General Mathematics Vol. 14, No. 3 (2006), 43–52
Angular estimates of analytic functions
defined by Carlson - Shaffer linear operator
B. A. Frasin
Abstract
The object of the present paper is to derive some argument properties of analytic functions defined by the Carlson - Shaffer linear
operator L(a, c)f (z) . Our results contain some interesting corollaries as the special cases.
2000 Mathematics Subject Classification: 30C45.
Keywords: Analytic functions, Argument, Hadamard product, linear
operator.
1
Received 12 July, 2006
Accepted for publication (in revised form) 15 September, 2006
43
1
44
B.A. Frasin
1
Introduction and definitions
Let A denote the class of functions of the form :
(1.1)
f (z) = z +
∞
X
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
(1.2)
f (z) = z +
∞
X
n
an z and g(z) = z +
∞
X
bn z n
n=2
n=2
their Hadamard product (or convolution) is defined by
(1.3)
(f ∗ g)(z) := z +
∞
X
an b n z n .
n=2
Define the function φ(a, c; z) by
(1.4)
φ(a, c; z)
:
=
∞
X
(a)n
n=0
(c)n
z n+1
−
(a ∈ R; c ∈ R\Z−
0 ; Z0 := {0, −1, −2, . . .}, z ∈ U),
where (λ)n is the Pochhammer symbol given, in terms of Gamma functions,

n = 0,
Γ(λ + n)  1,
(λ)n :=
=
 λ(λ + 1)(λ + 2) . . . (λ + n − 1), n ∈ N : {1, 2, . . .}.
Γ(λ)
Corresponding to the function φ(a, c; z), Carlson and Shaffer[1] intro-
duced a linear operator L(a, c) : A → A by
(1.5)
L(a, c)f (z) := φ(a, c; z) ∗ f (z),
45
Angular estimates of analytic functions
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
z(L(a, c)f (z))′ = aL(a + 1, c)f (z) − (a − 1)L(a, c)f (z),
(1.7)
and L(1, 1)f (z) = f (z), L(2, 1)f (z) = zf ′ (z) , L(3, 1)f (z) = zf ′ (z) +
1 2 ′′
z f (z).
2
Many properties of analytic functions defined by the Carlson-Shaffer
linear operator were studied by (among others) Owa and Srivastava [7],
Ding [3], Kim and Lee [4], Ravichandran et al.[6] and Shanmugam et al.
[5].
In this paper we shall derive some argument 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:
Lemma 1.1.( [2]). Let p(z) be analytic in U with p(0) = 1 and p(z) 6= 0
(z ∈ U) and suppose that
(1.8)
π
|arg(p(z) + βzp (z))| <
2
′
¶
µ
2
−1
α + tan αβ
π
then we have
(1.9)
|arg p(z)| <
π
α
2
(z ∈ U).
(α > 0, β > 0),
46
B.A. Frasin
2
Main Results
Theorem 2.1.
Let a + 1 > µ > 0, α > 0 , λ is any real number ,
L(a, c)f (z)/L(a + 1, c)f (z) 6= 0 (z ∈ U) and suppose that
¯ µ
·
¸¶
¯
L(a + 1, c)f (z)
L(a + 2, c)f (z)
(a + 1)L(a, c)f (z)
¯arg
λ
−µ
+1
¯
(a + 1 − µ)L(a + 1, c)f (z)
L(a, c)f (z)
L(a + 1, c)f (z)
(2.1)
µ
−
then we have
(2.2)
¶¯
µ
¶
2
µ
(a + 1)λ − aµ ¯¯ π
−1
¯ < 2 α + π tan a + 1 − µ α
a+1−µ
¯
µ
¶¯
¯
¯ π
L(a, c)f (z)
¯arg
¯< α
¯
L(a + 1, c)f (z) ¯ 2
(z ∈ U).
Proof. Define the function p(z) by
(2.3)
p(z) :=
L(a, c)f (z)
.
L(a + 1, c)f (z)
Then p(z) = 1 + b1 z + b2 z + · · · is analytic in U with p(0) = 1 and p(z) 6= 0
(z ∈ U) . Also, by a simple computation, we find from (2.3) that
(2.4)
zp′ (z)
=
p(z)
µ
z(L(a, c)f (z))′ z(L(a + 1, c)f (z))′
−
L(a, c)f (z)
L(a + 1, c)f (z)
¶
by making use of the familiar identity (1.7) in (2.4) , we get
·
¸
L(a + 1, c)f (z)
L(a + 2, c)f (z)
L(a, c)f (z)
λ
−µ
+1
L(a, c)f (z)
L(a + 1, c)f (z)
L(a + 1, c)f (z)
µ
¶
¸
·
′
a
µ
zp (z)
λ
1+
+ 1 p(z)
−
−
=
p(z) a + 1
p(z)
p(z)
1
=
[(a + 1)λ − aµ + (a + 1 − µ)p(z) + µzp′ (z)]
a+1
47
Angular estimates of analytic functions
or, equivalently,
·
¸
L(a + 1, c)f (z)
L(a + 2, c)f (z)
(a + 1)L(a, c)f (z)
λ
−µ
+1
(a + 1 − µ)L(a + 1, c)f (z)
L(a, c)f (z)
L(a + 1, c)f (z)
(2.5)
−
µ
(a + 1)λ − aµ
a+1−µ
¶
= p(z) +
µ
zp′ (z).
a+1−µ
The result of Theorem 2.1 now follows by an application of Lemma1.1.
Letting a = c = 1 in Theorem 2.1, we have
Corollary 2.2.
Let 2 > µ > 0, α > 0 , λ is any real number ,
f (z)/zf ′ (z) 6= 0 (z ∈ U) and suppose that
¯
h ′
³
¯
zf (z)
2f (z)
¯arg (2−µ)zf ′ (z) λ f (z) −
π
<
2
(2.6)
µ
µzf ′′ (z)
2f ′ (z)
µ
2
α
α + tan−1
π
2−µ
´´¯
i ³
¯
+ 1 − µ − 2λ−µ
¯
2−µ
¶
then we have
(2.7)
Theorem 2.3.
¯
¶¯
µ
¯ π
¯
f
(z)
¯< α
¯arg
¯
zf ′ (z) ¯ 2
(z ∈ U).
Let a 6= −1, λ 6= −µ, α, λ > 0, δ(a + 1)(λ + µ) > 0,
L(a + 1, c)f (z)/z 6= 0 (z ∈ U) and suppose that
¯
µ
¶¶¯
µ
¶ µ
¯
¯
1
L(a + 1, c)f (z)
L(a + 2, c)f (z)
¯arg
¯
+
µ
δ
λ
¯
¯
λ+µ
z
L(a + 1, c)f (z)
¶
µ
π
λ
2
(2.8) <
α
α + tan−1
2
π
δ(a + 1)(λ + µ)
then we have
(2.9)
¯
¶ ¯
µ
¯ π
¯
L(a
+
1,
c)f
(z)
¯< α
¯arg
δ
¯ 2
¯
z
(z ∈ U).
48
B.A. Frasin
Proof. Define the function p(z) by
(2.10)
p(z) :=
µ
L(a + 1, c)f (z)
z
¶
δ.
Then p(z) = 1 + b1 z + b2 z + · · · is analytic in U with p(0) = 1 and p(z) 6= 0
(z ∈ U) . Also, by a simple computation and by making use of the familiar
identity (1.7) we find from (2.10) that
L(a + 2, c)f (z)
1
zp′ (z)
=
+1
L(a + 1, c)f (z)
δ(a + 1) p(z)
(2.11)
by using (2.10) and (2.11), we get
1
λ+µ
(2.12)
µ
L(a + 1, c)f (z)
z
= p(z) +
¶
¶ µ
L(a + 2, c)f (z)
+µ
δ λ
L(a + 1, c)f (z)
λ
zp′ (z)
δ(a + 1)(λ + µ)
Using Lemma 1.1, we obtain the required result.
Letting a = c = 1 in Theorem 2.3, we have
Corollary 2.4. Let λ 6= −µ, α, λ > 0, 2δ(λ + µ) > 0, f ′ (z) 6= 0 (z ∈ U)
and suppose that
(2.13)
¯
µ
µ
¶¶¯
′′
¯
¯
λzf
(z)
′
¯arg (f (z)) δ 1 +
¯
¯
¯
2(λ + µ)f ′ (z)
¶
µ
π
λ
2
<
α
α + tan−1
2
π
2δ(λ + µ)
then we have
(2.14)
|arg (f ′ (z)) δ| <
π
α
2
(z ∈ U).
49
Angular estimates of analytic functions
Theorem 2.5. Let a 6= −1, α, λ, η, γ > 0, L(a + 1, c)f (z)/L(a, c)f (z) 6=
0 (z ∈ U) and suppose that
¯³
´ h ³
´
i¯
¯
¯
L(a+2,c)f (z)
L(a+1,c)f (z)
(z)
γη
γ
(a
+
1)
−
a
−
1
+
1
arg ¯ L(a+1,c)f
¯
L(a,c)f (z)
λ
L(a+1,c)f (z)
L(a,c)f (z)
π
<
2
(2.15)
µ
λ
2
α + tan−1 α
π
η
¶
then we have
¯
¶ ¯
µ
¯ π
¯
L(a
+
1,
c)f
(z)
¯< α
¯arg
γ
¯ 2
¯
L(a, c)f (z)
(2.16)
(z ∈ U).
Proof. Define the function p(z) by
(2.17)
p(z) :=
µ
L(a + 1, c)f (z)
L(a, c)f (z)
¶
γ.
Then p(z) = 1 + b1 z + b2 z + · · · is analytic in U with p(0) = 1 and p(z) 6= 0
(z ∈ U) . Also, by a simple computation and by making use of the familiar
identity (1.7), we find from (2.17) that
´ h ³
´
i
³
L(a+2,c)f (z)
L(a+1,c)f (z)
L(a+1,c)f (z)
γλ
γ η (a + 1) L(a+1,c)f (z) − a L(a,c)f (z) − 1 + 1
L(a,c)f (z)
λ
= p(z) + zp′ (z)
η
(2.18)
An application of Lemma 1.1, we obtain the required result.
Letting a = c = 1 in Theorem 2.5, we have
Corollary 2.6. Let α, λ, η, γ > 0, zf ′ (z)/ f (z) 6= 0 (z ∈ U) and
suppose that
(2.19)
¯
½µ ′ ¶ · µ
¶
¸¾¯
′′
′
¯
¯
zf
(z)
zf
(z)
γη
zf
(z)
¯arg
γ
1+ ′
+ 1 ¯¯
−
¯
f (z)
λ
f (z)
f (z)
µ
¶
π
2
λ
<
α + tan−1 α
2
π
η
50
B.A. Frasin
then we have
(2.20)
¯
µ ′ ¶ ¯
¯
¯
¯arg zf (z) γ ¯ < π α
¯
¯ 2
f (z)
(z ∈ U).
Letting λ = η = γ = 1 in Corollary 2.6, we have
Corollary 2.7. Let 0 < α ≤ 1, zf ′ (z)/ f (z) 6= 0 (z ∈ U) and suppose
that
¯
½µ ′ ¶ µ
¶
¶¾¯
µ
¯
¯ π
zf (z)
zf ′′ (z) zf ′ (z)
2
−1
¯
¯
(2.21) ¯arg
−
2+ ′
¯ < 2 α + π tan α
f (z)
f (z)
f (z)
then we have
(2.22)
¯
µ ′ ¶¯
¯
¯
¯arg zf (z) ¯ < π α
¯
f (z) ¯ 2
(z ∈ U),
that is, f (z) is strongly starlike function of order α in U.
Theorem 2.8. Let a 6= −1, α, λ, γ > 0, (γ − λ)(a + 1) > 0, z/L(a +
1, c)f (z) 6= 0 (z ∈ U) and suppose that
¯
µ
¶¸¯
·
¯
¯
L(a
+
2,
c)f
(z)
z
1
¯
¯arg
−
λz
γ
¯
γ−λ
L(a + 1, c)f (z)
[L(a + 1, c)f (z)]2 ¯
¶
µ
λ
2
π
−1
(2.23) <
α .
α + tan
2
π
(γ − λ)(a + 1)
then we have
(2.24)
¯
¶¯
µ
¯ π
¯
z
¯< α
¯arg
¯
L(a + 1, c)f (z) ¯ 2
Proof. Define the function p(z) by
(2.25)
p(z) :=
z
L(a + 1, c)f (z)
(z ∈ U).
51
Angular estimates of analytic functions
Then p(z) = 1 + b1 z + b2 z + · · · is analytic in U with p(0) = 1 and p(z) 6= 0
(z ∈ U). Also, by a simple computation and by making use of the familiar
identity (1.7), we find from (2.25) that
(2.26)
1
γ−λ
µ
L(a + 2, c)f (z)
z
− λz
γ
L(a + 1, c)f (z)
[L(a + 1, c)f (z)]2
¶
=
λ
zp′ (z).
(γ − λ)(a + 1)
The result of Theorem 2.8 now follows by an application of Lemma1.1.
= p(z) +
Letting a = c = 1 in Theorem 2.8, we have
Corollary 2.9. Let α, λ, γ > 0, (γ −λ)(a+1) > 0, 1/f ′ (z) 6= 0 (z ∈ U)
and suppose that
¯
¶
µ
¶¯
µ
′′
¯ π
¯
λ
λzf
(z)
2
1
−1
¯<
−
α
α + tan
(2.27) ¯¯arg
f ′ (z) 2(γ − λ)(f ′ (z))2 ¯ 2
π
2(γ − λ)
then we have
(2.28)
¯
µ
¶¯
¯
¯ π
1
¯arg
¯< α
¯
f ′ (z) ¯ 2
(z ∈ U).
References
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[3] S. Ding, Some properties of a class of analytic functions, J. Math. Ana.
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52
B.A. Frasin
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Department of Mathematics,
Al al-Bayt University,
P.O. Box: 130095
Mafraq, Jordan.
E-mail address: bafrasin@yahoo.com.