Volume 8 (2007), Issue 2, Article 53, 10 pp.
SOME PROPERTIES OF ANALYTIC FUNCTIONS DEFINED BY A LINEAR
OPERATOR
B.A. FRASIN
D EPARTMENT OF M ATHEMATICS
A L AL -BAYT U NIVERSITY
P.O. B OX : 130095 M AFRAQ , J ORDAN
bafrasin@yahoo.com
Received 29 August, 2006; accepted 03 April, 2007
Communicated by A. Sofo
A BSTRACT. The object of the present paper is to derive some properties of analytic functions
defined by the Carlson - Shaffer linear operator L(a, c)f (z) .
Key words and phrases: Analytic functions, Hadamard product, Linear operator.
2000 Mathematics Subject Classification. 30C45.
1. I NTRODUCTION AND D EFINITIONS
Let A denote the class of functions of the form:
∞
X
(1.1)
f (z) = z +
an z n ,
n=2
which are analytic in the open unit disc U = {z : |z| < 1} . For two functions f (z) and g(z)
given by
∞
∞
X
X
n
(1.2)
f (z) = z +
an z
and g(z) = z +
bn z n ,
n=2
n=2
their Hadamard product (or convolution) is defined by
∞
X
(1.3)
(f ∗ g)(z) := z +
an b n z n .
n=2
Define the function φ(a, c; z) by
φ(a, c; z) :=
(1.4)
∞
X
(a)n
n=0
(a ∈ R; c ∈
R\Z−
0;
Z−
0
(c)n
z n+1 ,
:= {0, −1, −2, . . .}, z ∈ U),
The author would like to thank the referee for his valuable suggestions.
228-06
2
B.A. F RASIN
where (λ)n is the Pochhammer symbol given in terms of Gamma functions,
Γ(λ + n)
Γ(λ)
1,
n = 0,
=
λ(λ + 1)(λ + 2) . . . (λ + n − 1), n ∈ N : {1, 2, . . .}.
(λ)n :=
Corresponding to the function φ(a, c; z), Carlson and Shaffer [1] introduced a linear operator
L(a, c) : A → A by
L(a, c)f (z) := φ(a, c; z) ∗ f (z),
(1.5)
or, equivalently, by
(1.6)
L(a, c)f (z) := z +
∞
X
(a)n
n=1
(c)n
an+1 z n+1
(z ∈ U).
It follows from (1.6) that
(1.7)
z(L(a, c)f (z))0 = aL(a + 1, c)f (z) − (a − 1)L(a, c)f (z),
and
L(2, 1)f (z) = zf 0 (z),
1
L(3, 1)f (z) = zf 0 (z) + z 2 f 00 (z).
2
Many properties of analytic functions defined by the Carlson-Shaffer linear operator were
studied by (among others) Owa and Srivastava [10], Ding [5], Kim and Lee [6], Ravichandran
et al. ([8, 9]), Shanmugam et al. [7] and Frasin ([2, 3]).
In this paper we shall derive some properties of analytic functions defined by the linear operator L(a, c)f (z).
In order to prove our main results, we recall the following lemma:
L(1, 1)f (z) = f (z),
Lemma 1.1 ([4]). Let Φ(u, v) be a complex valued function,
Φ : D → C,
(D ⊂ C2 ; C is the complex plane),
and let u = u1 + iu2 and v = v1 + iv2 . Suppose that the function Φ(u, v) satisfies:
(i) Φ(u, v) is continuous in D;
(ii) (1, 0) ∈ D and Re(Φ(1, 0)) > 0;
(iii) Re(Φ(iu2 , v1 )) ≤ 0 for all (iu2 , v1 ) ∈ D and such that v1 ≤ −(1 + u22 )/2.
Let p(z) = 1 + p1 z + p2 z 2 + · · · be regular in U such that (p(z), zp0 (z)) ∈ D for all z ∈ U.
If Re(Φ(p(z), zp0 (z))) > 0 (z ∈ U), then Re(p(z)) > 0 (z ∈ U).
2. M AIN R ESULTS
Theorem 2.1. Let α ∈ C, β ∈ C, (Re(β) ≥ 0), (α − β) ∈ R, and suppose that
L(a, c)f (z)
(a + 1)L(a + 2, c)f (z)
(2.1)
Re
α−β
+ aα > γ
L(a + 1, c)f (z)
L(a + 1, c)f (z)
for some γ(γ < (α − β)(a + 1)) and 2(α − β) + Re(β) 6= 0, then
L(a, c)f (z)
2γ − 2a(α − β) + Re(β)
(2.2)
Re
>
L(a + 1, c)f (z)
2(α − β) + Re(β)
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 53, 10 pp.
(z ∈ U).
http://jipam.vu.edu.au/
S OME P ROPERTIES OF A NALYTIC F UNCTIONS
3
Proof. Define the function p(z) by
L(a, c)f (z)
:= δ + (1 − δ)p(z),
L(a + 1, c)f (z)
(2.3)
where
δ=
(2.4)
2γ − 2a(α − β) + Re(β)
< 1.
2(α − β) + Re(β)
Then p(z) = 1 + b1 z + b2 z + · · · is regular in U. It follows from (2.3) that
(1 − δ)zp0 (z)
z(L(a, c)f (z))0 z(L(a + 1, c)f (z))0
−
=
L(a, c)f (z)
L(a + 1, c)f (z)
δ + (1 − δ)p(z)
(2.5)
by making use of the familiar identity (1.7) in (2.5), we get
(2.6)
(a + 1)L(a + 2, c)f (z)
a
(1 − δ)zp0 (z)
=1+
−
L(a + 1, c)f (z)
δ + (1 − δ)p(z) δ + (1 − δ)p(z)
or, equivalently,
(a + 1)L(a + 2, c)f (z)
L(a, c)f (z)
(2.7)
α−β
+ aα
L(a + 1, c)f (z)
L(a + 1, c)f (z)
= (δ + a)(α − β) + (1 − δ)(α − β)p(z) + β(1 − δ)zp0 (z).
Therefore, we have
L(a, c)f (z)
(a + 1)L(a + 2, c)f (z)
(2.8) Re
α−β
+ aα − γ
L(a + 1, c)f (z)
L(a + 1, c)f (z)
= Re {(δ + a)(α − β) − γ + (1 − δ)(α − β)p(z) + β(1 − δ)zp0 (z)} > 0.
If we define the function Φ(u, v) by
(2.9)
Φ(u, v) = (δ + a)(α − β) − γ + (1 − δ)(α − β)u + β(1 − δ)v
with u = u1 + iu2 and v = v1 + iv2 , then
(i) Φ(u, v) is continuous in D = C2 ;
(ii) (1, 0) ∈ D and Re(Φ(1, 0)) = (α − β)(a + 1) − γ > 0;
(iii) For all (iu2 , v1 ) ∈ D and such that v1 ≤ −(1 + u22 )/2,
Re(Φ(iu2 , v1 )) = Re{(δ + a)(α − β) − γ + β(1 − δ)v1 }
≤ (δ + a)(α − β) − γ −
(1 − δ)(1 + u22 ) Re(β)
2
(1 − δ)u22 Re(β)
≤ 0.
2
Therefore, the function Φ(u, v) satisfies the conditions in Lemma 1.1. Thus we have Re(p(z)) >
0 (z ∈ U), that is
L(a, c)f (z)
2γ − 2a(α − β) + Re(β)
Re
>
.
L(a + 1, c)f (z)
2(α − β) + Re(β)
=−
−
Letting β = −α in Theorem 2.1, we have
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 53, 10 pp.
http://jipam.vu.edu.au/
4
B.A. F RASIN
Corollary 2.2. Let α ∈ C, (Re(α) < 0), and suppose that
L(a, c)f (z)
− (a + 1)L(a + 2, c)f (z)
(2.10)
Re
α+α
+ aα > γ
L(a + 1, c)f (z)
L(a + 1, c)f (z)
for some γ (γ < 2(a + 1) Re(α)), then
2γ − (4a + 1) Re(α)
L(a, c)f (z)
>
(2.11)
Re
L(a + 1, c)f (z)
3 Re(α)
Letting a = c = 1 in Theorem 2.1, we have
(z ∈ U).
Corollary 2.3. Let α, β ∈ C, (Re(β) ≥ 0), (α − β) ∈ R, and suppose that
zf 00 (z)
f (z)
(2.12)
Re
α − 2β 1 +
+α >γ
zf 0 (z)
2f 0 (z)
for some γ(γ < 2(α − β)) and 2(α − β) + Re(β) 6= 0, then
f (z)
2γ − 2(α − β) + Re(β)
(2.13)
Re
>
zf 0 (z)
2(α − β) + Re(β)
(z ∈ U).
Theorem 2.4. Let α ∈ C, β ∈ C, (Re(β) ≥ 0), (α − β) ∈ R, and suppose that
L(a, c)f (z)
(a + 1)L(a + 2, c)f (z)
(2.14)
Re
α−β
+ aα < γ
L(a + 1, c)f (z)
L(a + 1, c)f (z)
for some γ(γ > (α − β)(a + 1)) and 2(α − β) + Re(β) 6= 0, then
L(a, c)f (z)
2γ − 2a(α − β) + Re(β)
(2.15)
Re
<
L(a + 1, c)f (z)
2(α − β) + Re(β)
(z ∈ U).
Proof. Define the function p(z) by
L(a, c)f (z)
(2.16)
:= δ + (1 − δ)p(z),
L(a + 1, c)f (z)
where
2γ − 2a(α − β) + Re(β)
(2.17)
δ=
> 1.
2(α − β) + Re(β)
Then p(z) = 1 + b1 z + b2 z + · · · is regular in U. It follows from (2.16) that
L(a, c)f (z)
(a + 1)L(a + 2, c)f (z)
(2.18) Re γ −
α−β
− aα
L(a + 1, c)f (z)
L(a + 1, c)f (z)
= Re {γ − (δ + a)(α − β) − (1 − δ)(α − β)p(z) − β(1 − δ)zp0 (z)} > 0.
If we define the function Φ(u, v) by
(2.19)
Φ(u, v) = γ − (δ + a)(α − β) − (1 − δ)(α − β)u − β(1 − δ)v
with u = u1 + iu2 and v = v1 + iv2 , then
(i) Φ(u, v) is continuous in D = C2 ;
(ii) (1, 0) ∈ D and Re(Φ(1, 0)) = γ − (α − β)(a + 1) > 0;
(iii) For all (iu2 , v1 ) ∈ D and such that v1 ≤ −(1 + u22 )/2,
Re(Φ(iu2 , v1 )) = Re{γ − (δ + a)(α − β) − β(1 − δ)v1 }
≤ γ − (δ + a)(α − β) +
=
(1 − δ)(1 + u22 ) Re(β)
2
(1 − δ)u22 Re(β)
≤ 0,
2
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 53, 10 pp.
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S OME P ROPERTIES OF A NALYTIC F UNCTIONS
5
applying Lemma 1.1, we have Re(p(z)) > 0 (z ∈ U), that is
L(a, c)f (z)
2γ − 2a(α − β) + Re(β)
Re
<
.
L(a + 1, c)f (z)
2(α − β) + Re(β)
Theorem 2.5. Let a > −1, µ > 0, α ∈ C, β ∈ C, (Re(β) ≥ 0), (α + β) ∈ R, and suppose
that
(
µ )
L(a + 1, c)f (z)
L(a + 2, c)f (z)
>γ
(2.20)
Re
α+β
z
L(a + 1, c)f (z)
for some γ(γ < α + β) and 2µ(α + β)(a + 1) + Re(β) 6= 0, then
(
µ )
2µγ(a + 1) + Re(β)
L(a + 1, c)f (z)
>
(2.21)
Re
z
2µ(α + β)(a + 1) + Re(β)
(z ∈ U).
Proof. Define the function p(z) by
µ
L(a + 1, c)f (z)
(2.22)
:= δ + (1 − δ)p(z),
z
where
2µγ(a + 1) + Re(β)
> 1.
2µ(α + β)(a + 1) + Re(β)
Then p(z) = 1 + b1 z + b2 z + · · · is regular in U. Also, by a simple computation and by making
use of the familiar identity (1.7) we find from (2.22) that
δ=
(2.23)
βL(a + 2, c)f (z)
β(1 − δ)zp0 (z)
=
+ β,
L(a + 1, c)f (z)
µ(a + 1)(δ + (1 − δ)p(z))
and by using (2.22) and (2.23), we get
µ L(a + 1, c)f (z)
βL(a + 2, c)f (z)
(2.24)
+α
z
L(a + 1, c)f (z)
β(1 − δ)zp0 (z)
=
+ (α + β)(δ + (1 − δ)p(z)).
µ(a + 1)
Therefore, we have
(
)
µ L(a + 1, c)f (z)
βL(a + 2, c)f (z)
(2.25) Re
+α −γ
z
L(a + 1, c)f (z)
β(1 − δ) 0
= Re (α + β)δ − γ + (1 − δ)(α + β)p(z) +
zp (z) > 0.
µ(a + 1)
If we define the function Φ(u, v) by
(2.26)
Φ(u, v) = (α + β)δ − γ + (1 − δ)(α + β)u +
β(1 − δ)
v
µ(a + 1)
with u = u1 + iu2 and v = v1 + iv2 , then
(i) Φ(u, v) is continuous in D = C2 ;
(ii) (1, 0) ∈ D and Re(Φ(1, 0)) = (α + β) − γ > 0;
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 53, 10 pp.
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6
B.A. F RASIN
(iii) For all (iu2 , v1 ) ∈ D and such that v1 ≤ −(1 + u22 )/2,
β(1 − δ)
v1 }
µ(a + 1)
(1 − δ)(1 + u22 ) Re(β)
≤ (α + β)δ − γ −
2µ(a + 1)
2
(1 − δ)u2 Re(β)
=−
≤ 0.
2µ(a + 1)
Re(Φ(iu2 , v1 )) = Re{(α + β)δ − γ +
Therefore, the function Φ(u, v) satisfies the conditions in Lemma 1.1. Thus we have Re(p(z)) >
0 (z ∈ U), that is,
(
µ )
2µγ(a + 1) + Re(β)
L(a + 1, c)f (z)
>
(z ∈ U).
Re
z
2µ(α + β)(a + 1) + Re(β)
−
Letting α = β in Theorem 2.5, we have
Corollary 2.6. Let a > −1, µ > 0, β ∈ C, (Re(β) > 0), and suppose that
µ L(a + 1, c)f (z)
L(a + 2, c)f (z)
(2.27)
Re
α+β
>γ
z
L(a + 1, c)f (z)
for some γ(γ < 2 Re(β)), then
(
µ )
L(a + 1, c)f (z)
2µγ(a + 1) + Re(β)
(2.28)
Re
>
z
(4µ(a + 1) + 1) Re(β)
(z ∈ U).
Letting a = c = 1 in Theorem 2.5, we have,
Corollary 2.7. Let µ > 0, α ∈ C, β ∈ C, (Re(β) ≥ 0), (α + β) ∈ R, and suppose that
zf 00 (z)
µ
0
(2.29)
Re (f (z)) α + β 1 +
>γ
2f 0 (z)
for some γ(γ < α + β) and 4µ(α + β) + Re(β) 6= 0, then
n
o
µ
4µγ + Re(β)
(2.30)
Re (f 0 (z)) >
4µ(α + β) + Re(β)
(z ∈ U).
Employing the same manner as in the proofs of Theorems 2.4 and 2.5, we have:
Theorem 2.8. Let a > −1, µ > 0, α ∈ C, β ∈ C, (Re(β) ≥ 0), (α + β) ∈ R, and suppose
that
(
µ )
L(a + 1, c)f (z)
L(a + 2, c)f (z)
(2.31)
Re
α+β
<γ
z
L(a + 1, c)f (z)
for some γ(γ > α + β) and 2µ(α + β)(a + 1) + Re(β) 6= 0, then
(
µ )
L(a + 1, c)f (z)
2µγ(a + 1) + Re(β)
(2.32)
Re
<
z
2µ(α + β)(a + 1) + Re(β)
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 53, 10 pp.
(z ∈ U).
http://jipam.vu.edu.au/
S OME P ROPERTIES OF A NALYTIC F UNCTIONS
7
Theorem 2.9. Let λ > 0, α ∈ C, β ∈ C, (Re(β) ≥ 0), (α + β) ∈ R, and suppose that
(
λ
L(a + 1, c)f (z)
(2.33) Re
L(a, c)f (z)
L(a + 2, c)f (z)
L(a + 1, c)f (z)
× λβ(a + 1)
− aλβ
+ λα
>γ
L(a + 1, c)f (z)
L(a, c)f (z)
for some γ(γ < λ(α + β)) and 2λ(α + β) + Re(β) 6= 0, then
λ
L(a + 1, c)f (z)
2γ + Re(β)
(2.34)
Re
>
L(a, c)f (z)
2λ(α + β) + Re(β)
(z ∈ U).
Proof. Define the function p(z) by
λ
L(a + 1, c)f (z)
(2.35)
:= δ + (1 − δ)p(z),
L(a, c)f (z)
where
δ=
(2.36)
2γ + Re(β)
< 1.
2λ(α + β) + Re(β)
Then p(z) = 1 + b1 z + b2 z + · · · is regular in U . Also, by a simple computation and by making
use of the familiar identity (1.7), we find from (2.35) that
(2.37)
L(a + 1, c)f (z)
L(a, c)f (z)
λ L(a + 2, c)f (z)
L(a + 1, c)f (z)
λβ(a + 1)
− aλβ
+ λα
L(a + 1, c)f (z)
L(a, c)f (z)
= λ(α + β)(δ + (1 − δ)p(z)) + β(1 − δ)zp0 (z)
Therefore, we have
(
λ
L(a + 1, c)f (z)
(2.38) Re
L(a, c)f (z)
L(a + 2, c)f (z)
L(a + 1, c)f (z)
× λβ(a + 1)
− aλβ
+ λα − γ
L(a + 1, c)f (z)
L(a, c)f (z)
= Re {λ(α + β)(δ + (1 − δ)p(z)) + β(1 − δ)zp0 (z) − γ} > 0.
If we define the function Φ(u, v) by
(2.39)
Φ(u, v) = λδ(α + β) − γ + λ(1 − δ)(α + β)u + β(1 − δ)v
with u = u1 + iu2 and v = v1 + iv2 , then
(i) Φ(u, v) is continuous in D = C2 ;
(ii) (1, 0) ∈ D and Re(Φ(1, 0)) = λ(α + β) − γ > 0;
(iii) For all (iu2 , v1 ) ∈ D and such that v1 ≤ −(1 + u22 )/2,
Re(Φ(iu2 , v1 )) = Re{λδ(α + β) − γ + β(1 − δ)v1 }
(1 − δ)(1 + u22 )
Re(β)
2
(1 − δ)u22 Re(β)
=−
≤ 0.
2
≤ λδ(α + β) − γ −
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 53, 10 pp.
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8
B.A. F RASIN
Therefore, the function Φ(u, v) satisfies the conditions in Lemma 1.1. Thus we have
λ
2γ + Re(β)
L(a + 1, c)f (z)
>
(z ∈ U).
Re
L(a, c)f (z)
2λ(α + β) + Re(β)
−
Letting α = β in Theorem 2.9, we have:
Corollary 2.10. Let λ > 0, α ∈ C, β ∈ C, (Re(β) > 0), (α + β) ∈ R, and suppose that
(
λ
L(a + 1, c)f (z)
(2.40) Re
L(a, c)f (z)
−
L(a + 2, c)f (z)
L(a + 1, c)f (z)
× λβ(a + 1)
− aλβ
+ λβ
>γ
L(a + 1, c)f (z)
L(a, c)f (z)
for some γ(γ < λ(α + β)), then
λ
2γ + Re(β)
L(a + 1, c)f (z)
(2.41)
Re
>
L(a, c)f (z)
(4λ + 1) Re(β)
(z ∈ U).
Letting a = c = λ = 1 in Theorem 2.9, we have:
Corollary 2.11. Let α ∈ C, β ∈ C, (Re(β) ≥ 0), (α + β) ∈ R, and suppose that:
00
0 f (z) f 0 (z)
zf (z)
(2.42)
Re
(2β + α) + βz
−
>γ
f (z)
f 0 (z)
f (z)
for some γ(γ < α + β) and 2(α + β) + Re(β) 6= 0, then f (z) is starlike of order δ, where
2γ+Re(β)
δ = 2(α+β)+Re(β)
.
Employing the same manner as in the proofs of Theorems 2.4 and 2.9, we have:
Theorem 2.12. Let λ > 0, α ∈ C, β ∈ C, (Re(β) ≥ 0), (α + β) ∈ R, and suppose that:
(
λ
L(a + 1, c)f (z)
(2.43) Re
L(a, c)f (z)
L(a + 2, c)f (z)
L(a + 1, c)f (z)
× λβ(a + 1)
− aλβ
+ λα
<γ
L(a + 1, c)f (z)
L(a, c)f (z)
for some γ(γ > λ(α + β)) and 2λ(α + β) + Re(β) 6= 0, then
λ
L(a + 1, c)f (z)
2γ + Re(β)
(2.44)
Re
<
L(a, c)f (z)
2λ(α + β) + Re(β)
(z ∈ U).
Theorem 2.13. Let a > −1, α ∈ C, β ∈ C, (Re(β) ≥ 0), (α − β) ∈ R, and suppose that
z
L(a + 2, c)f (z)
(2.45)
Re
(a + 1)α − β
>γ
L(a + 1, c)f (z)
L(a + 1, c)f (z)
for some γ(γ < (α − β)(a + 1)), then
z
2γ + Re(β)
(2.46)
Re
>
L(a + 1, c)f (z)
2(a + 1)(α − β) + Re(β)
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 53, 10 pp.
(z ∈ U).
http://jipam.vu.edu.au/
S OME P ROPERTIES OF A NALYTIC F UNCTIONS
9
Proof. Define the function p(z) by
z
:= δ + (1 − δ)p(z),
L(a + 1, c)f (z)
(2.47)
where
δ=
(2.48)
2γ + Re(β)
> 1.
2(a + 1)(α − β) + Re(β)
Then p(z) = 1 + b1 z + b2 z + · · · is regular in U. Also, by a simple computation and by making
use of the familiar identity (1.7), we find from (2.47) that
z
L(a + 2, c)f (z)
(2.49)
(a + 1)α − β
L(a + 1, c)f (z)
L(a + 1, c)f (z)
= (a + 1)(α − β)(δ + (1 − δ)p(z)) + β(1 − δ)zp0 (z).
Therefore, we have
z
L(a + 2, c)f (z)
(2.50) Re
(a + 1)α − β
−γ
L(a + 1, c)f (z)
L(a + 1, c)f (z)
= Re{(a + 1)(α − β)(δ + (1 − δ)p(z)) + β(1 − δ)zp0 (z) − γ} > 0.
If we define the function Φ(u, v) by
(2.51)
Φ(u, v) = δ(a + 1)(α − β) − γ + (a + 1)(α − β)(1 − δ)u + β(1 − δ)v
with u = u1 + iu2 and v = v1 + iv2 , then
(i) Φ(u, v) is continuous in D = C2 ;
(ii) (1, 0) ∈ D and Re(Φ(1, 0)) = (α − β)(a + 1) − γ > 0;
(iii) For all (iu2 , v1 ) ∈ D and such that v1 ≤ −(1 + u22 )/2,
Re(Φ(iu2 , v1 )) = Re{δ(a + 1)(α − β) − γ + β(1 − δ)v1 }
≤ δ(a + 1)(α − β) − γ −
(1 − δ)(1 + u22 )
Re(β)
2
(1 − δ)u22 Re(β)
≤ 0.
2
Therefore, the function Φ(u, v) satisfies the conditions in Lemma 1.1. Thus we have
z
2γ + Re(β)
Re
>
(z ∈ U).
L(a + 1, c)f (z)
2(a + 1)(α − β) + Re(β)
=−
−
Letting β = −α in Theorem 2.13, we have:
Corollary 2.14. Let a > −1, a 6= 0, α ∈ C, (Re(α) < 0), and suppose that
z
− L(a + 2, c)f (z)
(2.52)
Re
(a + 1)α + α
>γ
L(a + 1, c)f (z)
L(a + 1, c)f (z)
for some γ(γ < 2(a + 1) Re(α)), then
z
2γ − Re(α)
(2.53)
Re
>
L(a + 1, c)f (z)
(4a + 3) Re(α)
(z ∈ U).
Letting a = c = 1 in Theorem 2.13, we have:
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 53, 10 pp.
http://jipam.vu.edu.au/
10
B.A. F RASIN
Corollary 2.15. Let a > −1, α ∈ C, β ∈ C, (Re(β) ≥ 0), (α − β) ∈ R, and suppose that
1
zf 00 (z)
(2.54)
Re
2α − β −
β
>γ
f 0 (z)
2f 0 (z)
for some γ(γ < 2(α − β)) and 4(α − β) + Re(β) 6= 0, then
2γ + Re(β)
1
>
(2.55)
Re
0
f (z)
4(α − β) + Re(β)
(z ∈ U).
Employing the same manner as in the proofs of Theorem 2.4 and 2.13, we have:
Theorem 2.16. Let a > −1, α ∈ C, β ∈ C, (Re(β) ≥ 0), (α − β) ∈ R, and suppose that
z
L(a + 2, c)f (z)
(2.56)
Re
(a + 1)α − β
<γ
L(a + 1, c)f (z)
L(a + 1, c)f (z)
for some γ(γ > (α − β)(a + 1)) and 2(a + 1)(α − β) + Re(β) 6= 0, then
z
2γ + Re(β)
(2.57)
Re
<
(z ∈ U).
L(a + 1, c)f (z)
2(a + 1)(α − β) + Re(β)
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