Volume 9 (2008), Issue 2, Article 45, 8 pp.
CERTAIN SUBCLASSES OF p−VALENT MEROMORPHIC FUNCTIONS
INVOLVING CERTAIN OPERATOR
M.K. AOUF AND A.O. MOSTAFA
D EPARTMENT OF M ATHEMATICS
FACULTY OF S CIENCE
M ANSOURA U NIVERSITY
M ANSOURA 35516, E GYPT
mkaouf127@yahoo.com
adelaeg254@yahoo.com
Received 25 March, 2008; accepted 20 May, 2008
Communicated by N.K. Govil
Pα
A BSTRACT. In this paper, a new subclass p,β (η, δ, µ, λ) of p−valent meromorphic functions
defined by certain integral operator is introduced. Some interesting properties of this class are
obtained.
Key words and phrases: Analytic, p−valent, Meromorphic, Integral operator.
2000 Mathematics Subject Classification. 30C45.
1. I NTRODUCTION
Let
P
p
be the class of functions f of the form:
f (z) = z
(1.1)
−p
+
∞
X
ak−p z k−p
(p ∈ N = {1, 2, ...}),
k=1
which are analytic and p−valent in the punctured unit disc U ∗ = {z : z ∈ C and 0 < |z| <
1} = U \{0}.
P
P
Similar to [1], we define the following family of integral operators Qαβ,p : p → p (α ≥
0, β > −1; p ∈ N) as follows:
(i)
Z z
α+β−1
t
α
−(p+β)
(1.2)
Qβ,p f (z) =
αz
(1 − )α−1 tβ+p−1 f (t)dt
β−1
z
0
(1.3)
(α > 0; β > −1; p ∈ N; f ∈ Σp );
and
The authors would like to thank the referees of the paper for their helpful suggestions.
092-08
2
M.K. AOUF AND A.O. M OSTAFA
(ii)
(1.4)
Q0β,p f (z) = f (z) (for α = 0; β > −1; p ∈ N; f ∈ Σp ).
From (1.2) and (1.4), we have
∞
(1.5)
(1.6)
Qαβ,p f (z) = z −p +
Γ(α + β) X Γ(k + β)
ak−p z k−p
Γ(β) k=1 Γ(k + β + α)
(α ≥ 0; β > −1; p ∈ N; f ∈ Σp ).
Using the relation (1.5), it is easy to show that
α
z(Qαβ,p f (z))0 = (α + β − 1)Qα−1
β,p f (z) − (α + β + p − 1)Qβ,p f (z).
Pα
P
Definition 1.1. Let
p,β (η, δ, µ, λ) be the class of functions f ∈
p which satisfy:

!µ
!µ−1 
α−1


α
α
Qβ,p f (z) Qβ,p f (z)
Qβ,p f (z)
(1.8)
Re (1 − λ)
+
λ
> η,


Qαβ,p g(z)
Qαβ,p g(z)
Qα−1
β,p g(z)
P
where g ∈ p satisfies the following condition:
(
)
Qαβ,p g(z)
(1.9)
Re
> δ (0 ≤ δ < 1; z ∈ U ),
Qα−1
β,p g(z)
(1.7)
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 1.1 ([2]). Let Ω be a set in the complex plane C and let the function ψ : C2 → C satisfy
1+r2
the condition ψ(ir2 , s1 ) ∈
/ Ω for all real r2 , s1 ≤ − 2 2 . If q is analytic in U with q(0) = 1 and
ψ(q(z), zq 0 (z)) ∈ Ω, z ∈ U, then Re{q(z)} > 0 (z ∈ U ).
Lemma 1.2 ([3]). If q is analytic in U with q(0) = 1, and if λ ∈ C\{0} with Re{λ} ≥ 0, then
Re{q(z) + λzq 0 (z)} > α (0 ≤ α < 1) implies Re{q(z)} > α + (1 − α)(2γ − 1), where γ is
given by
Z 1
γ = γ(Re λ) =
(1 + tRe{λ} )−1 dt
0
which is increasing function of Re{λ} and
the bound cannot be improved.
1
2
≤ γ < 1. The estimate is sharp in the sense that
For real or complex numbers a, b and c (c 6= 0, −1, −2, . . . ), the Gauss hypergeometric
function is defined by
(1.10)
2 F1 (a, b; c; z) = 1 +
a·b z
a(a + 1) · b(b + 1) z 2
+
+ ··· .
c 1!
c(c + 1)
2!
We note that the series (1.10) converges absolutely for z ∈ U and hence represents an analytic
function in U (see, for details, [4, Ch. 14]). Each of the identities (asserted by Lemma 1.3
below) is fairly well known (cf., e.g., [4, Ch. 14]).
Lemma 1.3 ([4]). For real or complex numbers a, b and c (c 6= 0, −1, −2, . . . ),
Z 1
Γ(b)Γ(c − b)
(1.11)
tb−1 (1 − t)c−b−1 (1 − tz)−a dt =
2 F1 (a, b; c; z)
Γ(c)
0
(Re(c) > Re(b) > 0);
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 45, 8 pp.
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p−VALENT M EROMORPHIC F UNCTIONS
(1.12)
2 F1 (a, b; c; z)
−a
= (1 − z)
2 F1 (a, b; c; z)
(1.13)
2 F1
3
z
a, c − b; c;
z−1
;
=2 F1 (b, a; c; z);
and
2 F1
(1.14)
1
1, 1; 2;
2
= 2 ln 2.
2. M AIN R ESULTS
Unless otherwise mentioned, we assume throughout this paper that
α ≥ 0; β > −1; α + β 6= 1; µ > 0; 0 ≤ η < 1; p ∈ N and λ ≥ 0.
P
Theorem 2.1. Let f ∈ αp,β (η, δ, µ, λ). Then
!µ
Qαβ,p f (z)
2µη(α + β − 1) + λδ
(2.1)
Re
>
,
(z ∈ U ),
α
Qβ,p g(z)
2µ(α + β − 1) + λδ
P
where the function g ∈ p satisfies the condition (1.9).
Proof. Let γ =
2µη(α+β−1)+λδ
,
2µ(α+β−1)+λδ
and we define the function q by
"
!µ
#
Qαβ,p f (z)
1
q(z) =
−γ .
1−γ
Qαβ,p g(z)
(2.2)
Then q is analytic in U and q(0) = 1. If we set
h(z) =
(2.3)
Qαβ,p g(z)
Qα−1
β,p g(z)
,
then by the hypothesis (1.9), Re{h(z)} > δ. Differentiating (2.2) with respect to z and using
the identity (1.7), we have
(2.4) (1 − λ)
Qαβ,p f (z)
Qαβ,p g(z)
!µ
+λ
Qα−1
β,p f (z)
Qα−1
β,p g(z)
Qαβ,p f (z)
Qαβ,p g(z)
!µ−1
= [(1 − γ)q(z) + γ] +
λ(1 − γ)
zq 0 (z)h(z).
µ(α + β − 1)
Let us define the function ψ(r, s) by
(2.5)
ψ(r, s) = [(1 − γ)r + γ] +
Using (2.5) and the fact that f ∈
Pα
p,β (η, δ, µ, λ),
λ(1 − γ)
sh(z).
µ(α + β − 1)
we obtain
{ψ(q(z), zq 0 (z)); z ∈ U )} ⊂ Ω = {w ∈ C : Re(w) > η}.
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 45, 8 pp.
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4
M.K. AOUF AND A.O. M OSTAFA
Now for all real r2 , s1 ≤ −
1+r22
,
2
we have
λ(1 − γ)s1
Re h(z)
µ(α + β − 1)
λ(1 − γ)δ(1 + r22 )
≤γ−
2µ(α + β − 1)
λ(1 − γ)δ
≤γ−
= η.
2µ(α + β − 1)
Hence for each z ∈ U, ψ(ir2 , s1 ) ∈
/ Ω. Thus by Lemma 1.1, we have Re{q(z)} > 0 (z ∈ U )
and hence
!µ
Qαβ,p f (z)
Re
> γ (z ∈ U ).
Qαβ,p g(z)
Re{ψ(ir2 , s1 )} = γ +
This proves Theorem 2.1.
P
Corollary 2.2. Let the functions f and g be in p and let g satisfy the condition (1.9). If λ ≥ 1
and
)
(
Qα−1
Qαβ,p f (z)
β,p f (z)
(2.6)
Re (1 − λ) α
+ λ α−1
> η (0 ≤ η < 1; z ∈ U ),
Qβ,p g(z)
Qβ,p g(z)
then
(
Re
(2.7)
Qα−1
β,p f (z)
Qα−1
β,p g(z)
)
>γ=
η[2(α + β − 1) + δ] + δ(λ − 1)
2(α + β − 1) + δλ
(z ∈ U ).
Proof. We have
λ
Qα−1
β,p f (z)
Qα−1
β,p g(z)
(
=
Qα−1
Qαβ,p f (z)
β,p f (z)
+ λ α−1
(1 − λ) α
Qβ,p g(z)
Qβ,p g(z)
)
+ (λ − 1)
Qαβ,p f (z)
Qαβ,p g(z)
(z ∈ U ).
Since λ ≥ 1, making use of (2.6) and (2.1) (for µ = 1), we deduce that
)
( α−1
Qβ,p f (z)
η[2(α + β − 1) + δ] + δ(λ − 1)
Re
>γ=
(z ∈ U ).
α−1
2(α + β − 1) + δλ
Qβ,p g(z)
Corollary 2.3. Let λ ∈ C\{0} with Re{λ} ≥ 0. If f ∈
P
p
satisfies the following condition:
α−1
Re{(1 − λ)(z p Qαβ,p f (z))µ + λz p Qβ,p
f (z)(z p Qαβ,p f (z))µ−1 } > η
(z ∈ U ),
then
2µη(α + β − 1) + Re{λ}
Re{(z p Qαβ,p f (z))µ } >
2µ(α + β − 1) + Re{λ}
P
Further, if λ ≥ 1 and f ∈ p satisfies
(2.8)
(2.9)
(z ∈ U ).
Re{(1 − λ)z p Qαβ,p f (z) + λz p Qα−1
β,p f (z)} > η
(z ∈ U ),
2η(α + β − 1) + λ − 1
2(α + β − 1) + λ
(z ∈ U ).
then
(2.10)
Re{z p Qα−1
β,p f (z)} >
Proof. The results (2.8) and (2.10) follow by putting g(z) = z −p in Theorem 2.1 and Corollary
2.2, respectively.
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 45, 8 pp.
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p−VALENT M EROMORPHIC F UNCTIONS
5
Remark 1. Choosing α, λ and µ appropriately in Corollary 2.3, we have,
(i) For α = 0, β 6= 1 and λ = 1 in Corollary 2.3, we have:
1
zf 0 (z)
µ
p
Re
β+p−1+
(z f (z)) > η (z ∈ U ),
β−1
f (z)
which implies that
2µη(β − 1) + 1
Re {z p f (z)}µ >
(z ∈ U ).
2µ(β − 1) + 1
(ii) For α = 0, β 6= 1, µ = 1 and λ ∈ C\{0} with Re{λ} ≥ 0 in Corollary 2.3, we have
λ p+1 0
λp
p
Re (1 +
)z f (z) +
z f (z) > η,
β−1
β−1
which implies that
2η(β − 1) + Re{λ}
Re {z p f (z)} >
(z ∈ U ).
2(β − 1) + Re{λ}
0
(iii) Replacing f by − zfp in the result (ii), we have:
h
i z p+1 f 0 (z)
p+1 00
λ
λ
+ p(β−1) z f (z) > η (0 ≤ η < 1; z ∈ U ),
− Re 1 + β−1 (p + 1)
p
which implies that
p+1 0 z f (z)
2η(β − 1) + Re{λ}
− Re
>
(z ∈ U ).
p
2(β − 1) + Re{λ}
P
Theorem 2.4. Let λ ∈ C with Re{λ} > 0. If f ∈ p satisfies the following condition:
(2.11)
p α
µ−1
Re{(1 − λ)(z p Qαβ,p f (z))µ + λz p Qα−1
}>η
β,p f (z)(z Qβ,p f (z))
(z ∈ U ),
then
(2.12)
Re{(z p Qαβ,p f (z))µ } > η + (1 − η)(2ρ − 1),
where
(2.13)
1
µ(α + β − 1)
1
ρ = 2 F1 1, 1;
+ 1;
.
2
Re{λ}
2
Proof. Let
(2.14)
q(z) = (z p Qαβ,p f (z))µ .
Then q is analytic with q(0) = 1. Differentiating (2.14) with respect to z and using the identity
(1.7), we have
p α
µ−1
(1 − λ)(z p Qαβ,p f (z))µ + λz p Qα−1
β,p f (z)(z Qβ,p f (z))
= q(z) +
so that by the hypothesis (2.11), we have
λ
0
Re q(z) +
zq (z) > η
µ(α + β − 1)
In view of Lemma 1.2, this implies that
λ
zq 0 (z),
µ(α + β − 1)
(z ∈ U ).
Re{q(z)} > η + (1 − η)(2ρ − 1),
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 45, 8 pp.
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6
M.K. AOUF AND A.O. M OSTAFA
where
Z
1
ρ = ρ(Re{λ}) =
Re{λ}
1 + t µ(α+β−1)
−1
dt.
0
Putting Re{λ} = λ1 > 0, we have
Z 1
Z 1
−1
λ1
µ(α+β−1)
µ(α
+
β
−
1)
−1
ρ=
1 + t µ(α+β−1))
dt =
(1 + u)−1 u λ1
du
λ1
0
0
Using (1.11), (1.12), (1.13) and (1.14), we obtain
1
µ(α + β − 1)
1
ρ=
+ 1;
.
2 F1 1, 1;
2
λ1
2
This completes the proof of Theorem 2.1.
P
Corollary 2.5. Let λ ∈ R with λ ≥ 1. If f ∈ p satisfies
(2.15)
Re (1 − λ)z p Qαβ,p f (z) + λz p Qα−1
(z ∈ U ),
β,p f (z) > η
then
Re{z
p
1
(z ∈ U ),
> η + (1 − η)(2ρ1 − 1) 1 −
λ
1
(α + β − 1)
1
ρ1 =
+ 1;
.
2 F1 1, 1;
2
λ
2
Qα−1
β,p f (z)}
where
Proof. The result follows by using the identity
(2.16)
p α
p α−1
p α
λz p Qα−1
β,p f (z) = (1 − λ)z Qβ,p f (z) + λz Qβ,p f (z) + (λ − 1)z Qβ,p f (z).
Remark 2. We note that, for α = 0, β = 2 and λ = µ > 0 in Corollary 2.3, that is, if
(2.17)
Re (1 − λ)(z p f (z))λ + λ(z p+1 f (z))0 (z p f (z))λ−1 > η (z ∈ U ),
then (2.8) implies that
2η + 1
(z ∈ U ),
3
P
whereas, if f ∈ p satisfies the condition (2.17) then by using Theorem 2.4, we have
Re{(z p f (z))λ } >
(2.18)
Re{(z p f (z))λ } > 2(1 − ln 2)η + (2 ln 2 − 1) (z ∈ U ),
which is better than (2.18).
P
Theorem 2.6. Suppose that the functions f and g are in p and g satisfies the condition (1.9).
If
( α−1
)
Qβ,p f (z) Qαβ,p f (z)
(1 − η)δ
(2.19)
Re
− α
>−
(z ∈ U ),
α−1
Qβ,p g(z)
2(α + β − 1)
Qβ,p g(z)
for some η (0 ≤ η < 1), then
(
Re
(2.20)
Qαβ,p f (z)
Qαβ,p g(z)
)
>η
(z ∈ U )
and
(
(2.21)
Re
Qα−1
β,p f (z)
Qα−1
β,p g(z)
)
>
η[2(α + β − 1) + δ] − δ
2(α + β − 1)
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 45, 8 pp.
(z ∈ U ).
http://jipam.vu.edu.au/
p−VALENT M EROMORPHIC F UNCTIONS
7
Proof. Let
"
#
Qαβ,p f (z)
1
q(z) =
−η .
1 − η Qαβ,p g(z)
(2.22)
Then q is analytic in U with q(0) = 1. Setting
φ(z) =
(2.23)
Qαβ,p g(z)
Qα−1
β,p g(z)
(z ∈ U ),
we observe that from (1.9), we have Re{φ(z)} > δ (0 ≤ δ < 1) in U . A simple computation
shows that
Qα−1
Qαβ,p f (z)
(1 − η)zq 0 (z)
β,p f (z)
φ(z) = α−1
− α
α+β−1
Qβ,p g(z)
Qβ,p g(z)
= ψ(q(z), zq 0 (z)),
where
ψ(r, s) =
(1 − η)sφ(z)
.
α+β−1
Using the hypothesis (2.19), we obtain
0
{ψ(q(z), zq (z)); z ∈ U } ⊂ Ω =
Now, for all real r2 , s1 ≤ −
1+r22
,
2
(1 − η)δ
w ∈ C : Re w > −
2(α + β − 1)
.
we have
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)
Re {ψ(ir2 , s1 )} =
This shows that ψ(ir2 , s1 ) ∈
/ Ω for each z ∈ U. Hence by Lemma 1.1, 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:
( α−1
)
( α−1
)
(
)
Qβ,p f (z)
Qβ,p f (z) Qαβ,p f (z)
Qαβ,p f (z)
Re
= Re
− α
− Re
.
Qβ,p g(z)
Qαβ,p g(z)
Qα−1
Qα−1
β,p g(z)
β,p g(z)
This completes the proof of Theorem 2.6.
Remark 3.
(i) For α = 0 and g(z) = z −p in Theorem 2.6, we have
−(1 − η)δ
Re z p+1 f 0 (z) + pz p f (z) >
(z ∈ U ),
2
which implies that
Re {z p f (z)} > η (z ∈ U )
and
η[2(β − 1) + δ] − δ
Re z p+1 f 0 (z) + (p + β − 1)z p f (z) >
(z ∈ U ).
2
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 45, 8 pp.
http://jipam.vu.edu.au/
8
M.K. AOUF AND A.O. M OSTAFA
(ii) Putting α = 0, β = 2 in Theorem 2.6, we get that, if
0
zf (z) + (p + 1)f (z) f (z)
−(1 − η)δ
Re
−
>
(z ∈ U ),
0
zg (z) + (p + 1)g(z)
g(z)
2
then
f (z)
Re
> η (z ∈ U )
g(z)
and
0
η(2 + δ) − δ
zf (z) + (p + 1)f (z)
>
(z ∈ U ).
Re
0
zg (z) + (p + 1)g(z)
2
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J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 45, 8 pp.
http://jipam.vu.edu.au/