General Mathematics Vol. 17, No. 3 (2009), 105–124
Some analytic and multivalent functions
defined by subordination property 1
S. M. Khairnar and Meena More
Abstract
In this paper we introduce some functions which are multivalently analytic defined by the subordination property and the DziokSrivastava linear operator. We obtain characterizing property, growth
and distortion inequalities, closure theorem, extreme points, radius
of starlikeness, convexity, and close-to-convexity for the functions in
the class. We also discuss inclusion and neighbourhood properties,
region of p-valency and a class preserving linear operator for these
functions. Interesting consequences of the results obtained are also
indicated.
2000 Mathematics Subject Classification:Primary 30C45; Secondary
30C50, 26A33.
Key words and phrases: Multivalent functions, Radius of
close-to-convexity, Dziok-Srivastava operator, Subordination principle,
Maximum modulus theorem.
1
Received 12 September, 2008
Accepted for publication (in revised form) 29 September, 2008
105
106
1
S. M. Khairnar, Meena More
Introduction
Let A(p) denote the class of functions of the form
(1)
p
f (z) = z +
∞
X
ak z k
(ak ≥ 0, p, n ∈ IN )
k=p+n
which are analytic and p-valent in the open unit disc U = {z : |z| < 1}. If
f (z) ∈ A(p) is given by (1) and g(z) ∈ A(p) is given by
(2)
p
g(z) = z +
∞
X
bk z k (bk ≥ 0, p, n ∈ IN0 )
k=p+n
the convolution (f ∗ g)(z) of f and g is defined by
(3)
p
(f ∗ g)(z) := z +
∞
X
ak bk z k := (g ∗ f )(z).
k=p+n
A function f ∈ A(p) is said to be p-valently starlike of order ρ (0 ≤ ρ < p)
in U if and only if,
(4)
½ 0 ¾
f (z)
Re z
> ρ.
f (z)
Similarly, a function f (z) is p-valently convex of order ρ (0 ≤ ρ < p) in U if
½
¾
zf 00 (z)
(5)
Re 1 + 0
> ρ.
f (z)
It follows from expression (4) and (5) that f is convex if and only if, zf 0 is
starlike.A function f (z) ∈ A(p) is close-to-convex of order ρ if
½ 0 ¾
f (z)
(6)
Re
> ρ (0 ≤ ρ < p).
z p−1
Some analytic and multivalent functions...
107
For the two functions f and g, analytic in U , we say that the function f (z)
is subordinate to g(z) in U , and write f ≺ g, if there exists a Schwarz
function w(z), analytic in U with w(0) = 0 and |w(z)| < 1 (z ∈ U ), such
that f (z) = g(w(z)) (z ∈ U ). In particular, if the function g is univalent in
U , the above subordination is equivalent to f (0) = g(0) and f (U ) ⊂ g(U ).
The operator
(Hsq [a1 ]f )(z) := Hsq (a1 , · · · , aq ; b1 , · · · , bs )f (z)
= z p q Fs (a1 , · · · , aq ; b1 , · · · , bs ; z) ∗ f (z)
∞
X
(a1 )k−p · · · (aq )k−p ak
p
zk
= z +
(b
1 )k−p · · · (bs )k−p (k − p)!
k=p+n
(7)
p
= z +
∞
X
h(k)ak z k
k=p+n
where
(8)
Here
h(k) =
q Fs (z)
(a1 )k−p · · · (aq )k−p
(b1 )k−p · · · (bs )k−p (k − p)!
is the generalized hypergeometric function for aj ∈ C (j =
1, 2, · · · , q) and bj ∈ C (j = 1, 2, · · · , s) such that bj 6= 0, −1, −2, · · · (j =
1, 2, · · · , s) defined by
(9)
q Fs (z)
=
=
q Fs (a1 , · · · , aq ; b1 , · · ·
∞
X
(a1 )k · · · (aq )k k
k=0
(b1 )k · · · (bs )k k!
z
, bs ; z)
(q ≤ s + 1, q, s ∈ IN0 , z ∈ U )
where


(λ)k =
1
(k = 0)
Γ(λ + k)
=
 λ(λ + 1) · · · (λ + k − 1) (k ∈ IN )
Γ(λ)
108
The series
S. M. Khairnar, Meena More
q Fs (z)
in (9) converges absolutely for |z| < ∞ if q < s + 1 and
for |z| = 1 if q = s + 1.The linear operator defined in (7) is the DziokSrivastava operator (for details see [2], [3]) which contains the well-known
operators like the Hohlov linear operator [6], the Carlson-Shafer operator
[1], the Ruscheweyh derivative operator [11], the Srivastava-Owa fractional
derivative operator [9], the Saitoh generalized linear operator, the BernardiLibera-Livingston operator and many others. One may refer [9] for further
details and references for these operators.
Let T (p) denote the subclass of A(p) consisting of functions f of the
form
p
(10)
f (z) = z −
∞
X
ak z k (ak ≥ 0, p, n ∈ IN )
k=p+n
which are analytic and p-valent in U .
By applying the subordination definition we introduce a new class K(λ, µ, A, B)
of functions belonging to T (p) and satisfying
(11)
1 + Az
z(Hsq [a1 ]f )0 + λz 2 (Hsq [a1 ]f )00
≺p
L(z) =
q
q
0
(1 − µ)(Hs [a1 ]f ) + µz(Hs [a1 ]f )
1 + Bz
(0 ≤ µ ≤ λ ≤ 1, −1 ≤ A < B ≤ 1, aj ∈ C (j = 1, 2, · · · , q), bj ∈ C \
{0, −1, −2, · · · } (j = 1, 2, · · · , s), q ≤ s + 1, q, s ∈ IN, z ∈ U ).
Following the work of Goodman [5] and Ruscheweyh [11], we define the
(n, δ)-neighbourhood of a function f ∈ T (p) by
(12)
Nn,δ (f ) :=
(
p
g ∈ T (p) : g(z) = z −
∞
X
k=p+n
bk z
k
and
∞
X
k=p+n
)
k|ak − bk | ≤ δ
.
Some analytic and multivalent functions...
In particular, for the function e(z) = z p
(13)
(
Nn,δ (e) :=
g ∈ T (p) : g(z) = z p −
109
(p ∈ IN )
∞
X
bk z k and
k=p+n
∞
X
)
k|bk | ≤ δ
.
k=p+n
A function f (z) ∈ T (p) defined by (10) is said to be in the class K α (λ, µ, A, B)
if there exists a function g(z) ∈ K(λ, µ, A, B) such that
¯
¯
¯ f (z)
¯
¯
¯
(14)
¯ g(z) − 1¯ < p − α (z ∈ U, 0 ≤ α < p)
2
Main results and properties of the Class
K(λ, µ, A, B)
Theorem 1. Let the function f (z) be defined by (10). Then the function
f (z) belongs to the class K(λ, µ, A, B) if and only if
∞
X
(15)
M (λ, µ, A, B, k)ak ≤ 1
k=p+n
where
(16)
M (λ, µ, A, B, k) =
[k(1 + B)(1 + λ(k − 1)) − p(1 + A)(1 + µ(k − 1))]h(k)
p[(B − 1)(1 + λ(p − 1)) − (A − 1)(1 + µ(p − 1))]
for
(a1 )k−p · · · (aq )k−p
(b1 )k−p · · · (bs )k−p (k − p)!
(0 ≤ µ ≤ λ ≤ 1, −1 ≤ A < B ≤ 1, aj ∈ C (j = 1, 2, · · · , q), bj ∈ C \
h(k) =
{0, −1, −2, · · · } (j = 1, 2, · · · , s)). The result is sharp with the extremal
function f (z) given by
(17)
f (z) = z p −
1
z p+n (n ∈ IN ).
M (λ, µ, A, B, k)
110
S. M. Khairnar, Meena More
Proof. We suppose that f (z) ∈ K(λ, µ, A, B). Then by recalling the
condition (11), we have
(18)
˛
˛
∞
P
˛
˛
˛
˛
p[1+λ(p−1)−(1+µ(p−1))]−
[k(1+λ(k−1))−p(1+µ(k−1))]h(k)ak z k−p
˛
˛
k=p+n
˛
˛
˛
˛ ≤ 1 (z ∈ U ).
∞
P
˛
˛
˛p[B(1+λ(p−1))−A(1+µ(p−1))]−
[kB(1+λ(k−1))−Ap(1+µ(k−1))]h(k)ak z k−p ˛
˛
˛
k=p+n
Now choosing values of z on the real axis and allowing z → 1 from the
left through real values, the inequality (18) immediately yields the desired
condition in (15).
Conversely, by assuming the hypothesis (15) and |z| = 1, we note the
following by the subordination property:
˛
˛
˛ L(z) − p ˛
˛
˛
˛ L(z)B − pA ˛
˛
˛
∞
P
˛
˛
k−p
˛
˛
p[1
+
λ(p
−
1)
−
(1
+
µ(p
−
1))]
−
[k(1
+
λ(k
−
1))
−
p(1
+
µ(k
−
1))]h(k)a
z
k
˛
˛
k=p+n
˛
˛
≤˛
˛
∞
P
˛
˛
k−p
˛
˛ p[B(1 + λ(p − 1)) − A(1 + µ(p − 1))] −
[kB(1 + λ(k − 1)) − Ap(1 + µ(k − 1))]h(k)ak z
˛
˛
k=p+n
p[1 + λ(p − 1) − (1 + µ(p − 1))] +
≤
p[B(1 + λ(p − 1)) − A(1 + µ(p − 1))] −
∞
P
[k(1 + λ(k − 1) − p(1 + µ(k − 1))]h(k)ak
k=p+n
∞
P
.
[k(B(1 + λ(k − 1)) − Ap(1 + µ(k − 1))]h(k)ak
k=p+n
Hence by maximum modulus theorem f (z) ∈ K(λ, µ, A, B).Finally, it is
observed that the result is sharp and the extremal function is given by (17).
Theorem 1 immediately yields the following result.
Corollary 1. If a function f (z) of the form (10) is in T (p) and belongs to
the class K(λ, µ, A, B), then
(19)
ak ≤
p[(B − 1)(1 + λ(p − 1)) − (A − 1)(1 + µ(p − 1))]
(k ≥ p + n, n ∈ IN )
[k(1 + B)(1 + λ(k − 1)) − p(1 + A)(1 + µ(k − 1))]h(k)
where the equality holds true for the function (17).
Some analytic and multivalent functions...
111
Proof. The result (19) follows from the fact that the series in (15) converges.
Next we give some more interesting properties of the class K(λ, µ, A, B).
Theorem 2. Let 0 ≤ λ ≤ µ ≤ 1, −1 ≤ A < B ≤ 1, −1 ≤ A0 < B 0 ≤ 1.
Then
K(λ, µ, A, B) = K(λ, µ, A0 , B 0 )
(20)
if and only if
M (λ, µ, A, B, k) = M (λ, µ, A0 , B 0 , k).
(21)
Proof. Let f (z) ∈ K(λ, µ, A, B) and (21) hold true. Then by Theorem 1,
we have
∞
X
0
0
M (λ, µ, A , B , k)ak =
∞
X
M (λ, µ, A, B, k)ak ≤ 1.
k=p+n
k=p+n
This implies f (z) ∈ K(λ, µ, A0 , B 0 ). Similarly it can be shown that f (z) ∈
K(λ, µ, A0 , B 0 ) implies f (z) ∈ K(λ, µ, A, B). Hence (21) implies K(λ, µ, A, B) =
K(λ, µ, A0 , B 0 ). Conversely, suppose (20) holds true. Notice that a function
defined by (10) belonging to K(λ, µ, A, B) will belong to K(λ, µ, A0 , B 0 ) only
if
∞
X
0
0
M (λ, µ, A , B , k)ak ≤
∞
X
M (λ, µ, A, B, k)ak
k=p+n
k=p+n
that is if
(22)
M (λ, µ, A0 , B 0 , k) ≤ M (λ, µ, A, B, k).
Similarly, we can show that
(23)
M (λ, µ, A, B, k) ≤ M (λ, µ, A0 , B 0 , k).
112
S. M. Khairnar, Meena More
(22) and (23) together imply (21). Hence the result.
We state some more interesting deductions which follow using Theorem 1
and Theorem 2.
Theorem 3. Let 0 ≤ λ ≤ µ ≤ 1, −1 ≤ A < B1 ≤ B2 ≤ 1. Then
K(λ, µ, A, B1 ) ⊇ K(λ, µ, A, B2 ).
Proof. Notice that
(24)
M (λ, µ, A, B1 , k) ≤ M (λ, µ, A, B2 , k) for B1 ≤ B2 .
If f (z) ∈ K(λ, µ, A, B2 ) we have
∞
X
M (λ, µ, A, B1 , k)ak ≤
k=p+n
∞
X
M (λ, µ, A, B2 , k)ak ≤ 1
k=p+n
Thus by Theorem 1 it follows that f (z) ∈ K(λ, µ, A, B1 ). Hence the theorem
is proved.
Theorem 4. Let 0 ≤ λ ≤ µ ≤ 1, −1 ≤ A1 ≤ A2 < B ≤ 1. Then
K(λ, µ, A1 , B) ⊆ K(λ, µ, A2 , B).
Proof. The proof of the theorem is on the lines of Theorem 3 above.
Next we give a result which follows from Theorem 3 and Theorem 4.
Corollary 2. Let 0 ≤ λ ≤ µ ≤ 1, −1 ≤ A1 ≤ A2 < B1 ≤ B2 ≤ 1. Then
K(λ, µ, A1 , B2 ) ⊆ K(λ, µ, A2 , B2 ) ⊆ K(λ, µ, A2 , B1 ).
3
Growth and Distortion Theorem
Let us recall again the function h(k) given by (8)
h(k) =
(a1 )k−p · · · (aq )k−p
.
(b1 )k−p · · · (bs )k−p (k − p)!
Some analytic and multivalent functions...
113
We note that h(k) is a non-decreasing function of k for k ≥ p + n, n ∈ IN .
Thus
(25)
h(k) ≥ h(p + 1) =
a1 · · · aq
≥ 0.
b1 · · · bs
We now state the following growth and distortion inequalities for the class
K(λ, µ, A, B).
Theorem 5. If the function f (z) defined by (10) is in the class K(λ, µ, A, B),
then
(26)
p[(B − 1)(1 + λ(p − 1)) − (A − 1)(1 + µ(p − 1))]
|z|p+n , (n ∈ IN )
[(p + n)(1 + B)(1 + λ(p + n − 1)) − p(1 + A)(1 + µ(p + n − 1))]h(p + n)
||f (z)|−|z|p | ≤
and
(27)
||f 0 (z)|−p |z|p−1 | ≤
p(p + n)[(B − 1)(1 + λ(p − 1)) − (A − 1)(1 + µ(p − 1))]
|z|p+n−1 , (n ∈ IN ).
[(p + n)(1 + B)(1 + λ(p + n − 1)) − p(1 + A)(1 + µ(p + n − 1))]h(p + n)
The result in (26) and (27) are sharp with the extremal function
f (z) = z p −
p[(B − 1)(1 + λ(p − 1)) − (A − 1)(1 + µ(p − 1))]
z p+n , (n ∈ IN ).
[(p + n)(1 + B)(1 + λ(p + n − 1)) − p(1 + A)(1 + µ(p + n − 1))]h(p + n)
Proof. We have
p
f (z) = z +
∞
X
ak z k therefore
k=p+n
(28)
∞
X
|f (z)| ≤ |z|p +
ak |z|k ≤ |z|p +|z|p+n
k=p+n
≤ |z|p +
∞
X
ak
k=p+n
p[(B −1)(1+λ(p−1))−(A−1)(1+µ(p−1))]
|z|p+n .
[(p+n)(1+B)(1+λ(p+n−1))−p(1+A)(1+µ(p+n−1))]h(p+n)
Similarly
(29) |f (z)|
≥
|z|p −
∞
X
k=p+n
≥
ak |z|k ≥ |z|p − |z|p+n
∞
X
ak
k=p+n
p(p + n)[(B − 1)(1 + λ(p − 1)) − (A − 1)(1 + µ(p − 1))]
|z|p+n .
[(p + n)(1 + B)(1 + λ(p + n − 1)) − p(1 + A)(1 + µ(p + n − 1))]h(p + n)
Combining (28) and (29) we get the result (26). The next result in (27)
can be derived similarly.
114
S. M. Khairnar, Meena More
Remark. Let the function f (z) defined by (10) be in the class K(λ, µ, A, B).
Then f (z) is included in a disc with centre at the origin and radius
R1 = 1 +
p[(B − 1)(1 + λ(p − 1)) − (A − 1)(1 + µ(p − 1))]
, (n ∈ IN ).
[(p + n)(1 + B)(1 + λ(p + n − 1)) − p(1 + A)(1 + µ(p + n − 1))]h(p + n)
and f 0 (z) is included in a disc with centre at origin and radius
R2 = p +
p(p + n)[(B − 1)(1 + λ(p − 1)) − (A − 1)(1 + µ(p − 1))]
, (n ∈ IN ).
[(p + n)(1 + B)(1 + λ(p + n − 1)) − p(1 + A)(1 + µ(p + n − 1))]h(p + n)
Now we state a theorem of convex linear combinations of the functions in
the class K(λ, µ, A, B).
Theorem 6. Let the function
p
fj (z) = z −
∞
X
ak,j z k (ak,j ≥ 0, j = 1, 2, · · · , l)
k=p+n
be in the class K(λ, µ, A, B). Then
h(z) =
l
X
cj fj (z) ∈ K(λ, µ, A, B)
j=1
where
l
P
cj = 1 and cj ≥ 0 (j = 1, 2, · · · , l). Thus, we note that K(λ, µ, A, B)
j=1
is a convex set.
Proof. We have
(30)
h(z) =
X̀
Ã
cj
zp −
j=1
= z
p
cj −
j=1
= zp −
!
ak,j z k
k=p+n
X̀
= zp −
∞
X
∞
X̀ X
cj ak,j z k
j=1 k=p+n
∞
X
k=p+n
∞
X
k=p+n
Ã
X̀
j=1
ek z k
ak,j cj
!
zk
Some analytic and multivalent functions...
where ek =
P̀
115
ak,j cj .
j=1
Since fj ∈ K(λ, µ, A, B) by (15), we have
∞
X
(31)
M (λ, µ, A, B, k)ak,j ≤ 1.
k=p+n
In view of (30) h(z) ∈ K(λ, µ, A, B) if
∞
X
M (λ, µ, A, B, k)ek ≤ 1.
k=p+n
Now, we have
∞
X
M (λ, µ, A, B, k)ek =
k=p+n
∞
X
M (λ, µ, A, B, k)
X̀
j=1
≤
X̀
cj
ak,j cj
j=1
k=p+n
=
X̀
∞
X
M (λ, µ, A, B, k)ak,j
k=p+n
cj = 1.
j=1
Thus h(z) ∈ K(λ, µ, A, B).
4
Extreme Points
Theorem 7. Let fp (z) = z p and
fk (z) = z p −
p[(B − 1)(1 + λ(p − 1)) − (A − 1)(1 + µ(p − 1))]
zk
[k(1 + B)(1 + λ(k − 1)) − p(1 + A)(1 + µ(k − 1))]h(k)
(k ≥ p + n, n ∈ IN ). Then f (z) ∈ K(λ, µ, A, B) if and only if f (z) can be
expressed in the form
(32)
f (z) =
∞
X
k=p+n
dk fk (z)
116
S. M. Khairnar, Meena More
∞
P
where dk ≥ 0 and
dk = 1, n ∈ IN0 .
k=p+n
Proof. Let f (z) be expressible in the form
∞
X
f (z) =
λk fk (z) (n ∈ IN0 )
k=p+n
p
= z −
∞
X
1
dk z k (n ∈ IN ).
M (λ, µ, A, B, k)
k=p+n
Now
∞
X
M (λ, µ, A, B, k)
k=p+n
∞
X
1
dk =
dk = 1 − dp ≤ 1 (n ∈ IN ).
M (λ, µ, A, B, k)
k=p+n
Therefore, f (z) ∈ K(λ, µ, A, B) . Conversely, suppose that f (z) ∈ K(λ, µ, A, B).
Then setting
dk =
∞
X
1
ak and dp = 1 −
dk
M (λ, µ, A, B, k)
k=p+n
(n ∈ IN )
we notice that f (z) can be expressed in the form (32).
Remark. The extreme points of the class K(λ, µ, A, B) are fp (z) = z p and
fk (z) = z p −
5
p[(B − 1)(1 + λ(p − 1)) − (A − 1)(1 + µ(p − 1))]
z k , (k ≥ p + n, n ∈ IN ).
[k(1 + B)(1 + λ(k − 1)) − p(1 + A)(1 + µ(k − 1))h(k)
Inclusion Property
We now obtain an inclusion relation for the functions in the class K(λ, µ, A, B).
Theorem 8. If h(k) ≥ h(p + n) for k ≥ p + n, n ∈ IN and
(33)
δ :=
p(p + n)[(B − 1)(1 + λ(p − 1)) − (A − 1)(1 + µ(p − 1))]
[(p + n)(1 + B)(1 + λ(p + n − 1)) − p(1 + A)(1 + µ(p + n − 1))]h(p + n)
Some analytic and multivalent functions...
117
then
(34)
K(λ, µ, A, B) ⊆ Nn,δ (e).
Proof. Let f (z) ∈ K(λ, µ, A, B). Then in view of assertion (15) of Theorem
1 and the condition h(k) ≥ h(p + n) for k ≥ p + n, n ∈ IN , we get
∞
X
h(p + n)[(p + n)(1 + B)(1 + λ(p + n − 1)) − p(1 + A)(1 + µ(p + n − 1))]
ak
k=p+n
≤
∞
X
[k(1 + B)(1 + λ(k − 1)) − p(1 + A)(1 + µ(k − 1))]h(k)ak
k=p+n
(35)
≤ p[(B − 1)(1 + λ(p − 1)) − (A − 1)(1 + µ(p − 1))]
which implies
∞
X
(36)
k=p+n
ak ≤
p[(B − 1)(1 + λ(p − 1)) − (A − 1)(1 + µ(p − 1))]
.
[(p + n)(1 + B)(1 + λ(p + n − 1)) − p(1 + A)(1 + µ(p + n − 1))]h(p + n)
Applying the assertion (15) of Theorem 1 in conjunction with (36), we
obtain
(1 + B)(1 + λ(p + n − 1))h(p + n)
∞
X
kak
k=p+n
≤ p[(B − 1)(1 + λ(p − 1)) − (A − 1)(1 + µ(p − 1))] + p(1 + A)(1 + µ(p + n − 1))h(p + n)
∞
X
ak
k=p+n
≤
p(p + n)[(B − 1)(1 + λ(p − 1)) − (A − 1)(1 + µ(p − 1))]
:= δ
(p + n)(1 + B)(1 + λ(p + n − 1)) − p(1 + A)(1 + µ(p + n − 1))]h(p + n)
which by virtue of (12) establishes the inclusion relation (34).
6
Neighbourhood Property
In this section we determine the neighbourhood property for the class K α (λ, µ, A, B).
Theorem 9. If g(z) ∈ K(λ, µ, A, B) and
(37)
α=p−
δ
M (λ, µ, A, B, p + n)
p + n M (λ, µ, A, B, p + n) − 1
118
S. M. Khairnar, Meena More
then
Nn,δ (g) ⊂ K α (λ, µ, A, B).
Proof. Suppose that f (z) ∈ Nn,δ (g). We then find from (12) that
∞
X
k|ak − bk | ≤ δ
k=p+n
which readily implies the following coefficient inequality
∞
X
(38)
|ak − bk | ≤
k=p+n
δ
p+n
(n ∈ IN ).
Next, since g(z) ∈ K(λ, µ, A, B), in view of (36), we have
∞
X
(39)
bk ≤
k=p+n
1
M (λ, µ, A, B, p + n)
Using (38) and (39), we get
¯
¯
¯ f (z)
¯
¯
¯≤
−
1
¯ g(z)
¯
∞
P
|ak − bk |
k=p+n
1−
∞
P
≤
bk
M (λ, µ, A, B, p + n)
δ
p + n M (λ, µ, A, B, p + n) − 1
k=p+n
provided that α is given by (37). Thus by condition (14) f (z) ∈ K α (λ, µ, A, B)
where α is given by (37).
7
Radius of Starlikeness, Convexity and Closeto-convexity
Using the inequalities (4), (5) and (6) and Theorem 1 we can compute the
radius of starlikeness, convexity and close-to-convexity.
Some analytic and multivalent functions...
119
Theorem 10. Let a function f (z) ∈ K(λ, µ, A, B). Then f (z) is p-valently
starlike of order ρ (0 ≤ ρ < p) in the disc |z| < R3 where
½
R3 = inf
k
(p − ρ)
M (λ, µ, A, B, k)
(k − ρ)
1
¾ k−p
(k ≥ p + n, n ∈ IN )
for M (λ, µ, A, B, k) given by (16).
Proof : It is sufficient to show that
¯ 0
¯
¯ zf
¯
¯
¯
¯ f − p¯ ≤ p − ρ for 0 ≤ ρ < p and |z| < R3
(40)
¯
¯
∞
P
¯
¯
(k − p)ak z k−p ¯
¯ 0
¯ ¯−
¯ zf
¯
¯ ¯ k=p+n
¯
¯
¯=¯
−
p
∞
¯ f
¯
¯ ¯
P
¯ 1−
ak z k−p ¯¯
¯
k=p+n
(40) is bounded above by p − ρ if
(41)
∞
X
(k − ρ)
ak |z|k−p ≤ 1.
(p
−
ρ)
k=p+n
Also from Theorem 1, if f (z) ∈ K(λ, µ, A, B) then
(42)
∞
X
M (λ, µ, A, B, k)ak ≤ 1.
k=n+p
In view of (42) we notice that (41) holds true if
(k − ρ) k−p
|z|
≤ M (λ, µ, A, B, k).
(p − ρ)
That is if
½
|z| ≤
(p − ρ)M (λ, µ, A, B, k)
(k − ρ)
Setting |z| = R3 we get the desired result.
1
¾ k−p
.
120
S. M. Khairnar, Meena More
Theorem 11. Let a function f (z) ∈ K(λ, µ, A, B). Then f (z) is p-valently
convex of order ρ (0 ≤ ρ < p) in the disc |z| < R4 where
½
R4 = inf
k
1
¾ k−p
p(p − ρ)
M (λ, µ, A, B, k)
(k ≥ p + n, n ∈ IN )
k(k − ρ)
for M (λ, µ, A, B, k) given by (16).
Proof : It is sufficient to show that
¯ 00
¯
¯ zf
¯
¯
¯ ≤ p − ρ for 0 ≤ ρ < p and |z| < R4 .
+
1
−
p
¯ f
¯
Using arguments similar to the proof of Theorem 10, we get the result.
Theorem 12. Let a function f (z) ∈ K(λ, µ, A, B). Then f (z) is p-valently
close-to-convex of order ρ (0 ≤ ρ < p) in the disc |z| < R5 where
½
R5 = inf
k
1
¾ k−p
(p − ρ)
M (λ, µ, A, B, k)
k
(k ≥ p + n, n ∈ IN )
for M (λ, µ, A, B, k) given by (16).
Proof : It is sufficient to show that
¯ 0
¯
¯ f
¯
¯
¯
¯ z p−1 − p¯ ≤ p − ρ for 0 ≤ ρ < p and |z| < R5 .
The result follows by application of arguments similar to the proof of Theorem 10.
Some analytic and multivalent functions...
8
121
Application of Class Preserving Integral
Operator
In this Section we give a class preserving integral operator due to JungKim-Srivastava, please refer [8].
(43)

I(z) =
Qαβ,p f (z)
=

α+β+p−1
β+p−1
 α
zβ
Z
z
0
t
tβ−1 (1 − )α−1 f (t)dt
z
(α > 0, β > −p, z ∈ U ).
It can be easily verified that
(44)
I(z) =
Qαβ,p f (z)
∞
X
Γ(β + k)Γ(α + β + k)
=z −
ak z k .
Γ(α
+
β
+
k)Γ(β
+
p)
k=p+1
p
A function I(z) is said to be close-to-convex and p-valent in the disc |z| < R6
if
¯
¯ 0
¯
¯ I (z)
¯ ≤ p in |z| < R6
¯
−
p
¯
¯ z p−1
(45)
and (α > 0, β > −p, z ∈ U ).
Theorem 13. Let α > 0, β > −p and f (z) belong to the class K(λ, µ, A, B).
Then the function I(z) defined by (43) is close-to-convex and p-valent in the
disc |z| < R6 , where
(46)
½
R6 = inf
k
pΓ(α + β + k)Γ(β + p)M (λ, µ, A, B, k)
kΓ(β + k)Γ(α + β + k)
Proof : We show that
(47)
¯
¯ 0
¯
¯ I (z)
¯ ≤ p in |z| < R6
¯
−
p
¯
¯ z p−1
1
¾ k−p
.
122
S. M. Khairnar, Meena More
R6 is given by (46).
In view of (44), we have
¯
¯
¯ 0
¯
∞
¯
¯ X
¯ I (z)
¯
kΓ(β
+
k)Γ(α
+
β
+
k)
k−p ¯
¯
¯ = ¯¯−
a
z
−
p
¯
k
¯ z p−1
¯
¯
¯
Γ(α + β + k)Γ(β + p)
k=p+n
∞
X
kΓ(β + k)Γ(α + β + k)
≤
ak |z|k−p .
Γ(α + β + k)Γ(β + p)
k=p+n
The last inequality is bounded above by p if
∞
X
kΓ(β + k)Γ(α + β + k)
ak |z|k−p ≤ 1.
pΓ(α
+
β
+
k)Γ(β
+
p)
k=p+n
(48)
Also, since f (z) ∈ K(λ, µ, A, B) by Theorem 1, we have
∞
X
(49)
M (λ, µ, A, B, k)ak ≤ 1
k=p+n
where M (λ, µ, A, B, k) is given in (19). Thus (48) and consequently (47)
will hold if
kΓ(β + k)Γ(α + β + k)
ak |z|k−p ≤ M (λ, µ, A, B, k)ak .
pΓ(α + β + k)Γ(β + p)
That is, if
½
|z| ≤
pΓ(α + β + k)Γ(β + p)M (λ, µ, A, B, k)
kΓ(β + k)Γ(α + β + k)
1
¾ k−p
for k ≥ p + n, n ∈ IN . The result follows by setting |z| = R6 .
References
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Some analytic and multivalent functions...
123
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with the generalized hypergeometric function, Applied Mathematics and
Computation, 103(1)(1993), 1-13.
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S. M. Khairnar, Meena More
[10] J. K. Prajapat and R. K. Raina, Some new inclusion and neighbourhood properties for certain multivalent function classes associated with
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S.M.Khairnar and Meena More*
Department of Mathematics,
Maharashtra Academy of Engineering,
Alandi, Pune - 412105, Maharashtra, India
Email: smkhairnar2007@gmail.com
Email*: meenamores@gmail.com