ISSN1842-6298 (electronic), 1843-7265 (print) Volume4(2009), 77 – 88
APPLICATIONS OF GENERALIZED
RUSCHEWEYH DERIVATIVE TO UNIVALENT
FUNCTIONS WITH FINITELY MANY
COEFFICIENTS
Abdul Rahman S. Juma and S. R. Kulkarni
Abstract. By making use of the generalized Ruscheweyh derivative, the authors investigate several interesting properties of certain subclasses of univalent functions having the form
f(z) =z− m ❳
n=2
(1−β)en (1−nβ+α(1−n))B
λ,µ
1 (n)
zn− ∞ ❳
k=m+1
akzk.
1
Introduction
LetA(n, p) denote the class of functions f normalized byf(z) =zp− ∞
P
k=n+p
akzk,(p, n∈
IN ={1,2,3,4, ....}),which are analytic and multivalent in the unit disk U ={z :
|z|<1}. The function f(z) is said to be starlike of order δ(0≤δ < p) if and only if Rezff(′z(z)) > δ,(z ∈ U). On the other hand f(z) is said to be convex of order
δ(0≤δ < p) if and only if Re1 +zff′′′((zz))
> δ,(z∈ U).
Now for α≥0,0≤β <1 and λ >−1, µ, ν ∈IR, the following classes
Mλ,µp (α, β) =
(
f(z)∈A(n, p) :Re (
Jpλ,µf(z)
z(Jpλ,µf(z))′
)
> α
Jpλ,µf(z)
z(Jpλ,µf(z))′
−p
+β )
(1.1)
Mλ,µ1,1(α, β) =
(
f(z)∈A(1,1) :Re (
J1λ,µf(z)
z(J1λ,µf(z))′
)
> α
J1λ,µf(z)
z(J1λ,µf(z))′ −1
+β )
(1.2)
2000 Mathematics Subject Classification: 30C45
whereJ1λ,µf(z) is the generalized Ruscheweyh derivative off defined by
J1λ,µf(z) = Γ(µ−λ+ν+ 2) Γ(ν+ 2)Γ(µ+ 1)zJ
λ,µ,ν
0,z (zµ
−1f(z)) (1.3)
whereJ0λ,µ,ν,z f(z) generalized fractional derivative operator of orderλ(see eg.[1], [5], [7], [8]).
We can write (1.3) as
J1λ,µf(z) =z− ∞
X
k=n+1
akB1λ,µ(k)zk (1.4)
where
B1λ,µ(k) = Γ(k+µ)Γ(ν+ 2 +µ−λ)Γ(k+ν+ 1)
Γ(k)Γ(k+ν+ 1 +µ−λ)Γ(ν+ 2)Γ(1 +µ). (1.5)
Consider the class Mλ,µ1,1(α, β, em) is subclass of Mλ,µ1,1(α, β) consisting of func-tions of the form
f(z) =z−
m
X
n=2
(1−β)en
(1−nβ+α(1−n))B1λ,µ(n)z n−
∞
X
k=m+1
akzk, (1.6)
where
en=
[1−nβ+α(1−n)]Bλ,µ1 (n) 1−β an.
Many authors have studied the different cases (e.g. [2], [4], [6]), and for λ =
µ, ν = 1, we get the case was studied by [3].
2
Coefficient Bounds
In order to prove our results, we need the following Lemma due to A.R.S.Juma and S. R. Kulkarni [2].
Lemma 1. Let f(z) =zp− ∞
P
k=m+p
akzk. Then f(z)∈ Mλ,µp (α, β) if and only if
∞
X
k=m+p
[(1 +α)−k(αp+β)] [(1 +α)−p(αp+β)]B
λ,µ
p (k)ak<1. (2.1)
Theorem 2. Let the function f(z) defined by (1.6). Then f ∈ Mλ,µ1,1(α, β, em) if and only if
∞
X
k=m+1
(1−kβ) +α(1−k) 1−β B
λ,µ
1 (k)ak <1− m
X
n=2
en. (2.2)
Proof. Let
an=
(1−β)en
(1−nβ+α(1−n))Bλ,µ1 (n). (2.3)
We can say that f ∈ Mλ,µ1,1(α, β, em)⊂ Mλ,µ1,1(α, β) if and only if
m
X
n=2
[(1−nβ) +α(1−n)]Bλ,µ1 (n) 1−β an+
∞
X
k=m+1
[1−kβ+α(1−k)]B1λ,µ(k) 1−β ak<1
or
∞
X
k=m+1
[1−kβ+α(1−k)]B1λ,µ(k)
1−β ak<1−
m
X
n=2
en.
Corollary 3. Let f(z) defined by (1.6) be in Mλ,µ1,1(α, β, em). Then for k≥m+ 1 we have
ak≤
(1−β)(1− ∞
P
n=2
en)
(1−kβ+α(1−k))B1λ,µ(k), (2.4)
this result is sharp for
h(z) =z−
m
X
n=2
(1−β)en
[(1−nβ) +α(1−n)]B1λ,µ(n)z n−
(1−β)(1− ∞
P
n=2
en)
[(1−kβ) +α(1−k)]B1λ,µ(k)z k.
(2.5)
Corollary 4. Let f defined by (1.6). Then f ∈ M10,,01(0, β, em), that is,
f(z) =z−
m
X
n=2
en(1−β) 1−nβ z
n−
∞
X
k=m+1
akzk,
if and only if
∞
X
k=m+1
1−kβ
1−β ak<1−
m
X
n=2
Corollary 5. Let f defined by (1.6). Then f ∈ M11,,01(0, β, em), that is,
f(z) =z−
m
X
n=2
(1−β)(ν+ 1)en (1−nβ)Γ(n+ν) =z
n−
∞
X
k=m+1
akzk
if and only if
∞
X
k=m+1
(1−kβ)Γ(k+ν)
(1−β)(ν+ 1) ak<1− m
X
n=2
en.
Corollary 6. Let f defined by (1.6). Then f ∈ M10,,01(α,0, em), that is,
f(z) =z−
m
X
n=2
en 1 +α(1−n)z
n−
∞
X
k=m+1
akzk
if and only if
∞
X
k=m+1
(1 +α(1−k))ak<1− m
X
n=2
en.
Corollary 7. Let f defined by (1.6). Then f ∈ M11,,01(α,0, em) , that is,
f(z) =z−
m
X
n=2
en(ν+ 1)
(1 +α(1−n))Γ(n+ν)z n−
∞
X
k=m+1
akzk
if and only if
∞
X
k=m+1
[1 +α(1−k)]Γ(k+ν)
(ν+ 1) ak<1− m
X
n=2
en.
We claim that all these results are entirely new.
3
Extreme points and other results
Theorem 8. Let f1(z), f2(z), ..., fℓ(z) defined by
fi(z) =z− m
X
n=2
(1−β)en
[1−nβ+α(1−n)]B1λ,µ(n)z n−
∞
X
k=m+1
ak,izk, (3.1)
(i= 1,· · · , ℓ) be in the class Mλ,µ1,1(α, β, em). Then G(z) defined by
G(z) = ℓ
X
i=1
λifi and ℓ
X
i=1
λi = 1,0≤ m
X
n=2
en≤1,0≤en≤1
Proof. In view of Theorem2, we have
Proof. It is clear that
by Theorem 2, we have the last expression is less than 1ℓ Pℓ
The next Theorem is very useful to obtain the extreme points of the class
Mλ,µ1,1(α, β, em).
Theorem 12. Let
= z−
therefore at the end we can write
Theorem 14. Let f(z) defined by (1.6) be in the class Mλ,µ1,1(α, β, em). Then f(z)
Proof. It is sufficient to show that
Theorem 15. [Alexander’s Theorem]Let f be analytic in U withf(0) =f′(0)−
1 = 0. Then f is convex function if and only if zf′ is starlike function.
Corollary 16. Let f ∈ Mλ,µ1,1(α, β, em). Then f(z) is convex of order δ(0≤δ < 1) in |z|< r where r is the largest value such that
m
X
n=2
nen
(1−nβ+α(1−n))B1λ,µ(n)r n−1+
k(1−
m
P
n=2
en)
(1−kβ+α(1−k))B1λ,µ(k)r
k−1 < 1
1−β.
Theorem 17. Let f ∈ Mλ,µ1,1(α, β, em) andλ >0, if
dn=
(1−β)e2n
(1−nβ+α(1−n))B1λ,µ(n) (2≤n≤m), (3.6)
then the function
H(z) =z−
m
X
n=2
(1−β)dn
((1−nβ) +α(1−n))B1λ,µ(n)z n−
∞
X
k=m+1
akzk
is also in the classMλ,µ1,1(α, β, em).
Proof. By assumption we have (1−nβ+α(1−n))Bλ,µ1 (n)>1 therefore,
dn=
(1−β)e2n
(1−nβ+α(1−n))Bλ,µ1 (n) < en≤1.
So, 0≤
m
P
n=2
dn< m
P
n=2
en≤1, thus
∞
X
k=m+1
(1−kβ+α(1−k))B1λ,µ(k)
(1−β)(1−
m
P
n=2
dn)
ak<
∞
X
k=m+1
(1−kβ+α(1−k))B1λ,µ(k)
(1−β)(1−
m
P
n=2
en)
<1.
Theorem 18. Let f, g ∈ Mλ,µ1,1(α, β, em) and λ >0. Then
(f ∗g)(z) =z−
m
X
n=2
(1−β)2e2n
(1−nβ+α(1−n))2(Bλ,µ 1 (n))2
zn− ∞
X
k=m+1
akbkzk
is also in the classMλ,µ1,1(α, β, dm)ifλ1 ≤inf
k
(Bλ,µ1 (k)) 2
(1− m
P
n=2
dn)−1
, wheredn(2≤n≤m)
Proof. By making use of (3.6), we have
(f∗g)(z) =z−
m
X
n=2
(1−β)dn
(1−nβ+α(1−n))B1λ,µ(n)z n−
∞
X
k=m+1
akbkzk.
By Theorem 17, we have
∞
X
k=m+1
(1−kβ+α(1−k))B1λ,µ(k)
(1−β)(1− Pm
n=2
dn)
ak<1
and
∞
X
k=m+1
(1−kβ+α(1−k))B1λ,µ(k)
(1−β)(1− ∞
P
n=2
dn)
bk<1.
Therefore, by Cauchy-Schwarz inequality we obtain
∞
X
k=m+1
(1−kβ+α(1−k))B1λ,µ(k)
(1−β)(1−
m
P
n=2
dn)
p
akbk <1. (3.7)
Now, we want to prove that
∞
X
k=m+1
(1−kβ+α(1−k))B1λ,µ(k)
(1−β)(1− Pm
n=2
dn)
akbk<1. (3.8)
In view of (3.7) the inequality (3.8) holds true if
p akbk
Bλ1,µ
1 (k)
B1λ,µ(k) <1. (3.9)
But we have
B1λ,µ(k)
1−
m
P
n=2
dn
p akbk<
(1−kβ+α(1−k))B1λ,µ(k)
(1−β)(1−
m
P
n=2
dn)
p
akbk<1.
Therefore (3.9) holds true if
1− Pm
n=2
dn
Bλ,µ1 (k) <
B1λ,µ(k)
Bλ1,µ
1 (k)
, or Bλ1,µ
1 (k)<
(Bλ,µ1 (k))2
1− Pm
n=2
dn
.
Then λ1 < (B λ,µ
1 (k)) 2
1− m
P
n=2
Theorem 19. Let f(z)∈ Mλ,µ1,1(α, β, em). Then
J1λ,µf(z) = exp
Z z
0
E(t)−α
(βE(t)−α)tdt
,|E(z)|<1, z∈ U.
Proof. The case α = 0 is obvious. Therefore, suppose that α 6= 0. Then for f ∈
Mλ,µ1,1(α, β, em) and let w =
J1λ,µf(z)
z(J1λ,µf(z))′ we have Re(w) > α|w−1|+β, therefore,
w−1
w−β
<
1
α or w−1
w−β = E(z)
α , where|E(z)|<1, z∈ U. This yields
J1λ,µf(z)
z(J1λ,µf(z))′ −1 J1λ,µf(z)
z(J1λ,µf(z))′ −β
= E(z)
α or
J1λ,µf(z)−z(J1λ,µf(z))′ J1λ,µf(z)−βz(J1λ,µf(z))′ =
E(z)
α .
Thus (J λ,µ
1 f(z))′
(J1λ,µf(z)) =
E(z)−α
(βE(z)−α)z.
By the integration we get the result.
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Abdul Rahman S. Juma S. R. Kulkarni
Department of Mathematics Department of Mathematics Al-Anbar University, Fergusson College,
Ramadi, Iraq. Pune 411004, India
