Mathematica Aeterna, Vol. 3, 2013, no. 8, 617-624 Harmonic Mappings for which Second Dilatation is Janowski Functions Emel Yavuz Duman İstanbul Kültür University Department of Mathematics and Computer Science Ataköy Campus 34156, İstanbul, Turkey Yaşar Polatoğlu İstanbul Kültür University Department of Mathematics and Computer Science Ataköy Campus 34156, İstanbul, Turkey Yasemin Kahramaner İstanbul Commerce University Department of Mathematics Üsküdar Campus 34672, İstanbul, Turkey Maslina Darus Universiti Kebangsaan Malaysia School of Mathematical Sceinces 43600 angi, Selangor D. Ehsan Kuala Lumpur, Malaysia Abstract In the present paper we extend the fundamental property that if h(z) and g(z) are regular functions in the open unit disc D with the properties h(0) = g(0) = 0, h maps D onto many-sheeted region which is starlike with respect to the origin, and Re g (z) > 0, then Reg(z) h(z) > 0, introduced by R.J. Libera [5] to the Janowski functions and give some applications of this to the harmonic functions.
618 Emel Yavuz Duman,Yaşar Polatoğlu,Yasemin Kahramaner and Maslina Darus Mathematics Subject Classification: 30C45; 30C55 Keywords: Harmonic mappings, Janowski functions, growth theorem, distortion theorem, coefficient inequality 1 Introduction Let Ω be the class of functions φ(z) regular in the open unit disc D = {z C z < 1} and satisfy the conditions φ(0) = 0, φ(z) < 1 for all z D. For arbitrary fixed numbers A,, 1 < A 1, 1 < A we denote by P(A,) the family of functions p(z) = 1 + p 1 z + p 2 z 2 + regular in D such that p(z) is in P(A,) if and only if p(z) = 1+Aφ(z) 1+φ(z) for some function φ(z) Ω and every z D. Let S (A,) denote the family of functions s(z) = z +c 2 z 2 + regular in D such that s(z) in S (A,) if and only if z s (z) s(z) (1) = p(z) (2) for some p(z) in P(A,) and all z D. We note that every function in this family maps the unit disc univalently onto a region wihich is starlike with respect to the origin. Let s 1 (z) = z+d 2 z 2 + and s 2 (z) = z+e 2 z 2 + be analytic functions in D. If there exists φ(z) Ω such that s 1 (z) = s 2 (φ(z)), then we say that s 1 (z) is subordinate to s 2 (z) and write s 1 (z) s 2 (z) so that s 1 (D) s 2 (D). Finally, univalent harmonic functions are generalization of univalent analytic functions the point of the departure is the canonical representation f = h(z)+g(z), g(0) = 0 (3) of a harmonic function f in the unit disc D as the sum of an analytic function h(z) and conjugate of an analytic function g(z). With the convention that g(0) = 0, the representation is unique. The power series expansions of h(z) and g(z) are denoted by h(z) = a n z n, g(z) = b n z n. (4) n=0 If f is sense-preserving harmonic mapping of D onto some other region, then by Lewy s theorem its Jacobian is strictly positive, i.e, n=0 J f(z) = 2 g (z) 2 > 0.
Harmonic Mappings for which Second Dilatation is Janowski Functions 619 Equivalently, the inequality g (z) < holds for all z D. This shows, in particular, that h (0) 0 and h (0) = 1. The class of all sense-preserving harmonic mapping of the disc with a 0 = b 0 = 0, a 1 = 1 will be denoted by S H. Thus S H contains the standard class S of analytic univalent functions. Although the analytic part h(z) of a function f S H is locally univalent, it will be come apparent that it not be univalent. The class of functions f S H with g (0) = 0 will be denoted by SH 0. At the same time, we note that S H is a normal family and SH 0 is a compact normal family. For details, see [2]. Now, we consider the following class of harmonic mappings in the plane { SHPST = f = h(z)+g(z) f S H, h(z) S (A,), w(z) = g (z) } b 1 P(A,), 1 < A 1. InthispaperwewillinvestigatethesubclassS HPST. Wewillneedthefollowing theorems in the sequel. Theorem 1.1 [4] Let h(z) be an element of S (A,), then where C(r, A, ) h(z) C(r,A,), (5) C(r,A,) = {r(1+r) A re Ar, = 0. These bounds are sharp, being attained at the point z = re iθ, 0 θ 2π, by respectively, where f 0 (z;a,) = f = zf 0 (z; A, ), (6) f = zf 0 (z;a,), (7) {(1+e iθ z) A re Ae iθz, = 0. Theorem 1.2 [6] If h(z) = z +a 2 z 2 + belongs to S (A,), then a n { n 2 (A )+k k=0 n 2 k=0 k+1, 0, A k+1 0. These bound are sharp because the extremal function is {(1+z) A f (z) = re Az, = 0. Lemma 1.3 (Jack s Lemma, [3]) Let φ(z) be regular in the unit disc D, with φ(0) = 0. Then if φ(z) attains its maximum value on the circle z = r at the point z 1, one has z 1 φ (z 1 ) = kφ(z 1 ) for some k 1.
620 Emel Yavuz Duman,Yaşar Polatoğlu,Yasemin Kahramaner and Maslina Darus 2 Main Results Theorem 2.1 If h(z) and g(z) are regular in D such that h(0) = g(0) = 0. If h(z) S (A,) and g (z) g(z) b 1 P(A,), then P(A,). b 1 h(z) Proof. Since the linear transformation 1+Az maps z = r onto the disc with ( ) 1+z 1 Ar the center C(r) = 2,0 and the radius ρ(r) = (A )r and using the 1 2 r 2 1 2 r 2 subordination principle we can write so thus 1 g (z) b 1 1+Az 1+z g (z) b 1+Az 1 1+z 1 g (z) b 1 1 Ar2 1 2 r 2 (A )r 1 2 r 2 g (z) b 1(1 Ar 2 ) 1 2 r 2 b 1 (A )r. (8) 1 2 r 2 The inequality (8) shows that the values of g (z)/ are in the disc { } g (z) h D r (b 1 ) = (z) g (z) b 1(1 Ar 2 ) 1 2 r b 1 (A )r, 0, 2 1 2 r { } 2 g (z) g (z) b 1 b 1 Ar, = 0. Now we define a function φ(z) by g(z) h(z) = b 1+Aφ(z) 1 1+φ(z). Then φ(z) is analytic in D, φ(0) = 0. On the other hand since h(z) S (A,) then z h (z) h(z) D r = 1 Ar2 1 2 r (A )r, 0, 2 1 2 r 2 z h (z) 1 h(z) Ar, = 0 for all z = r < 1. Thus, for a point z 1 on the bound of this disc we have or z 1 h (z 1 ) h(z 1 ) 1 Ar2 1 2 r 2 = (A )r 1 2 r 2eiθ, 0, z 1 h (z 1 ) h(z 1 ) 1 = Areiθ, 0, h(z 1 ) z 1 h(z 1 ) = 1 2 r 2 [1 Ar 2 ]+(A )re D r, 0, iθ h(z 1 ) z 1 h(z 1 ) = 1 1+Are D r, 0, iθ
Harmonic Mappings for which Second Dilatation is Janowski Functions 621 where D r is the boundary of the disc D r. Therefore, by Jack s lemma, z 1 φ (z 1 ) = kφ(z 1 ) and k 1, we have that { 1+Aφ(z1 w(z 1 ) = g ) (z 1 ) b 1 h (z 1 ) = / w(d (1 Ar 2 )+(A )re iθ r (b 1 )), 0, 1 1+Aφ(z 1 )+Akφ(z 1 ) / w(d 1+Are iθ r (b 1 )), = 0, (9) because φ(z 1 ) = 1 and k 1. ut this is a contradiction to the condition g (z) b 1 1+Az 1+z 1+φ(z 1 ) + (A )kφ(z 1) (1+φ(z 1 )) 2 1 2 r 2 and so we have φ(z) < 1 for all z D. Lemma 2.2 Let f = h(z) + g(z) S H, then for a function defined by ω(z) = g (z) we have b and for all z = r < 1. 1 r 1 b 1 r ω(z) b 1 +r 1+ b 1 r, (10) (1 r 2 )(1 b 1 2 ) (1+ b 1 r) 2 1 ω(z) 2 (1 r2 )(1 b 1 2 ) (1 b 1 r) 2, (11) (1 r)(1+ b 1 ) 1 b 1 r 1+ ω(z) (1+r)(1+ b 1 ) 1+ b 1 r Proof. Since f = h(z)+g(z) S H, it follows that (12) ω(z) = g (z) = b 1 +2b 2 z + 1+2a 2 z + ω(0) = b 1 ω(z) < 1 ω(0) = b 1 < 1. So the function φ(z) = ω(z) ω(0) 1 ω(0)ω(z) = ω(z) b 1 1 b 1 ω(z) satisfies the conditions of Schwarz lemma. Therefore we have ω(z) = b 1 +φ(z) 1+b 1 φ(z) if and only if ω(z) b 1 +z 1+b 1 z ( ) b 1 +z 1+b 1 z) 1 b 1 2 r 2 (z D). (13) On the other hand, the linear transformation maps z = r onto the ( disc with the center C(r) = (1 r 2 )Re(b 1 ) 1 b 1, (1 r2 )Im(b 1 ) with the radius ρ(r) = 2 r 2 (1 b 1 2 )r 1 b 1. Then, we have 2 r 2 ω(z) b 1(1 r 2 ) 1 b 1 2 (1 b 1 2 )r 1 b 1 2 r, (14) 2 which gives (10), (11) and (13).
622 Emel Yavuz Duman,Yaşar Polatoğlu,Yasemin Kahramaner and Maslina Darus Corollary 2.3 Let f = h(z)+g(z) be an element of S HPST, then b 1 (1 Ar) 2 (1 r) A 3 and g (z) b 1 (1+Ar) 2 (1+r) A 3 b 1 (1 Ar) 2 e Ar g (z) b 1 (1+Ar) 2 e Ar, = 0, (15) b 1 r(1 Ar)(1 r) A 2 g(z) b 1 r(1+ar)(1+r) A 2 b 1 (1 Ar)e Ar g(z) b 1 (1+Ar)e Ar, = 0, (16) for all z = r < 1. Proof. Since h(z) S (A,), then we have z h (z) 1 Ar2 h(z) 1 2 r (A )r 1 Ar z 2 1 2 r 2 1 r h(z) 1+Ar, 0, 1+r z h (z) z 1 h(z) Ar 1 Ar h(z) 1+Ar, = 0, for all z D. Using Theorem 1.1 and after simple calculations we get (1 Ar)(1 r) A 2 (1+Ar)(1+r) A 2 (1 Ar)e Ar (1+Ar)e Ar, = 0. On the other hand, if we use Theorem 2.1, then we can write F(r, A,, b 1 ) g(z) h(z) F(r,A,, b 1 ), 0, F(r, A, b 1 ) g(z) F(r,A, b 1 ), = 0, and F(r, A, ) F(r, A) h(z) g (z) g (z) F(r,A,), 0, F(r,A), = 0, (17) (18) (19) (20) for all z = r < 1, where F(r,A,, b 1 ) = b 1 (1+Ar) 1+r and F 1 (r,a, b 1 ) = b 1 (1+Ar). Considering Theorem 1.1, and equations (18), (19) and (20), we get (15) and (16). Corollary 2.4 Let f = h(z)+g(z) be an element of SHPST (A,), then b 1 (1 Ar) 2 (1 r) 2(A 2) (1 r 2 )(1 b 1 2 ) (1+ b 1 r) 2 J f(z) b 1 (1+Ar) 2 (1 r) 2(A 2) b 1 (1 Ar) 2 e 2Ar(1 r2 )(1 b 1 2 ) (1+ b 1 r) 2 J f(z) (1 r 2 )(1 b 1 2 ) (1 b 1 r) 2, 0, b 1 (1+Ar) 2 e 2Ar(1 r2 )(1 b 1 2 ) (1 b 1 r) 2, = 0, (21)
Harmonic Mappings for which Second Dilatation is Janowski Functions 623 for all z = r < 1, where Proof. This is a consequence of Lemma 2.2 and the inequalities in (18). Corollary 2.5 Let f = h(z)+g(z) be an element of SHPST (A,), then ( ) 1 A [ ] b 1 2F 1 2 A,2 A,3 A, + b 1 b 1 A 2 b 1 1 2 (1+A)r2 F 1 [2,2 A ],1,3,r, b 1 r + b 1 1 3 Ar3 F 1 [3,2 A ],1,4,r, b 1 r b 1 (1 r) A + b 1 ( ) A [ ] b1 ( 1+r) (1+ b 1 r) 2F 1 2 A,2 A,3 A, + b 1 (1+ b 1 r) (A 2)(1+ b 1 r) 2 ( ) 1 A [ ] b1 2F 1 2 A,2 A,3 A,1 b 1 b 1 A 2 + b 1 1 2 (1+A)r2 F 1 [2,2 A ],1,3, r, b 1 r + b 1 1 3 Ar3 F 1 [3,2 A ],1,4, r, b 1 r b 1 (1 r) A + b 1 ( ) A [ ] b1 (1+r) (1+ b 1 r) 2F 1 2 A,2 A,3 A, b 1 (1+ b 1 r), (A 2)(1+ b 1 r) 2 f (22) for all z = r < 1, where 2 F 1 and F 1 are denote the Gauss and Appel hypergeometric functions, respectively [1]. Proof. Using Lemma 2.2 and the inequalities in (18), after the simple calculations, we get ( g (z) ) dz df ( + g (z) ) dz (1 w(z)) dz df (1+w(z)) dz b 1 (1 Ar)(1 r) A 2 (1 b 1 )(1 r) dr df 1+ b 1 r b 1 (1+Ar)(1+r) A 2 for all z = r < 1. Integrating the inequality (23), we obtain (22). (1+ b 1 )(1+r) dr 1+ b 1 r (23)
624 Emel Yavuz Duman,Yaşar Polatoğlu,Yasemin Kahramaner and Maslina Darus Theorem 2.6 Let f = h(z)+g(z) be an element of SHPST (A,), then b n+1 b 1 n+1 n+1 k=1 k(a ) k 2 m=0 Proof. Since f = h(z)+g(z) SHPST (A,), then we have Therefore, we can write p(z) = g (z), p(z) P(A,). b 1 (A )+m. (24) m+1 1+p 1 z +p 2 z 2 + +p n z n + = b 1 +2b 2 z + b 1 (1+2a 2 z + ) (1+p 1 z +p 2 z 2 + +p n z n + )(b 1 (1+2a 2 z + )) = b 1 +2b 2 z + b n+1 = b n+1 1 ka k p n k+1 (25) n+1 where a 1 1, p 0 1. Using Theorem 1.2 in (25) we obtain (24). References k=1 [1] Abramowitz, M., Stegun, I., Handbook of Mathematical Functions, Dover Pub., New York, 1970. [2] Duren, P., Harmonic mappings in the plane. Cambridge Tracts in Mathematics, 156. Cambridge University Press, Cambridge, 2004. [3] Jack, I.S., Functions starlike and convex of order α, J. London Math. Soc. (2) 3 (1971), 469-474. [4] Janowski, W., Some extremal problems for certain families of analytic functions, I. Ann. Polon. Math. 28 (1973), 297-326. [5] Libera, R.J., Some classes of regular univalent functions, Proc. Amer. Math. Soc. 16 (1965) 755-758. [6] Polatoğlu, Y., olcal, M., A coefficient inequality for the class of analytic functions in the unit disc, Int. J. Math. Math. Sci. 59 (2003), 3753-3759. Received: September, 2013