Hindawi Publishing Corporation International Journal of Mathematics and Mathematical Sciences Volume 2010, Article ID 369078, 15 pages doi:10.1155/2010/369078 Research Article Differential Subordination Defined by New Generalised Derivative Operator for Analytic Functions Ma’moun Harayzeh Al-Abbadi and Maslina Darus School of Mathematical Sciences, Faculty of Science and Technology, Universiti Kebangsaan Malaysia, 43600 Bangi Selangor (Darul Ehsan), Malaysia Correspondence should be addressed to Maslina Darus, maslina@ukm.my Received 9 June 2009; Revised 20 December 2009; Accepted 9 March 2010 Academic Editor: Dorothy Wallace Copyright q 2010 M. Harayzeh Al-Abbadi and M. Darus. 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. A new generalised derivative operator μn,m is introduced. This operator generalised many λ1 ,λ2 well-known operators studied earlier by many authors. Using the technique of differential subordination, we will study some of the properties of differential subordination. In addition we investigate several interesting properties of the new generalised derivative operator. 1. Introduction and Preliminaries Let A denote the class of functions of the form fz z ∞ ak zk , where ak is complex number, 1.1 k2 which are analytic in the open unit disc U {z ∈ C : |z| < 1} on the complex plane C. Let S, S∗ α, Cα 0 ≤ α < 1 denote the subclasses of A consisting of functions that are univalent, starlike of order α, and convex of order α in U, respectively. In particular, the classes S∗ 0 S∗ and C0 C are the familiar classes of starlike and convex functions in U, respectively. A function f ∈ Cα if Re1 zf /f > α. Furthermore a function f analytic in U is said to be convex if it is univalent and fU is convex. Let HU be the class of holomorphic function in unit disc U {z ∈ C : |z| < 1}. Let 1.2 An f ∈ HU : fz z an1 zn1 · · · , z ∈ U , with A1 A. 2 International Journal of Mathematics and Mathematical Sciences For a ∈ C and n ∈ N {1, 2, 3, . . .} we let Ha, n f ∈ HU : fz z an zn an1 zn1 · · · , z ∈ U . 1.3 ∞ k k Let fz z ∞ k2 ak z and gz z k2 bk z be analytic in the open unit disc U {z ∈ C : |z| < 1}. Then the Hadamard product or convolution f ∗ g of the two functions f, g is defined by ∞ fz ∗ gz f ∗ g z z ak bk zk . 1.4 k2 Next, we state basic ideas on subordination. If f and g are analytic in U, then the function f is said to be subordinate to g, and can be written as f ≺ g, fz ≺ gz, z ∈ U, 1.5 if and only if there exists the Schwarz function w, analytic in U, with w0 0 and |wz| < 1 such that fz gwz, z ∈ U . Furthermore if g is univalent in U, then f ≺ g if and only if f0 g0 and fU ⊂ gU see 1, page 36. Let ψ : C3 × U → C and let h be univalent in U. If p is analytic in U and satisfies the second-order differential subordination ψ pz, zp z, z2 p z; z ≺ hz, z ∈ U, 1.6 then p is called a solution of the differential subordination. The univalent function q is called a dominant of the solutions of the differential subordination, or more simply a dominant, if p ≺ q for all p satisfying 1.6. A dominant q that satisfies q ≺ q for all dominants q of 1.6 is said to be the best dominant of 1.6. Note that the best dominant is unique up to a rotation of U. Now, xk denotes the Pochhammer symbol or the shifted factorial defined by xk ⎧ ⎨1 for k 0, x ∈ C \ {0}, ⎩xx 1x 2 · · · x k − 1 for k ∈ N {1, 2, 3, . . .}, x ∈ C. 1.7 To prove our results, we need the following equation throughout the paper: n,m n,m μn,m1 μ λ fz − λ fz ∗ φ z μ fz ∗ φ , 1 z z 1 λ 1 λ 2 2 λ1 ,λ2 λ1 ,λ2 λ1 ,λ2 z ∈ U, 1.8 where n, m ∈ N0 {0, 1, 2, . . .}, λ2 ≥ λ1 ≥ 0, and φλ2 z is analytic function given by φλ2 z z ∞ zk . 1 λ2 k − 1 k2 1.9 International Journal of Mathematics and Mathematical Sciences 3 is the generalized derivative operator which we shall introduce later in the paper. Here μn,m λ1 , λ2 Moreover, we need the following lemmas in proving our main results. Lemma 1.1 see 2, page 71. Let h be analytic, univalent, and convex in U, with h0 a, γ / 0 and, Re γ ≥ 0. If p ∈ Ha, n and pz zp z ≺ hz, γ z ∈ U, 1.10 then pz ≺ qz ≺ hz, where qz γ/nzγ/n z 0 z ∈ U, 1.11 ht tγ/n−1 dt, z ∈ U. The function q is convex and is the best a, n-dominant. Lemma 1.2 see 3. Let g be a convex function in U and let hz gz nαzg z, 1.12 where α > 0 and n is a positive integer. If pz g0 pn zn pn1 zn1 · · · , z ∈ U, 1.13 is holomorphic in U and pz αzp z ≺ hz, z ∈ U, 1.14 then pz ≺ gz, 1.15 zf z 1 >− , Re 1 f z 2 1.16 and this result is sharp. Lemma 1.3 see 4. Let f ∈ A, if then 2 z z ftdt, 0 belongs to the class of convex functions. z ∈ U, z / 0, 1.17 4 International Journal of Mathematics and Mathematical Sciences 2. Main Results In the present paper, we will use the method of differential subordination to derive certain fz. Note that differential subordination properties of generalised derivative operator μn,m λ1 ,λ2 has been studied by various authors, and here we follow similar works done by Oros 5 and G. Oros and G. I. Oros 6. In order to derive our new generalised derivative operator, we define the analytic function Fλm1 ,λ2 z z ∞ 1 λ1 k − 1m k2 1 λ2 k − 1 m−1 zk , 2.1 where m ∈ N0 {0, 1, 2, . . .} and λ2 ≥ λ1 ≥ 0. Now, we introduce the new generalised as follows. derivative operator μn,m λ1 ,λ2 is defined by μn,m :A → A Definition 2.1. For f ∈ A the operator μn,m λ1 ,λ2 λ1 ,λ2 μn,m fz Fλm1 ,λ2 z ∗ Rn fz, λ1 ,λ2 z ∈ U, 2.2 where n, m ∈ N0 N ∪ {0}, λ2 ≥ λ1 ≥ 0, and Rn fz denotes the Ruscheweyh derivative operator 7, given by Rn fz z ∞ cn, kak zk , n ∈ N0 , z ∈ U, 2.3 k2 where cn, k n 1k−1 /1k−1 . If f is given by 1.1, then we easily find from equality 2.2 that fz z μn,m λ1 ,λ2 ∞ 1 λ1 k − 1m k2 1 λ2 k − 1m−1 cn, kak zk , z ∈ U, 2.4 nk−1 where n, m ∈ N0 {0, 1, 2 . . .}, λ2 ≥ λ1 ≥ 0, and cn, k n n 1k−1 /1k−1 . Remark 2.2. Special cases of this operator include the Ruscheweyh derivative operator in two n ≡ Rn and μn,0 ≡ Rn 7, the Salagean derivative operator μ0,m cases μn,1 1,0 ≡ S 8, the 0,λ2 λ1 ,0 generalised Ruscheweyh derivative operator in two cases μn,1 ≡ Rnλ and μn,0 ≡ Rnλ 9, λ1 ,λ2 λ1 ,λ2 the generalised Salagean derivative operator μ0,m ≡ Snβ introduced by Al-Oboudi 10, and λ1 ,0 n ≡ Dλ,β that can be found in the generalised Al-Shaqsi and Darus derivative operator μn,m λ1 ,0 11. Now, we remind the well-known Carlson-Shaffer operator La, c 12 associated with the incomplete beta function φa, c; z, defined by La, c : A → A, La, cfz : φa, c; z ∗ fz, z ∈ U, 2.5 International Journal of Mathematics and Mathematical Sciences 5 where φa, c; z z ∞ ak−1 ck−1 k2 zk , 2.6 a is any real number, and c / ∈ z−0 ; z−0 {0, −1, −2, . . .}. It is easily seen that 0,1 1,2 fz μ0,m μ0,0 0,0 fz μ0,λ2 fz μ0,1 fz La, afz fz, λ1 ,0 2.7 1,1 0,0 μ1,0 fz μ1,m 0,0 fz μ0,λ2 fz μλ1 ,1 fz L2, 1fz zf z, λ1 ,0 and also fz μa−1,1 fz μa−1,m fz La, 1fz, μa−1,0 0,0 λ1 ,0 0,λ2 2.8 where a 1, 2, 3, . . . . Next, we give the following. α denote the class of Definition 2.3. For n, m ∈ N0 , λ2 ≥ λ1 ≥ 0, and 0 ≤ α < 1, let Rn,m λ1 ,λ2 functions f ∈ A which satisfy the condition Re μn,m fz > α, λ1 ,λ2 z ∈ U. 2.9 Also let Kλn,m δ denote the class of functions f ∈ A which satisfy the condition 1 ,λ2 Re μn,m fz ∗ φλ2 z > δ, λ1 ,λ2 z ∈ U. 2.10 Remark 2.4. It is clear that R0,1 α ≡ Rλ1 , α, and the class of functions f ∈ A satisfying λ1 ,0 Re λ1 zf z f z > α, z ∈ U 2.11 is studied by Ponnusamy 13 and others. Now we begin with the first result as follows. Theorem 2.5. Let hz 1 2α − 1z , 1z z ∈ U, 2.12 be convex in U, with h(0)=1 and 0 ≤ α < 1. If n, m ∈ N0 , λ2 ≥ λ1 ≥ 0, and the differential subordination n m1 μλ1, ,λ2 fz ≺ hz, z ∈ U, 2.13 6 International Journal of Mathematics and Mathematical Sciences holds, then μn,m fz ∗ φλ2 z λ1 ,λ2 ≺ qz 2α − 1 21 − α 1 σ , λ1 λ1 z1/λ1 2.14 where σ is given by σx z 0 tx−1 dt, 1t z ∈ U. 2.15 The function q is convex and is the best dominant. Proof. By differentiating 1.8, with respect to z, we obtain μn,m1 fz λ1 ,λ2 μn,m fz ∗ φλ2 z λ1 z μn,m fz ∗ φλ2 z . λ1 ,λ2 λ1 ,λ2 2.16 Using 2.16 in 2.13, differential subordination 2.13 becomes 1 2α − 1z n,m μn,m . fz ∗ φ z λ z μ fz ∗ φ z ≺ hz λ 1 λ 2 2 λ1 ,λ2 λ1 ,λ2 1z 2.17 Let ∞ 1 λ1 k − 1m n,m k pz μλ1 ,λ2 fz ∗ φλ2 z z cn, kak z 1 λ2 k − 1m k2 1 p1 z p2 z2 · · · , 2.18 p ∈ H1, 1, z ∈ U . Using 2.18 in 2.17, the differential subordination becomes pz λ1 zp z ≺ hz 1 2α − 1z . 1z 2.19 By using Lemma 1.1, we have pz ≺ qz 1 λ1 z1/λ1 z ht t1/λ1 −1 dt 0 z 1 2α − 1t t1/λ1 −1 dt 1t 0 21 − α 1 2α − 1 σ , 1/λ 1 λ λ1 z 1 1 λ1 z1/λ1 2.20 International Journal of Mathematics and Mathematical Sciences 7 where σ is given by 2.15, so we get 1 21 − α n,m μλ1 ,λ2 fz ∗ φλ2 z ≺ qz 2α − 1 σ . 1/λ 1 λ1 λ1 z 2.21 The function q is convex and is the best dominant. The proof is complete. Theorem 2.6. If n, m ∈ N0 , λ2 ≥ λ1 ≥ 0, and 0 ≤ α < 1, then one has Rn,m1 ⊂ Kλn,m δ, λ1 ,λ2 α 1 ,λ2 2.22 21 − α 1 δ 2α − 1 σ , λ1 λ1 2.23 where and σ is given by 2.15. α, then from 2.9 we have Proof. Let f ∈ Rn,m1 λ1 ,λ2 Re μn,m1 fz > α, λ1 ,λ2 z ∈ U, 2.24 1 2α − 1z . 1z 2.25 which is equivalent to μn,m1 fz λ1 ,λ2 ≺ hz Using Theorem 2.5, we have 21 − α 1 μn,m fz ∗ φ z ≺ qz 2α − 1 σ . λ 2 λ1 ,λ2 1/λ λ1 λ1 z 1 2.26 Since q is convex and qU is symmetric with respect to the real axis, we deduce that Re μn,m fz ∗ φ > Re q1 δ δα, λ1 z λ 2 λ1 ,λ2 2α − 1 21 − α 1 σ , λ1 λ1 2.27 α ⊂ Kλn,m δ. This completes the proof of Theorem 2.6. from which we deduce Rn,m1 λ1 ,λ2 1 ,λ2 Remark 2.7. Special case of Theorem 2.6 with λ2 0 was given earlier in 11. Theorem 2.8. Let q be a convex function in U, with q0 1, and let hz qz λ1 zq z, z ∈ U. 2.28 8 International Journal of Mathematics and Mathematical Sciences If n, m ∈ N0 , λ2 ≥ λ1 ≥ 0,and f ∈ A and satisfies the differential subordination μn,m1 fz λ1 ,λ2 ≺ hz, z ∈ U, 2.29 then μn,m fz ∗ φλ2 z ≺ qz, λ1 ,λ2 z ∈ U, 2.30 and this result is sharp. Proof. Using 2.18 in 2.16, differential subordination 2.29 becomes pz λ1 zp z ≺ hz qz λ1 zq z, z ∈ U. 2.31 Using Lemma 1.2, we obtain pz ≺ qz, z ∈ U. 2.32 Hence μn,m fz ∗ φλ2 z ≺ qz, λ1 ,λ2 z ∈ U. 2.33 And the result is sharp. This completes the proof of the theorem. We give a simple application for Theorem 2.8. Example 2.9. For n 0, m 1, λ2 ≥ λ1 ≥ 0, qz 1 z/1 − z, f ∈ A, and z ∈ U and applying Theorem 2.8, we have hz 1z 1 z 1 2λ1 z − z2 λ1 z . 1−z 1−z 1 − z2 2.34 By using 1.8 we find that μ0,1 fz 1 − λ1 fz λ1 zf z. λ1 ,λ2 2.35 Now, μ0,1 fz λ1 ,λ2 ∞ ak zk ak kzk λ1 z . ∗ φλ2 z 1 − λ1 z 1 λ2 k − 1 1 λ2 k − 1 k2 k2 ∞ 2.36 International Journal of Mathematics and Mathematical Sciences 9 A straightforward calculation gives the following: ∞ 1 λ1 k − 1kak k−1 z fz ∗ φλ2 z 1 μ0,1 λ1 ,λ2 1 λ2 k − 1 k2 ∞ λ1 k − 1/1 λ2 k − 1zk z ∞ fz ∗ φλ2 z ∗ z k2 k1 λ1 k − 1zk . z z k2 kak 1 2.37 Similarly, using 1.8, we find that μ0,2 fz 1 − λ1 μ0,1 fz ∗ φλ2 z λ1 z μ0,1 fz ∗ φλ2 z , λ1 ,λ2 λ1 ,λ2 λ1 ,λ2 2.38 then fz μ0,2 λ1 ,λ2 μ0,1 fz ∗ φλ2 z λ1 z μ0,1 fz ∗ φλ2 z . λ1 ,λ2 λ1 ,λ2 2.39 By using 2.37 we obtain ∞ kk − 1ak 1 λ1 k − 1 k−2 μ0,1 z . fz ∗ φλ2 z λ1 ,λ2 1 λ2 k − 1 k2 2.40 We get ∞ ∞ kak 1 λ1 k − 1 k−1 kk − 1ak 1 λ1 k − 1 k−1 μ0,2 fz 1 λ z z 1 λ1 ,λ2 1 λ2 k − 1 1 λ2 k − 1 k2 k2 1 ∞ kak 1 λ1 k − 12 zk−1 1 λ − 1 k 2 k2 2 k fz ∗ φλ2 z ∗ z ∞ k2 k1 λ1 k − 1 z z 2.41 . From Theorem 2.8 we deduce that 2 k fz ∗ φλ2 z ∗ z ∞ k2 k1 λ1 k − 1 z z ≺ 1 2λ1 z − z2 1 − z2 , 2.42 implies that k fz ∗ φλ2 z ∗ z ∞ 1z k2 k1 λ1 k − 1z ≺ , z 1−z z ∈ U. 2.43 10 International Journal of Mathematics and Mathematical Sciences Theorem 2.10. Let q be a convex function in U, with q0 1 and let hz qz zq z, z ∈ U. 2.44 If n, m ∈ N0 , λ2 ≥ λ1 ≥ 0, and f ∈ A and satisfies the differential subordination μn,m fz λ1 ,λ2 ≺ hz, 2.45 then μn,m fz λ1 ,λ2 z ≺ qz, z ∈ U. 2.46 And the result is sharp. Proof. Let pz μn,m fz λ1 ,λ2 z z ∞ k2 1 λ1 k − 1m /1 λ2 k − 1m−1 cn, kak zk 1 p1 z p2 z2 · · · , 2.47 z p ∈ H1, 1, z ∈ U . Differentiating 2.47, with respect to z, we obtain μn,m fz λ1 ,λ2 pz zp z, z ∈ U. 2.48 Using 2.48, 2.45 becomes pz zp z ≺ hz qz zq z. 2.49 Using Lemma 1.2, we deduce that pz ≺ qz, z ∈ U, 2.50 and using 2.47, we have μn,m fz λ1 ,λ2 z ≺ qz, z ∈ U. This proves Theorem 2.10. We give a simple application for Theorem 2.10. 2.51 International Journal of Mathematics and Mathematical Sciences 11 Example 2.11. For n 0, m 1, λ2 ≥ λ1 ≥ 0, qz 1/1 − z, f ∈ A, and z ∈ U and applying Theorem 2.10, we have 1 1 1 z . 1−z 1−z 1 − z2 2.52 fz 1 − λ1 fz λ1 zf z, μ0,1 λ1 ,λ2 2.53 hz From Example 2.9, we have so fz μ0,1 λ1 ,λ2 f z λ1 zf z. 2.54 Now, from Theorem 2.10 we deduce that f z λ1 zf z ≺ 1 1 − z2 2.55 implies that 1 − λ1 fz λ1 zf z 1 ≺ . z 1−z 2.56 Theorem 2.12. Let 1 2α − 1z , 1z hz z ∈ U, 2.57 be convex in U, with h0 1 and 0 ≤ α < 1. If n, m ∈ N0 , λ2 ≥ λ1 ≥ 0, f ∈ A, and the differential subordination holds as μn,m fz λ1 ,λ2 ≺ hz, 2.58 then μn,m fz λ1 ,λ2 z ≺ qz 2α − 1 The function q is convex and is the best dominant. 21 − α ln1 z . z 2.59 12 International Journal of Mathematics and Mathematical Sciences Proof. Let pz fz μn,m λ1 ,λ2 z z ∞ k2 1 λ1 k − 1m /1 λ2 k − 1m−1 cn, k ak zk 2.60 z p ∈ H1, 1, z ∈ U . 1 p1 z p2 z2 · · · , Differentiating 2.60, with respect to z, we obtain fz μn,m λ1 ,λ2 pz zp z, z ∈ U. 2.61 Using 2.61, the differential subordination 2.58 becomes pz zp z ≺ hz 1 2α − 1z , 1z z ∈ U. 2.62 From Lemma 1.1, we deduce that pz ≺ qz 1 z 1 z 1 z z htdt 0 z 0 z 0 1 2α − 1t dt 1t 1 dt 2α − 1 1t 2α − 1 z 0 2.63 t dt 1t 21 − α ln1 z . z Using 2.60, we have μn,m fz λ1 ,λ2 z ≺ qz 2α − 1 21 − α ln1 z . z 2.64 The proof is complete. From Theorem 2.12, we deduce the following corollary. Corollary 2.13. If f ∈ Rn,m α, then λ1 ,λ2 Re μn,m fz λ1 ,λ2 z > 2α − 1 21 − α ln 2, z ∈ U. 2.65 International Journal of Mathematics and Mathematical Sciences 13 α, from Definition 2.3 we have Proof. Since f ∈ Rn,m λ1 ,λ2 Re μn,m fz > α, λ1 ,λ2 z ∈ U, 2.66 1 2α − 1z . 1z 2.67 which is equivalent to μn,m fz λ1 ,λ2 ≺ hz Using Theorem 2.12, we have μn,m fz λ1 ,λ2 ≺ qz 2α − 1 21 − α z ln1 z . z 2.68 Since q is convex and qU is symmetric with respect to the real axis, we deduce that Re μn,m fz λ1 ,λ2 > Re q1 2α − 1 21 − α ln 2, z z ∈ U. 2.69 Theorem 2.14. Let h ∈ HU, with h0 1, h 0 / 0 which satisfy the inequality zh z Re 1 h z 1 >− , 2 z ∈ U. 2.70 If n, m ∈ N0 , λ2 ≥ λ1 ≥ 0,and f ∈ A and satisfies the differential subordination μn,m fz λ1 ,λ2 ≺ hz, z ∈ U, 2.71 then μn,m fz λ1 ,λ2 z ≺ qz 1 z z htdt. 2.72 0 Proof. Let pz fz μn,m λ1 ,λ2 z z ∞ k2 1 λ1 k − 1m /1 λ2 k − 1m−1 cn, kak zk 1 p1 z p2 z2 · · · , z p ∈ H1, 1, z ∈ U . 2.73 14 International Journal of Mathematics and Mathematical Sciences Differentiating 2.73, with respect to z, we have μn,m fz λ1 ,λ2 pz zp z, z ∈ U. 2.74 Using 2.74, the differential subordination 2.71 becomes pz zp z ≺ hz, z ∈ U. 2.75 From Lemma 1.1, we deduce that pz ≺ qz 1 z z htdt, 2.76 0 and using 2.73, we obtain μn,m fz λ1 ,λ2 z ≺ qz 1 z z htdt. 2.77 0 From Lemma 1.3, we have that the function q is convex, and from Lemma 1.1, q is the best dominant for subordination 2.71. This completes the proof of Theorem 2.14. 3. Conclusion We remark that several subclasses of analytic univalent functions can be derived and studied . using the operator μn,m λ1 ,λ2 Acknowledgment This work is fully supported by UKM-ST-06-FRGS0107-2009, MOHE, Malaysia. References 1 K. S. Padmanabhan and R. 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