Research Article A Subclass of Analytic Functions Related to k-uniformly Convex and Starlike Functions

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1 Hindawi Function Spaces Volume 2017, Article ID , 7 pages Research Article A Subclass of Analytic Functions Related to k-uniformly Convex and Starlike Functions Saqib Hussain, 1 Akhter Rasheed, 2 Muhammad Asad Zaighum, 2 and Maslina Darus 3 1 Department of Mathematics, COMSATS Institute of Information Technology, Abbottabad, Pakistan 2 Department of Mathematics & Statistics, Riphah International University, Islamabad, Pakistan 3 School of Mathematical Sciences, Faculty of Science and Technology, Universiti Kebangsaan Malaysia, 43600Bangi,Selangor,Malaysia Correspondence should be addressed to Saqib Hussain; saqib_math@yahoo.com Received 26 January 2017; Accepted 20 April 2017; Published 23 May 2017 Academic Editor: Maria Alessandra Ragusa Copyright 2017 Saqib Hussain et al. 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. We investigate some subclasses of k-uniformly convex and k-uniformly starlike functions in open unit disc, which is generalization of class of convex and starlike functions. Some coefficient inequalities, a distortion theorem, the radii of close-to-convexity, and starlikeness and convexity for these classes of functions are studied. The behavior of these classes under a certain modified convolutionoperatorisalsodiscussed. 1. Introduction Let A be the class of all analytic functions f in open unit disc Δ={z: z <1},normalizedbyf(0) = 0 and f (0) = 1. Thus, any f Ahas the following Maclaurin s series: f (z) =z+ a n z n. (1) Afunctionf is said to be univalent if it never takes same value twice. By S we mean the subclass of A which is composed of univalent functions. By ST and CV we mean the well-known subclasses of A that are, respectively, starlike and convex. In 1991, Goodman [1, 2] introduced the classes UCV and UST of uniformly convex and uniformly starlike functions, respectively. A function f S is uniformly convex if f(z) maps every circular arc γ contained in Δ with center ζ Δ onto a convex arc. The function f Sis uniformly starlike if f(z) maps every circular arc γ contained in Δ with center ζ Δontoastarlike arc with respect to f(ζ). Amoreuseful representation of UCV and UST was given in [3 6] as f UCV f A, Re ( (zf (z)) zf (z) f )> (z) f, (z) f UST f A, Re ( zf (z) f (z) )> zf (z) 1, f (z) z Δ. z Δ. In 1999, for k 0, Kanas and Wisniowska [7] introduced the class k UCV and k UST as Re ( (zf (z)) f (z) f k UCV zf k UST f A, zf (z) )>k f, (z) z Δ. Observe that 0 UCV CV, 0 UST UST and 1 UCV UCV, 1 UST UST. For fixed k 0, these classes have a nice geometrical representation; for detail see [7 9]. (2) (3)

2 2 Function Spaces A lot of authors obtain very useful properties of UCV and UST and their generalization in several direction; for example, see [1, 2, 7, 8, 10, 11] and reference cited therein. For (0 α<1), in [4] (see also [12]), Ronning introduced thefollowingtwo importantsubclasses k UST(α) and k UCV(α) as Re { zf (z) f (z) f k UST (α) zf (z) α}>k 1, f (z) f k UCV (α) zf k UST (α). z Δ. Recently in [13] El-Ashwah et al. introduced two important subclass k UCV(α, β) and k UST(α, β) of kuniformly convex starlike functions as f k UCV (α, β) Re { (zf (z)) { f α } (zf (z)) (z) } >k f β, (z) { } Re { zf (z) f (z) f k UST (α, β) zf (z) α}>k β, f (z) where (0 α<β 1)and k(1 β) < 1 α. Let f j (j=1,2,...)be defined by z Δ, z Δ, (4) (5) f j (z) =z+ a n,j z n, a n,j 0; j N; (6) then the modified Hadmard product of f 1 (z) and f 2 (z) is defined by (f 1 f 2 ) (z) =z a n,1 a n,2 z n. (7) We denote T by subclass of S consisting of functions having all negative coefficients in their Maclaurin s series expansions, so any f T has a series of the form: f (z) =z a n zn, z Δ. (8) Let V η be the class of functions f S given in (1) for which arg(a n ) = π + (n 1)η, n 2.NotethatV 0 = T [11]. In recent years, more and more researchers are interested in the above defined classes (see [9, 11, 14 22]). In this paper, by taking inspiration from the above cited paper, we introduce some new subclasses of analytic functions and obtain some interesting results. Definition 1. For (0 α<β 1), 0 δ<1, k(1 β) < 1 α, and 0 λ<1,afunctionf S is in class k U(α,β,λ,δ)if and only if Re { (1 δ) zf +δ(zf + (1+2λ) z 2 f +λz 3 f ) (1 δ) f+δ(zf +λz 2 f ) α} (1 δ) zf +δ(zf + (1+2λ) z 2 f +λz 3 f ) k (1 δ) f+δ(zf +λz 2 f ) β, z Δ. Also (9) k VU η (α,β,λ,δ)=k U (α,β,λ,δ) V η. (10) It is worth mentioning that, for special values of parameters, these classes were extensively studied by many authors; here we mention few of them. (1) k U(α,β,λ,1)=k U(λ,β,α)[21]. (2) k VU 0 (α,β,λ,1)=k VU η (λ, β, α) [21]. (3) 0 VU 0 (α,1,0,1)=cv(α) [11]. (4) k VU 0 (α,1,0,1)=k UCV(α) [23]. (5) 1 U(α,1,0,1)=UCV(α) [4]. (6) k U(α,β,0,0)=k UST(α, β) [13]. Throughout the paper 1 α β 1, 0 λ < 1, 1 α>k(1 β), and z Δ, unless otherwise stated. 2. Main Results Theorem 2. Afunctionf(z) given by (1) is in class k U(α,β,λ,δ)if where [ ] a n, Π n = (n+δn(n 1)(1+λn)), Ω n =1 δ+δn(1+λ(n 1)). (11) (12) Proof. It is sufficient to prove that inequality (9) holds. As we know Re (w) >k w β +α iff Re ((1 + ke iθ )w βke iθ ) α; (13)

3 Function Spaces 3 then inequality (9) can be written as From (18) and (19), we have A (z) + (1 α) B (z) A (z) (1+α) B (z) Re ((1 + ke iθ ) 2[(1 α) k(1 β)] z (1 δ) zf +δ(zf + (1+2λ) z 2 f +λz 3 f ) (1 δ) f+δ(zf +λz 2 f ) (14) 2[(Π n βω n )k+(π n αω n )] a n z n (20) βke iθ ) α. =2[[(1 α) k(1 β)] z This is, where A (z) A (z) Re ( ) α, (15) B (z) =(1+ke iθ )(zf +δ((1+2λ) z 2 f +λz 3 f )) βke iθ ((1 δ) f+δ(zf +λz 2 f )), B (z) = (1 δ) f+δ(zf +λz 2 f ); then we have Now (16) A (z) + (1 α) B (z) A (z) (1+α) B (z) 0. (17) A (z) + (1 α) B (z) = and also (1 β) ke iθ + (2 α) z [(βω n Π n )ke iθ (1 α) Ω n Π n n z n ((1 β)k+(2 α)) z [(βω n Π n )k+(1 α) Ω n +Π n ] a n z n, A (z) (1+α) B (z) = ((1 β) ke iθ α)z + [(Π n βω n )ke iθ (1+α) Ω n +Π n n z n ((1 β)k+α) z + [(Π n βω n )k (1+α) Ω n +Π n ] a n zn. (18) (19) [ ] a n z n ]. The last expression is bounded below by 0 if [ ] a n, which completes the proof. (21) In the next theorem, we prove that condition (11) is also necessary for function f k U(α,β,λ,δ). Theorem 3. Let f(z) be given by (1) and in V η ;thenf k VU η (α,β,λ,δ)if and only if [ ] a n. (22) Proof. From Theorem 2, we need only to show that f k VU η (α,β,λ,δ) satisfies inequality (22). If f k VU η (α,β,λ,δ), then by definition, we have Re ( (1 α) + (Π n αω n )a n z n 1 1+ Ω na n z n 1 ) (1 β) + k (Π n βω n )a n z n 1 1+ Ω. na n z n 1 (23) Since f is function of form (1) with the argument property giveninclassv η and letting z=re θ in the above inequality, we have (1 α) (Π n αω n ) a n rn 1 1 Ω n a n rn 1 k( (1 β) (Π n βω n ) a n rn 1 1 Ω n a ) n rn 1 for r 1, and (24) leads to require inequality [ ] a n. (24) (25)

4 4 Function Spaces The function f n,η (z) =z ()ei(1 n)η z n, is extremal function. 0 η 2π, n 2 (26) Corollary 4. Let f(z) given in (1) be in class k VU η (α, β, λ, δ).then a n, n 2. (27) Inequality (27) is attained for the function given in (26). Theorem 5. Let the function f(z) given in (1) be in class k VU η (α,β,λ,δ).then for z <r=1 f (z) r f (z) r+ r 2, r 2. (28) The results in (28) are attained for the function given in (26) for z=±r. Proof. As we know from Theorem 3 [ ] a n Theorem 6. Let the function f(z) given in (1) be in class k VU η (α,β,λ,δ).then for z <r=1 1 2() r f (z) 1+ 2() r. Proof. For f(z) given by (1), we have f (z) 1 n a n z n 1 1 rn a n, f (z) 1+ n a n z n 1 1+rn a n. In view of Theorem 3, [ ] 2 or, equivalently, n a n [ ] a n, (32) (33) (34) n a n 2() [ ]. (35) A substitution from (35) into (33) yields inequality (32), which is required. Theorem 7. Let f k VU η (α,β,λ,δ) with argument property as in class V η.definef j (z) = z and As similarly [ ] a n. f (z) z r f (z) z + r+ a n z n r r 2 a n r 2, This completes the proof. a n z n r+r 2 a n r 2. (29) (30) (31) f n,η =z ei(1 n)η [ ] zn, (36) where 0 η 2π, n 2. Then function f(z) is in class k VU η (α,β,λ,δ) if and only if it can be expressed as f (z) = n=1 where μ n 0 (n 1) and n=1 μ n =1. Proof. Assume that f (z) =μ 1 f 1 (z) = μ n f n,η, (37) + μ n [z ei(1 n)η z n ] n=1 μ n z [ ei(1 n)η ]μ n z n. (38)

5 Function Spaces 5 Then it follows that e i(1 n)η μ n [ ] = μ n [] (1 μ 1 ) [], (39) by Theorem 3, f k VU η (α,β,λ,δ). Conversely, assume that the function f(z) defined by (1) belongs to class k VU η (α,β,λ,δ),andthen Set a n, n 2. (40) μ n = a n, n 2, (41) and μ 1 =1 μ n, n 2.Thenf(z) = n=1 μ nf n,η and this completes the proof. Theorem 8. Let f k VU η (α,β,λ,δ).thenf(z) is close to convex of order σ(0 σ < 1) in the disc z < r 1,where r 1 = inf [ (1 σ)(π 1/(n 1) n (1+k) (kβ+α)ω n ) ], n() n 2. (42) Proof. As f V η,wheref is close to convex of order σ, we have as f (z) 1 <1 σ, (43) f (z) 1 n a n z n 1, (44) this expression is less than 1 σif n 1 σ a n z n 1 <1. (45) By the fact that f k VU η (α,β,λ,δ)if and only if inequality (43) is true if [ () n 1; (46) n 1 σ zn 1 ; (47) () or, equivalently, z n 1 =[ (1 σ)( ) ]. (48) n() Theorem 9. Let f k VU η (α,β,λ,δ).thenf(z) is close to convex of order σ(0 σ<1)in the disc z < r 2,where r 2 = inf [ (1 σ)(π 1/(n 1) n (1+k) (kβ+α)ω n ) ], (n σ)() n 2. Proof. As f V η and f is starlike of order σ,thenwehave as zf (z) f (z) zf (z) f (z) 1 The last expression is less than 1 σif (49) 1 <1 σ, (50) (n 1) a n z n 1 1 a. (51) n z n 1 n σ 1 σ a n z n 1 <1. (52) Using the fact that f k VU η (α,β,λ,δ)if and only if (50) is true if [ () n 1 (53) n σ 1 σ z n 1 <. (54) () Or equivalently z n 1 = (1 σ)( ), (55) (n σ)() which is required. Theorem 10. Let f k VU η (α,β,λ,δ).thenf(z) is convex of order σ(0 σ<1)in the disc z < r 3,where r 3 = inf [ (1 σ)[π 1/(n 1) n (1+k) (kβ+α)ω n ] ], n (n σ)(1 α k (1 β)) n 2. (56) Proof. Using the fact that f is convex if and only if zf is starlike, following the lines of Theorem 9, we have the required results. Theorem 11. Let f j (z) (j = 1, 2,...) given by (6) be in class k VU η (α,β,λ,δ).then(f 1 f 2 ) k VU η (φ 1,λ,δ),for

6 6 Function Spaces φ 1 = ( ) 2 (Π n (1+k) kβω n )() 2 ( ) 2 Ω n () 2. (57) Proof. We need to prove the largest φ 1 such that ( ) a (1 φ 1 k(1 β)) n,1 a n,2 1. (58) From Theorem 3, we have [ () n,1 1, [ () n,2 1. By Cauchy-Schwarz inequality, we have (59) For n 2 a n,1 a n,2 ( )(1 φ 1 k(1 β)) (62) (Π n (1+k) (kβ+φ 1 )Ω n )(). Note that a n,1 a n,2 (1 α k (1 β)) ( ). (63) [ ] a (1 α k (1 β)) n,1 a n,2 1. (60) Thus, it is sufficient to show [ Π n (1+k) (kβ+φ 1 )Ω n (1 φ 1 k(1 β)) n,1 a n,2 [ ] a (1 α k (1 β)) n,1 a n,2, n 2. (61) We need to show () ( ) ( )(1 φ 1 k(1 β)) (Π n (1+k) (kβ+φ 1 )Ω n )(), or equivalently (64) φ 1 ( ) 2 (Π n (1+k) kβω n )() 2 ( ) 2 Ω n () 2 =ω(n). (65) ω(n) is an increasing function for n 2.For in (65), φ 1 ω(2) = ( ) 2 (Π 2 (1+k) kβω 2 )() 2 ( ) 2 Ω 2 () 2, (66) which proves main assertion of Theorem 11. Conflicts of Interest The authors declare that they have no conflicts of interest. Authors Contributions All authors jointly work on the results, and they read and approved the final manuscript. Acknowledgments The work here is supported by MOHE Grant: FRGS/1/2016/ STG06/UKM/01/1. References [1] A. W. Goodman, On uniformly convex functions, Annales Polonici Mathematici,vol.56,no.1,pp.87 92,1991. [2] A. W. Goodman, On uniformly starlike functions, Journal of Mathematical Analysis and Applications, vol.155,no.2,pp , 1991.

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