Open Access

Properties of multivalent functions associated with the integral operator defined by the hypergeometric function

Journal of Inequalities and Applications20132013:458

https://doi.org/10.1186/1029-242X-2013-458

Received: 15 February 2013

Accepted: 6 September 2013

Published: 7 November 2013

Abstract

In this paper, we introduce a new class of multivalent functions by using a generalized integral operator defined by the hypergeometric function. Some properties such as inclusion, radius problem and integral preserving are considered.

MSC:30C45, 30C50.

Keywords

Bazilevic functionsmultivalent functionsintegral operatorhypergeometric functions

1 Introduction and preliminaries

Let A p denote the class of functions f ( z ) of the form
f ( z ) = z p + n = p + 1 a n z n ( p N = { 1 , 2 , 3 , } ) ,
(1.1)
which are analytic in the open unit disc E. Also A 1 = A , the usual class of analytic functions defined in the open unit disc E = { z : | z | < 1 } . A function f A p is a p-valent starlike function of order ρ if and only if
Re z f ( z ) f ( z ) > ρ , 0 ρ < p , z E .
This class of functions is denoted by S p ( ρ ) . It is noted that S p ( 0 ) = S p . Let f ( z ) and g ( z ) be analytic in E, we say f ( z ) is subordinate to g ( z ) , written f g or f ( z ) g ( z ) if there exists a Schwarz function w ( z ) , w ( 0 ) = 0 and | w ( z ) | < 1 in E, then f ( z ) = g ( w ( z ) ) . In particular, if g is univalent in E, then we have the following equivalence
f ( z ) g ( z ) f ( 0 ) = g ( 0 ) and f ( E ) g ( E ) .
For any two analytic functions f ( z ) and g ( z ) with
f ( z ) = n = 0 b n z n + 1 and g ( z ) = n = 0 c n z n + 1 , z E ,
the convolution (the Hadamard product) is given by
( f g ) ( z ) = n = 0 b n c n z n + 1 , z E .
A function f A is said to be in the class, denoted by SD ( k , δ ) ( 0 δ < 1 ), if and only if
Re { z f ( z ) f ( z ) } > k | z f ( z ) f ( z ) 1 | + δ , k 0 , z E .
(1.2)
Similarly, a function f A is said to be in the class, denoted by CD ( k , δ ) of k-uniformly convex of order δ ( 0 δ < 1 ), if
Re { 1 + z f ( z ) f ( z ) } > k | z f ( z ) f ( z ) | + δ , k 0 , z E .
(1.3)
Geometric interpretation The functions f SD ( k , δ ) and f CD ( k , δ ) if and only if z f ( z ) f ( z ) and z f ( z ) f ( z ) + 1 , respectively, take all the values in the conic domain Ω k , δ defined by
Ω k , δ = { u + i v : u > k ( u 1 ) 2 + v 2 + δ }
with p ( z ) = z f ( z ) f ( z ) or p ( z ) = z f ( z ) f ( z ) + 1 and considering the functions which map E onto the conic domain Ω k , δ such that 1 Ω k , δ . One may rewrite the conditions (1.2) or (1.3) in the form
p ( z ) q k , δ ( z ) .
The function q k , δ ( z ) plays the role of extremal for these classes and is given by
q k , δ ( z ) = { 1 + ( 1 2 δ ) z 1 z , k = 0 , 1 + 2 δ γ π 2 ( log 1 + z 1 z ) 2 , k = 1 , 1 + 2 δ 1 k 2 sinh 2 [ ( 2 π arccos k ) arctan h z ] , 0 < k < 1 , 1 + δ k 2 1 sin ( π 2 R ( t ) 0 u ( z ) t 1 1 x 2 1 ( t x ) 2 d x ) + δ k 2 1 , k > 1 .
(1.4)
For f ( z ) in A p , the operator D μ + p 1 : A p A p is defined by
D μ + p 1 f ( z ) = z p ( 1 z ) μ + p f ( z ) ( μ > p ) ,
or equivalently
D μ + p 1 f ( z ) = z p ( z μ 1 f ( z ) ) μ + p 1 ( μ + p 1 ) ! ,
(1.5)
where μ is any integer greater than −p. If f ( z ) is given by (1.1), then it follows that
D μ + p 1 f ( z ) = z p + n = p + 1 ( μ + n 1 ) ! ( n p ) ! ( μ + p 1 ) ! a n z n .
The symbol D μ + p 1 when p = 1 , was introduced by Ruscheweyh [1] and D μ + p 1 is called the ( μ + p 1 ) th order Ruscheweyh derivative. We now introduce a function ( z 2 p F 1 ( a , b , c ; z ) ) 1 given by
( z p 2 F 1 ( a , b , c ; z ) ) ( z p 2 F 1 ( a , b , c ; z ) ) 1 = z p ( 1 z ) μ + p ( μ > p ) ,
and the following linear operator
I μ , p ( a , b , c ) f ( z ) = ( z p 2 F 1 ( a , b , c ; z ) ) 1 f ( z ) ,
(1.6)

where a, b, c are real or complex numbers other than 0 , 1 , 2 ,  , μ > p , z E and f ( z ) A p . This operator was recently introduced in [2]. In particular, for p = 1 , this operator is studied by Noor [3]. For b = 1 , this operator reduces to the well-known Cho-Kwon-Srivastava operator I μ , p ( a , c ) , which was studied by Cho et al. [4], and for μ = 1 , b = c , a = n + p , see [5]. For a = n + p , b = c = 1 , this operator was investigated by Liu [6] and Liu and Noor [7].

Simple computations yield
I μ , p ( a , b , c ) f ( z ) = z p + n = p + 1 ( c ) n ( μ + p ) n ( a ) n ( b ) n a n z n .
From (1.6), we note that
I μ , 1 ( a , b , c ) f ( z ) = I μ ( a , b , c ) f ( z ) (see [3]) , I 0 , p ( a , p , a ) f ( z ) = f ( z ) , I 1 , p ( a , p , a ) f ( z ) = z f ( z ) p .
Also, it can be easily seen that
z ( I μ , p ( a , b , c ) f ( z ) ) = ( μ + p ) I μ + 1 , p ( a , b , c ) f ( z ) μ I μ , p ( a , b , c ) f ( z ) ,
(1.7)
and
z ( I μ , p ( a + 1 , b , c ) f ( z ) ) = a I μ , p ( a , b , c ) f ( z ) ( a p ) I μ , p ( a , b , c ) f ( z ) .

We define the following class of multivalent analytic functions by using the operator I μ , p ( a , b , c ) f ( z ) above.

Definition 1.1 Let f A p for p N . Then f U B μ , p α ( a , b , c ; γ , k , δ ) for a , b , c R Z 0 , μ > p , α > 0 , k 0 , 0 δ < 1 and γ > 0 if and only if
( 1 γ ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α + γ I μ + 1 , p ( a , b , c ) f ( z ) I μ + 1 , p ( a , b , c ) g ( z ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α 1 q k , δ ( z ) ,
(1.8)
where g A p is such that
q ( z ) = I μ + 1 , p ( a , b , c ) g ( z ) I μ , p ( a , b , c ) g ( z ) P ( ρ ) , ρ = k + δ k + 1 , z E .
(1.9)
Furthermore, for different choices of parameters being involved, we obtain many other well-known subclasses of the class A p and A as special cases.
  1. (i)

    a = c , b = 1 , k = 0 , μ = m N 0 , we have B m , p α ( γ , δ ) studied in [8].

     
  2. (ii)
    a = c = b = p = γ = 1 , k = μ = 0 , g ( z ) = z , the class U B μ , p α ( a , b , c ; γ , k , δ ) reduces to the class
    B α ( δ ) = { f A ( 1 ) : z f ( z ) f ( z ) ( f ( z ) z ) α P ( δ ) }

    studied in [9].

     
  3. (iii)

    a = c = b = p = γ = 1 , k = μ = 0 , g ( z ) = z , the U B μ , p α ( a , b , c ; γ , k , δ ) reduces to B ( α ) is the class of Bazilevich functions investigated by Singh [10].

     
  4. (iv)
    a = c = b = p = α = 1 , γ = 0 , k = μ = 0 , g ( z ) = z , the class U B μ , p α ( a , b , c ; γ , k , δ ) reduces to the class
    B δ = { f A ( 1 ) : f ( z ) z P ( δ ) } ,
     

the class studied by Chen [11].

Let f A p and F η , p : A p A p be defined by
F η , p ( z ) = ( η + p ) z η 0 z t η 1 f ( t ) d t , η > p .
(1.10)

We need the following lemmas which will be used in our main results.

Lemma 1.2 [12]

Let u = u 1 + i u 2 and v = v 1 + i v 2 , and let ψ : D C 2 C be a complex-valued function satisfying the conditions:
  1. (i)

    ψ ( u , v ) is continuous in a domain D C 2 ,

     
  2. (ii)

    ( 1 , 0 ) D and ψ ( 1 , 0 ) > 0 ,

     
  3. (iii)

    Re ψ ( i u 2 , v 1 ) 0 , whenever ( i u 2 , v 1 ) D and v 1 1 2 ( 1 + u 2 2 ) .

     

If h ( z ) = 1 + c 1 z + c 2 z 2 + is analytic in E such that ( h , z h ) D and Re ψ ( h ( z ) , z h ( z ) ) > 0 for z E , then Re h ( z ) > 0 .

Lemma 1.3 [13]

Let h be convex in the unit disc E, and let A 0 . Suppose that B ( z ) is analytic in E with Re B ( z ) A . If g is analytic in E and g ( 0 ) = h ( 0 ) . Then
A z 2 g ( z ) + B ( z ) z g ( z ) + g ( z ) h ( z ) implies that g ( z ) h ( z ) .

Lemma 1.4 [14]

Let F be analytic and convex in E. If f , g A p and f , g F . Then
σ f + ( 1 σ ) g F , 0 σ 1 .

Lemma 1.5 [15]

Let h be convex in E with h ( 0 ) = a and β C such that Re β 0 . If p H [ a , n ] and
p ( z ) + z p ( z ) β h ( z ) ,
then p ( z ) q ( z ) h ( z ) , where
q ( z ) = β n z β / n 0 z h ( t ) t β / n 1 d t

and q ( z ) is the best dominant.

2 Main results

Theorem 2.1 Let f U B μ , p α ( a , b , c ; γ , k , δ ) for a , b , c R Z 0 , μ > p , p N , α > 0 , k 0 , 0 δ < 1 and γ > 0 . Then f U B μ , p α ( a , b , c ; 0 , k , δ ) .

Proof Consider
h ( z ) = ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α ,
(2.1)
where h is analytic in E with h ( 0 ) = 1 , and g A p satisfies condition (1.9). Differentiating (2.1) logarithmically and using (1.7), we have
z h ( z ) h ( z ) = α ( μ + p ) { ( I μ + 1 , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) f ( z ) I μ + 1 , p ( a , b , c ) g ( z ) I μ , p ( a , b , c ) g ( z ) ) } .
Using (2.1) and simplifying, we obtain
h ( z ) + γ z h ( z ) α ( μ + p ) q ( z ) = ( 1 γ ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α + γ I μ + 1 , p ( a , b , c ) f ( z ) I μ + 1 , p ( a , b , c ) g ( z ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α 1 .
Since f U B μ , p α ( a , b , c ; γ , k , δ ) , therefore, we can write
h ( z ) + γ z h ( z ) α ( μ + p ) q ( z ) q k , δ ( z ) , z E .

Now using Lemma 1.3 for A = 0 and B ( z ) = γ α ( μ + p ) q ( z ) with Re q ( z ) > 0 , we have Re B ( z ) 0 , therefore, h ( z ) q k , δ ( z ) . Hence f U B μ , p α ( a , b , c ; 0 , k , δ ) . □

Theorem 2.2 Let f U B μ , p α ( a , b , c ; γ , 0 , δ ) for a , b , c R Z 0 , μ > p , p N . Then f U B μ , p α ( a , b , c ; 0 , 0 , δ 1 ) , where
δ 1 = 2 α δ ( p + μ ) | q ( z ) | 2 + γ ρ 2 α ( p + μ ) | q ( z ) | 2 + γ ρ .
Proof Consider
h ( z ) = 1 ( 1 δ 1 ) { ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α δ 1 } ,
(2.2)
where h is analytic in E with h ( 0 ) = 1 , and g A p satisfies condition (1.9). Differentiating (2.2), we have
( 1 δ 1 ) α h ( z ) = ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α 1 × { ( I μ , p ( a , b , c ) f ( z ) ) I μ , p ( a , b , c ) g ( z ) I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ( I μ , p ( a , b , c ) g ( z ) ) I μ , p ( a , b , c ) g ( z ) } .
Using (1.7) and simplifying, we obtain
( 1 γ ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α + γ I μ + 1 , p ( a , b , c ) f ( z ) I μ + 1 , p ( a , b , c ) g ( z ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α 1 = ( 1 δ 1 ) h ( z ) + δ 1 + γ ( 1 δ 1 ) z h ( z ) α ( p + μ ) q ( z ) .
Since f U B μ , p α ( a , b , c ; γ , 0 , δ ) , therefore we have
( 1 δ 1 ) h ( z ) + δ 1 + γ ( 1 δ 1 ) z h ( z ) α ( p + μ ) q ( z ) 1 + ( 1 2 δ ) z 1 z , 0 δ < 1 , z E .
This implies that
1 1 δ { ( 1 δ 1 ) h ( z ) + δ 1 δ + γ ( 1 δ 1 ) z h ( z ) α ( p + μ ) q ( z ) } q 0 , 0 ( E ) = P .
To obtain our desired result, we show that h P , for z E . Let u = u 1 + i u 2 , v = v 1 + i v 2 , and let Ψ : D C 2 C be a complex-valued function such that u = h ( z ) , v = z h ( z ) . Then
Ψ ( u , v ) = ( 1 δ 1 ) u + δ 1 δ + γ ( 1 δ 1 ) v α ( p + μ ) q ( z ) .
The first two conditions of Lemma 1.2 are easily verified. To verify the third condition, we consider
Re Ψ ( i u 2 , v 1 ) = Re { ( 1 δ 1 ) i u 2 + δ 1 δ + γ ( 1 δ 1 ) v 1 α ( p + μ ) q ( z ) } = δ 1 δ + Re γ ( 1 δ 1 ) v 1 α ( p + μ ) q ( z ) δ 1 δ Re γ ( 1 δ 1 ) ( 1 + u 2 2 ) q ( z ) ¯ 2 α ( p + μ ) | q ( z ) | 2 δ 1 δ γ ( 1 δ 1 ) ( 1 + u 2 2 ) ρ 2 α ( p + μ ) | q ( z ) | 2 = A + B u 2 C ,

where A = 2 α ( p + μ ) ( δ 1 δ ) | q ( z ) | 2 γ ρ ( 1 δ 1 ) , B = γ ρ ( 1 δ 1 ) 0 if 0 δ 1 < 1 and C = 2 α ( p + μ ) | q ( z ) | 2 > 0 . From the relation δ 1 = 2 α δ ( p + μ ) | q ( z ) | 2 + γ ρ 2 α ( p + μ ) | q ( z ) | 2 + γ ρ , we have A 0 . This implies that Re Ψ ( i u 2 , v 1 ) 0 . Using Lemma 1.2, we have h P for z E . This completes the proof. □

Theorem 2.3 Let a , b , c R Z 0 , μ > p , p N , α > 0 , k 0 and 0 δ < 1 . Then
U B μ , p α ( a , b , c ; γ 2 , k , δ ) U B μ , p α ( a , b , c ; γ 1 , k , δ ) , 0 γ 1 < γ 2 , z E .
Proof Since f U B μ , p α ( a , b , c ; γ 2 , k , δ ) , therefore, we have
( 1 γ 2 ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α + γ 2 I μ + 1 , p ( a , b , c ) f ( z ) I μ + 1 , p ( a , b , c ) g ( z ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α 1 = h 1 ( z ) q k , δ ( z ) ,
(2.3)
where g A p satisfies condition (1.9). From Theorem 2.1, we write
( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α = h 2 ( z ) q k , δ ( z ) , z E .
(2.4)
Now, for γ 1 0 , we obtain
( 1 γ 1 ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α + γ 1 I μ + 1 , p ( a , b , c ) f ( z ) I μ + 1 , p ( a , b , c ) g ( z ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α 1 = ( 1 γ 1 γ 2 ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α + γ 1 γ 2 { ( 1 γ 2 ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α + γ 2 I μ + 1 , p ( a , b , c ) f ( z ) I μ + 1 , p ( a , b , c ) g ( z ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α 1 } = γ 1 γ 2 h 1 ( z ) + ( 1 γ 1 γ 2 ) h 2 ( z ) .
Using the convexity of the class of the function q k , δ ( z ) and Lemma 1.4, we write
γ 1 γ 2 h 1 ( z ) + ( 1 γ 1 γ 2 ) h 2 ( z ) q k , δ ( z ) , z E ,

where h 1 and h 2 are given by (2.3) and (2.4), respectively. This implies that f U B μ , p α ( a , b , c ; γ 1 , k , δ ) . Hence the proof of the theorem is completed. □

Theorem 2.4 Let f U B μ , p 1 ( a , b , c ; γ , k , δ ) , a , b , c R Z 0 , μ > p , p N , α > 0 , k 0 , γ 1 and 0 δ < 1 . Then f U B μ + 1 , p 1 ( a , b , c ; 1 , k , δ ) .

Proof Since f U B μ , p 1 ( a , b , c ; γ , k , δ ) , therefore, we have
( 1 γ ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) + γ I μ + 1 , p ( a , b , c ) f ( z ) I μ + 1 , p ( a , b , c ) g ( z ) q k , δ ( z ) .
Now, consider
γ I μ + 1 , p ( a , b , c ) f ( z ) I μ + 1 , p ( a , b , c ) g ( z ) = ( 1 γ ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) + γ I μ + 1 , p ( a , b , c ) f ( z ) I μ + 1 , p ( a , b , c ) g ( z ) + ( γ 1 ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) .
This implies that
I μ + 1 , p ( a , b , c ) f ( z ) I μ + 1 , p ( a , b , c ) g ( z ) = 1 γ { ( 1 γ ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) + γ I μ + 1 , p ( a , b , c ) f ( z ) I μ + 1 , p ( a , b , c ) g ( z ) } + ( 1 1 γ ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) .

Using Theorem 2.1, Lemma 1.4 and the convexity of q k , δ ( z ) , we have the required result. □

Now, using the operators I μ , p ( a , b , c ) and F η , p defined by (1.7) and (1.10), respectively, we have
z ( I p μ ( a , b , c ) F η , p ( f ) ( z ) ) = ( p + η ) I p μ ( a , b , c ) f ( z ) η I p μ ( a , b , c ) F η , p ( f ) ( z ) , η > p .
(2.5)
Theorem 2.5 Let f A p and F η , p be given by (1.10). If
( 1 γ ) I μ , p ( a , b , c ) F η , p ( f ) ( z ) z p + γ I μ , p ( a , b , c ) ( f ( z ) ) z p q k , δ ( z ) , z E ,
(2.6)
with a , b , c R Z 0 , μ , η > p , p N , γ > 0 , then
I μ , p ( a , b , c ) F η , p ( f ( z ) ) z p h ( z ) q k , δ ( z ) , z E ,
where
h ( z ) = p + η γ z ( p + η ) / γ 0 z q k , δ ( z ) t ( p + η ) / γ 1 d t .
Proof Let
I μ , p ( a , b , c ) F η , p ( f ( z ) ) z p = h 1 ( z ) , z E ,
where h 1 is analytic in E with h 1 ( 0 ) = 1 . Then
z ( I μ , p ( a , b , c ) F η , p ( f ( z ) ) ) = p z p h 1 ( z ) + z p + 1 h 1 ( z ) .
Using (2.5), we have
γ ( p + η ) I μ , p ( a , b , c ) f ( z ) z p γ η I μ , p ( a , b , c ) F η , p ( f ) ( z ) z p = p γ h 1 ( z ) + γ z h 1 ( z ) .
Thus,
( 1 γ ) I μ , p ( a , b , c ) F η , p ( f ( z ) ) z p + γ I μ , p ( a , b , c ) ( f ( z ) ) z p = h 1 ( z ) + γ z h 1 ( z ) p + η .
(2.7)
From (2.7), it follows that
h 1 ( z ) + γ z h 1 ( z ) p + η q k , δ ( z ) , z E .

Using Lemma 1.5, for β 1 = p + η γ , n = 1 and a = 1 , we obtain h 1 ( z ) h ( z ) q k , δ ( z ) . That is, I μ , p ( a , b , c ) F η , p ( f ( z ) ) z p q k , δ ( z ) . □

Theorem 2.6 Let f U B μ , p α ( a , b , c ; 0 , 0 , δ ) for a , b , c R Z 0 , μ > p , p N , α , γ > 0 , 0 δ < 1 . Then f U B μ , p α ( a , b , c ; γ , 0 , δ ) , for | z | < r 0 , where
r 0 = α ( p + μ ) + γ γ 2 + 2 α γ ( p + μ ) α ( p + μ ) .
(2.8)
Proof Let f U B μ , p α ( a , b , c ; 0 , 0 , δ ) . Then we have
( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α = ( 1 δ ) h ( z ) + δ ,
(2.9)
where g A p satisfies the condition
q ( z ) = I μ + 1 , p ( a , b , c ) g ( z ) I μ , p ( a , b , c ) g ( z ) P , z E
and h P . Differentiating (2.9) and then using (1.7), we obtain
γ I μ + 1 , p ( a , b , c ) f ( z ) I μ + 1 , p ( a , b , c ) g ( z ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α 1 γ ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α = γ ( p δ ) z h ( z ) α ( p + μ ) q ( z ) .
This implies that
1 1 δ { ( 1 γ ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α + γ I μ + 1 , p ( a , b , c ) f ( z ) I μ + 1 , p ( a , b , c ) g ( z ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α 1 δ } = h ( z ) + γ z h ( z ) α ( p + μ ) q ( z ) , z E .
(2.10)
Now, using the well-known distortion result for class P, we have
| z h ( z ) | 2 r Re h ( z ) 1 r 2 and Re h ( z ) 1 r 1 + r , | z | < r < 1 , z E .
Thus, due to the applications of these inequalities, we have
Re ( h ( z ) + γ z h ( z ) α ( p + μ ) q ( z ) ) Re h ( z ) γ | z h ( z ) | α ( p + μ ) | q ( z ) | Re h ( z ) ( 1 2 γ r α ( p + μ ) ( 1 r ) 2 ) = Re h ( z ) ( α ( p + μ ) ( 1 r ) 2 2 γ r α ( p + μ ) ( 1 r ) 2 ) .

For | z | < r 0 , where r 0 is given in (2.8), the inequality above is positive. Sharpness of the result follows by taking h ( z ) = 1 + z 1 z . Hence from (2.10), we have the required result. □

Declarations

Authors’ Affiliations

(1)
Department of Mathematics, GC University
(2)
Department of Mathematics, COMSATS Institute of Information Technology

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