J. Aust. Math. Soc. 100 (2016), 1–20

Lemma 1.1 [21]. Let p(z) = 1 + P∞ n=1 cnzn be an analytic function with Re p (z) > 0 in D and µ be a complex number. Then |c2 −µc2 1| ≤2 max{1, |2µ −1|}. (1.4) The result is sharp for the functions given by p(z) = (1 + z2)/(1 −z2) and p(z) = (1 + z)/(1 −z).

Lemma 1.2 [7]. Let h(z) be convex in D with h(0) = a. If p(z) is analytic in D, with p(0) = a and p(z) + zp′(z) ≺h(z), then p(z) ≺1 z Z z 0 h(t) dt. https://doi.org/10.1017/S1446788715000336 Published online by Cambridge University Press

Lemma 1.3 [8]. Let f ∈A be given by f(z) = z + P∞ n=2 anzn. Then the inverse function F(w) of the function f(z) is analytic in |w| < ρ for some ρ > 0. Also, suppose that  z f(z) t = 1 + ∞ X n=1 a(−t) n zn and  w

Theorem 2.1. Let f(z) be of the form (1.1) and f ∈M( β) for some 1 < β ≤2. Then |an| ≤2( β −1) n −1 for n ≥2. Equality is attained for the function fn(z) = z(1 −zn−1)(2( β−1))/(n−1) for n ≥2.

Theorem 2.2. Let f ∈N( β) be of the form (1.1) for some 1 < β ≤2. Then |an| ≤2( β −1) n(n −1) for n ≥2. Equality is attained for the function fn(z) given by f ′ n(z) = (1 −zn−1)(2( β−1))/(n−1), n ≥2.

Theorem 2.3. (i) If f ∈M( β) for some β > 1, then z f(z) ≺(1 −z)−2( β−1). (2.4) (ii) If f ∈N( β) for some β > 1, then f ′(z) ≺(1 −z)2( β−1) (2.5) and f(z) z ≺1 −(1 −z)2β−1 (2β −1)z

Corollary 3.1d.1, page 76]) z f(z) = g(z) ≺exp Z z 0 2( β −1) 1 −t dt  = (1 −z)−2( β−1). (ii) Let f ∈N( β) for some β > 1. Then, by the definition of the class N( β), Re β −1 −zf ′′(z) f ′(z) β −1

Corollary 3.1d.1, page 76]) f ′(z) ≺exp  − Z z 0 2( β −1) 1 −t dt  = (1 −z)2( β−1). (2.10) Next, suppose that h(z) = f(z)/z and so zh′(z) + h(z) = f ′(z). Therefore, (2.10) becomes

Corollary 2.4. For f ∈M( β) for some β > 1, the following hold. (i) r(1 −r)2( β−1) ≤|f(z)| ≤r(1 + r)2( β−1), |z| = r < 1. Equality holds for the function f(z) = z(1 −z)2( β−1) or its rotation. https://doi.org/10.1017/S1446788715000336 Published online by Cambridge University Press

Corollary 2.5. (i) If f ∈N( β) for some β > 1, then, for each z = reiθ in D, (1 −r)2( β−1) ≤|f ′(z)| ≤(1 + r)2( β−1). (2.11) Equality holds for the function f(z) given by f ′(z) = (1 −z)2( β−1) or its rotation. (ii) For each f ∈N( β) ( β > 1), | arg f ′(z)| ≤2( β −1) sin−1 r, |z| = r < 1. Equality holds for the function f(z) given by f ′(z) = (1 −z)2( β−1) or its rotation. (iii) If f ∈N( β) and β > 1, then, for each z = reiθ in D, |f(z)| ≤(1 + r)2β−1 −1 2β −1

Theorem 2.6. Let f ∈N( β) for some β > 1. Then, for every positive number r ≤ 1/(2β −1), the function f maps the disk |z| < r onto a convex domain. The result is best possible, that is, the radius of convexity for the class N( β) is 1/(2β −1). https://doi.org/10.1017/S1446788715000336 Published online by Cambridge University Press

Theorem 3.1. Let f(z) be of the form f(z) = z + P∞ n=2 anzn and f ∈M( β) for some β > 1. Then, for any λ ∈C, |a3 −λa2 2| ≤  β −1 for λ −2β −3 4(β −1) ≤ 1 4(β −1), 4(β −1)2 λ −2β −3

Theorem 3.2. Let f(z) be of the form f(z) = z + P∞ n=2 anzn and f ∈M( β) for some β > 1. Then, for any λ ∈C, |a3 −λa2 2| ≤  β −1 3 for λ −2β −3 3(β −1) ≤ 1 3(β −1), (β −1)2

Lemma 4.1. Let f(z) be of the form (1.1) and f ∈M( β) for some β > 1. Also, for a fixed n ∈N, let (f(z)/z)−n have an expansion of the form  f(z) z −n = 1 + ∞ X k=1 a(−n) k zk. Then, for each k ≥1, |a(−n) k

Theorem 4.2. Let f ∈M( β) for some fixed β > 1. Then, for 0 < r ≤1, max f∈M( β) ∆  r, z f(z)  = 4πr2(β −1)2 F(2β −1, 2β −1; 2; r2). The maximum is attained for the function f0(z) = z(1 −z)2(β−1).

Theorem 4.3. Let f ∈M( β) for some β > 1 and F(w) be the inverse function of f(z) having the expansion F(w) = w + P∞ n=2 Anwn, which is valid in some neighborhood of the origin. Then |An| ≤1 n

Theorem 4.4. Let the function f(z) be in N( β) for some β > 1 and F(w) be the inverse function of f(z), with the following expansion: F(w) = w + ∞ X n=2 Anwn, (4.4) which is valid in some neighborhood of the origin. Then |An| ≤(−1)n+1(2β −1)n  1 2β −1 n 

Theorem 4.5. Let g ∈MΣ( β) (β > 1) be of the form g(z) = z(1 + P∞ n=1 bnz−n) for z ∈∆. Then, for each n ≥1, |bn| ≤

Theorem 4.6. Let g ∈NΣ( β) (β > 1) be given by g(z) = z(1 + P∞ n=1 bnz−n) for z ∈∆. Then, for n ≥2, |bn| ≤ 1 n −1

Theorem 4.8. Let g ∈MΣ( β) (β > 1) and G(w) be the inverse of g(z) and suppose that G(w) has the following expansion: G(w) = w  1 + ∞ X n=1 Bnw−n in some neighborhood of the point at infinity. Then: (i) |B1| ≤2(β −1); (ii) for n ≥2, |Bn| ≤