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p-Valent Meromorphic Functions M.K. Aouf and A.O. Mostafa

vol. 9, iss. 2, art. 45, 2008

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CERTAIN SUBCLASSES OF p-VALENT MEROMORPHIC FUNCTIONS INVOLVING

CERTAIN OPERATOR

M.K. AOUF AND A.O. MOSTAFA

Department of Mathematics Faculty of Science

Mansoura University Mansoura 35516, Egypt.

EMail:mkaouf127@yahoo.com adelaeg254@yahoo.com

Received: 25 March, 2008

Accepted: 20 May, 2008

Communicated by: N.K. Govil 2000 AMS Sub. Class.: 30C45.

Key words: Analytic,p−valent, Meromorphic, Integral operator.

Abstract: In this paper, a new subclassPα

p,β(η, δ, µ, λ)ofp−valent meromorphic func- tions defined by certain integral operator is introduced. Some interesting proper- ties of this class are obtained.

Acknowledgements: The authors would like to thank the referees of the paper for their helpful sug- gestions.

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p-Valent Meromorphic Functions M.K. Aouf and A.O. Mostafa

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Contents

1 Introduction 3

2 Main Results 6

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1. Introduction

LetP

p be the class of functionsf of the form:

(1.1) f(z) =z−p+

X

k=1

ak−pzk−p (p∈N={1,2, ...}),

which are analytic and p−valent in the punctured unit discU = {z : z ∈ C and 0<|z|<1}=U\{0}.

Similar to [1], we define the following family of integral operatorsQαβ,p:P

p → P

p (α≥0, β >−1;p∈N)as follows:

(i)

Qαβ,pf(z) =

α+β−1 β−1

αz−(p+β) Z z

0

(1− t

z)α−1tβ+p−1f(t)dt (1.2)

(α >0; β >−1; p∈N; f ∈Σp);

(1.3) and (ii)

(1.4) Q0β,pf(z) =f(z) (forα= 0; β >−1; p∈N; f ∈Σp).

From(1.2)and(1.4),we have

Qαβ,pf(z) =z−p+ Γ(α+β) Γ(β)

X

k=1

Γ(k+β)

Γ(k+β+α)ak−pzk−p (1.5)

(α≥0; β >−1; p∈N; f ∈Σp).

(1.6)

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Using the relation(1.5),it is easy to show that

(1.7) z(Qαβ,pf(z))0 = (α+β−1)Qα−1β,p f(z)−(α+β+p−1)Qαβ,pf(z).

Definition 1.1. Let Pα

p,β(η, δ, µ, λ)be the class of functionsf ∈P

p which satisfy:

(1.8) Re

(1−λ) Qαβ,pf(z) Qαβ,pg(z)

!µ

+λQα−1β,p f(z) Qα−1β,p g(z)

Qαβ,pf(z) Qαβ,pg(z)

!µ−1

> η,

whereg ∈P

p satisfies the following condition:

(1.9) Re

(Qαβ,pg(z) Qα−1β,p g(z)

)

> δ (0≤δ <1;z ∈U),

and η and µ are real numbers such that 0 ≤ η < 1, µ > 0 and λ ∈ C with Re{λ}>0.

To establish our main results we need the following lemmas.

Lemma 1.2 ([2]). Letbe a set in the complex plane C and let the function ψ : C2 → C satisfy the condition ψ(ir2, s1) ∈/ Ω for all real r2, s1 ≤ −1+r222. If q is analytic in U with q(0) = 1 and ψ(q(z), zq0(z)) ∈ Ω, z ∈ U, then Re{q(z)} >

0 (z ∈U).

Lemma 1.3 ([3]). If q is analytic in U with q(0) = 1, and if λ ∈ C\{0} with Re{λ} ≥ 0, then Re{q(z) +λzq0(z)} > α (0 ≤ α < 1) implies Re{q(z)} >

α+ (1−α)(2γ−1),whereγis given by

γ =γ(Reλ) = Z 1

0

(1 +tRe{λ})−1dt

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which is increasing function ofRe{λ}and 12 ≤γ < 1.The estimate is sharp in the sense that the bound cannot be improved.

For real or complex numbersa, bandc(c6= 0,−1,−2, . . .), the Gauss hyperge- ometric function is defined by

(1.10) 2F1(a, b;c;z) = 1 + a·b c

z

1! +a(a+ 1)·b(b+ 1) c(c+ 1)

z2

2! +· · · .

We note that the series(1.10)converges absolutely forz ∈ U and hence represents an analytic function inU (see, for details, [4, Ch. 14]). Each of the identities (as- serted by Lemma1.4below) is fairly well known (cf., e.g., [4, Ch. 14]).

Lemma 1.4 ([4]). For real or complex numbersa, bandc(c6= 0,−1,−2, . . .),

Z 1 0

tb−1(1−t)c−b−1(1−tz)−adt = Γ(b)Γ(c−b)

Γ(c) 2F1(a, b;c;z) (1.11)

(Re(c)>Re(b)>0);

(1.12) 2F1(a, b;c;z) = (1−z)−a 2F1

a, c−b;c; z z−1

;

(1.13) 2F1(a, b;c;z) =2 F1(b, a;c;z);

and

(1.14) 2F1

1,1; 2;1 2

= 2 ln 2.

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2. Main Results

Unless otherwise mentioned, we assume throughout this paper that

α≥0; β >−1; α+β 6= 1; µ >0; 0≤η <1; p∈Nandλ≥0.

Theorem 2.1. Letf ∈Pα

p,β(η, δ, µ, λ).Then

(2.1) Re Qαβ,pf(z) Qαβ,pg(z)

!µ

> 2µη(α+β−1) +λδ

2µ(α+β−1) +λδ , (z ∈U),

where the functiong ∈P

p satisfies the condition(1.9).

Proof. Letγ = 2µη(α+β−1)+λδ

2µ(α+β−1)+λδ,and we define the functionqby

(2.2) q(z) = 1

1−γ

"

Qαβ,pf(z) Qαβ,pg(z)

!µ

−γ

# .

Thenqis analytic inU andq(0) = 1. If we set

(2.3) h(z) = Qαβ,pg(z)

Qα−1β,p g(z),

then by the hypothesis(1.9), Re{h(z)} > δ.Differentiating(2.2)with respect toz and using the identity(1.7), we have

(2.4) (1−λ) Qαβ,pf(z) Qαβ,pg(z)

!µ

+λQα−1β,p f(z) Qα−1β,p g(z)

Qαβ,pf(z) Qαβ,pg(z)

!µ−1

= [(1−γ)q(z) +γ] + λ(1−γ)

µ(α+β−1)zq0(z)h(z).

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Let us define the functionψ(r, s)by

(2.5) ψ(r, s) = [(1−γ)r+γ] + λ(1−γ)

µ(α+β−1)sh(z).

Using(2.5)and the fact thatf ∈Pα

p,β(η, δ, µ, λ), we obtain

{ψ(q(z), zq0(z));z ∈U)} ⊂Ω ={w∈C : Re(w)> η}.

Now for all realr2, s1 ≤ −1+r222,we have

Re{ψ(ir2, s1)}=γ+ λ(1−γ)s1

µ(α+β−1)Reh(z)

≤γ− λ(1−γ)δ(1 +r22) 2µ(α+β−1)

≤γ− λ(1−γ)δ

2µ(α+β−1) =η.

Hence for eachz ∈ U, ψ(ir2, s1) ∈/ Ω.Thus by Lemma1.2, we haveRe{q(z)} >

0 (z ∈U)and hence

Re Qαβ,pf(z) Qαβ,pg(z)

!µ

> γ (z ∈U).

This proves Theorem2.1.

Corollary 2.2. Let the functionsf and g be in P

p and let g satisfy the condition (1.9). Ifλ ≥1and

(2.6) Re

(

(1−λ)Qαβ,pf(z)

Qαβ,pg(z) +λQα−1β,p f(z) Qα−1β,p g(z)

)

> η (0≤η <1;z ∈U),

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then

(2.7) Re

(Qα−1β,p f(z) Qα−1β,p g(z)

)

> γ = η[2(α+β−1) +δ] +δ(λ−1)

2(α+β−1) +δλ (z ∈U).

Proof. We have

λQα−1β,p f(z) Qα−1β,p g(z) =

(

(1−λ)Qαβ,pf(z)

Qαβ,pg(z) +λQα−1β,p f(z) Qα−1β,p g(z)

)

+ (λ−1)Qαβ,pf(z)

Qαβ,pg(z) (z ∈U).

Sinceλ≥1,making use of(2.6)and(2.1) (forµ= 1), we deduce that Re

(Qα−1β,p f(z) Qα−1β,p g(z)

)

> γ = η[2(α+β−1) +δ] +δ(λ−1)

2(α+β−1) +δλ (z ∈U).

Corollary 2.3. Letλ ∈ C\{0}with Re{λ} ≥ 0.Iff ∈ P

p satisfies the following condition:

Re{(1−λ)(zpQαβ,pf(z))µ+λzpQα−1β,p f(z)(zpQαβ,pf(z))µ−1}> η (z ∈U),

then

(2.8) Re{(zpQαβ,pf(z))µ}> 2µη(α+β−1) + Re{λ}

2µ(α+β−1) + Re{λ} (z ∈U).

Further, ifλ≥1andf ∈P

p satisfies

(2.9) Re{(1−λ)zpQαβ,pf(z) +λzpQα−1β,p f(z)}> η (z ∈U),

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then

(2.10) Re{zpQα−1β,p f(z)}> 2η(α+β−1) +λ−1

2(α+β−1) +λ (z ∈U).

Proof. The results (2.8)and (2.10) follow by puttingg(z) = z−p in Theorem 2.1 and Corollary2.2, respectively.

Remark 1. Choosingα, λandµappropriately in Corollary2.3, we have, (i) Forα= 0, β 6= 1andλ= 1in Corollary2.3, we have:

Re 1

β−1

β+p−1 + zf0(z) f(z)

(zpf(z))µ

> η (z ∈U),

which implies that

Re{zpf(z)}µ> 2µη(β−1) + 1

2µ(β−1) + 1 (z∈U).

(ii) Forα= 0, β6= 1, µ= 1andλ∈C\{0}withRe{λ} ≥0in Corollary2.3, we have

Re

(1 + λp

β−1)zpf(z) + λ

β−1zp+1f0(z)

> η,

which implies that

Re{zpf(z)}> 2η(β−1) + Re{λ}

2(β−1) + Re{λ} (z ∈U).

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(iii) Replacingf byzfp0 in the result(ii), we have:

−Re h

1 + β−1λ (p+ 1)izp+1f0(z)

p + p(β−1)λ zp+1f00(z)

> η (0≤η <1;z ∈U),

which implies that

−Re

zp+1f0(z) p

> 2η(β−1) + Re{λ}

2(β−1) + Re{λ} (z ∈U).

Theorem 2.4. Let λ ∈ C with Re{λ} > 0. If f ∈ P

p satisfies the following condition:

(2.11) Re{(1−λ)(zpQαβ,pf(z))µ+λzpQα−1β,p f(z)(zpQαβ,pf(z))µ−1}> η (z ∈U), then

(2.12) Re{(zpQαβ,pf(z))µ}> η+ (1−η)(2ρ−1), where

(2.13) ρ= 1

2 2F1

1,1;µ(α+β−1) Re{λ} + 1;1

2

.

Proof. Let

(2.14) q(z) = (zpQαβ,pf(z))µ.

Thenqis analytic withq(0) = 1.Differentiating(2.14)with respect toz and using the identity(1.7), we have

(1−λ)(zpQαβ,pf(z))µ+λzpQα−1β,p f(z)(zpQαβ,pf(z))µ−1

=q(z) + λ

µ(α+β−1)zq0(z),

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so that by the hypothesis(2.11), we have Re

q(z) + λ

µ(α+β−1)zq0(z)

> η (z ∈U).

In view of Lemma1.3, this implies that

Re{q(z)}> η+ (1−η)(2ρ−1),

where

ρ=ρ(Re{λ}) = Z 1

0

1 +t

Re{λ}

µ(α+β−1)

−1 dt.

PuttingRe{λ}=λ1 >0,we have ρ=

Z 1 0

1 +t

λ1 µ(α+β−1))

−1

dt= µ(α+β−1) λ1

Z 1 0

(1 +u)−1u

µ(α+β−1) λ1 −1

du

Using(1.11),(1.12),(1.13)and(1.14),we obtain ρ= 1

2 2F1

1,1;µ(α+β−1)

λ1 + 1;1 2

.

This completes the proof of Theorem2.1.

Corollary 2.5. Letλ ∈Rwithλ ≥1.Iff ∈P

psatisfies

(2.15) Re

(1−λ)zpQαβ,pf(z) +λzpQα−1β,p f(z) > η (z ∈U),

then

Re{zpQα−1β,p f(z)}> η+ (1−η)(2ρ1−1)

1− 1 λ

(z ∈U),

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where

ρ1 = 1 2 2F1

1,1;(α+β−1)

λ + 1;1 2

.

Proof. The result follows by using the identity

(2.16) λzpQα−1β,p f(z) = (1−λ)zpQαβ,pf(z) +λzpQα−1β,p f(z) + (λ−1)zpQαβ,pf(z).

Remark 2. We note that, forα= 0, β = 2andλ=µ >0in Corollary2.3, that is, if

(2.17) Re

(1−λ)(zpf(z))λ+λ(zp+1f(z))0(zpf(z))λ−1 > η (z ∈U),

then(2.8)implies that

(2.18) Re{(zpf(z))λ}> 2η+ 1

3 (z ∈U), whereas, if f ∈ P

p satisfies the condition (2.17) then by using Theorem2.4, we have

Re{(zpf(z))λ}>2(1−ln 2)η+ (2 ln 2−1) (z ∈U), which is better than(2.18).

Theorem 2.6. Suppose that the functions f and g are in P

p and g satisfies the condition(1.9). If

(2.19) Re

(Qα−1β,p f(z)

Qα−1β,p g(z) −Qαβ,pf(z) Qαβ,pg(z)

)

>− (1−η)δ

2(α+β−1) (z ∈U),

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for someη(0≤η <1),then

(2.20) Re

(Qαβ,pf(z) Qαβ,pg(z)

)

> η (z ∈U)

and

(2.21) Re

(Qα−1β,p f(z) Qα−1β,p g(z)

)

> η[2(α+β−1) +δ]−δ

2(α+β−1) (z ∈U).

Proof. Let

(2.22) q(z) = 1

1−η

"

Qαβ,pf(z) Qαβ,pg(z) −η

# .

Thenqis analytic inU withq(0) = 1. Setting

(2.23) φ(z) = Qαβ,pg(z)

Qα−1β,p g(z) (z ∈U),

we observe that from(1.9), we haveRe{φ(z)} > δ (0 ≤ δ < 1)inU. A simple computation shows that

(1−η)zq0(z)

α+β−1 φ(z) = Qα−1β,p f(z)

Qα−1β,p g(z) −Qαβ,pf(z) Qαβ,pg(z)

=ψ(q(z), zq0(z)), where

ψ(r, s) = (1−η)sφ(z) α+β−1 .

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Using the hypothesis(2.19), we obtain {ψ(q(z), zq0(z));z ∈U} ⊂Ω =

w∈C: Rew >− (1−η)δ 2(α+β−1)

.

Now, for all realr2, s1 ≤ −1+r222, we have

Re{ψ(ir2, s1)}= s1(1−η) Re{φ(z)}

α+β−1

≤ −(1−η)δ(1 +r22) 2(α+β−1)

≤ −(1−η)δ 2(α+β−1).

This shows that ψ(ir2, s1) ∈/ Ω for each z ∈ U. Hence by Lemma 1.2, we have Re{q(z)} > 0 (z ∈ U).This proves (2.20). The proof of(2.21)follows by using (2.20)and(2.21)in the identity:

Re

(Qα−1β,p f(z) Qα−1β,p g(z)

)

= Re

(Qα−1β,p f(z)

Qα−1β,p g(z) −Qαβ,pf(z) Qαβ,pg(z)

)

−Re

(Qαβ,pf(z) Qαβ,pg(z)

) .

This completes the proof of Theorem2.6.

Remark 3.

(i) Forα = 0andg(z) =z−p in Theorem2.6, we have Re

zp+1f0(z) +pzpf(z) > −(1−η)δ

2 (z ∈U),

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which implies that

Re{zpf(z)}> η (z ∈U) and

Re

zp+1f0(z) + (p+β−1)zpf(z) > η[2(β−1) +δ]−δ

2 (z ∈U).

(ii) Puttingα = 0, β = 2in Theorem2.6, we get that, if

Re

zf0(z) + (p+ 1)f(z)

zg0(z) + (p+ 1)g(z) − f(z) g(z)

> −(1−η)δ

2 (z ∈U), then

Re

f(z) g(z)

> η (z ∈U)

and

Re

zf0(z) + (p+ 1)f(z) zg0(z) + (p+ 1)g(z)

> η(2 +δ)−δ

2 (z ∈U).

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References

[1] E. AQLAN, J.M. JAHANGIRI AND S.R. KULKARNI, Certain integral oper- ators applied to meromorphic p-valent functions, J. of Nat. Geom., 24 (2003), 111–120.

[2] S.S. MILLER AND P.T. MOCANU, Second order differential equations in the complex plane, J. Math. Anal. Appl., 65 (1978), 289–305.

[3] S. PONNUSAMY, Differential subordination and Bazilevic functions, Proc. In- dian Acad. Sci. (Math. Sci.), 105 (1995), 169–186.

[4] E.T. WHITTAKER AND G.N. WATSON, A Course of Modern Analysis: An Introduction to the General Theory of Infinite Processes and of Analytic Func- tions; With an Account of the Principal Transcendental Functions, Fourth Edi- tion, Cambridge University Press, Cambridge, 1927.

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