Mathematics & Statistics

Theorem 2.1. Let p be an analytic function in D with p(0) = 1. Let p(z) + βzp′(z) ≺ez, β > 0. Then the following are true: (a) p(z) ≺ez. (b) If β ≥βL ≃2.35, then p(z) ≺√1 + z where βL is the unique root of √ 2 − ∞ X k=0 1 k!(1 + βk) = 0 in (0, ∞). (c) If β ≥βs ≃0.73, then p(z) ≺φs(z) where βs is the unique root of sin 1 −

Lemma 2.2 ([19, Theorem 3.4h., p. 132]). Let q be analytic in D and let θ and φ be analytic in domain U containing q(D), with φ(w) ̸= 0, when w ∈q(D). Set Q(z) = zq′(z)φ(q(z)), h(z) := θ(q(z)) + Q(z) and suppose that either h is convex or Q is starlike. In addition, assume that Re zh′(z)/Q(z) > 0. If p is analytic in D, with p(0) = q(0), p(D) ⊂D and θ(p(z)) + zp′(z)φ(p(z)) ≺θ(q(z)) + zq′(z)φ(q(z)) = h(z) then p ≺q, and q is the best dominant.

Theorem 2.4. Let a, b and c be nonzero real numbers such that F(a, b, c; z) has no zeros in D. Then zF(a, b, c; z) is starlike if (1) c ≥max{1 + a + b −ab, 2 + 2ab −(a + b)} and (2) (c −1)(c −2) ≥a2 + b2 −ab −a −b.

Theorem 2.5. Let p be an analytic function in D with p(0) = 1. Let the subordination p(z) + βzp′(z) ≺ √ 1 + z, β > 0 holds. Let χ(β) denotes the infinite series given by χ(β) = 1 Γ(−1 2) ∞ X k=0 Γ(−1 2 + k)

Theorem 2.7. If the analytic function p : D →C satisfies p(0) = 1 and the subordination p(z) + βzp′(z) ≺1 + z, β > 0, then the following are true: (a) If β ≥1/(e −1), then p(z) ≺ez. (b) If β ≥(1 −sin 1)/ sin 1, then p(z) ≺φs(z) = 1 + sin z. (c) If β ≥2(1 + √ 2), then p(z) ≺φ0(z), where φ0(z) is given by (1.1). (d) If β ≥1/2, then p(z) ≺φc(z) = 1 + 4z/3 + 2z2/3. Estimates on β are sharp.

Theorem 3.1. Let p be the analytic function in D with p(0) = 1 and satisfies the subor- dination 1 + βzp′(z) ≺φ0(z). Then the following are true: (a) If β ≥(1 + √ 2)  −1 + 2k log  1 + 1 k  ≃1.62574, then p(z) ≺φ0(z). (b) If β ≥

Theorem 3.3. Let p be an analytic function in D with p(0) = 1. Let 1 + βzp′(z) p(z) ≺φ0(z). (a) If β ≥−  −1 + 2k log  1 + 1 k  /k log(2( √ 2 + 1)) ≃1.4819, then p(z) ≺φ0(z).

Theorem 3.5. Let p be an analytic function in D with p(0) = 1. Let 1 + βzp′(z) p2(z) ≺φ0(z). (a) If β ≥2  −1 + 2k log  1 + 1 k  ≃1.34681, then p(z) ≺φ0(z). (b) If β ≥−(1+sin 1) k sin 1 

Theorem 4.1. Let p be an analytic function in D with p(0) = 1. Let 1 + βzp′(z) ≺φc(z). Then (a) If β ≥e/(e −1), then p(z) ≺ez. (b) If β ≥3 + 2 √ 2, then p(z) ≺φ0(z). (c) If β ≥5/(3 sin 1), then p(z) ≺φs(z). (d) If β ≥5/(3 √ 2 −3), then p(z) ≺√1 + z. (e) If β ≥max {(1 −B)/(A −B), 5(1 + B)/3(A −B)}, then p(z) ≺(1+Az)/(1+Bz) The results are sharp.

Theorem 4.3. Let p be an analytic function in D with p(0) = 1. Let 1 + βzp′(z) p(z) ≺φc(z). Then (a) If β ≥5/3, then p(z) ≺ez. (b) If β ≥(log(1/2( √ 2 −1)))−1 ≃5.31275, then p(z) ≺φ0(z). (c) If β ≥5/(3 log(1 + sin 1)), then p(z) ≺φs(z). (d) If β ≥5/3 log √ 2 , then p(z) ≺√1 + z. (e) If β ≥max 5(3 log((1 + A)/(1 + B)))−1, (log((1 −B)/(1 −A)))−1 , then p(z) ≺

Theorem 4.5. Let p be an analytic function in D with p(0) = 1. Let 1 + βzp′(z) p2(z) ≺φc(z). (a) If β ≥5e/3(e −1), then p(z) ≺ez. (b) If β ≥2(1 + √ 2), then p(z) ≺φ0(z). (c) If β ≥5(1 + sin 1)/3 sin 1, then p(z) ≺φs(z). (d) If β ≥5 √ 2/3( √ 2 −1), then p(z) ≺√1 + z. (e) If β ≥max {(1 −A)/(A −B), 5(1 + A)/(3(A −B))}, then p(z) ≺(1 + Az)/(1 +

Theorem 5.1. Let p be an analytic function in D with p(0) = 1. Let 1 + βzp′(z) ≺φs(z). (a) If β ≥ 1 sin 1 P∞ n=0 (−1)n (2n+1)(2n+1)! ≃1.12432, then p(z) ≺φs(z). (b) If β ≥3 2 P∞ n=0 (−1)n (2n+1)(2n+1)! ≃1.41912, then p(z) ≺φc(z).

Theorem 5.2. Let p be an analytic function in D with p(0) = 1. Let 1 + βzp′(z) p(z) ≺φs(z). (a) If β ≥ 1 log(1+sin 1) P∞ n=0 (−1)n (2n+1)(2n+1)! ≃1.54952, then p(z) ≺φs(z). (b) If β ≥ 1 log 3 P∞

Theorem 5.3. Let p be an analytic function in D with p(0) = 1. Let 1 + βzp′(z) p2(z) ≺φs(z). (a) If β ≥1+sin 1 sin 1 P∞ n=0 (−1)n (2n+1)(2n+1)! ≃2.0704, then p(z) ≺φs(z). (b) If β ≥3 2 P∞ n=0 (−1)n

Theorem 6.1. Let p be an analytic function in D with p(0) = 1. Let 1 + βzp′(z) ≺ez. (a) If β ≥ e e−1 P∞ n=1 (−1)n+1 n!n ≃1.2602, then p(z) ≺ez. (b) If β ≥ 1 sin 1 P∞ n=1

Theorem 6.2. Let p be an analytic function in D with p(0) = 1. Let 1 + βzp′(z) p(z) ≺ez. (a) If β ≥P∞ n=1 1 n!n ≃1.3179, then p(z) ≺ez. (b) If β ≥ 1 log(1+sin 1) P∞ n=1 1 n!n ≃2.1585, then p(z) ≺φs(z). (c) If β ≥

Theorem 6.3. Let p be an analytic function in D with p(0) = 1. Let 1 + βzp′(z) p2(z) ≺ez. (a) If β ≥ e e−1 P∞ n=1 1 n!n ≃2.08489, then p(z) ≺ez.