# A coefficient inequality for the class of analytic functions in the unit disc

The aim of this paper is to give a coeï¬cient inequality for the class of analytic functions in the unit disc D = {z | |z| < 1}. 2000 Mathematics Subject Classiï¬cation: 30C45. 1. Introduction. Let â¦ be the family of functions Ï(z) regular in the disc D and satisfying the conditions Ï(0) = 0 and |Ï(z)| < 1 for z â D. Next, for arbitrary ï¬xed numbers A and B, â1 < A â¤ 1, â1 â¤ B < A, denote by P (A, B) the family of functions p(z) = 1 + p1 z + p2 z2 + Â· Â· Â· regular in D such that p(z) is in P (A, B) if and only if p(z) = 1 + AÏ(z) 1 + BÏ(z) (1.2) (1.1) for some function Ï(z) â â¦ and every z â D. The class P (A, B) was introduced by Janowski [3]. Moreover, let S â (A, B, b) (b â 0, complex) denote the family of functions f (z) = z + a2 z2 + Â· Â· Â· + an zn + Â· Â· Â· regular in D and such that f (z) is in S â (A, B, b) if and only if 1+ 1 f (z) â 1 = p(z) z b f (z) (1.3) (1.4) for some p(z) in P (A, B) and all z in D. For the aim of this paper we need Jackâs lemma [2]. âLet Ï(z) be a regular in the unit disc with Ï(0) = 0, then if |Ï(z)| attains its maximum value on the circle |z| = r at a point z1 , we can write z1 Ï (z1 ) = kÏ(z1 ), where k is real and k â¥ 1.â 2. Coeï¬cient inequality. The main purpose of this paper is to give sharp upper bound of the modulus of the coeï¬cient an . Therefore, we need the following lemma. Lemma 2.1. The necessary and suï¬cient condition for g(z) = z +a2 z2 +Â· Â· Â· belongs to S â (A, B, b) is ï£± ï£²z Â· 1 + BÏ(z) â g(z) â S â (A, B, b) â g(z) = ï£³z Â· ebAÏ(z) , where Ï(z) â â¦. Proof. The proof of this lemma is in four steps. Step 1. Let B â 0 and g(z) = z Â· 1 + BÏ(z) b(AâB)/B b(AâB)/B B â 0, B = 0, (2.1) (2.2) If we take the logarithmic derivative from equality (2.2), we obtain g (z) z Â· Ï (z) 1 zÂ· â 1 = (A â B) . b g(z) 1 + BÏ(z) If we use Jackâs lemma [2] in equality (2.3), we get (A â B)Ï(z) 1 g (z) â1 = . z b g(z) 1 + BÏ(z) After the simple calculations from (2.4), we see that 1+ g (z) 1 + AÏ(z) 1 zÂ· â1 = . b g(z) 1 + BÏ(z) (2.3) (2.4) (2.5) Equality (2.5) shows that g(z) â S â (A, B, b). Step 2. Let B = 0 and g(z) = z Â· ebAÏ(z) . Similarly, we obtain 1+ g (z) 1 + AÏ(z) 1 z â1 = = 1 + AÏ(z). b g(z) 1 + BÏ(z) (2.6) (2.7) This shows that g(z) â S â (A, B, b). Step 3. Let g(z) â S â (A, B, b) and B â 0, then we have 1+ 1 g (z) 1 + AÏ(z) z â1 = . b g(z) 1 + BÏ(z) (2.8) Equality (2.8) can be written in the from b(A â B) Ï(z)/z 1 g (z) = + . g(z) 1 + BÏ(z) z If we use Jackâs lemma (2.9), we obtain b(A â B)Ï (z) 1 g (z) = + . g(z) 1 + BÏ(z) z Integrating both sides of equality (2.10), we get g(z) = z Â· 1 + BÏ(z) b(AâB)/B (2.9) (2.10) (2.11) Step 4. Let g(z) â S â (A, B, b) and B = 0. Similarly, we obtain g(z) = z Â· ebAÏ(z) which ends the proof. We note that we choose the branch of (1 + Bw(z))b(AâB)/B such that 1 + Bw(0) b(AâB)/B (2.12) =1 at z = 0. (2.13) Theorem 2.2. If f (z) = z + a2 z2 + Â· Â· Â· + an zn + Â· Â· Â· belongs to S â (A, B, b), then b(A â B) + kB k+1 |bA| k+1 if B = 0. if B = 0, (2.14) These bounds are sharp because the extremal function is ï£± ï£´ ï£² z , (1 â BÎ´z)âb(AâB)/B ï£´ bAz ï£³ze , |Î´| = 1, if B = 0, if B = 0. fâ (z) = (2.15) Proof. Let B â 0. If we use the deï¬nition of the class S â (A, B, b), then we write 1+ f (z) 1 z â 1 = p(z). b f (z) (2.16) Equality (2.16) can be written by using the Taylor expansion of f (z) and p(z) in the form z + 2a2 z2 + 3a3 z3 + Â· Â· Â· + nan zn + Â· Â· Â· = z + a2 z2 + Â· Â· Â· + an zn + Â· Â· Â· 1 + bp1 z + bp2 z2 + Â· Â· Â· + bpn zn + Â· Â· Â· . (2.17) Evaluating the coeï¬cient of zn in both sides of (2.17), we get nan = an + bp1 a + bp2 a + Â· Â· Â· + bp . on the other hand, pn â¤ (A â B). (2.19) (2.18) Inequality (2.19) was proved by Aouf [1]. If we consider the relations (2.18) and (2.19) together, then we obtain (n â 1) |b||A â B| 1 + a2 + a3 + Â· Â· Â· + a which can be written in the form 1 (n â 1) (2.20) |b||A â B| ak , a1 = 1. (2.21) To prove (2.14), we will use the induction principle. Now, we consider inequalities (2.21) and b(A â B) + kB . k+1 (2.22) The right-hand sides of these inequalities are the same because (i) for n = 2, |b||A â B| (n â 1) ak , a1 = 1 â a2 â¤ |b||A â B|, (2.23) â a2 â¤ |b||A â B|; b(A â B) + kB = b(A â B) k+1 (ii) for n = 3, |b||A â B| (n â 1) a3 â¤ â a3 a3 â¤ ak = 1 |b||A â B| 1 + a2 2 1 1 2 â¤ |b| |A â B|2 + |b||A â B|, 2 2 b(A â B) + B b(A â B) + kB = |b||A â B| k+1 2 (2.24) â a3 â¤ â a3 1 1 |b||A â B| |b||A â B| + |B| â¤ |b||A â B| |b||A â B| + 1 2 2 1 1 2 â¤ |b| |A â B|2 + |b||A + B|. 2 2 Suppose that this result is true for n = p, then we have |b||A â B| (n â 1) ak , a1 = 1 â ap |b||A â B| 1 + a2 + a3 + Â· Â· Â· + apâ1 â¤ (p â 1) (2.25) , b(A â B) + kB k+1 pâ2 â ap â¤ b(A â B) + kB k+1 (2.26) â ap â¤ 1 |b||A â B| |b||A â B| + 1 |b||A â B| + 2 (p â 1)! Â· |b||A â B| + 3 Â· Â· Â· |b||A â B| + (p â 2) from (2.25), (2.26), and induction hypothesis, we have |b||A â B| 1 + a2 + a3 + Â· Â· Â· + apâ1 (p â 1) 1 |b||A â B| |b||A â B| + 1 = (p â 1)! Â· |b||A â B| + 2 Â· Â· Â· |b||A â B| + (p â 2) . If we write x = |b||A â B| > 0, equality (2.27) can be written in the form. x 1 + a2 + a3 + Â· Â· Â· + apâ1 (p â 1) 1 = x(x + 1)(x + 2) Â· Â· Â· x + (p â 2) . (p â 1)! (2.27) (2.28) After the simple calculation from equality (2.28), we get 1 1 x + (p â 1) 1 + a2 + a3 + Â· Â· Â· + apâ1 p (p â 1) 1 (x + 1)(x + 2)(x + 3) Â· Â· Â· x + (p â 2) x + (p â 1) = p! 1 x â 1 + a2 + a3 + Â· Â· Â· + apâ1 p p â1 1 + 1 + a2 + a3 + Â· Â· Â· + apâ1 p 1 = (x + 1)(x + 2)(x + 3) Â· Â· Â· x + (p â 2) x + (p â 1) p! 1 1 â ap + 1 + a2 + a3 + Â· Â· Â· + apâ1 p p 1 (x + 1)(x + 2)(x + 3) Â· Â· Â· x + (p â 2) x + (p â 1) = p! x 1 + a2 + a3 + Â· Â· Â· + apâ1 + ap â p 1 x(x + 1)(x + 2)(x + 3) Â· Â· Â· x + (p â 2) x + (p â 1) . = p! Equality (2.29) shows that the result is valid for n = p + 1. Therefore, we have (2.14). Corollary 2.3. The ï¬rst inequality of (2.14) can be rewritten in the form (2.29) b(A â B) + kb k+1 = B(A â B) 1 b(A â B) + B 2 1 1 Â· b(A â B) + 2B Â· Â· Â· b(A â B) + (n â 2)B 3 (n â 1) 1 b(A â B) Â· b(A â B) + B = (n â 1)! Â· b(A â B) + 2B Â· Â· Â· b(A â B) + (n â 2)B â¤ 1 b(A â B) b(A â B) + |B| (n â 1)! Â· b(A â B) + 2|B| Â· Â· Â· b(A â B) + (n â 2)|B| . (2.30) If A = 1, B = â1, and b = 1, then n! 1 2 Â· (2 + 1) Â· (2 + 2) Â· Â· Â· n = = n. (n â 1)! (n â 1)! (2.31) This is the coeï¬cient inequality for the starlike function which is well known. Corollary 2.4. If A = 1, B = â1, an < 1 (n â 1)! |2b + k|. (2.32) This inequality was obtained by Aouf [1]. Therefore, by giving the special value to A, B, and b, we obtain the coeï¬cient inequality for the classes S â (1, â1, Î²), S â (1, â1, eâiÎ» Cos Î»), S â (1, â1, (1â Î²)eâiÎ» Cos Î»), S â (1, 0, b), S â (Î², 0, b), S â (Î², âÎ², b), S â (1, (â1 + 1/M), b), and S â (1 â 2Î², â1, b), where 0 â¤ Î² < 1, |Î»| < Ï /2, and M > 1. References [1] [2] [3] M. K. Aouf, On a class of p-valent starlike functions of order Î±, Int. J. Math. Math. Sci. 10 (1987), no. 4, 733â744. I. S. Jack, Functions starlike and convex of order Î±, J. London Math. Soc. (2) 3 (1971), 469â474. W. Janowski, Some extremal problems for certain families of analytic functions. I, Ann. Polon. Math. 28 (1973), 297â326. YaÂ¸ar PolatoËlu: Department of Mathematics and Computer Science, Faculty of Scis g ence and Letters, Istanbul KÃ¼ltÃ¼r University, Sirinevler 34510, Istanbul, Turkey E-mail address: y.polatoglu@iku.edu.tr Metin Bolcal: Department of Mathematics and Computer Science, Faculty of Science and Letters, Istanbul KÃ¼ltÃ¼r University, Sirinevler 34510, Istanbul, Turkey E-mail address: m.bolcal@iku.edu.tr http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png

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