Results & Lemmas (6)
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Lemma 1.
Lemma 1. [14] Let Ω⊂C and let ψ: C3 × D →C be such that ψ(r, s, t; z) /∈Ωwhenever r = eeiθ, s = meiθeeiθ, and ℜ (s + t)e−iθe−eiθ ≥0, for…
Lemma 1. [14] Let Ω⊂C and let ψ : C3 × D →C be such that ψ(r, s, t; z) /∈Ωwhenever r = eeiθ, s = meiθeeiθ, and ℜ (s + t)e−iθe−eiθ ≥0, for θ ∈[0, 2π), z ∈D, and m ≥1. If q is analytic in D with q(0) = 1 and ψ(q(z), zq′(z), z2q′′(z); z) ∈Ω, for all z ∈D, then q ∈Pe. It is worth observing that the admissibility requirement in Lemma 1 is verified whenever ψ(r, s, t; z) /∈Ωfor r = eeiθ, s = meiθeeiθ, and ℜ (s + t)e−iθe−eiθ ≥0, where θ ∈[0, 2π), z ∈D, and m ≥1. In the special case ψ : C2×D →C, t
Lemma 2.
Lemma 2. [15] Suppose Ω⊂C and let ψ: C3 × D →C satisfy ψ(iρ, σ, µ + iv; z) /∈Ω whenever z ∈D, ρ ∈R, σ ≤−1+ρ2 2, and σ + µ ≤0. If q is…
Lemma 2. [15] Suppose Ω⊂C and let ψ : C3 × D →C satisfy ψ(iρ, σ, µ + iv; z) /∈Ω whenever z ∈D, ρ ∈R, σ ≤−1+ρ2 2 , and σ + µ ≤0. If q is analytic in D with q(0) = 1 and ψ(q(z), zq′(z), z2q′′(z); z) ∈Ω, then ℜ(q(z)) > 0 for all z ∈D. For the reduced case ψ : C2 × D →C, the admissibility condition in Lemma 2 becomes ψ(iρ, σ; z) /∈Ω, where ρ ∈R and σ ≤−1+ρ2 2 .
Lemma 3.
Lemma 3. [12] Assume q belongs to the normalized class of analytic functions and satisfies q(z) ̸= 1. Let Ω⊂C and let ψ: C3 × D →C be such…
Lemma 3. [12] Assume q belongs to the normalized class of analytic functions and satisfies q(z) ̸= 1. Let Ω⊂C and let ψ : C3 × D →C be such that ψ(r, s, t; z) /∈Ωwhenever r = √ 2 cos 2θ, eiθ, s = me3iθ 2 √ 2 cos 2θ, and ℜ (s + t)e−3iθ ≥ 3m2 8 √ 2 cos 2θ, for −π
Theorem 1.
Theorem 1. Assume that α > −1, n ∈N ∪0 and hold the condition 1 e[ℜ(α −1) −n] > 0, then Mn,α(z) ∈Pe.
Theorem 1. Assume that α > −1, n ∈N ∪0 and hold the condition 1 e[ℜ(α −1) −n] > 0, then Mn,α(z) ∈Pe.
Theorem 2.
Theorem 2. Let α > −1, n ∈N, h1 = h −(C −D)(1 + D) −(n + 1)(1 + D)2ih −(C −D)(1 −D) + (n + 1)(1 −D)2i > 0, h2 = h (C −D) + (C −D)2 + (C…
Theorem 2. Let α > −1, n ∈N, h1 = h −(C −D)(1 + D) −(n + 1)(1 + D)2ih −(C −D)(1 −D) + (n + 1)(1 −D)2i > 0, h2 = h (C −D) + (C −D)2 + (C −D)(1 + D)(α + 1) ih −(C −D)(1 −D) + (n + 1)(1 −D)2i + h −(C −D)(1 + D) −(n + 1)(1 + D)2ih (C −D) −(C −D)2 + (C −D)(1 −D)(α + 1)
Corollary 1.
Corollary 1. Let us consider the relation, we have zMn,α(z) = − α n + 1zM′ n+1,α−1(z), and consequently z (zMn,α(z))′ zMn,α(z) = 1 + z M′′…
Corollary 1. Let us consider the relation, we have zMn,α(z) = − α n + 1zM′ n+1,α−1(z), and consequently z (zMn,α(z))′ zMn,α(z) = 1 + z M′′ n+1,α−1(z) M′ n+1,α−1(z) Simultaneously with the Theorem 2 and substituting n = n −1, α = α + 1, yields the following reuslt as zMn,α(z) ∈S∗[C, D] (Janowski starlike). 5. Conclusion
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