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Journal of Applied Mathematics
Volume 2012 (2012), Article ID 982321, 10 pages
doi:10.1155/2012/982321
Research Article

On Generalized Bazilevic Functions Related with Conic Regions

Department of Mathematics, COMSATS Institute of Information Technology, Park Road, Islamabad, Pakistan

Received 12 March 2012; Accepted 18 March 2012

Academic Editor: Yonghong Yao

Copyright © 2012 Khalida Inayat Noor and Kamran Yousaf. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We define and study some generalized classes of Bazilevic functions associated with convex domains. These convex domains are formed by conic regions which are included in the right half plane. Such results as inclusion relationships and integral-preserving properties are proved. Some interesting special cases of the main results are also pointed out.

1. Introduction

Let 𝐴 denote the class of analytic functions 𝑓 ( 𝑧 ) defined in the unit disc 𝐸 = { 𝑧 | 𝑧 | < 1 } and satisfying the conditions 𝑓 ( 0 ) = 0 , 𝑓 ( 0 ) = 1 . Let 𝑆 denote the subclass of 𝐴 consisting of univalent functions in 𝐸 , and let 𝑆 and 𝐶 be the subclasses of 𝑆 which contains, respectively, star-like and convex in Bazilevič [1] introduced the class 𝐵 ( 𝛼 , 𝛽 , , 𝑔 ) as follows.

Let 𝑓 𝐴 . Then, 𝑓 𝐵 ( 𝛼 , 𝛽 , , 𝑔 ) , 𝛼 , 𝛽 real and 𝛼 > 0 if 𝑓 ( 𝑧 ) = ( 𝛼 + 𝑖 𝛽 ) 𝑧 0 ( 𝑧 ) 𝑔 𝛼 ( 𝑡 ) 𝑡 𝑖 𝛽 1 𝑑 𝑡 1 / ( 𝛼 + 𝑖 𝛽 ) , ( 1 . 1 ) for some 𝑔 𝑆 and R e ( 𝑧 ) > 0 , 𝑧 𝐸 .

The powers appearing in (1.1) are meant as principle values. The functions 𝑓 in the class 𝐵 ( 𝛼 , 𝛽 , , 𝑔 ) are shown to be analytic and univalent, see [1]. 𝐵 ( 𝛼 , 𝛽 , , 𝑔 ) is the largest known subclass of univalent functions defined by an explicit formula and contains many of the heavily researched subclasses of 𝑆 . We note the following:(i) 𝐵 ( 1 , 0 , 1 , 𝑔 ) = 𝐶 , (ii) 𝐵 ( 1 , 0 , 𝑧 𝑔 / 𝑔 , 𝑔 ) = 𝑆 , (iii) 𝐵 ( 1 , 0 , , 𝑔 ) = 𝐾 , where 𝐾 is the class of close-to-convex functions introduced by Kaplan [2],(iv) 𝐵 ( c o s 𝛾 , s i n 𝛾 , c o s ( 𝑧 𝑔 / ( 𝑔 + 𝑖 s i n 𝛾 ) ) , 𝑔 ) is the class of 𝛾 -spiral like functions which are univalent for | 𝛾 | < 𝜋 / 2 .

For analytic functions 𝑓 ( 𝑧 ) = 𝑛 = 0 𝑎 𝑛 𝑧 𝑛 and 𝑔 ( 𝑧 ) = 𝑛 = 0 𝑏 𝑛 𝑧 𝑛 , by 𝑓 𝑔 we denote the Hadamard product (convolution) of 𝑓 and 𝑔 , defined by ( 𝑓 𝑔 ) ( 𝑧 ) = 𝑛 = 0 𝑎 𝑛 𝑏 𝑛 𝑧 𝑛 . ( 1 . 2 ) For 𝑘 [ 0 , ) , the conic domain Ω 𝑘 is defined in [3] as follows: Ω 𝑘 = 𝑢 + 𝑖 𝑣 𝑢 > 𝑘 ( 𝑢 1 ) 2 + 𝑣 2 . ( 1 . 3 ) For fixed 𝑘 , Ω 𝑘 represents the conic region bounded successively by the imaginary axis ( 𝑘 = 0 ) , the right branch of hyperbola ( 0 < 𝑘 < 1 ) , a parabola ( 𝑘 = 1 ) and an ellipse ( 𝑘 > 1 ) .

The following univalent functions, defined by 𝑝 𝑘 ( 𝑧 ) with 𝑝 𝑘 ( 0 ) = 1 and 𝑝 𝑘 ( 0 ) > 0 , map the unit disc 𝐸 onto Ω 𝑘 𝑝 𝑘 ( 𝑧 ) = 1 + 𝑧 2 1 𝑧 , ( 𝑘 = 0 ) , 1 + 𝜋 2 l o g 1 + 𝑧 1 𝑧 2 2 , ( 𝑘 = 1 ) , 1 + 1 𝑘 2 s i n h 2 𝐴 ( 𝑘 ) a r c t a n h 𝑧 2 , ( 0 < 𝑘 < 1 ) , 1 + 𝑘 2 1 s i n 2 𝜋 𝐹 2 𝐾 ( 𝑡 ) 𝑧 𝑡 , 𝑡 , ( 𝑘 > 1 ) , ( 1 . 4 ) where 𝐴 ( 𝑘 ) = ( 2 / 𝜋 ) a r c c o s 𝑘 , 𝐹 ( 𝑤 , 𝑡 ) is the Jacobi elliptic integral of the first kind: 𝐹 ( 𝑤 , 𝑡 ) = 𝑤 0 𝑑 𝑥 1 𝑥 2 1 𝑡 2 𝑥 2 , ( 1 . 5 ) and 𝑡 ( 0 , 1 ) is chosen such that 𝑘 = c o s h ( 𝜋 𝐾 ( 𝑡 ) / 2 𝐾 ( 𝑡 ) ) , where 𝐾 ( 𝑡 ) is the complete elliptic integral of the first kind, 𝐾 ( 𝑡 ) = 𝐹 ( 1 , 𝑡 ) , 𝐾 ( 𝑡 ) = 𝐾 ( 1 𝑡 2 ) .

It is known that 𝑝 𝑘 ( 𝑧 ) are continuous as regards to 𝑘 and have real coefficients for 𝑘 [ 0 , ) .

Let 𝑃 ( 𝑝 𝑘 ) be the subclass of the class 𝑃 of Caratheodory functions 𝑝 ( 𝑧 ) , analytic in 𝐸 with 𝑝 ( 0 ) = 1 and such that 𝑝 ( 𝑧 ) is subordinate to 𝑝 𝑘 ( 𝑧 ) , written as 𝑝 ( 𝑧 ) 𝑝 𝑘 ( 𝑧 ) in 𝐸 .

We define the following.

Definition 1.1. Let ( 𝑧 ) be analytic in 𝐸 with ( 0 ) = 1 . Then, 𝑃 𝑚 ( 𝑝 𝑘 ) if and only if, for 𝑚 2 , 𝑘 [ 0 , ) , 1 , 2 𝑃 ( 𝑝 𝑘 ) we can write 𝑚 ( 𝑧 ) = 4 + 1 2 1 𝑚 ( 𝑧 ) 4 1 2 2 ( 𝑧 ) , 𝑧 𝐸 . ( 1 . 6 ) We note that 𝑃 2 ( 𝑝 𝑘 ) = 𝑃 ( 𝑝 𝑘 ) , and 𝑃 𝑚 ( 𝑝 0 ) = 𝑃 𝑚 , see [4].

Definition 1.2. Let 𝑓 𝐴 . Then, 𝑓 ( 𝑧 ) is said to belong to the class 𝑘 𝑅 𝑚 if and only if 𝑧 𝑓 / 𝑓 𝑃 𝑚 ( 𝑝 𝑘 ) for 𝑘 [ 0 , ) , 𝑚 2 , and 𝑧 𝐸 .
For 𝑚 = 2 , 𝑘 = 0 , the class 0 𝑅 2 = 𝑅 2 coincides with the class 𝑆 of starlike functions, and 0 𝑅 𝑚 = 𝑅 𝑚 consists of analytic functions with bounded radius rotation, see [5, 6]. Also 𝑘 𝑅 2 is the class 𝑆 𝑇 studied by several authors, see [7, 8].

Definition 1.3. Let 𝑓 𝐴 . Then, 𝑓 𝑘 𝐵 𝑚 ( 𝛼 , 𝛽 , , 𝑔 ) if and only if 𝑓 ( 𝑧 ) is as given by (1.1) for some 𝑔 𝑘 𝑅 2 , 𝑃 𝑚 ( 𝑝 𝑘 ) in 𝐸 with 𝑘 [ 0 , ) , 𝑚 2 , 𝛼 > 0 and 𝛽 real.
When 𝑚 = 2 and 𝑘 = 0 , we obtain the class 𝐵 ( 𝛼 , 𝛽 , , 𝑔 ) of Bazilevic functions.

We shall assume throughout, unless otherwise stated, that 𝑘 [ 0 , ) , 𝑚 2 , 𝛼 > 0 , 𝛽 real and 𝑧 𝐸 .

2. Preliminary Results

Lemma 2.1 (see [3]). Let 0 𝑘 < , and let 𝛽 0 , 𝛿 be any complex numbers with 𝛽 0 0 and R e ( 𝛽 0 𝑘 / ( 𝑘 + 1 ) + 𝛿 ) > 0 . If ( 𝑧 ) is analytic in 𝐸 , ( 0 ) = 1 and satisfies ( 𝑧 ) + 𝑧 ( 𝑧 ) 𝛽 0 ( 𝑧 ) + 𝛿 𝑝 𝑘 ( 𝑧 ) ( 2 . 1 ) and 𝑞 𝑘 ( 𝑧 ) is analytic solution of 𝑞 𝑘 ( 𝑧 ) + 𝑧 𝑞 𝑘 ( 𝑧 ) 𝛽 0 𝑞 𝑘 ( 𝑧 ) + 𝛿 = 𝑝 𝑘 ( 𝑧 ) , ( 2 . 2 ) then 𝑞 𝑘 ( 𝑧 ) is univalent, ( 𝑧 ) 𝑞 𝑘 ( 𝑧 ) 𝑝 𝑘 ( 𝑧 ) , ( 2 . 3 ) and 𝑞 𝑘 ( 𝑧 ) is the best dominant of (2.1).

Lemma 2.2 (see [9]). Let 𝑞 ( 𝑧 ) be convex in 𝐸 and 𝑗 𝐸 with R e 𝑗 ( 𝑧 ) > 0 , 𝑧 𝐸 . If 𝑝 ( 𝑧 ) is analytic in 𝐸 with 𝑝 ( 0 ) = 1 and satisfies { 𝑝 ( 𝑧 ) + 𝑗 ( 𝑧 ) 𝑧 𝑝 ( 𝑧 ) } 𝑞 ( 𝑧 ) , then 𝑝 ( 𝑧 ) 𝑞 ( 𝑧 ) .

Lemma 2.3 (see [9]). Let 𝑢 = 𝑢 1 + 𝑖 𝑢 2 , 𝑣 = 𝑣 1 + 𝑖 𝑣 2 , and let 𝜓 ( 𝑢 , 𝑣 ) be a complex-valued function satisfying the conditions(i) 𝜓 ( 𝑢 , 𝑣 ) is continuous in a domain 𝐷 2 ,(ii) ( 0 , 1 ) 𝐷 and R e 𝜓 ( 1 , 0 ) > 0 ,(iii) R e 𝜓 ( 𝑖 𝑢 2 , 𝑣 1 ) 0 , whenever ( 𝑖 𝑢 2 , 𝑣 1 ) 𝐷 and 𝑣 1 ( 1 / 2 ) ( 1 + 𝑢 2 2 ) .If ( 𝑧 ) = 1 + 𝑐 1 𝑧 + 𝑐 2 𝑧 2 + is a function analytic in 𝐸 such that ( ( 𝑧 ) , 𝑧 ( 𝑧 ) ) 𝐷 and R e 𝜓 { ( 𝑧 ) , 𝑧 ( 𝑧 ) } > 0 for 𝑧 𝐸 , then R e ( 𝑧 ) > 0 in 𝐸 .

3. Main Results

Theorem 3.1. Let ( 1 / ( 1 𝛾 ) ) { 𝑧 𝑓 ( 𝑧 ) / 𝑓 ( 𝑧 ) 𝛾 } 𝑃 𝑚 ( 𝑝 𝑘 ) , for 𝑧 𝐸 and 𝛾 [ 0 , 1 ] . Define 𝑔 ( 𝑧 ) = ( 𝑐 + 1 ) 𝑧 𝑐 𝑧 0 𝑡 𝑐 1 𝑓 𝛼 ( 𝑡 ) 𝑑 𝑡 1 / 𝛼 , 𝛼 > 0 , 𝑐 , R e 𝑐 0 . ( 3 . 1 ) Then, ( 1 / ( 1 𝛾 ) ) { 𝑧 𝑔 ( 𝑧 ) / 𝑔 ( 𝑧 ) 𝛾 } 𝑃 𝑚 ( 𝑝 𝑘 ) in 𝐸 . In particular 𝑔 𝑘 𝑅 𝑚 in 𝐸 .

Proof. Let 𝑧 𝑔 ( 𝑧 ) 𝑔 ( 𝑧 ) = ( 1 𝛾 ) 𝑝 ( 𝑧 ) + 𝛾 , ( 3 . 2 ) where 𝑝 ( 𝑧 ) is analytic in 𝐸 with 𝑝 ( 0 ) = 1 , and let 𝑚 𝑝 ( 𝑧 ) = 4 + 1 2 𝑝 1 𝑚 ( 𝑧 ) 4 1 2 𝑝 2 ( 𝑧 ) . ( 3 . 3 ) From (3.1) and (3.2), we have 𝑔 𝛼 [ ] ( 𝑧 ) = 𝛼 ( 1 𝛾 ) 𝑝 ( 𝑧 ) + 𝑐 + 𝛼 𝛾 = 𝑓 𝛼 ( 𝑧 ) . ( 3 . 4 ) Logarithmic differentiation of (3.4) and some computation yield 𝑝 ( 𝑧 ) + 𝑧 𝑝 ( 𝑧 ) = 1 𝛼 ( 1 𝛾 ) 𝑝 ( 𝑧 ) + ( 𝑐 + 𝛼 𝛾 ) 1 𝛾 𝑧 𝑓 ( 𝑧 ) 𝑓 ( 𝑧 ) 𝛾 . ( 3 . 5 ) That is 𝑝 ( 𝑧 ) + 𝑧 𝑝 ( 𝑧 ) 𝛼 ( 1 𝛾 ) 𝑝 ( 𝑧 ) + ( 𝑐 + 𝛼 𝛾 ) 𝑃 𝑚 𝑝 𝑘 i n 𝐸 . ( 3 . 6 ) Let 𝜙 𝑎 , 𝑏 ( 𝑧 ) = 𝑧 + 𝑛 = 2 𝑧 𝑛 / ( ( 𝑛 1 ) 𝑎 + 𝑏 ) . Then, 𝜙 𝑝 ( 𝑧 ) 𝑎 , 𝑏 ( 𝑧 ) 𝑧 𝑎 = 𝑝 ( 𝑧 ) + 𝑧 𝑝 ( 𝑧 ) 𝑝 ( 𝑧 ) + 𝑏 . ( 3 . 7 ) Using convolution technique (3.7) with 𝑎 = 1 / 𝛼 ( 1 𝛾 ) , 𝑏 = ( 𝑐 + 𝛼 𝛾 ) / 𝛼 ( 1 𝛾 ) , we obtain, from (3.3) and (3.6), 𝑝 𝑖 ( 𝑧 ) + 𝑧 𝑝 𝑖 ( 𝑧 ) 𝛼 ( 1 𝛾 ) 𝑝 𝑖 ( 𝑧 ) + ( 𝑐 + 𝛼 𝛾 ) 𝑝 𝑘 ( 𝑧 ) i n 𝐸 , 𝑖 = 1 , 2 . ( 3 . 8 ) Since R e { ( 𝛼 ( 1 𝛾 ) 𝑘 / ( 𝑘 + 1 ) ) + 𝑐 + 𝛼 𝛾 } 0 , we apply Lemma 2.1 with 𝛽 0 = 𝛼 ( 1 𝛾 ) , 𝛿 = 𝑐 + 𝛼 𝛾 to obtain 𝑝 𝑖 ( 𝑧 ) 𝑞 𝑘 ( 𝑧 ) 𝑝 𝑘 ( 𝑧 ) , where 𝑞 𝑘 ( 𝑧 ) is the best dominant and is given as 𝑞 𝑘 𝛽 ( 𝑧 ) = 0 1 0 𝑡 𝛽 0 + 𝛿 1 e x p 𝑧 𝑡 𝑧 𝑝 𝑘 ( 𝑢 ) 1 𝑢 𝑑 𝑢 𝛽 0 𝑑 𝑡 1 𝛿 𝛽 0 . ( 3 . 9 )
Consequently, 𝑝 𝑃 𝑚 ( 𝑝 𝑘 ) in 𝐸 , and this completes the result.
As a special case, we prove the following.

Corollary 3.2. Let 𝑘 = 0 and let ( 1 / ( 1 𝛾 1 ) ) { 𝑧 𝑓 ( 𝑧 ) / 𝑓 ( 𝑧 ) 𝛾 1 } 𝑃 𝑚 in 𝐸 . Then, for 𝑔 defined by (3.1), 1 / ( 1 𝛾 ) { 𝑧 𝑔 ( 𝑧 ) / 𝑔 ( 𝑧 ) 𝛾 } 𝑃 𝑚 in 𝐸 where

2 𝛾 = 2 𝑐 2 𝛼 𝛾 1 + + 1 2 𝑐 2 𝛼 𝛾 1 + 1 2 + 8 𝛼 . ( 3 . 1 0 )

Proof. We can write 𝑧 𝑓 ( 𝑧 ) = 𝑓 ( 𝑧 ) 1 𝛾 1 ( 𝑧 ) + 𝛾 1 , ( 3 . 1 1 ) where 𝑃 𝑚 in 𝐸 .
Now proceeding as before, we have, with 𝑧 𝑔 ( 𝑧 ) 𝑚 𝑔 ( 𝑧 ) = ( 1 𝛾 ) 𝑝 ( 𝑧 ) + 𝛾 = 4 + 1 2 ( 1 𝛾 ) 𝑝 1 𝑚 ( 𝑧 ) + 𝛾 4 1 2 ( 1 𝛾 ) 𝑝 2 ( 𝑧 ) + 𝛾 ( 3 . 1 2 ) ( 1 𝛾 ) 𝑝 ( 𝑧 ) + 𝛾 + ( 1 𝛾 ) 𝑧 𝑝 ( 𝑧 ) = 𝛼 ( 1 𝛾 ) 𝑝 ( 𝑧 ) + ( 𝑐 + 𝛼 𝛾 ) 𝑧 𝑓 ( 𝑧 ) . 𝑓 ( 𝑧 ) ( 3 . 1 3 ) Using convolution technique together with (3.11), we obtain R e ( 1 𝛾 ) 𝑝 𝑖 ( 𝑧 ) + 𝛾 𝛾 1 + ( 1 𝛾 ) 𝑧 𝑝 𝑖 ( 𝑧 ) 𝛼 ( 1 𝛾 ) 𝑝 𝑖 ( 𝑧 ) + ( 𝑐 + 𝛼 𝛾 ) > 0 , ( 3 . 1 4 ) for 𝑖 = 1 , 2 .
We construct the functional 𝜓 ( 𝑢 , 𝑣 ) by taking 𝑢 = 𝑝 𝑖 ( 𝑧 ) , 𝑣 = 𝑧 𝑝 𝑖 ( 𝑧 ) as 𝜓 ( 𝑢 , 𝑣 ) = ( 1 𝛾 ) 𝑢 + 𝛾 𝛾 1 + ( 1 𝛾 ) 𝑣 𝛼 ( 1 𝛾 ) 𝑢 + ( 𝑐 + 𝛼 𝛾 ) . ( 3 . 1 5 ) The first two conditions of Lemma 2.3 are clearly satisfied. We verify condition (iii) as follows. R e 𝜓 𝑖 𝑢 2 , 𝑣 1 = 𝛾 𝛾 1 + R e ( 1 𝛾 ) 𝑣 1 𝑖 𝛼 ( 1 𝛾 ) 𝑢 2 , = + ( 𝑐 + 𝛼 𝛾 ) 𝛾 𝛾 1 + ( 𝑐 + 𝛼 𝛾 ) ( 1 𝛾 ) 𝑣 1 ( 𝑐 + 𝛼 𝛾 ) 2 + 𝛼 2 ( 1 𝛾 ) 2 𝑢 2 2 , 𝛾 𝛾 1 + ( 𝑐 + 𝛼 𝛾 ) ( 1 𝛾 ) 1 + 𝑢 2 2 2 ( 𝑐 + 𝛼 𝛾 ) 2 + 𝛼 2 ( 1 𝛾 ) 2 𝑢 2 2 , 𝑣 1 1 + 𝑢 2 2 2 , = 𝐴 + 𝐵 𝑢 2 2 , 2 𝐶 ( 3 . 1 6 ) where
𝐴 = 2 ( 𝛾 𝛾 1 ) ( 𝑐 + 𝛼 𝛾 ) 2 ( 1 𝛾 ) ( 𝑐 + 𝛼 𝛾 ) , 𝐵 = 2 𝛼 2 ( 𝛾 𝛾 1 ) ( 1 𝛾 ) 2 ( 1 𝛾 ) ( 𝑐 + 𝛼 𝛾 ) , 𝐶 = ( 𝑐 + 𝛼 𝛾 ) 2 + 𝛼 2 ( 1 𝛾 ) 2 𝑢 2 2 > 0 .
R e 𝜓 ( 𝑖 𝑢 2 , 𝑣 1 ) 0 if and only if 𝐴 0 , 𝐵 0 . From 𝐴 0 , we obtain 𝛾 as given by (3.10) and 𝐵 0 ensures that 𝛾 [ 0 , 1 ) .
Now proceeding as before, it follows from (3.12) that 𝑝 𝑃 𝑚 , and this proves our result.

By assigning certain permissible values to different parameters, we obtain several new and some known result.

Corollary 3.3. Let 𝑓 𝑘 𝑅 2 = 𝑘 𝑆 𝑇 . Then, it is known that 𝑓 𝑆 ( 𝛾 1 ) , 𝛾 1 = 𝑘 / ( 𝑘 + 1 ) and, form Corollary 3.2, it follows that 𝑔 𝑆 ( 𝛾 ) where 𝛾 is given by (3.10). Also a starlike function is 𝑘 -uniformly convex for | 𝑧 | < 𝑟 𝑘 , 𝑟 𝑘 = 1 2 ( 𝑘 + 1 ) + 4 𝑘 2 [ 8 ] + 6 𝑘 + 3 , s e e . ( 3 . 1 7 ) Therefore, for 𝑓 𝑘 𝑅 2 , it follows that ( 1 / ( 1 𝛾 ) ) { ( 𝑧 𝑔 ( 𝑧 ) ) / 𝑔 ( 𝑧 ) 𝛾 } 𝑝 𝑘 for | 𝑧 | < 𝑟 𝑘 , where 𝛾 is given by (3.10).
As special cases we note the following.
(i)For 𝑘 = 0 , we have 𝑟 0 = 1 / ( 2 + 3 ) and 𝑓 𝑆 ( 0 ) implies that 𝑔 𝐶 ( 𝛾 ) , with 𝛾 = 2 2 ( 𝑐 + 1 ) + 2 ( 𝑐 + 1 ) 2 + 8 𝛼 . ( 3 . 1 8 ) (ii)When 𝑘 = 1 , we have 𝛾 1 = 1 / 2 , 𝛾 = 2 / ( ( 2 𝑐 𝛼 + 1 ) + ( 2 𝑐 𝛼 + 1 ) 2 + 8 𝛼 ) and 𝑟 1 = 1 / ( 4 + 1 3 ) .

Theorem 3.4. Let 𝐹 𝑘 𝐵 𝑚 ( 𝛼 , 𝛽 , 𝑝 , 𝑓 ) , 𝑓 𝑘 𝑅 2 , 𝑝 𝑃 𝑚 ( 𝑝 𝑘 ) . Define, for R e [ 𝛼 𝑘 / ( 𝑘 + 1 ) + ( 𝑐 + 𝑖 𝛽 ) ] > 0 , 𝐺 ( 𝑧 ) = ( 𝑐 + 1 ) 𝑧 𝑐 𝑧 0 𝑡 𝑐 1 𝐹 𝛼 + 𝑖 𝛽 ( 𝑡 ) 𝑑 𝑡 1 / ( 𝛼 + 𝑖 𝛽 ) . ( 3 . 1 9 ) Then, 𝐺 𝑘 𝐵 𝑚 ( 𝛼 , 𝛽 , , 𝑔 ) in 𝐸 , where 𝑔 ( 𝑧 ) is given by (3.1), and ( 𝑧 ) is analytic in 𝐸 with ( 0 ) = 1 .

Proof. Set 𝑧 𝐺 ( 𝑧 ) 𝐺 𝛼 + 𝑖 𝛽 1 ( 𝑧 ) 𝑧 𝑖 𝛽 𝑔 𝛼 𝑚 ( 𝑧 ) = ( 𝑧 ) = 4 + 1 2 1 𝑚 ( 𝑧 ) 4 1 2 2 ( 𝑧 ) . ( 3 . 2 0 ) We note that ( 𝑧 ) is analytic in 𝐸 with ( 0 ) = 1 . From (3.20), we have 𝑧 𝑖 𝛽 𝑔 𝛼 ( 𝑧 ) 𝑧 𝛼 ( 𝑧 ) + ( 𝑧 ) 𝑧 𝑔 ( 𝑧 ) 𝑔 ( 𝑧 ) + 𝑐 + 𝑖 𝛽 = 𝑧 𝐹 ( 𝑧 ) 𝐹 𝛼 + 𝑖 𝛽 1 ( 𝑧 ) , ( 3 . 2 1 ) using (3.1), we note that 𝑓 ( 𝑧 ) 𝑔 ( 𝑧 ) 𝛼 = 𝛼 𝑧 𝑔 ( 𝑧 ) 𝑔 ( 𝑧 ) + 𝑐 + 𝑖 𝛽 . ( 3 . 2 2 ) From (3.21) and (3.22), it follows that ( 𝑧 ) + 𝑧 ( 𝑧 ) 𝛼 0 ( 𝑧 ) + ( 𝑐 + 𝑖 𝛽 ) 𝑃 𝑚 𝑝 𝑘 , ( 3 . 2 3 ) where 0 ( 𝑧 ) = 𝑧 𝑔 ( 𝑧 ) / 𝑔 ( 𝑧 ) 𝑃 ( 𝑝 𝑘 ) since 𝑔 𝑘 𝑅 2 by Theorem 3.1.
It can easily be seen that 𝑔 𝑆 ( 𝑘 / ( 𝑘 + 1 ) ) and R e { 𝛼 𝑧 𝑔 ( 𝑧 ) / 𝑔 ( 𝑧 ) + 𝑐 + 𝑖 𝛽 } > 0 .
Now, using (3.8), we can easily derive 𝑖 ( 𝑧 ) + 𝑗 ( 𝑧 ) 𝑧 𝑖 ( 𝑧 ) 𝑝 𝑘 ( 𝑧 ) i n 𝐸 , 𝑖 = 1 , 2 , ( 3 . 2 4 ) where 1 / 𝑗 ( 𝑧 ) = { 𝛼 𝑧 𝑔 ( 𝑧 ) / 𝑔 ( 𝑧 ) + 𝑐 + 𝑖 𝛽 } and R e 𝑗 ( 𝑧 ) > 0 .
Applying Lemma 2.2, it follows from (3.24) 𝑖 ( 𝑧 ) 𝑝 𝑘 ( 𝑧 ) in 𝐸 and therefore 𝑃 𝑚 ( 𝑝 𝑘 ) in 𝐸 . This completes the proof.

Theorem 3.5. Let 𝑓 ( 𝑧 ) be given by (1.1) with ( 𝑧 ) = 1 , { ( 𝛼 2 + 𝛽 2 ) 1 / 2 𝑒 𝑖 𝛾 ( 𝑧 𝑔 / 𝑔 ) } 𝑃 𝑚 ( 𝑝 𝑘 ) ( 𝛼 2 + 𝛽 2 ) 1 / 2 𝑒 𝑖 𝛾 = 𝛼 + 𝑖 𝛽 , | 𝛾 | < 𝜋 / 2 . Then, for 𝑧 𝐸 (i) 𝑒 𝑖 𝛾 ( 𝑧 𝑓 ( 𝑧 ) / 𝑓 ( 𝑧 ) ) = c o s 𝛾 ( 𝑝 ( 𝑧 ) ) + 𝑖 s i n 𝛾 , 𝑝 𝑃 𝑚 ( 𝑝 𝑘 ) , (ii)For 𝛼 + 𝑖 𝛽 = 𝑡 ( 𝛼 + 𝑖 𝛽 ) , 𝑡 1 , 𝑘 𝐵 𝑚 ( 𝛼 , 𝛽 , 1 , 𝑔 ) 𝑘 𝑡 𝐵 𝑚 𝛼 , 𝛽 , 1 , 𝑔 . ( 3 . 2 5 )

Proof. (i) From (1.1), we have 1 + 𝑧 𝑓 ( 𝑧 ) 𝑓 ( 𝑧 ) + ( 𝛼 1 + 𝑖 𝛽 ) 𝑧 𝑓 ( 𝑧 ) = 𝛼 𝑓 ( 𝑧 ) 𝑧 𝑔 ( 𝑧 ) 𝑔 ( 𝑧 ) + 𝑖 𝛽 = 𝐻 2 ( 𝑧 ) , 𝐻 2 𝑃 𝑚 𝑝 𝑘 i n 𝐸 . ( 3 . 2 6 ) Define a function 𝑝 ( 𝑧 ) analytic in 𝐸 by 𝑒 𝑖 𝛾 𝑧 𝑓 ( 𝑧 ) 𝑓 ( 𝑧 ) = c o s 𝛾 ( 𝑝 ( 𝑧 ) ) + 𝑖 s i n 𝛾 , 𝛾 = t a n 1 𝛽 𝛼 . ( 3 . 2 7 ) We can easily check that 𝑝 ( 0 ) = 1 .
Now, from (3.26) and (3.27), we have 𝑧 𝑝 ( 𝑧 ) 𝑝 ( 𝑧 ) + 𝑖 t a n 𝛾 + 𝛼 𝑝 ( 𝑧 ) + 𝑖 𝛽 𝑃 𝑚 𝑝 𝑘 i n 𝐸 . ( 3 . 2 8 ) That is 𝛼 𝑧 𝑝 ( 𝑧 ) 𝛼 𝑝 ( 𝑧 ) + 𝑖 𝛽 + 𝛼 𝑝 ( 𝑧 ) + 𝑖 𝛽 𝑃 𝑚 𝑝 𝑘 , ( 3 . 2 9 ) and, with ( 𝑧 ) = 𝛼 𝑝 ( 𝑧 ) + 𝑖 𝛽 = ( 𝑚 / 4 + 1 / 2 ) 1 ( 𝑧 ) ( 𝑚 / 4 1 / 2 ) 2 ( 𝑧 ) , we apply convolution technique used before to have 𝑖 ( 𝑧 ) + 𝑧 𝑖 ( 𝑧 ) 𝑖 ( 𝑧 ) 𝑝 𝑘 ( 𝑧 ) i n 𝐸 . ( 3 . 3 0 ) Applying Lemma, it follows that 𝑖 ( 𝑧 ) 𝑞 𝑘 ( 𝑧 ) 𝑝 𝑘 ( 𝑧 ) , 𝑧 𝐸 , ( 3 . 3 1 ) where 𝑞 𝑘 ( 𝑧 ) is the best dominant and is given by 𝑞 𝑘 ( 𝑧 ) = 1 0 e x p 0 𝑡 𝑧 𝑝 𝑘 ( 𝑢 ) 1 𝑢 𝑑 𝑢 1 . ( 3 . 3 2 ) From (3.31), we have ( 𝑧 ) = ( 𝛼 𝑝 ( 𝑧 ) + 𝑖 𝛽 ) 𝑃 𝑚 ( 𝑝 𝑘 ) in 𝐸 , and this proves part (i).
(ii) From part (i), we have 𝛼 2 + 𝛽 2 1 / 2 𝑒 𝑖 𝛾 𝑧 𝑓 ( 𝑧 ) 𝑓 ( 𝑧 ) = 𝐻 1 ( 𝑧 ) , 𝐻 1 𝑃 𝑚 𝑝 𝑘 i n 𝐸 . ( 3 . 3 3 ) Now, 1 + 𝑧 𝑓 ( 𝑧 ) 𝑓 + 𝛼 ( 𝑧 ) 1 + 𝑖 𝛽 𝑧 𝑓 ( 𝑧 ) = 𝑓 ( 𝑧 ) 1 + 𝑧 𝑓 ( 𝑧 ) 𝑓 ( 𝑧 ) + ( 𝛼 1 + 𝑖 𝛽 ) 𝑧 𝑓 ( 𝑧 ) 𝑓 𝛼 ( 𝑧 ) + ( 𝑡 1 ) 2 + 𝛽 2 1 / 2 𝑒 𝑖 𝛾 𝑧 𝑓 ( 𝑧 ) , 𝑓 ( 𝑧 ) = 𝐻 2 ( 𝑧 ) + ( 𝑡 1 ) 𝐻 1 ( 𝑧 ) , 𝐻 𝑖 𝑃 𝑚 𝑝 𝑘 1 , 𝑖 = 1 , 2 , = 𝑡 1 𝑡 𝐻 1 1 ( 𝑧 ) + 𝑡 𝐻 2 , ( 𝑧 ) = 𝑡 𝐻 , 𝑡 1 , ( 3 . 3 4 ) 𝐻 𝑃 𝑚 ( 𝑝 𝑘 ) , since 𝑃 𝑚 ( 𝑝 𝑘 ) is convex set, see [8].
Therefore, 𝑓 𝑘 𝑡 𝐵 𝑚 ( 𝛼 , 𝛽 , 1 , 𝑔 ) for 𝑧 𝐸 . This completes the proof.

As a special case, with 𝑚 = 2 , 𝑘 = 0 , we obtain a result proved in [10].

By assigning certain permissible values to the parameters 𝛼 , 𝛽 and 𝑚 , we have several other new results.

Acknowledgment

The authors are grateful to Dr. S. M. Junaid Zaidi, Rector, COMSATS Institute of Information Technology, Islamabad, Pakistan, for providing excellent research facilities and environment

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