On the Janowski convexity and starlikeness of the confluent hypergeometric function

Transcription

1 Reprint from the Bulletin of the Belgian Mathematical Society Simon Stevin On the Janowski convexity and starlikeness of the confluent hypergeometric function Rosihan M. Ali Saiful R. Mondal V. Ravichandran Bull. Belg. Math. Soc. Simon Stevin 015, 7 50 The Bulletin of the Belgian Mathematical Society - Simon Stevin is published by The Belgian Mathematical Society, with financial support from the Universitaire Stichting van Belgie Fondation Universitaire de Belgique and the Fonds National de la Recherche Scientifique FNRS. It appears quarterly and is indexed and/or abstracted in Current Contents, Current Mathematical Publications, Mathematical Reviews, Science Citation Index Expanded and Zentralblatt für Mathematik. The Bulletin of the Belgian Mathematical Society - Simon Stevin is part of Project Euclid Cornell University Library, an aggregation of electronic journals. It is available online to subscribers to Project Euclid For more informations about the Belgian Mathematical Society - Simon Stevin, see our web site at

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3 On the Janowski convexity and starlikeness of the confluent hypergeometric function Rosihan M. Ali Saiful R. Mondal V. Ravichandran Abstract For 1 B < A 1, conditions on A, B, a, c are determined that ensure the confluent hypergeometric function Φa; c; z satisfies the subordination Φa; c; z 1 + Az/1 + Bz. This gives rise to conditions for c/aφa; c; z 1 to be close-to-convex, Φa; c; z to be Janowski convex, and zφa; c; z to be Janowski starlike. 1 Introduction Let A denote the class of analytic functions f defined in the open unit disk D = {z : z < 1} normalized by the conditions f0 = 0 = f 0 1. If f and g are analytic in D, then f is subordinate to g, written fz gz, if there is an analytic self-map w of D satisfying w0 = 0 and fz = gwz. For 1 B < A 1, let P[A, B] be the class consisting of normalized analytic functions pz = 1+ c 1 z+ in D satisfying pz 1+ Az 1+ Bz. For instance, if 0 β < 1, then P[1 β, 1] is the class of functions pz = 1+ c 1 z+ satisfying Re pz > β in D. Received by the editors in May In revised form in September 014. Communicated by H. De Schepper. 010 Mathematics Subject Classification : 30C45, 30C80, 40G05. Key words and phrases : Subordination, Janowski starlike and convex functions, confluent hypergeometric function. Bull. Belg. Math. Soc. Simon Stevin 015, 7 50

4 8 R. M. Ali S. R. Mondal V. Ravichandran The class S [A, B] of Janowski starlike functions [6] consists of f A satisfying z f z P[A, B]. fz For 0 β < 1, S [1 β, 1] := S β is the usual class of starlike functions of order β; S [1 β, 0] := Sβ = { f A : z f z/ fz 1 < 1 β}, and S [β, β] := S [β] = { f A : z f z/ fz 1 < β z f z/ fz + 1 }. These classes have been studied, for example, in [, 3]. A function f A is said to be close-to-convex of order β [5, 9] if Rez f z/gz > β for some g S := S 0. This paper studies the confluent Kummer hypergeometric function Φa; c; z given by Φa; c; z = 1 F 1 a; c; z = Γc Γa Γa+n z n n=0 Γc+n n! = a n z n n=0 c n n!, 1.1 where a, c C, c = 0, 1,,, and λ n denotes the Pochhammer symbol given by λ 0 = 1, λ n = λλ + 1 n 1. The function Φ is a solution of the differential equation zφ a; c; z+c zφ a; c; z aφa; c; z = 0 1. introduced by Kummer in 1837 [1]. The Kummer hypergeometric function is an entire analytic function in C and satisfies the relation c Φ a; c; z = a Φa+ 1; c+1; z. 1.3 When Re c > Re a > 0, Φ can be expressed in the integral form Φa; c; z = Γc 1 t a 1 1 t c a 1 e tz dt. ΓaΓc a 0 Several important functions are special cases of the confluent hypergeometric function. These include the Bessel functions, Laguerre polynomials, and the error function. An exposition on Φ can be found in the book by Temme [1]. Several works [1, 8, 10, 11] have studied geometric properties of the function Φa; c; z, which include studies on its close-to-convexity, starlikeness, and convexity. Miller and Mocanu [8] proved that Re Φa; c; z > 0 in D for real a and c satisfying either a > 0 and c a, or a 0 and c a. Ponnusamy and Vuorinen [10, Theorem 1.9, p. 77] obtained sufficient conditions for Re Φa; c; z > β, 0 β < 1/. They also determined, as an application of 1.3, conditions that ensure c/a Φa; c; z 1 is close-to-convex of positive order with respect to the identity function. Additionally they derived conditions for close-to-convexity of zφa; c; z with respect to the starlike function z/1 z, as well as close-to-convexity of zφa; c; z with respect to the starlike function z/1 z. Constraints on a and c so that Φa; c; z is convex of positive order are also found in [10, Theorem 5.1, p. 88]. For positive a, Acharya [1, Theorem 4.1.1, p. 50] proved that zφa; c; z is starlike and close-to-convex

5 On the confluent hypergeometric function 9 with respect to z and z/1 z if respectively c a 1+ a + a+ 1 and 6c 3 + 3c a+ 6+3c 6a 3a+4+a5a In Section of this paper, sufficient conditions on A, B, a, c are determined that ensure Φ satisfies the subordination Φa; c; z 1+ Az/1 + Bz. Set in this general framework, our findings are obtained through a computationallyintensive methodology with shrewd manipulations. The benefits of such general results are that they not only extend and improved earlier known works, but by judicious choices of the parameters A and B, they give rise to several interesting applications. Examples involving the Laguerre polynomials and the modified Bessel functions are given in Section 3 to illustrate this significance. Sufficient conditions are also obtained for c/aφ a; c; z P[A, B], which readily yields conditions for c/aφa; c; z 1 to be close-to-convex. Section 4 gives emphasis to the investigation of Φa; c; z to be Janowski convex as well as of zφa; c; z to be Janowski starlike. The following lemma is needed in the sequel. Lemma 1.1. [8, 9] Let Ω C, and Ψ : C 3 D C satisfy Ψiρ, σ, µ+iν; z Ω whenever z D, ρ real, σ 1+ρ / and σ+µ 0. If p is analytic in D with p0 = 1, and Ψpz, zp z, z p z; z Ω for z D, then Re pz > 0 in D. In the case Ψ : C D C, then the condition in Lemma 1.1 reduces to ρ real and σ 1+ρ /. Ψiρ, σ; z Ω, Close-to-convexity of the confluent hypergeometric function Here is one main result in a general form that has several interesting ramifications. Theorem.1. Let 1 B Suppose B < A 1, and a, c R satisfy { } c 1 max 1, 1+ B1+ A 1+ a..1 Further let A, B, c and a satisfy either the inequality c 1 aa+b c whenever + c 11+B 1 B + 1+ B 1 B 1 A 1 B A1+B a a A+B a 1. c 11 BaA+B+ +1+ B+1+ A1+Ba

6 30 R. M. Ali S. R. Mondal V. Ravichandran or the inequality 1 B A1+ a B1 a,.3 c 1 A1+a B1 a + 1+ B 1 B 4 a + A+B a+ 1 whenever c A1+B a c 11+ B 1 B a 1 AB.4 c 11 BaA + B+ +1+ B+1+ A1+Ba < 1 B A1+a B1 a..5 If 1+ BΦa; c; z = 1+ A, then Φa; c; z P[A, B]. Proof. Define the analytic function p : D C by where Φz := Φa; c; z. Then and 1 A 1 BΦz pz = 1+ A 1+ BΦz, Φz = Φ z = 1 A+1+ Apz 1 B+1+ Bpz,.6 p z 1 B+1+ Bpz,.7 Φ z = 1 B+1+ Bpzp z 41+ Bp z 1 B+1+ Bpz 3..8 Using.6.8, the Kummer differential equation 1. yields z p 1+ B z 1 B+1+ Bpz zp z +c zzp z 1 B+1+ Bpz1 A+1+ Apz az = 0..9 With Ω = {0}, define Ψr, s, t; z by 1+ B Ψr, s, t; z := t 1 B+1+ Br s +c zs 1 B+1+ Br1 A+1+ Ar az..10

7 On the confluent hypergeometric function 31 It follows from.9 that Ψpz, zp z, z p z; z Ω. To show Re pz > 0 for z D, from Lemma 1.1, it is sufficient to establish Re Ψiρ, σ, µ+iν; z < 0 in D for any real ρ, σ 1+ρ /, and σ+µ 0. With z = x + iy D, it readily follows from.10 that 1 B Re Ψiρ, σ, µ+iν; z = µ ρ1 AB 1 B +1+ B ρ σ +c xσ+ ay 1 B1 A 1+ B1+ Aρ ax..11 Since σ 1+ρ /, and B [ 1, 3 ], Thus 1 B 1 B +1+ B ρ σ 1 B 1+ρ 1 B +1+ B ρ 4 Re Ψiρ, σ, µ+iν; z µ+σ+c x 1σ+ where + ρ1 AB 1+ B1+ Aρ ax 1+ B 1 B. 1 B1 A ay ax 1+ B 1 B 1 1+ B1+ c x 11+ρ Aρ ρ1 AB + ax+ ay 1 B1 A ax 1+ B 1 B = p 1 ρ + q 1 ρ+r 1 := Qρ, p 1 = 1 1+ B1+ Aax c x 1+, 1 AB q 1 = ay, 1 B1 A r 1 = ax 1 1+ B c x 1 1 B. Condition.1 shows that p 1 = 1 1+ B1+ Aax c x 1+ < 1 c 1 1+ B1+ Aa 1+ < 0. Since max {p 1ρ + q 1 ρ+r 1 } = 4p 1 r 1 q 1 /4p 1 for p 1 ρ R Qρ < 0 when 1 AB a y < c x 1 1+ B1+ A ax < 0, it is clear that

8 3 R. M. Ali S. R. Mondal V. Ravichandran c x 1+ 1 B1 A x, y < 1. As y < 1 x, the above condition holds whenever 1 AB a 1 x 1+ B1+ Aax c x 1 that is, when a + a A+B 1+ B 1+ a 1 B + 1 x 1+ A1+B c x 1+ c 1 a A+B B1 A + ax+ 1+ B, 1 B ax+ 1+ B, 1 B x+c 1 +c 1 1+B AB a1 1 B 0..1 To establish inequality.1, consider the polynomial R given by Rx := mx + nx+r, x < 1, where m := a + a A+B + 1 = A1+ a B1 a, n := c 1 a A+B B 1+ A1+B 1+ a 1 B r := c 1 +c 1 1+B 1 B AB a1. The constraint.3 yields n m, and thus Rx m + r n. inequality. readily implies that, Now Rx m+r n = a + a A+ B + 1+c 1 + c 11+B a 1 AB 1 B c 1 a A+B B 1+ A1+B 1+ a 1 B = c 1 c 1 a A+B B 1+ A1+B 1+ a 1 B + c 11+ B 1 B 1 A 1 B a + A+ B a+1 0. Now considers the case of the constraint.5, which is equivalent to n < m. Then the minimum of R occurs at x = n/m, and.4 yields Rx 4mr n 4m 0.

9 On the confluent hypergeometric function 33 is, Evidently Ψ satisfies the hypothesis of Lemma 1.1, and thus Re pz > 0, that 1 A 1 BΦz 1+ A 1+ BΦz 1+z 1 z. Hence there exists an analytic self-map w of D with w0 = 0 such that 1 A 1 BΦz 1+ A 1+ BΦz = 1+wz 1 wz, which implies that Φz 1+ Az/1+Bz. Theorem.1 gives rise to simple conditions on a and c to ensure Φa; c; z maps D into a half-plane. Corollary.1. Let a 0 and c 1+a. Then Re Φa; c; z > a/a 1. Proof. The result follows from Theorem.1 by choosing A = a+1/a 1, and B = 1. In this case, condition.1 reduces to c which clearly holds for c 1+ a. Also.3 and. respectively reduces to and Since c +a, and 1+ a c 1+a 0,.13 c 1 c 1a + 1 a a c 1+a 1+a 1+a = a 4 0, c 1 c 1a + 1+a + 1 = c c a 0, which establishes both.13 and.14. Remark.1. It is noteworthy to compare the result obtained in Corollary.1 with a result of Ponnusamy and Vuorinen [10]. With β := a/a 1, a 1, 0], the result of Ponnusamy and Vuorinen [10, Theorem 1.9, p.77] gives Re Φa; c; z > a/a 1 provided c a /1 a. Since a /1 a +a, it is clear that Corollary.1 extends the range of c. Further, since a 0, Corollary.1 extends the range of β to the whole interval β [0, 1. Corollary.. Let a > 0 and c {, a 0, 1 1+ a, a 1. Then Re Φa; c; z > 1+a 1+a 1.

10 34 R. M. Ali S. R. Mondal V. Ravichandran Proof. Let A = 1 a /1+ a, and B = 1 in Theorem.1. In this case,.1 reduces to c, which clearly holds. Now condition.3 is trivially true. Thus the result follows from Theorem.1 provided. holds, or equivalently, if Let a 0, 1 and c. Then c 1 c 1 1 a a..15 c 1 c 1 1 a = c 1c +c 1a a. If a 1, c 1+ a, then c 1 c 1 1 a = c 1c a a. Thus.15 holds in either case. From Theorem.1, it is evident that Φa; c; z P[A, 1], or equivalently, Re Φa; c; z > 1 A = 1+ a 1+ a. Remark.. Corollary. improves a result in [10, Corollary 1.11, p.77] to the wider range a 0, a 0 ], where a 0 = 1.70 is a root of a 4 a 1 = 0. It also improves [10, Corollary 1.1, p.78] for a 0, 1. Corollary.3. Let a, c be real such that a, a 0 c, 0 a 1 a, a 1. Then Re Φa; c; z > 1/. Proof. Put A = 0 and B = 1 in Theorem.1. The condition.1 reduces to c, which holds in all cases. It is sufficient to establish conditions.3 and., or equivalently, and c 1 1 a 0,.16 c 1 c 1 1 a a Consider the case when a 0 and c a. Then c 1 1 a = c + a a 0, and c 1 c 1 1 a a+1 = c c +a 0. Thus both.16 and.17 hold. For a [0, 1] and c, then c 1 1 a = c + a a 0, and c 1 c 1 1 a a+1 = c c +a 0. Finally it is readily established for a 1 and c a that c 1 1 a = c a a 0, and c 1 c 1 1 a a+ 1 = cc a 0. Remark.3. Corollary.3 encompasses for a 1 a result in [10, Theorem 1.13, p.78]. It also expands the range of a to include all real numbers.

11 On the confluent hypergeometric function 35 Corollary.4. Let a 1/, a = 0, and {, a [ 1/, 0 c a+3, a > 0. Then1+ a Φa; c; z 1 < a. In particular, z z ez/ 3 zi 0 +z 1I 1 3 < 3, where I n z is the modified Bessel function of order n. Proof. Choose A = a /a + 1 and B = 0 in Theorem.1. First consider the case when a > 0. The condition.1 reduces to c a+3, and condition., which is equivalent to c 1c a 3 0, clearly holds. Now.3 reduces to c 1+a, which holds because c a+3, a > 0. From Theorem.1 it is evident that Φa; c; z 1+ az/1+a, that is, Φa; c; z 1 < a/1+a. Now consider the case when a [ 1/, 0. Then.1 gives c. Conditions. and.3 respectively become and Since c and a [ 1/, 0, cc 1 c 11+a > 0,.18 ac 1 +a1+a > cc 1 c 11+ a = cc 1 c 11+ a = c 1c a > 0, and ac 1 +a1+a = ac 1+a+ a = ac +a > 0, which established both.18 and.19. Again from Theorem.1 it follows that An application of the identity Φ 1 ; ; z = ez/ 3 establishes the final assertion. 1+ a Φa; c; z 1 < a. z z 3 zi 0 +z 1I 1 Theorem.. Let 3 < B < A 1, and a, c R satisfy { } c 1 max 1, 1+ B1+ A 1+ a..0 Suppose A, B, c and a satisfy c 1 A+B c 1 a+1 + 8B1 B 1+ B A1+B a

12 36 R. M. Ali S. R. Mondal V. Ravichandran B1 B + 16c 1 1+ B 3 1 A 1 B a A+B a 1.1 whenever A+ B c 1 a+ 1 8B1 B 1+ A1+B + 1+ B 3 1+ a A1+a B1 a,. or the inequality aa+b c B1 B + 1+ B 3 1+a A+ B + a when A+ B c 1 a A1+B a c 1 +c 1 16B1 B 1+ B 3 a 8B1 B 1+ B A1+B a 1 AB.3 < A1+a B1 a..4 If 1+ BΦa; c; z = 1+ A, then Φa; c; z P[A, B]. Proof. Proceeding similarly as in the proof of Theorem.1, consider Re Ψiρ, σ, µ+iν; z as given in.11. For σ 1+ρ /, ρ R, and B 3, 1 B 1 B +1+ B ρ σ 1 B 1+ρ 1 B +1+ B ρ 4 8B1 B 1+ B 3. With z = x+ iy D, and µ+σ < 0, it follows that Re Ψiρ, σ, µ+iν; z 1 1+ B1+ c x 11+ρ Aρ + ax ρ1 AB 1 B1 A 8B1 B + ay ax 1+ B 3 where = p ρ + q ρ+r := Q 1 ρ, p = 1 1+ B1+ Aax c x 1+, 1 ABay q =, 1 B1 A r = ax 1 8B1 B c x 1 1+ B 3.

13 On the confluent hypergeometric function 37 As per the proof of Theorem.1, observe that the constraint.0 implies that p < 0. Thus Q 1 ρ < 0 for all ρ R provided q 4p r, that is, 1 AB a y c x 1 1+ B1+ A ax c x 1+ x, y < 1. With y < 1 x, it is sufficient to show 1 AB a 1 x c x 1 c x 1+ 1 B1 A ax+ 1+ B1+ Aax 1 B1 A ax+ for x < 1. The above inequality is equivalent to showing where m 1 := 1+a A+B + a, aa+b n 1 := c B1 B 1+ B 3, 16B1 B 1+ B 3 R 1 x := m 1 x + n 1 x+r 1 0,.5 8B1 B 1+ B 3 1+ r 1 := c 1 16B1 B +c 1 1+ B 3 a 1 AB. 1+ A1+B a, If. holds, then n 1 m 1. Since R 1 is increasing, then R 1 x m 1 + r 1 n 1, which is nonnegative from.1. On the other hand, if.4 holds, then n 1 < m 1, R 1 x 4m 1 r 1 n 1 /4m 1, and.3 implies R 1 x 0. Either case establishes.5. The following results are immediate consequences from relation 1.3 and respectively Theorem.1 and Theorem.. Theorem.3. Let 1 B Suppose B < A 1, a, c R with a = 0 and satisfying { } c max 1, 1+ B1+ A 1+ a+1. Further let A, B, c and a satisfy either c c a+1 A+B B 1+ 1 B 1+ A1+B a+ 1 c1+ B + 1 B 1 A 1 B a+ 1 A+B a+ 1 1

14 38 R. M. Ali S. R. Mondal V. Ravichandran whenever c1 Ba+ 1A+B+ +1+ B+1+ A1+Ba+ 1 1 B A+a Ba, or the inequality A+ a+ba c when 4 a B 1+ 1 B A+ B a A1+B a+ 1 c + 1+ B 1 B 1 AB c a+1 c1 Ba+ 1A+B+ +1+ B+1+ A1+Ba+ 1 < 1 B A+ a Ba. If 1+ BΦa+ 1; c+1; z = 1+ A, thenc/aφ a; c; z P[A, B]. Theorem.4. Let 3 < B < A 1. Suppose a, c R, a = 0, such that { } c max 1, 1+ B1+ A a. Further let A, B, c and a satisfy c A+ B c a B1 B + 16c 8B1 B 1+ B 3 1+ B 3 1 A 1 B when A+ B c a+1+1 8B1 B + 1+ B 3 A+a Ba, 1+ or the inequality A+B c a+1+1 8B1 B + 1+ B 3 1+a+ 1 A+ B + a when A+ B c a < A+a Ba. 8B1 B 1+ B A1+B a+1 a+ 1 A+B a A1+B a A1+B a+1 c 16B1 B + c 1+ B 3 a+1 1 AB A1+B a+1 If 1+ BΦa+ 1; c+1; z = 1+ A, thenc/aφ a; c; z P[A, B].

15 On the confluent hypergeometric function 39 With B = 1, the following results are easily deduced from Theorem.3 by choosing respectively A = a+ /a, A = a /+a and A = 0. Corollary.5. Let a 1, and c 1+1+a. Then c/aφa; c; z 1 is close-to-convex of ordera+1/a with respect to the identity function. Corollary.6. Let a > 1, a = 0, and c { 1, a 1, 0 a, a > 0. Then c/aφa; c; z 1 is close-to-convex of order + a + a / + a with respect to the identity function. Corollary.7. Let a be a nonzero real number, and 1 a, a 1 c 1, 1 a < 0 1+a, a > 0. Then Rec/aΦ a; c; z > 1/. 3 Examples The associated Laguerre polynomial L α n [4] is given by the series L α nz := n k=0 1 k Γn+α+1 n k!γα+ k+ 1k! zk. It relates to the confluent hypergeometric function via the identity Thus α+ 1 n Φ n; α+ 1; z = n!l α nz. m! 1+ m4 4 m m 4 L 4 m z = Φ m; 1+ m4 4 ; z. The associated Laguerre polynomial L α n is used to illustrate the significance of Theorem.1. Example 3.1. For every m =, 3,..., m! 1+ m4 4 m m 4 L 4 m z 1+ z 1 m 1+ z 1+m, 3.1 provided1+m m+m 1+m 4 /4 m = m! 1 m +m+m L m4 /4 m z.

16 40 R. M. Ali S. R. Mondal V. Ravichandran To verify the subordination given in3.1, consider p : D C given by Φ 1 m pz := 1+m 1 z Φ 1 m 1+m 1 z, where Φ 1 z := Φ m; 1+m 4 /4; z. Proceeding as in the proof of the Theorem.1 with A = 1/1 m and B = 1/1+m, then.11 reduces to mm+m Re Ψiρ, σ, µ+ iν; z = µ + m+ m m+ +m + m+ ρ σ +1+ m4 4 xσ ρ 4 m m y+ 8 1m m 4 m ρ x. Now σ 1+ρ / yields mm+m + m+ m m+ +m + m+ ρ σ x 1 8 mm+m + m+ 1+ρ m m+ +m + m+ ρ 4 3. := hρ. Taking logarithmic derivatives, it is evident that h is increasing for ρ 0, and thus hρ m + m+/mm+. Now as µ+σ < 0, 3. yields Re Ψiρ, σ, µ+iν; z µ+σ m + m+ mm+ 1 m x 11+ρ ρ 4 m m y+ 1 8 m m 4 x 1 m ρ x 8 where < p 1 ρ + q 1 ρ+r 1, p 1 = m4 1+ x < 0, 8 q 1 = 1 4 m m y, r 1 = m + m+ mm+ m m x. 8 As x < 1, y < 1 and y < 1 x, it follows that q 1 4p 1r 1 = 1 16 m4 m y + m4 1+ x 1 8 m x m4 8 m + m+ mm+ = 1 16 m4 m x + y m4 m x

17 On the confluent hypergeometric function 41 m4 m 6 < m m + m+ mm+ 4 + m 3 m+ 1+ x 1+ x < 0. Since p 1 < 0, the above inequality implies that maxp 1 ρ + q 1 ρ + r 1 = 4p 1 r 1 q 1 /4p 1 < 0 over ρ R. Hence Re Ψiρ, σ, µ + iν; z < 0. The final assertion now follows readily by using the same arguments as in the proof of Theorem.1. The next example illustrates both Theorem.1 and Theorem. in the sense that the real parameters satisfy the conditions of each theorem within different constraint intervals. Example 3.. If a, 1 and c max{, 1+ a 4 /4}, then Φa; c; z 1+ z 1+a 1+ z 1 a, z D, 3.3 provided1 a +a+a = 1+a a a Φa; c; z. To establish the subordination3.3, define the analytic function p : D C by pz := a 1+a 1 1 Φa; c; z Φa; c; z 1 a 1 a Proceeding similarly as in the proof of the Theorem.1 with A = 1/1 + a and B = 1/1 a, the equivalent form for.11 is aa a a+ Re Ψiρ, σ, µ+ iν; z = µ a a +a a+ ρ σ +c xσ ρ 4 a a y+ 1 8 a a 4 x 1 a ρ x Now σ 1+ρ / yields a a a +a a+ ρ σ a ρ a a +a a+ ρ 4 := aa a a+ hρ, where hρ = 1+ρ / a a +a a+ ρ > 0. By taking logarithmic derivatives, it follows that h ρ = a4 4a 3 + 8a 4+a a+ ρ 1+ρ a a +a a+ ρρhρ. 3.5

18 4 R. M. Ali S. R. Mondal V. Ravichandran We next consider two cases. First is when a, ], that is, B 3. In this case, it follows from 3.5 that h is increasing for ρ 0, and hence hρ 1/ a a. Thus Re Ψiρ, σ, µ+iν; z µ+σ a a+ aa +c x 1σ ρ 4 a a y+ 8 1 a a 4 x 1 8 where = p 1 ρ + q 1 ρ+r 1, p 1 = 1 a4 c 1 8 x < 1 a4 c , q 1 = 1 4 a a y, r 1 = a a+ a a 1 c x a a 4 x 1 = aa c Since x < 1, y < 1 and y < 1 x, a computation gives q 1 4p 1r 1 = 1 16 a4 a y c 1+ a4 4 x aa + c 1 4 a ρ x a x. < 1 16 a4 a 1 4 c 1a x+ a4 1 4 aa + c 1 c 1 aa + c < 1 16 a4 a 1 4 c 1a + a4 1 4 aa + c 1 c 1 aa + c = 1 a aa + c a 4 4c 1 0. a x Since maxp 1 ρ + q 1 ρ+r 1 = 4p 1 r 1 q 1 /4p 1 < 0 for ρ R, it follows that Re Ψiρ, σ, µ+ iν; z < 0. The final assertion for the case a, ] follows by adopting the same arguments used in the proof of Theorem.1. Next consider the case when a, 1, that is, the case B 3. For this range of a, the expression a 4 4a 3 + 8a 4 is negative, and hence h given in3.5 is non-negative when ρ > 0 and ρ a 4 4a 3 + 8a 4/a a+. Together these imply that hρ 16a 1 a a+ 4. x

19 On the confluent hypergeometric function 43 It is immediate from 3.4 that 8aa a 1 Re Ψiρ, σ, µ+iν; z µ+σ a a+ 3 +c x 1σ ρ 4 a a y+ 1 8 a a 4 x 1 a ρ x 8 where = p ρ + q ρ+r, p = 1 a4 c 1 8 x < 1 a4 c 1+ 8 < 0, q = 1 4 a a y, 8aa a 1 r = a a+ 3 c a x. 8 We need to show that q 4p r < 0. Since x < 1, y < 1 and y < 1 x, a computation yields q 4p r = a4 16 a x + y c 1 16aa a 1 c 1 a a+ 3 a 4 4 c 1+ 4a5 a a 1 a a+ 3 c 1a x 4 a < a 4 4c 1 + c 1 4aa a a a+ 3. Evidently, the conclusion follows if the expression M a = a a+ 3 a 4 a + +3aa a 1 is positive, which clearly holds for a, 1. 4 Janowski starlikeness of the confluent hypergeometric function This section looks at finding conditions to ensure a normalized confluent hypergeometric function zφa; c; z is Janowski starlike. For this purpose, we first derive sufficient conditions for Φa; c; z to be Janowski convex, after which an application of relation 1.3 yields conditions for zφa; c; z S [A, B]. Theorem 4.1. Let a, c R be such that A BΦ a; c, z = 1+BzΦ a; c; z, 1 B < A 1. Suppose c1+ B 1+ B 1+a1+B

20 44 R. M. Ali S. R. Mondal V. Ravichandran Further let A, B, c and a satisfy either 1 B+B 1+B 1+ a + c1 B 1+ B1+a 1+ a1 B +1+ c1+ B1 A+ B+c1 B B 4. whenever 1 B+B 1+B 1+a + c1 B 1+ B1+ a 1, 4.3 or the inequality 1+a1 B c1+ B1 A+B+c1 B B 1 B+B 1+B 1+a + c1 B1+ a B when 1 B+B 1+B 1+a + c1 B 1+ B1+ a If 0 / Φ D, 0 / Φ D, then 1+ zφ a; c, z Φ a; c, z 1+ Az 1+ Bz. Proof. For convenience, write Φz := Φa; c; z, and define an analytic function p : D C by pz := Φ z+1 BzΦ z Φ z 1+ BzΦ z. Then and z Φ z+zφ z zφ z zφ z Φ z = pz 1 pz+ 1+Bpz 1, 4.6 zφ z Φ z = zp z pz 1 1+ Bzp z pz+ 1+Bpz 1 zp z pz+ 1+ Bpz 1 1+ Bpz 1 =. pz 1pz+ 1+Bpz 1

21 On the confluent hypergeometric function 45 Thus zφ z Φ z = and together with 4.6 imply that zφ z Φ z zφ z Φ = z zp z pz 1pz+ 1+ Bpz 1 1+ zφ z Φ z, A differentiation of1. leads to It follows that zφ z Φ z Using4.6 and4.7, 4.8 yields that is, pz 1zp z pz 1 pz+ 1+Bpz 1 pz 1 pz+ 1+Bpz 1 + pz 1. pz+ 1+ Bpz 1 zφ z+c z+1φ z 1+ aφ z = zφ z Φ +c z+1 zφ z z Φ 1+ az = z zp z + pz 1 pz+ 1+Bpz 1 pz+ 1+ Bpz 1 + pz 1c z pz+ 1+ Bpz 1 zp z+ + c z1+b 1+az = 0, pz pz az1+b +c zb + 1+ a1 B z c z 1+B 1+az1 B Denote by Ψpz, zp z, z := zp z+f 1 pz + F pz+f 3, where F 1 = + c z1+b 1+ az1+b, F = c zb 1+ a1 B z, F 3 = + c z 1+B 1+ az1 B. =

22 46 R. M. Ali S. R. Mondal V. Ravichandran With Ω = {0}, 4.9 yields Ψpz, zp z, z Ω. Now with z = x+iy D, let G 1 := ReF 1 = = 1 + c x1+b +c1+ B x1+ B G := ReiF = yb+ 1+a1 B y, G 3 := ReF 3 = + c x 1+B = 1 c1 B+x1 B For σ 1+ρ /, ρ R, 1+ ax1+b a1+B 1+ ax1 B 1 1+a1 B Re Ψiρ, σ, z = σ G 1 ρ + G ρ+g G 1 ρ G ρ G := Qρ.,. Note that condition4.1 implies1+g 1 / > 0. In this case, Q has a maximum at ρ = G /1+G 1. Thus Qρ < 0 for all real ρ provided Since y < 1 x, it is left to show that G 1+G 11 G 3, x, y < 1. 1+a1 B B 1 x 1+ a1+b 1+ +c1+ B x1+ B 1+ 1 A+B+c1 B x1 B 1 1+a1 B, x < 1. The above inequality is equivalent to where Hx := x h A, Bx+ h 3 A, B 0, 4.10 h A, B = 1 BA+ B + B 1+B 1+a B1+ a B c, h 3 A, B = 1+ +c1+ B1 A+ B+c1 B 1+ a1 B B.

23 On the confluent hypergeometric function 47 To establish 4.10, first consider the case when 4.3 holds. Then h 1, and hence H x = 0 at x = h 1, 1. Since H x > 0, H has a minimum at x = h and from 4., it follows that Hx Hh = h h + h 3 = h + h 3 0. Now consider the case when 4.5 holds. In this case h 1, and H x = x h 1 h 0. Hence H is decreasing. From 4.4, it is evident that Hx H1 = 1 h + h 3 1 h +h 3 0. Thus Ψ satisfies the hypothesis of Lemma 1.1, and hence Re pz > 0, or equivalently Φ +1 BzΦ Φ 1+ BzΦ 1+z 1 z. By definition of subordination, there exists an analytic self-map w of D with w0 = 0 and A simple computation shows that Φ z+1 BzΦ z Φ z 1+ BzΦ z = 1+ wz 1 wz. 1+ zφ z Φ z = 1+ Awz 1+ Bwz, and hence 1+ zφ z Φ z The relation 1.3 also shows that 1+ Az 1+ Bz. z zφa; c; z zφa; c; z = 1+ zφ a 1; c 1; z Φ a 1; c 1; z. Together with Theorem 4.1, it immediately yields the following result for zφa; c; z S [A, B]. Theorem 4.. Let a and c be real numbers such that A BΦ a 1; c 1, z = 1+ BzΦ a 1; c 1; z, 1 B < A 1. Suppose c1+ B 1+ B a1+b Further let A, B, c and a satisfy either 1 B+B 1+B a +c 11 B 1+ Ba

24 48 R. M. Ali S. R. Mondal V. Ravichandran +c1+ BB A+c1 B whenever 1 B+B A a B1+B + or c 11 B a1 B B Ba 1, 4.13 a1 B 1+A B+c1+ BB A+c1 B B 1 a B+B 1+B + 1+ Ba 1 B c when 1 B+B A a B1+B + 1+ Ba 1 B c Then zφa; c; z S [A, B]. Remark 4.1. Choosing A = 1 β, β [0, 1, and B = 1, then Theorem 4.1 gives [10, Theorem 5.1, p. 88], while Theorem 4. gives [10, Theorem 5.3, p. 88]. For a = 1, c = 1+ δ, δ 1, Theorem 4.1 gives sufficient conditions for 1 Φ1; 1+δ; z = δ 1 t δ 1 e tz dt 0 to be Janowski convex. This is a generalization of a result in [10, Corollary 5.8, p. 89]. Writing fz = zφa; c; z, and gz = fz /z, it follows that zg z gz = f z z fz Now 4.16 and Theorem 4. immediately yield the following result. Corollary 4.1. Let a and c be real numbers such that A BΦ a + 1; c+1; z = 1+ BzΦ a+ 1; c+1; z and 4.11 holds for 1 B < 1. Further let A, B, c and a satisfy either4.1 when4.13 holds, or4.14 when4.15 holds. Then zφa; c; z S [, B].

25 On the confluent hypergeometric function 49 Remark 4.. Particular choice of A and B yield the following ramifications. i For A = 1 β and B = 1, Corollary 4.1 reduces to [10, Corollary 5.13, p. 93]. ii For a = 1/, Corollary 4.1 is useful in obtaining conditions for the generalized error function zφ1/; c, z S [, B]. Acknowledgment. The work presented here was supported by a FRGS and RU research grants from Universiti Sains Malaysia. The authors are thankful to the referee for the insightful comments that helped improve the clarity of this manuscript. References [1] A. P. Acharya, Univalence criteria for analytic functions and applications to hypergeometric functions, Ph.D diss., University of Würzburg, [] R. M. Ali, V. Ravichandran and N. Seenivasagan, Sufficient conditions for Janowski starlikeness, Int. J. Math. Math. Sci. 007, Art. ID 695, 7 pp. [3] R. M. Ali, R. Chandrashekar and V. Ravichandran, Janowski starlikeness for a class of analytic functions, Appl. Math. Lett , no. 4, [4] W. W. Bell, Special functions for scientists and engineers, D. Van Nostrand Co., Ltd., London, [5] A. W. Goodman, Univalent functions. Vol. I & II, Mariner, Tampa, FL, [6] W. Janowski, Some extremal problems for certain families of analytic functions. I, Ann. Polon. Math , [7] S. S. Miller and P. T. Mocanu, Univalence of Gaussian and confluent hypergeometric functions, Proc. Amer. Math. Soc , no., [8] S. S. Miller and P. T. Mocanu, Differential subordinations and inequalities in the complex plane, J. Differential Equations , no., [9] S. S. Miller and P. T. Mocanu, Differential subordinations, Monographs and Textbooks in Pure and Applied Mathematics, 5, Dekker, New York, 000. [10] S. Ponnusamy and M. Vuorinen, Univalence and convexity properties for confluent hypergeometric functions, Complex Variables Theory Appl , no. 1,

26 50 R. M. Ali S. R. Mondal V. Ravichandran [11] St. Ruscheweyh and V. Singh, On the order of starlikeness of hypergeometric functions, J. Math. Anal. Appl , no. 1, [1] N. M. Temme, Special functions, A Wiley-Interscience Publication, Wiley, New York, School of Mathematical Sciences, Universiti Sains Malaysia, USM Penang Malaysia rosihan@usm.my Department of Mathematics, King Faisal University, Ahsaa 3198, Saudi Arabia smondal@kfu.edu.sa Department of Mathematics, University of Delhi, Delhi , India vravi@maths.du.ac.in