Subclasses of Analytic Functions with Respect. to Symmetric and Conjugate Points

Transcription

1 Theoretical Mathematics & Applications, vol.3, no.3, 2013, 1-11 ISSN: (print), (online) Scienpress Ltd, 2013 Subclasses of Analytic Functions with Respect to Symmetric and Conjugate Points B. Srutha Keerthi 1 and S. Chinthamani 2 Abstract In this paper, we introduce new subclasses of analytic functions with respect to other points. The coefficient estimates for these classes are obtained. Mathematics Subject Classification: 30C45 Keywords: Analytic; univalent; starlike with respect to symmetric points; coefficient estimates 1 Introduction Let U be the class of functions which are analytic and univalent in the open unit disc D z : z < 1} given by w(z) z n b k z k 1 Department of Mathematics, School of Advanced Sciences, VIT Chennai Campus, Chennai , India. 2 Research Scholar, Anna University. Article Info: Received : December 14, Revised : March 20, 2013 Published online : September 1, 2013

2 2 Subclasses of Analytic Functions and satisfying the conditions w(0) 0, w(z) < 1, z D. Let S denote the class of functions f which are analytic and univalent in D of the form Also let S S f(z) z a n z n. (1) n2 be the subclass of S consisting of functions given by (1) satisfying } zf (z) Re > 0, z D. f(z) f( z) These functions are called starlike with respect to symmetric points and were introduced by Sakaguchi in Ashwah and Thomas in 2 introduced another class namely the class SC consisting of functions starlike with respect to conjugate points. Let SC be the subclass of S consisting of functions given by (1) and satisfying the condition } zf (z) Re > 0, z D. f(z) f(z) Motivated by S S, many authors discussed the following class C S of functions convex with respect to symmetric points and its subclasses. Let C S be the subclass of S consisting of functions given by (1) and satisfying the condition } (zf (z)) Re (f(z) f( z)) > 0, z D. In terms of subordination, Goel and Mehrok in 1982 introduced a subclass of SS, denoted by S S (A, B). Let SS (A, B) be the class of functions of the form (1) and satisfying the condition 2zf (z) f(z) f( z) 1 Az, 1 B < A 1, z D. 1 Bz Also let SC (A, B) be the class of functions of the form (1) and satisfying the condition 2zf (z) (f(z) f(z)) 1 Az, 1 B < A 1, z D. 1 Bz

3 B. Srutha Keerthi and S. Chinthamani 3 Let C S (A, B) be the class of functions of the form (1) and satisfying the condition 2(zf (z)) 1 Az, 1 B < A 1, z D. (f(z) f( z)) 1 Bz Also let C C (A, B) be the class of functions of the form (1) and satisfying the condition 2(zf (z)) 1 Az, 1 B < A 1, z D. (f(z) f(z)) 1 Bz In this paper, we introduce the class M S (ρ, µ, A, B) consisting of analytic functions f of the form (1) and satisfying 2ρµz 3 f (z) (2ρµ ρ µ)z 2 f (z) zf (z) ρµz 2 f (z) f ( z) (ρ µ)zf (z) f ( z) (1 ρ µ)f(z) f( z) 1 Az 1 Bz 1 B < A 1, 0 µ ρ 1, z D. We note that M S (0, 0, A, B) S S (A, B) and M S(1, 0, A, B) C S (A, B). Also introduce the class M C (ρ, µ, A, B) consisting of analytic functions f of the form (1) and satisfying 2ρµz 3 f (z) (2ρµ ρ µ)z 2 f (z) zf (z) ρµz 2 (f(z) f(z)) (ρ µ)z(f(z) f(z)) (1 ρ µ)(f(z) f(z)) 1 Az 1 Bz 1 B < A 1, 0 µ ρ 1, z D. Note that M C (0, 0, A, B) S C (A, B) and M C(1, 0, A, B) C C (A, B). By definition of subordination it follows that f M S (ρ, µ, A, B) if and only if 2ρµz 3 f (z) (2ρµ ρ µ)z 2 f (z) zf (z) ρµz 2 f (z) f ( z) (ρ µ)zf (z) f ( z) (1 ρ µ)f(z) f( z) w U and that f M C (ρ, µ, A, B) if and only if 2ρµz 3 f (z) (2ρµ ρ µ)z 2 f (z) zf (z) 1 Aw(z) p(z), (2) 1 Bw(z) ρµz 2 (f(z) f(z)) (ρ µ)z(f(z) f(z)) (1 ρ µ)(f(z) f(z)) 1 Aw(z) p(z), w U (3) 1 Bw(z)

4 4 Subclasses of Analytic Functions where p(z) 1 p n z n (4) We study the classes M S (ρ, µ, A, B) and M C (ρ, µ, A, B), the coefficient estimates for functions belonging to these classes are obtained. We also need the following lemma for proving our results. Lemma If p(z) is given by (4) then p n A B, n 1, 2, 3,.... (5) n1 2 Main Results In this section, we give the coefficient inequalities for the classes M S (ρ, µ, A, B) and M C (ρ, µ, A, B). Theorem 2.1. Let f M S (ρ, µ, A, B). Then for n 1, 0 µ ρ 1 a 2n1 n 1 (A B 2j) (6) 2 n n!(2n 1)(2n)ρµ (2n 1)(ρ µ) 1 n 1 (A B 2j) (7) 2 n n!(2n 1)(2n)ρµ (2n)(ρ µ) 1 Proof. From (2) and (4), we have ρµ6a3 z 3 24a 4 z 4 (2n)(2n 1)(2n 2)a 2n z 2n (2ρµ ρ µ)2a 2 z 2 6a 3 z 3 (2n 1)(2n)a 2n z 2n z 2a 2 z 2 3a 3 z 3 2na 2n z 2n } ρµ 6a 3 z 3 20a 5 z 5 (2n 1)(2n 2)a 2n 1 z 2n 1 (2n 1)(2n)a 2n1 z 2n1 (ρ µ)z 3a 3 z 3 (2n 1)a 2n 1 z 2n 1 (2n 1)a 2n1 z 2n1 (1 ρ µ)z a 3 z 3 a 2n 1 z 2n 1 a 2n1 z 2n1 } 1 p1 z p 2 z 2 p 2n 1 z 2n 1 p 2n z 2n }

5 B. Srutha Keerthi and S. Chinthamani 5 Equating the coefficients of like powers of z, we have 2a 2 2ρµ (ρ µ) 1 p 1, 2a 3 6ρµ 2(ρ µ) 1 p 2 (8) } 4a 4 12ρµ 3(ρ µ) 1 p 3 a 3 p 1 6ρµ 2(ρ µ) 1 4a 5 20ρµ 4(ρ µ) 1 p 4 a 3 p 2 6ρµ 2(ρ µ) 1 (9) 2na 2n (2n 1)(2n)ρµ (2n 1)(ρ µ) 1 p 2n 1 p 2n 3 a 3 6ρµ 2(ρ µ) 1 p 1 a 2n 1 (2n 1)(2n 2)ρµ (2n 2)(ρ µ) 1 (10) (2n)a 2n1 (2n 1)(2n)ρµ 2n(ρ µ) 1 p 2n p 2n 2 a 3 6ρµ 2(ρ µ) 1 p 2 a 2n 1 (2n 1)(2n 2)ρµ (2n 2)(ρ µ) 1 (11) Using Lemma 1.1 and (8), we get a 2 22ρµ (ρ µ) 1, a 3 26ρµ 2(ρ µ) 1 (12) Again by applying (11) and followed by Lemma 1.1, we get from (9) a 4 (A B 2) (2)(4)12ρµ 3(ρ µ) 1, a 5 (A B 2) (2)(4)20ρµ 4(ρ µ) 1 It follows that (6) and (7) hold for n 1, 2. We prove (6) using induction. Equation (10) in conjunction with Lemma 1.1 yield 2n(2n 1)(2n)ρµ (2n 1)(ρ µ) 1 1 (2k 1)(2k)ρµ 2k(ρ µ) 1 a 2k1 (13) We assume that (6) holds for k 3, 4,..., (n 1). Then from (13), we obtain 2n(2n 1)(2n)ρµ (2n 1)(ρ µ) 1 k 1 1 (A B 2j) 2 k k! (14)

6 6 Subclasses of Analytic Functions In order to complete the proof, it is sufficient to show that 2m(2m 1)(2m)ρµ (2m 1)(ρ µ) 1 m 1 k 1 1 (A B 2j) 2 k k! m 3, 4,..., n. m 1 (A B 2j),(15) 2 m m!(2m 1)(2m)ρµ (2m 1)(ρ µ) 1 (15) is valid for m 3. (14) Let us suppose that (15) is true for all m, 3 < m (n 1). Then from 2n(2n 1)(2n)ρµ (2n 1)(ρ µ) 1 k 1 1 (A B 2j) 2 k k! (n 1) n 2(n 1)(2n 1)(2n)ρµ (2n 1)(ρ µ) 1 n 2 k 1 1 (A B 2j) 2 k k! 2n(2n 1)(2n)ρµ (2n 1)(ρ µ) 1 n 2 (A B 2j) (n 1) n n 2 2 n 1 (n 1)! 2 n 1 (n 1)!(2n 1)(2n)ρµ (2n 1)(ρ µ) 1 (A B 2j) 2n(2n 1)(2n)ρµ (2n 1)(ρ µ) 1 n 2 (A B 2j) 2 n 1 (n 1)!

7 B. Srutha Keerthi and S. Chinthamani 7 2 n 1 (n 1)!(2n 1)(2n)ρµ (2n 1)(ρ µ) 1 n 2 (A B 2j)(A B 2(n 1)) n 1 (A B 2j) 2 n n!(2n 1)(2n)ρµ (2n 1)(ρ µ) 1 Thus (15) holds for m n and hence (6) follows. Similarly we can prove (7). Theorem 2.2. Let f M C (ρ, µ, A, B). Then for n 1, 0 µ ρ 1 2n 2 (16) (2n 1)!(2n 1)(2n)ρµ (2n 1)(ρ µ) 1 a 2n1 2n 1 (17) (2n)!(2n 1)(2n)ρµ (2n)(ρ µ) 1 Proof. From (3) and (4), we have ρµ6a3 z 3 24a 4 z 4 (2n)(2n 1)(2n 2)a 2n z 2n (2ρµ ρ µ)2a 2 z 2 6a 3 z 3 (2n 1)(2n)a 2n z 2n z 2a 2 z 2 3a 3 z 3 2na 2n z 2n } ρµ2a 2 z 2 6a 3 z 3 (2n 1)(2n)a 2n z 2n (ρ µ)z 2a 2 z 2 2na 2n z 2n (1 ρ µ)z a 2 z 2 a 2n z 2n } 1 p1 z p 2 z 2 p 2n z 2n } Equating the coefficients of like powers of z, we have a 2 (2ρµ(ρ µ)1) p 1, 2a 3 (6ρµ2(ρ µ)1) p 2 a 2 p 1 (2ρµ(ρ µ)1) (18) 3a 4 (12ρµ3(ρ µ)1) p 2 a 2 p 2 (2ρµ(ρ µ)1)a 3 p 1 (6ρµ2(ρ µ)1) (19)

8 8 Subclasses of Analytic Functions 4a 5 (20ρµ 4(ρ µ) 1) p 4 a 2 p 3 (2ρµ (ρ µ) 1) a 3 p 2 (6ρµ 2(ρ µ) 1) (2n 1)a 2n ((2n 1)(2n)ρµ (2n 1)(ρ µ) 1) p 2n 1 a 2 p 2n 2 (2ρµ (ρ µ) 1) a 4 p 1 (12ρµ 3(ρ µ) 1) (20) a 2n 1 p 1 ((2n 2)(2n 1)ρµ (2n 2)(ρ µ) 1) (2n)a 2n1 ((2n 1)(2n)ρµ (2n)(ρ µ) 1) By using Lemma 1.1 and (18), we get a 2 p 2n a 2 p 2n 1 (2ρµ (ρ µ) 1) (21) a 2n p 1 ((2n)(2n 1)ρµ (2n 1)(ρ µ) 1) (22) 2ρµ (ρ µ) 1, a 3 (A B 1) 2(6ρµ 2(ρ µ) 1) (23) Again by applying (23) and followed by Lemma 5, we get from (19) and (20), we have (A B 1)(A B 2) a 4 (2)(3)(12ρµ 3(ρ µ) 1) (A B 1)(A B 2)(A B 3) a 5 (2)(3)(4)(20ρµ 4(ρ µ) 1) It follows that (16) hold for n 1, 2. We now prove (16) using induction. Equation (21) in conjunction with Lemma 1.1 yield (2n 1)(2n 1)(2n)ρµ (2n 1)(ρ µ) 1 1 a 2k a 2k1 (24) We assume that (16) holds for k 3, 4,..., (n 1). Then from (24), we obtain (2n 1)(2n 1)(2n)ρµ (2n 1)(ρ µ) 1 2k 2 A B 1 (2k 1)! (2k)! 2k 1 (25)

9 B. Srutha Keerthi and S. Chinthamani 9 In order to complete the proof, it is sufficient to show that (2m 1)(2m 1)(2m)ρµ (2m 1)(ρ µ) 1 m 1 2k 2 A B m 1 1 (2k 1)! (2k)! (2m 1)!((2m 1)(2m)ρµ (2m 1)(ρ µ) 1) 2m 2 2k 1, (26) m 3, 4, 5,..., n. (3.21) is valid for m 3. Let us suppose that (3.21) is true for all m, 3 < m (n 1). Then from (25) (2n 1)(2n 1)(2n)ρµ (2n 1)(ρ µ) 1 2k 2 A B 1 (2k 1)! (2k)! 2k 1 (2n 3) (2n 1) (2(n 1) 1)((2n 1)(2n)ρµ (2n 1)(ρ µ) 1) n 2 2k 2 A B n 2 2k 1 1 (2k 1)! (2k)! (2n 1)((2n 1)(2n)ρµ (2n 1)(ρ µ) 1) 2n 4 A B (2(n 1) 1)! 2n 3 A B (2n 1)((2n 1)(2n)ρµ (2n 1)(ρ µ) 1) (2(n 1))! 2n 4 (2n 3) (2n 1) (2(n 1) 1)!((2n 1)(2n)ρµ (2n 1)(ρ µ) 1) (2n 1)((2n 1)(2n)ρµ (2n 1)(ρ µ) 1) 2n 4 A B (2(n 1) 1)!

10 10 Subclasses of Analytic Functions (2n 1)(2(n 1) 1)!((2n 1)(2n)ρµ (2n 1)(ρ µ) 1) 2n 3 A B (2n 1)((2n 1)(2n)ρµ (2n 1)(ρ µ) 1) (2(n 1))! 2n 3 2n 2 (2n 1)!((2n 1)(2n)ρµ (2n 1)(ρ µ) 1) Thus (26) holds for m n and hence (16) follows. Similarly we can prove (17). On specializing the values of ρ, µ in Theorem 2.1 and 2.2, we get the following. Remark 2.3. In Theorem 2.1, if we set µ 0 and ρ 0, we get starlike functions with respect to symmetric points and if we set µ 0 and ρ 1, we get convex functions with respect to symmetric points. Remark 2.4. In Theorem 2.2, if we set µ 0 and ρ 0, we get starlike functions with respect to conjugate points and if we set µ 0 and ρ 1, we get convex functions with respect to conjugate points. For other values of µ and ρ, the transition is smooth. Acknowledgements. The present investigation was supported by Science and Engineering Research Board, New Delhi project no: SR/S4/MS:716/10 with titled On Classes of Certain Analytic Univalent Functions and Sakaguchi Type Functions. References 1 R.N. Das and P. Singh, On subclass of Schlicht mapping, Indian J. Pure Appl. Math., 8, (1977),

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