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A class of analytic functions involving in the Dziok-Srivastava operator

Abstract

Let A be a class of functions f(z) of the form

f(z)=z+ n = 2 a n z n
(0.1)

which are analytic in the open unit disk U. By means of the Dziok-Srivastava operator, we introduce a new subclass

S m l ( α 1 ,α,μ) ( l m + 1 , l , m N { 0 } , π 2 < α < π 2 , μ > cos α )

of A. In particular, S 0 1 (2,0,0) coincides with the class of uniformly convex functions introduced by Goodman. The order of starlikeness and the radius of α-spirallikeness of order β (β<1) are computed. Inclusion relations and convolution properties for the class S m l ( α 1 ,α,μ) are obtained. A special member of S m l ( α 1 ,α,μ) is also given. The results presented here not only generalize the corresponding known results, but also give rise to several other new results.

MSC:30C45.

1 Introduction

Let A be a class of functions f(z) of the form

f(z)=z+ n = 2 a n z n
(1.1)

which are analytic in the open unit disk U={z:|z|<1}. For β<1, a function f(z)A is said to be starlike of order β in U if

z f ( z ) f ( z ) >β(zU).
(1.2)

This class is denoted by S (β) (β<1). For π 2 <α< π 2 and β<1, a function f(z)A is said to be α-spirallike of order β in U if

{ e i α z f ( z ) f ( z ) } >βcosα(zU).
(1.3)

When 0β<1, it is well known that all the starlike functions of order β and α-spirallike functions of order β are univalent in U. A function f(z)A is said to be convex univalent in U if

{ 1 + z f ( z ) f ( z ) } >0(zU).
(1.4)

We denote this class by K. Also, let UCV(K) be the class of uniformly convex functions in U introduced by Goodman [1]. It was shown in [2] that f(z)A is in UCV if and only if

{ 1 + z f ( z ) f ( z ) } >| z f ( z ) f ( z ) |(zU).
(1.5)

In [2], Rønning investigated the class S p defined by

S p = { f ( z ) S ( 0 ) : f ( z ) = z g ( z ) , g ( z ) UCV } .
(1.6)

The uniformly convex and related functions have been studied by many authors (see, e.g., [110] and the references therein).

If

f(z)=z+ n = 2 a n z n Aandg(z)=z+ n = 2 b n z n A,

then the Hadamard product (or convolution) of f(z) and g(z) is given by

(fg)(z)=z+ n = 2 a n b n z n .

For

α j C(j=1,2,,l)and β j C{0,1,2,}(j=1,2,,m),

the generalized hypergeometric function

F m l ( α 1 ,, α l ; β 1 ,, β m ;z)

is defined by the following infinite series:

F m l ( α 1 , , α l ; β 1 , , β m ; z ) = n = 0 ( α 1 ) n ( α l ) n ( β 1 ) n ( β m ) n z n n ! ( l m + 1 ; l , m N 0 = N { 0 } ; z U ) ,

where ( c ) n is the Pochhammer symbol defined by

( c ) n = { 1 ( n = 0 ) , c ( c + 1 ) ( c + n 1 ) ( n N ) .

Corresponding to the function

z l F m ( α 1 ,, α l ; β 1 ,, β m ;z),

the Dziok-Srivastava operator (see [11])

H( α 1 ,, α l ; β 1 ,, β m ):AA

is defined by the following Hadamard product:

H ( α 1 , , α l ; β 1 , , β m ) f ( z ) = ( z l F m ( α 1 , , α l ; β 1 , , β m ; z ) ) f ( z ) ( l m + 1 ; l , m N 0 ; z U ) .

If f(z)A is given by (1.1), then we have

H( α 1 ,, α l ; β 1 ,, β m )f(z)=z+ n = 1 ( α 1 ) n ( α l ) n ( β 1 ) n ( β m ) n a n + 1 n ! z n + 1 (zU).
(1.7)

In order to make the notation simple, we write

H m l ( α 1 )=H( α 1 ,, α l ; β 1 ,, β m )(lm+1;l,m N 0 ).
(1.8)

It should also be remarked that the Dziok-Srivastava operator H m l ( α 1 ) is a generalization of several linear operators considered in earlier investigations (see [1219], also see [20]).

In this paper we introduce and investigate the following subclass of A.

Definition A function f(z)A is said to be in S m l ( α 1 ,α,μ) if it satisfies the condition

{ e i α z ( H m l ( α 1 ) f ( z ) ) H m l ( α 1 ) f ( z ) } +μ>| z ( H m l ( α 1 ) f ( z ) ) H m l ( α 1 ) f ( z ) 1|(zU),
(1.9)

where

lm+1,l,m N 0 , π 2 <α< π 2 andμ>cosα.
(1.10)

Note that f(z)=z S m l ( α 1 ,α,μ) and that

S 0 1 (1,α,0)= { f ( z ) A : { e i α z f ( z ) f ( z ) } > | z f ( z ) f ( z ) 1 | ( z U ) } .
(1.11)

Also,

S 0 1 (1,0,0)= S p and S 0 1 (2,0,0)=UCV.
(1.12)

Throughout this paper we assume, unless otherwise stated, that l, m, α and μ satisfy (1.10).

2 Subordination theorem

Let f(z) and g(z) be analytic in U. We say that the function f(z) is subordinate to g(z) in U, and we write f(z)g(z), if there exists an analytic function w(z) in U such that

| w ( z ) | |z|andf(z)=g ( w ( z ) ) (zU).

If g(z) is univalent in U, then

f(z)g(z)f(0)=g(0)andf(U)g(U).

Theorem 1 A function f(z)A is in S m l ( α 1 ,α,μ) if and only if

e i α z ( H m l ( α 1 ) f ( z ) ) H m l ( α 1 ) f ( z ) h(z)cosα+isinα,
(2.1)

where

h ( z ) = 1 + 2 π 2 ( 1 + μ cos α ) ( log 1 + z 1 z ) 2 = 1 + 8 π 2 ( 1 + μ cos α ) { z + 2 3 z 2 + 23 45 z 3 + } ( z U ) .
(2.2)

Proof Let us define w(z)=u+iv by

e i α z ( H m l ( α 1 ) f ( z ) ) H m l ( α 1 ) f ( z ) =w(z)cosα+isinα(zU).
(2.3)

Then w(0)=1 and the inequality (1.9) can be rewritten as

u> cos α 2 ( cos α + μ ) v 2 + 1 2 ( 1 μ cos α ) .
(2.4)

Thus

w(U)Ω= { w = u + i v : u  and  v  satisfy  ( 2.4 ) } .

It follows from (2.2) that h(0)=1. In order to prove the theorem, it suffices to show that the function w=h(z) given by (2.2) maps U conformally onto the parabolic region Ω.

Note that 1 2 (1 μ cos α )<1. Consider the transformations

w 1 = w 1 , w 2 =exp ( π w 1 2 cos α cos α + μ ) ,t= 1 2 ( w 2 + 1 w 2 ) .
(2.5)

It is easy to verify that the composite function

t=ch ( π 2 cos α ( w 1 ) cos α + μ ) =g(w)(say)

maps Ω + =Ω{w=u+iv:v>0} conformally onto the upper half-plane Im(t)>0 so that w=(w)[ 1 2 (1 μ cos α ),+) corresponds to t=(t)[1,+) and w=1 to t=1. With the help of the symmetry principle, the function t=g(w) maps Ω conformally onto the region G={t:|arg(t+1)|<π}. Since

t=2 ( 1 + z 1 z ) 2 1
(2.6)

maps U onto G, we see that

w = g 1 ( t ) = 1 + 1 2 π 2 ( 1 + μ cos α ) ( log ( t + t 2 1 ) ) 2 = 1 + 2 π 2 ( 1 + μ cos α ) ( log 1 + z 1 z ) 2 = h ( z )

maps U conformally onto Ω. The proof of the theorem is now completed. □

Corollary 1 Let f(z) S m l ( α 1 ,α,μ). Then for zU,

| ( H m l ( α 1 ) f ( z ) z ) sec α e i α |exp { 2 π 2 ( 1 + μ cos α ) 0 1 1 ρ ( log 1 + ρ | z | 1 ρ | z | ) 2 d ρ }
(2.7)

and

| ( H m l ( α 1 ) f ( z ) z ) sec α e i α |exp { 8 π 2 ( 1 + μ cos α ) 0 1 1 ρ ( arctan ρ | z | ) 2 d ρ } .
(2.8)

The results are sharp.

Proof

From Theorem 1 we have

e i α cos α ( z ( H m l ( α 1 ) f ( z ) ) H m l ( α 1 ) f ( z ) 1 ) h(z)1

for f(z) S m l ( α 1 ,α,μ) and h(z) given by (2.2). Since the function h(z)1 is univalent and starlike (with respect to the origin) in U, using the result of Suffridge [[21], Theorem 3], we get

e i α cos α 0 z ( ( H m l ( α 1 ) f ( t ) ) H m l ( α 1 ) f ( t ) 1 t ) dt 0 z h ( t ) 1 t dt.

This implies that

e i α cos α log H m l ( α 1 ) f ( z ) z = 0 1 h ( ρ w ( z ) ) 1 ρ dρ(zU),
(2.9)

where w(z) is analytic and |w(z)||z| in U.

Noting that h(z) maps the disk |z|<ρ (0<ρ1) onto a region which is convex and symmetric with respect to the real axis, we know that

h ( ρ | z | ) { h ( ρ w ( z ) ) } h ( ρ | z | ) (zU).
(2.10)

Now (2.2), (2.9) and (2.10) lead to

log| ( H m l ( α 1 ) f ( z ) z ) sec α e i α | 2 π 2 ( 1 + μ cos α ) 0 1 1 ρ ( log 1 + ρ | z | 1 ρ | z | ) 2 dρ

and

log | ( H m l ( α 1 ) f ( z ) z ) sec α e i α | 2 π 2 ( 1 + μ cos α ) 0 1 1 ρ ( log 1 + i ρ | z | 1 i ρ | z | ) 2 d ρ = 8 π 2 ( 1 + μ cos α ) 0 1 1 ρ ( arctan ρ | z | ) 2 d ρ

for zU. Hence we have (2.7) and (2.8).

Furthermore, for

α j C{0,1,2,}(j=1,,l),

it is easy to see that the function f 0 (z) in S m l ( α 1 ,α,μ), defined by

H m l ( α 1 ) f 0 ( z ) = z exp { 2 π 2 ( 1 + μ cos α ) cos α e i α 0 z 1 t ( log 1 + t 1 t ) 2 d t } ( z U ) ,
(2.11)

shows that the estimates (2.7) and (2.8) are sharp. □

Corollary 2 Let f(z) S m l ( α 1 ,α,μ), where

α j C{0,1,2,}(j=1,2,,l).

Then

f ( z ) = z exp { 2 cos α e i α π 2 ( 1 + μ cos α ) 0 1 1 ρ ( log 1 + ρ w ( z ) 1 ρ w ( z ) ) 2 d ρ } { z + n = 1 n ! ( β 1 ) n ( β m ) n ( α 1 ) n ( α l ) n z n + 1 } ( z U ) ,
(2.12)

where w(z) is analytic in U with w(0)=0 and |w(z)|<1 (zU).

Proof From (2.9) and (2.2), we have

H m l ( α 1 ) f ( z ) = z exp { 2 cos α e i α π 2 ( 1 + μ cos α ) 0 1 1 ρ ( log 1 + ρ w ( z ) 1 ρ w ( z ) ) 2 d ρ } ( z U ) .
(2.13)

For

α j C{0,1,2,}(j=1,2,,l),

from (2.13) and (1.7), we obtain (2.12). □

3 Properties of the class S m l ( α 1 ,α,μ)

Theorem 2 Let f(z) S m l ( α 1 ,α,μ). Then

H m l ( α 1 )f(z) S ( 1 2 ( 1 μ cos α ) )
(3.1)

and the order 1 2 (1 μ cos α ) is sharp.

Proof Let h(z) be given by (2.2). It follows from the proof of Theorem 1 that

h(U)= { w = u + i v : u = cos α 2 ( cos α + μ ) v 2 + 1 2 ( 1 μ cos α ) } .
(3.2)

By using (3.2), we find that

min | z | = 1 ( z 1 ) { e i α ( h ( z ) cos α + i sin α ) } = min v ( , + ) g(v)cosα+ sin 2 α,

where

g(v)= cos 2 α 2 ( cos α + μ ) v 2 + cos α μ 2 +vsinα(<v<+).

Since

g (v)= cos 2 α cos α + μ v+sinα, g (v)>0,

the function g(v) attains its minimum value at

v 0 = ( cos α + μ ) sin α cos 2 α .

Thus

min | z | = 1 ( z 1 ) { e i α ( h ( z ) cos α + i sin α ) } = g ( v 0 ) cos α + sin 2 α = sin 2 α ( cos α + μ ) 2 cos α + cos α ( cos α μ ) 2 + sin 2 α = 1 2 ( 1 μ cos α ) .
(3.3)

If f(z) S m l ( α 1 ,α,μ), then we deduce from Theorem 1 and (3.3) that

z ( H m l ( α 1 ) f ( z ) ) H m l ( α 1 ) f ( z ) > 1 2 ( 1 μ cos α ) (zU)

and the order 1 2 (1 μ cos α ) in (3.1) is sharp for the function f 0 (z) defined by (2.11). □

Theorem 3 Let f(z) S m l ( α 1 ,α,μ) and 1 2 (1 μ cos α )β<1. Then H m l ( α 1 )f(z) is α-spirallike of order β in |z|<ρ, where

ρ=ρ(β,α,μ)= ( tan ( π 4 2 cos α ( 1 β ) cos α + μ ) ) 2 .
(3.4)

The result is sharp.

Proof From (3.4) and (2.2) we have

0<ρ1 ( 1 2 ( 1 μ cos α ) β < 1 )

and

h ( ρ ) = 1 + 2 π 2 ( 1 + μ cos α ) ( log 1 + i ρ 1 i ρ ) 2 = 1 8 π 2 ( 1 + μ cos α ) ( arctan ρ ) 2 = β .

Hence

inf | z | < ρ h(z)=h(ρ)=β.
(3.5)

Let f(z) S m l ( α 1 ,α,μ). Then it follows from Theorem 1 and (3.5) that

{ e i α z ( H m l ( α 1 ) f ( z ) ) H m l ( α 1 ) f ( z ) } >βcosα ( | z | < ρ ) ,

that is, H m l ( α 1 )f(z) is α-spirallike of order β in |z|<ρ. Also, the result is sharp for the function f 0 (z) defined by (2.11). □

Setting β= 1 2 (1 μ cos α ), Theorem 3 reduces to the following.

Corollary 3 Let f(z) S m l ( α 1 ,α,μ). Then H m l ( α 1 )f(z) is α-spirallike of order 1 2 (1 μ cos α ) in U. The result is sharp.

For β1, a function f(z)A is said to be prestarlike of order β in U if

{ z ( 1 z ) 2 ( 1 β ) f ( z ) S ( β ) , β < 1 , f ( z ) z > 1 2 , β = 1 ,
(3.6)

(see [20]). We denote this class by R(β) (β1). The following lemma is due to Ruscheweyh [[20], p.54].

Lemma 1 Let β1, f(z)R(β) and g(z) S (β). Then, for any analytic function F(z) in U,

f ( F g ) f g (U) co ¯ ( F ( U ) ) ,

where co ¯ (F(U)) denotes the convex hull of F(U).

Applying the lemma, we derive Theorems 4 and 5 below.

Theorem 4 Let

α 1 >0and α 1 max { α 1 , 1 + μ cos α } .
(3.7)

Then

S m l ( α 1 , α , μ ) S m l ( α 1 ,α,μ).
(3.8)

Proof

Define

ϕ(z)=z+ n = 1 ( α 1 ) n ( α 1 ) n z n + 1 (zU)

for α 1 and α 1 satisfying (3.7). Then ϕ(z)A and

z ( 1 z ) α 1 ϕ(z)= z ( 1 z ) α 1 (zU).
(3.9)

In view of α 1 α 1 >0, it follows from (3.9) that

z ( 1 z ) α 1 ϕ(z) S ( 1 α 1 2 ) S ( 1 α 1 2 ) ,

which implies that

ϕ(z)R ( 1 α 1 2 ) .
(3.10)

Also, for f(z)A, (3.9) leads to

{ H m l ( α 1 ) f ( z ) = ϕ ( z ) H m l ( α 1 ) f ( z ) , z ( H m l ( α 1 ) f ( z ) ) = ϕ ( z ) ( z ( H m l ( α 1 ) f ( z ) ) ) .
(3.11)

Let f(z) S m l ( α 1 ,α,μ). Then, by Theorems 1 and 2, we have

{ F ( z ) = z ( H m l ( α 1 ) f ( z ) ) H m l ( α 1 ) f ( z ) e i α ( h ( z ) cos α + i sin α ) , H m l ( α 1 ) f ( z ) S ( 1 2 ( 1 μ cos α ) ) S ( 1 α 1 2 )
(3.12)

for h(z) given by (2.2) and α 1 1+ μ cos α . Since the function e i α (h(z)cosα+isinα) is convex univalent in U, from (3.10), (3.11), (3.12) and the lemma, we deduce that

z ( H m l ( α 1 ) f ( z ) ) H m l ( α 1 ) f ( z ) = ϕ ( z ) ( z ( H m l ( α 1 ) f ( z ) ) ) ϕ ( z ) H m l ( α 1 ) f ( z ) = ϕ ( z ) ( F ( z ) H m l ( α 1 ) f ( z ) ϕ ( z ) H m l ( α 1 ) f ( z ) e i α ( h ( z ) cos α + i sin α ) .

Therefore, by Theorem 1, f(z) S m l ( α 1 ,α,μ) and (3.8) is proved. □

Theorem 5 Let f(z) S m l ( α 1 ,α,μ) and g(z)R( 1 2 (1 μ cos α )). Then

(fg)(z) S m l ( α 1 ,α,μ).
(3.13)

Proof Let f(z) S m l ( α 1 ,α,μ). According to Theorems 1 and 2, we have

F(z)= z ( H m l ( α 1 ) f ( z ) ) H m l ( α 1 ) f ( z ) e i α ( h ( z ) cos α + i sin α )

and

H m l ( α 1 )f(z) S ( 1 2 ( 1 μ cos α ) ) .
(3.14)

If we put ϕ(z)=(fg)(z), then

z ( H m l ( α 1 ) ϕ ( z ) ) H m l ( α 1 ) ϕ ( z ) = g ( z ) ( z ( H m l ( α 1 ) f ( z ) ) ) g ( z ) H m l ( α 1 ) f ( z ) = g ( z ) ( F ( z ) H m l ( α 1 ) f ( z ) ) g ( z ) H m l ( α 1 ) f ( z ) ( z U )
(3.15)

for g(z)R( 1 2 (1 μ cos α )).

In view of (3.14) and (3.15), an application of the lemma leads to

z ( H m l ( α 1 ) ϕ ( z ) ) H m l ( α 1 ) ϕ ( z ) e i α ( h ( z ) cos α + i sin α ) .

Consequently, by applying Theorem 1, ϕ(z) S m l ( α 1 ,α,μ) and the proof of (3.13) is completed. □

Note that R( 1 2 )= S ( 1 2 ). Since R( β 1 )R( β 2 ) for β 1 β 2 1 (see [[15], p.49], we have

K=R(0)R ( 1 2 ( 1 μ cos α ) ) (cosα<μcosα).

Thus Theorem 5 yields the following.

Corollary 4

  1. (i)

    If f(z) S m l ( α 1 ,α,0) and g(z) S ( 1 2 ), then

    (fg)(z) S m l ( α 1 ,α,0).
  2. (ii)

    If f(z) S m l ( α 1 ,α,μ) with cosα<μcosα and g(z)K, then

    (fg)(z) S m l ( α 1 ,α,μ).

Theorem 6 The function f(z)A defined by

H m l ( α 1 )f(z)= z ( 1 b z ) 2 cos α e i α (zU)
(3.16)

belongs to the class S m l ( α 1 ,α,μ), where

α j C{0,1,2,}(j=1,2,,l),

b is complex and

|b| { cos α + μ 3 cos α μ ( cos α < μ < cos α 3 ) , μ cos α + μ ( μ cos α 3 ) .
(3.17)

The result is sharp, that is, |b| cannot be increased.

Proof For f(z)A defined by (3.16) and

α j C{0,1,2,}(j=1,2,,l),

we easily have

e i α z ( H m l ( α 1 ) f ( z ) ) H m l ( α 1 ) f ( z ) = 1 + b z 1 b z cosα+isinα(zU).
(3.18)

Hence, by Theorem 1, f(z) S m l ( α 1 ,α,μ) if and only if

1 + b z 1 b z h(z),
(3.19)

where h(z) is given by (2.2). Clearly, (3.19) is equivalent to

{ w : | w 1 + | b | 2 1 | b | 2 | < 2 | b | 1 | b | 2 } h(U)
(3.20)

for 0<|b|<1. Let

δ=min { | w 1 + | b | 2 1 | b | 2 | : w h ( U ) } ,
(3.21)

where h(U) is given by (3.2). Then we have

{ δ = min { g ( u ) : u 1 2 ( 1 μ cos α ) } , g ( u ) = ( u 1 + | b | 2 1 | b | 2 ) 2 + 2 ( 1 + μ cos α ) ( u cos α μ 2 cos α ) ( u cos α μ 2 cos α ) .
(3.22)

Note that

1 2 ( 1 μ cos α ) < 1 + | b | 2 1 | b | 2 , g (u)=2 ( u ( 2 | b | 2 1 | b | 2 μ cos α ) ) .
(3.23)

(i)If

cosα<μ< cos α 3 and|b|= cos α + μ 3 cos α μ ,
(3.24)

then

1 | b | 1 + | b | = 1 2 ( 1 μ cos α ) , | b | 2 = ( cos α + μ 3 cos α μ ) 2 < cos α + μ 5 cos α + μ ,

and so

2 | b | 2 1 | b | 2 μ cos α < 1 2 ( 1 μ cos α ) .
(3.25)

From (3.22), (3.23) and (3.25), we have g (u)>0(u 1 2 (1 μ cos α )), and hence

δ= g ( 1 2 ( 1 μ cos α ) ) = 1 + | b | 2 1 | b | 2 1 2 ( 1 μ cos α ) = 2 | b | 1 | b | 2 .
(3.26)

(ii) If

cosα<μ< cos α 3 and cos α + μ 3 cos α μ <|b|< cos α + μ 5 cos α + μ ,
(3.27)

then

1 | b | 1 + | b | < 1 2 ( 1 μ cos α ) and g (u)>0 ( u 1 2 ( 1 μ cos α ) ) .

Hence

δ= g ( 1 2 ( 1 μ cos α ) ) < 2 | b | 1 | b | 2 .
(3.28)

(iii) If

μ cos α 3 and|b|= μ cos α + μ ,
(3.29)

then

| b | 2 = μ cos α + μ cos α + μ 5 cos α + μ ,

and so

2 | b | 2 1 | b | 2 μ cos α 1 2 ( 1 μ cos α ) .

Thus g(u) attains its minimum value at

u 0 = 2 | b | 2 1 | b | 2 μ cos α

and

δ= g ( u 0 ) =2|b| cos α + μ cos α ( 1 | b | 2 ) = 2 | b | 1 | b | 2 .
(3.30)

(iv) If

μ cos α 3 and μ cos α + μ <|b|<1,
(3.31)

then from (iii) we easily have

δ= g ( u 0 ) < 2 | b | 1 | b | 2 .
(3.32)

Now, by virtue of (3.19), (3.20), (3.21), and (i)-(iv), we have proved the theorem. □

Theorem 7 Let

f(z)=z+ n = 2 a n z n S m l ( α 1 ,α,μ),

where

α j C{0,1,2,3,}(j=1,2,,l).

Then

| a 2 | 8 ( cos α + μ ) π 2 | β 1 β m α 1 α l |.
(3.33)

The result is sharp.

Proof It can be easily verified that, for zU,

z ( H m l ( α 1 ) f ( z ) ) H m l ( α 1 ) f ( z ) =1+ α 1 α l β 1 β m a 2 z+
(3.34)

and

h ( z ) = 1 + 8 z π 2 ( 1 + μ cos α ) ( n = 1 z n 1 2 n 1 ) 2 = 1 + 8 π 2 ( 1 + μ cos α ) n = 1 ( 1 n ν = 0 n 1 1 2 ν + 1 ) z n = 1 + 8 π 2 ( 1 + μ cos α ) z + ,
(3.35)

where

f(z)=z+ a 2 z 2 + S m l ( α 1 ,α,μ)

and h(z) is given by (2.2). From (3.34), (3.35) and Theorem 1, we obtain

π 2 e i α 8 ( cos α + μ ) ( z ( H m l ( α 1 ) f ( z ) ) H m l ( α 1 ) f ( z ) 1 ) = π 2 e i α α 1 α l 8 ( cos α + μ ) β 1 β m a 2 z + π 2 cos α 8 ( cos α + μ ) ( h ( z ) 1 ) K .
(3.36)

It is the well-known Rogosinski result (cf. [[22], p.195]) that if

g(z)= n = 1 b n z n

is analytic in U, g(z)ϕ(z) and ϕ(z)K, then | b n |1 (nN). Hence (3.33) follows from (3.36) at once. □

The estimate (3.33) is sharp since equality is attained for the function f 0 (z) defined by (2.11).

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Acknowledgements

Dedicated to Professor Hari M Srivastava.

This work was partially supported by the National Natural Science Foundation of China (Grant No. 11171045).

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Correspondence to Neng Xu.

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Xu, N., Yang, DG. & Sokół, J. A class of analytic functions involving in the Dziok-Srivastava operator. J Inequal Appl 2013, 138 (2013). https://doi.org/10.1186/1029-242X-2013-138

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Keywords

  • uniformly convex functions
  • convex univalent functions
  • starlike functions
  • α-spirallike functions
  • convolution
  • Dziok-Srivastava operator
  • subordination