β)Γ0

.

The result is sharp for the functionsfi(z)(i= 1,2) given by

fi(z) = 1 zp +

2βγ(p−α) p(1 + 2βγ−β)Γ0

zp,(i= 1,2, p∈N)

proof: Sincefi(z)∈Np,q,s(α1;α, β, γ)(i= 1,2).

Then from Theorem 2.1, we have: ∞

X

k=0

(p+k)(1 + 2βγ−β)Γk

2βγ(p−α) ap+k,i≤1(i= 1,2).

Thus by the Cauchy- Schwarz inequality, we obtain ∞

X

k=0

(p+k)(1 + 2βγ−β)Γk 2βγ(p−α)

√

ap+k,1ap+k,2≤1 (10)

To prove the theorem we need to find the largestη such that ∞

X

k=0

(p+k)(1 + 2βγ−β)Γk

2βγ(p−η) ap+k,1ap+k,2≤1,

or we must get:

ap+k,1ap+k,2

p−η ≤

√

ap+k,1ap+k,2

p−α (k≥0;p∈N), which is equivalent to

√

ap+k,1ap+k,2≤

p−η

p−α(k≥0;p∈N). From (10), we have

≤ p−η

p−α.

By simplifying it, we get:

η≤p− 2βγ(p−α) 2

(p+k)(1 + 2βγ−β)Γk

.

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Now, defining the functionφ(k) by

φ(k) =p− 2βγ(p−α) 2

(p+k)(1 + 2βγ−β)Γk

.

This function is an increasing function ofk. Thus, we have

η≤φ(0) =p− 2βγ(p−α) 2

p(1 + 2βγ−β)Γ0

.

Theorem 5.2: Let the functionf1(z) defined by (9) be in the classNp,q,s(α1;α2, β, γ)

and the function f2(z) defined by (9) be in the class Np,q,s(α1;α3, β, γ). Then

(f1∗f2)(z)∈Np,q,s(α1;ζ, β, γ), where

ζ=p−2βγ(p−α2)(p−α3)

p(1 + 2βγ−β)Γ0

f1(z) =

1 zp +

2βγ(p−α2)

p(1 + 2βγ−β)Γ0

zp(p∈N),

and

f2(z) =

1 zp +

2βγ(p−α3)

p(1 + 2βγ−β)Γ0

zp(p∈N).

proof : By using the same technique of Theorem 5.1 we prove the theorem, hence it is omitted.

Theorem 5.3 : Let f1(z) = _z1p + P∞

k=0ap+k,1zp+k ∈ Np,q,s(α1;α, β, γ) and

f2(z) = _z1p + P∞

k=0ap+k,2zp+k ∈Np,q,s(α1;α, β, γ) with|ap+k,2| ≤1, k ≥0, p∈N. Then (f1∗f2)(z)∈Np,q,s(α1;α, β, γ).

proof: By using Theorem 2.1 it is enough to show that: ∞

X

k=0

(p+k)(1 + 2βγ−β)Γk

2βγ(p−α) ap+k,1ap+k,2≤1,

Since

∞

X

k=0

(p+k)(1 + 2βγ−β)Γk

2βγ(p−α) |ap+k,1ap+k,2|,

= ∞

X

k=0

(p+k)(1 + 2βγ−β)Γk

2βγ(p−α) ap+k,1|ap+k,2|,

≤

∞

X

k=0

(p+k)(1 + 2βγ−β)Γk 2βγ(p−α) ap+k,1,

≤1.

Thus (f1∗f2)(z)∈Np,q,s(α1;α, β, γ).

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Corollary 5.1 : Let f1(z) = _z1p + P∞

k=0ap+k,1z

p+k _∈ _N

p,q,s(α1;α, β, γ) and

f2(z) =_z1p+ P∞

k=0ap+k,2zp+k∈Np,q,s(α1;α, β, γ) with 0≤ |ap+k,2| ≤1, k≥0, p∈

N. Then (f1∗f2)(z)∈Np,q,s(α1;α, β, γ).

Theorem 5.4 : If the functions fi(z)(i = 1,2) defined by (9) are in the class

Np,q,s(α1;α, β, γ) and

p(1 + 2βγ−β)Γ0−4βγ(p−α)≥0,

then the function h(z) defined by

h(z) = 1 zp +

∞

X

k=0

(a2_p₊_k,₁+a_p2₊_k,₂)zp+k,

X

k=0

(p+k)(1 + 2βγ−β)Γk

2βγ(p−α) ap+k,1≤1,

and

∞

X

k=0

(p+k)(1 + 2βγ−β)Γk

2βγ(p−α) ap+k,2≤1. Then

∞

X

k=0

[(p+k)(1 + 2βγ−β)Γk 2βγ(p−α) ]

2_a2

p+k,1≤1,

and

∞

X

k=0

[(p+k)(1 + 2βγ−β)Γk 2βγ(p−α) ]

2

a2p+k,2≤1.

Hence

∞

X

k=0

1 2[

(p+k)(1 + 2βγ−β)Γk 2βγ(p−α) ]

2₍_a2

p+k,1+a 2

p+k,2)≤1.

To proof the theorem it is sufficient to show that ∞

X

k=0

[(p+k)(1 + 2βγ−β)Γk 2βγ(p−α) ](a

2

p+k,1+a 2

p+k,2)≤1. (11)

Thus the inequality (11) will be satisfied if (p+k)(1 + 2βγ−β)Γk

2βγ(p−α) ≤ 1 2[

(p+k)(1 + 2βγ−β)Γk 2βγ(p−α) ]

2_{, k}_{= 0}_,₁_,₂_{, ...}

or

(p+k)(1 + 2β−β)Γk−4βγ(p−α)≥0, k= 0,1,2, ....

The left hand side of the above inequality is an increasing function of k, so it satisfied for allk if

p(1 + 2β−β)Γ0−4βγ(p−α)≥0,

which is given by our hypothesis.

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Theorem 5.5: If the functionsfi(z)(i= 1,2) defined by

fi(z) = 1 zp +

∞

X

k=0

ap+k,izp+k,(ap+k,i≥0;i= 1,2), (8)

are in the classNp,q,s(α1;α, β, γ), then the function h(z) defined by

h(z) = 1 zp +

∞

X

k=0

(a2_p₊_k,₁+a_p2₊_k,₂)zp+k,

is in the classNp,q,s(α1;ζ, β, γ), where

ζ=p− 4βγ(p−α) 2

p(1 + 2βγ−β)Γ0

fi(z) = 1 zp +

2βγ(p−α) p(1 + 2βγ−β)Γ0

zp,(i= 1,2, p∈N)

X

k=0

(p+k)(1 + 2βγ−β)Γk

2βγ(p−α) ap+k,i≤1(i= 1,2).

Now

∞

X

k=0

[(p+k)(1 + 2βγ−β)Γk 2βγ(p−α) ]

2

a2p+k,i

≤

∞

X

k=0

[(p+k)(1 + 2βγ−β)Γk 2βγ(p−α) ap+k,i]

2

≤1(i= 1,2),

hence

∞

X

k=0

1 2[

(p+k)(1 + 2βγ−β)Γk 2βγ(p−α) ]

2₍_a2

p+k,1+a 2

p+k,2)≤1. (12)

To prove the theorem, it is sufficient to show that: ∞

X

k=0

(p+k)(1 + 2βγ−β)Γk 2βγ(p−ζ) (a

2

p+k,1+a 2

p+k,2)≤1. (13)

From (12)(13), we must find the largest value ofζ such that 1

p−ζ ≤

(p+k)(1 + 2βγ−β)Γk 4βγ(p−α)2 ,

or

ζ≤p− 4βγ(p−α) 2

(p+k)(1 + 2βγ−β)Γk

,

since the right hand side of the above inequality is an increasing function of k, we get

ζ=p− 4βγ(p−α) 2

p(1 + 2βγ−β)Γ0

.

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International Journal of Basic & Applied Sciences IJBAS-IJENS Vol:13 No:05 43

6. _{Radii of Meromorphically p-Valent Starlikness and Convexity}

Then f(z) is meromorphically p-valent starlike of order δ(0 ≤δ < p) in the disk

|z|< r1, where

r1= inf

k≥0{

(p+k)(1 + 2βγ−β)Γk(p−δ) 2βγ(p−α)(p+k−δ) }

1 2p+k

The result is sharp .

proof: From Theorem 2. 1, we have: ∞

X

k=0

(p+k)(1 + 2βγ−β)Γkap+k ≤2βγ(p−α),

andf(z) is said to be meromorphically p-valent starlike of orderδ(0≤δ < p), if

<{−zf

0₍_z₎

f(z) }> δ, or

|zf

0₍_z_{) +}_pf₍_z₎

f(z) | ≤p−δ(0≤δ < p). Now

|zf

0₍_z_{) +}_pf₍_z₎

f(z) |=|

P∞

k=0(2p+k)ap+kz

p+k

z−p₊P∞

k=0ap+kzp+k

|,

≤

P∞

k=0(2p+k)ap+k|z|2p+k 1−P∞

k=0ap+k|z|2p+k

.

To prove the theorem the above inequality must be less than or equal to p−δ, so ∞

X

k=0

(p+k−δ) (p−δ) ap+k|z|

2p+k

≤p−δ. (14)

Then by Corollary 2. 1 the inequality (14) will be true if

|z|2p+k≤ (p+k)(1 + 2βγ−β)Γk(p−δ)

2βγ(p−α)(p+k−δ) , that is,

|z| ≤ {(p+k)(1 + 2βγ−β)Γk(p−δ)

2βγ(p−α)(p+k−δ) }

1 2p+k,

The infimum of the above quantity is the radii of starlikeness of the functionf(z) in the classNp,q,s(α1;α, β, γ).

The sharpness follows by choosing the same extremal function (5). This completes the proof of the theorem.

Then f(z) is meromorphically p-valent convex of order µ(0 ≤µ < p) in the disk

|z|< r2, where

r2= inf

k≥0{

p(p−µ)(1 + 2βγ−β)Γk 2βγ(p−α)(3p+k−µ) }

1 2p+k_.

The result is sharp .

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International Journal of Basic & Applied Sciences IJBAS-IJENS Vol:13 No:05 44

proof: It is enough to show that

<{−1−zf

00₍_z₎

f0₍_z₎ }> µ(0≤µ < p);|z|< r2;p∈N), or

|(zf

0₍_z₎₎0₊_pf0₍_z₎

f0₍_z₎ |=|

P∞

k=0(p+k)(2p+k)ap+kz

p+k

−pz−(p+1)₊P∞

k=0(p+k)ap+kzp+k−1

|,

≤

P∞

k=0(p+k)(2p+k)ap+k|z|2p+k

p−P∞

k=0(p+k)ap+k|z|2p+k

.

To prove the theorem the above inequality must be less than or equal to p−µ, or ∞

X

k=0

(p+k)(3p+k−µ) p(p−µ) ap+k|z|

2p+k_≤₁_.

By using Theorem 2.1,we obtain

|z|2p+k_≤ p(p−µ)(1 + 2βγ−β)Γk 2βγ(p−α)(3p+k−µ) . Thus

|z| ≤ {p(p−µ)(1 + 2βγ−β)Γk

2βγ(p−α)(3p+k−µ) }

1

2p+k₍_k≥₀_{, p}∈_N_.

By choosingr2 to be the infimum of the above quantity we get the result.

The sharpness follows by choosing the same extremal function (5). This completes the proof of the theorem.

7. Weighted Mean and Arithmetic Mean

Definition 1.1: If the functions f(z) and g(z) defined by (1) are in the class Np,q,s(α1;α, β, γ), then the weighted mean hi(z) of the two functions is defined as follows

hi(z) = 1

2[(1−i)f(z) + (1 +i)g(z)].

Theorem 7.1 : If the functions f(z) and g(z) defined by (1) are in the class Np,q,s(α1;α, β, γ). Then their weighted mean is also in the classNp,q,s(α1;α, β, γ).

proof: The weighted mean off(z) andg(z) is:

hi(z) = 1

2[(1−i)f(z) + (1 +i)g(z)],

=1

2[(1−i)( 1 zp +

∞

X

k=0

ap+kzp+k) + (1 +i)( 1 zp +

∞

X

k=0

bp+kzp+k)],

= 1 zp +

∞

X

k=0

1

2((1−i)ap+k+ (1 +i)bp+k)z p+k_.

By using Theorem 2.1, it is sufficient to show that ∞

X

k=0

(p+k)(1 + 2βγ−β)Γk[ 1

2((1−i)ap+k+ (1 +i)bp+k)],

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International Journal of Basic & Applied Sciences IJBAS-IJENS Vol:13 No:05 45

= 1 2(1−i)

∞

X

k=0

(p+k)(1 + 2βγ−β)Γkap+k+ 1 2(1 +i)

∞

X

k=0

(p+k)(1 + 2βγ−β)Γkbp+k,

≤1

2(1−i)(2βγ(p−α)) + 1

2(1 +i)(2βγ(p−α)), = 2βγ(p−α).

Hence hi(z)∈Np,q,s(α1;α, β, γ).

Theorem 7.2: If the functionsfi(z)(i= 1, ..., d) defined by

fi(z) = 1 zp +

∞

X

k=0

ap+k,izp+k,(ap+k,i≥0, k≥0, i= 1, ..., d),

belongs to the classNp,q,s(α1;α, β, γ), then their arithmetic mean defined by

h(z) =1 d

d

X

i=1

fi(z),

proof: Since

h(z) = 1 zp +

∞

X

k=0

(1 d

d

X

i=1

ap+k,i)zp+k.

Then by using Theorem 2. 1, we must show that ∞

X

k=0

(p+k)(1 + 2βγ−β)Γk( 1 d

d

X

i=1

ap+k,i),

= 1 d

d

X

i=1

∞

X

k=0

(p+k)(1 + 2βγ−β)Γkap+k,i,

≤ 1

d

X

i=1

2βγ(p−α) = 2βγ(p−α).

Thereforeh(z)∈Np,q,s(α1;α, β, γ).

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