Results & Lemmas (8)
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Theorem 2.1.
Theorem 2.1. Let f be a holomorphic function in D such that f(0) = 1. Then f ∈GN if and only if there exists a function h ∈S ∗such that…
Theorem 2.1. Let f be a holomorphic function in D such that f(0) = 1. Then f ∈GN if and only if there exists a function h ∈S ∗such that f(z) = r h(z) z exp 1 2 −z + z3 9 , z ∈D, (2.1)
Theorem 2.3.
Theorem 2.3. Let f be a holomorphic function in D such that f(0) = 1. Then f ∈GN if and only if there exists a function h ∈S ∗(1/2) such…
Theorem 2.3. Let f be a holomorphic function in D such that f(0) = 1. Then f ∈GN if and only if there exists a function h ∈S ∗(1/2) such that f(z) = h(z) z exp 1 2 −z + z3 9 , z ∈D. (2.5)
Theorem 2.5.
Theorem 2.5. If f ∈GN, then the following sharp estimate 1 1 + |z| exp −1 2ℜ z −z3 9 ≤|f(z)| ≤ 1 1 −|z| exp −1
Theorem 2.5. If f ∈GN , then the following sharp estimate 1 1 + |z| exp −1 2ℜ z −z3 9 ≤|f(z)| ≤ 1 1 −|z| exp −1
Theorem 2.6.
Theorem 2.6. If f(z) = 1 + P∞ n=1 dnzn ∈GN, then the coefficients dn satisfy the following sharp coefficient inequalities |d1| ≤3 2 (2.8) |2d1…
Theorem 2.6. If f(z) = 1 + P∞ n=1 dnzn ∈GN , then the coefficients dn satisfy the following sharp coefficient inequalities |d1| ≤3 2 (2.8) |2d1 + 1| ≤2 (2.9) 2d2 −d2 1 ≤1 (2.10) 6(3d3 −3d1d2 + d3 1) −1 ≤6
Theorem 2.7.
Theorem 2.7. The region of values of the coefficient d1, i.e. d1: g ∈GN, g(z) = 1 + d1z + · · · has the form w ∈C: w + 1 2 ≤1 . In this…
Theorem 2.7. The region of values of the coefficient d1, i.e. {d1 : g ∈GN , g(z) = 1 + d1z + · · · } has the form w ∈C : w + 1 2 ≤1 . In this section, we establish specific differential subordination findings related to the class GN. To prove differential subordination results, we recall the following lemma (see [18, Theorem 3.4h, p. 132]).
Lemma 2.8.
Lemma 2.8. Let q be univalent in D, θ and ϕ be holomorphic in a domain D containing q(D) with ϕ(w) ̸= 0 when w ∈q(D). Let Q(z):=…
Lemma 2.8. Let q be univalent in D, θ and ϕ be holomorphic in a domain D containing q(D) with ϕ(w) ̸= 0 when w ∈q(D). Let Q(z) := zq′(z)ϕ(q(z)) and h(z) := θ(q(z))+ Q(z) for z ∈D. Suppose that either (i) Q is starlike univalent in D, or (ii) h is convex univalent in D. Assume also that (iii) ℜzh′(z) Q(z) > 0, z ∈D. If p ∈H 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)), z ∈D, then p ≺q and q is the best dominant.
Theorem 2.9.
Theorem 2.9. Let f ∈H and f(0) = 1. If f satisfies the subordination condition 2zf ′(z) f(z) + 1 + z −z3 3 ≺1 + z 1 −z, z ∈D, (2.16)
Theorem 2.9. Let f ∈H and f(0) = 1. If f satisfies the subordination condition 2zf ′(z) f(z) + 1 + z −z3 3 ≺1 + z 1 −z , z ∈D, (2.16)
Theorem 2.10.
Theorem 2.10. Let f ∈H with f(0) = 1. If f satisfies 2zf ′(z) f(z) + 1 + z −z3 3 ≺1 + z 1 −z, z ∈D, (2.20) then p(z):= z f(z) 1 −z 2 Z z…
Theorem 2.10. Let f ∈H with f(0) = 1. If f satisfies 2zf ′(z) f(z) + 1 + z −z3 3 ≺1 + z 1 −z , z ∈D, (2.20) then p(z) := z f(z) 1 −z 2 Z z 0 f(ζ)
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