Abstract
The aim of a study of the presented paper is the differential subordination
involving harmonic means of the expressions p(z), p(z) + zp′(z), and p(z) + zp′(z)
p(z)
when p is an analytic function in the unit disk, such that p(0) = 1, p(z) ̸≡1. Several
applications in the geometric functions theory are given.
Results & Lemmas (20)
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Lemma 1.1
Lemma 1.1 [10, p. 24] Let q ∈Q, with q(0) = a, and let p(z) = a + anzn + · · · be analytic in D with p(z) ̸≡a and n ≥1. If p is not…
Lemma 1.1 [10, p. 24] Let q ∈Q, with q(0) = a, and let p(z) = a + anzn + · · · be analytic in D with p(z) ̸≡a and n ≥1. If p is not subordinate to q, then there exist points z0 = r0eiθ0 ∈D and ζ0 ∈∂D \ E(q) and an m ≥n ≥1 for which p(Dr0) ⊂q(D), (1) p(z0) = q(ζ0), (2) z0 p′(z0) = mζ0q′(ζ0), (3) ℜz0 p′′(z0) p′(z0) + 1 ≥mℜ ζ0q′′(ζ0) q′(ζ0) + 1 . 2 Harmonic Mean
Theorem 2.1
Theorem 2.1 Let p(z) = 1 + a1z + · · · be analytic in D with p(z) ̸≡1. Then ℜ
Theorem 2.1 Let p(z) = 1 + a1z + · · · be analytic in D with p(z) ̸≡1. Then ℜ
Theorem 2.1
Theorem 2.1 is complete. ⊓⊔
Theorem 2.1 is complete. ⊓⊔
Corollary 2.1
Corollary 2.1 Let f (z) = z + a2z2 + · · · be analytic in D. Then ℜ2 f (z) f ′(z) f (z) + zf ′(z) > 0 ⇒ℜf (z) z > 0.
Corollary 2.1 Let f (z) = z + a2z2 + · · · be analytic in D. Then ℜ2 f (z) f ′(z) f (z) + zf ′(z) > 0 ⇒ℜf (z) z > 0.
Theorem 2.2
Theorem 2.2 Let p(z) = 1 + a1z + · · · be analytic in D with p(z) ̸≡1. Then ℜ 2p(z) + 2zp′(z) 1 + p2(z) + zp(z)p′(z) > 0 ⇒ℜp(z) > 0.
Theorem 2.2 Let p(z) = 1 + a1z + · · · be analytic in D with p(z) ̸≡1. Then ℜ 2p(z) + 2zp′(z) 1 + p2(z) + zp(z)p′(z) > 0 ⇒ℜp(z) > 0.
Corollary 2.2
Corollary 2.2 Let f (z) = z + a2z2 + · · · be analytic in D. Then ℜ ⎡ ⎢⎣ 2 z f (z) f ′(z) z f (z) + f ′(z) ⎤ ⎥⎦> 0 ⇒ℜf (z) z > 0. 123
Corollary 2.2 Let f (z) = z + a2z2 + · · · be analytic in D. Then ℜ ⎡ ⎢⎣ 2 z f (z) f ′(z) z f (z) + f ′(z) ⎤ ⎥⎦> 0 ⇒ℜf (z) z > 0. 123
Theorem 2.3
Theorem 2.3 Let p(z) = 1 + a1z + · · · be analytic in D with p(z) ̸≡1. Then ℜ ⎧ ⎪⎪⎨ ⎪⎪⎩ 2 p(z) + zp′(z) p(z) 2 + zp′(z) p2(z) ⎫ ⎪⎪⎬ ⎪⎪⎭
Theorem 2.3 Let p(z) = 1 + a1z + · · · be analytic in D with p(z) ̸≡1. Then ℜ ⎧ ⎪⎪⎨ ⎪⎪⎩ 2 p(z) + zp′(z) p(z) 2 + zp′(z) p2(z) ⎫ ⎪⎪⎬ ⎪⎪⎭
Corollary 2.3
Corollary 2.3 Let f (z) = z + a2z2 + · · · be analytic in D. Then ℜ ⎧ ⎪⎪⎨ ⎪⎪⎩ 2zf ′(z) f (z) 1 + zf ′′(z) f ′(z) 1 + zf ′(z) f (z) + zf…
Corollary 2.3 Let f (z) = z + a2z2 + · · · be analytic in D. Then ℜ ⎧ ⎪⎪⎨ ⎪⎪⎩ 2zf ′(z) f (z) 1 + zf ′′(z) f ′(z) 1 + zf ′(z) f (z) + zf ′′(z) f ′(z) ⎫
Theorem 2.4
Theorem 2.4 Let p(z) = 1 + a1z + · · · be analytic in D with p(z) ̸≡1. Then ℜ ⎧ ⎪⎪⎨ ⎪⎪⎩ 2 p(z) + zp′(z) p(z) 1 + p2(z) + zp′(z) ⎫ ⎪⎪⎬…
Theorem 2.4 Let p(z) = 1 + a1z + · · · be analytic in D with p(z) ̸≡1. Then ℜ ⎧ ⎪⎪⎨ ⎪⎪⎩ 2 p(z) + zp′(z) p(z) 1 + p2(z) + zp′(z) ⎫ ⎪⎪⎬ ⎪⎪⎭ > 0 ⇒ℜp(z) > 0.
Corollary 2.4
Corollary 2.4 Let f (z) = z + a2z2 + · · · be analytic in D. Then ℜ ⎧ ⎪⎪⎨ ⎪⎪⎩ 2 f (z) zf ′(z) 1 + zf ′′(z) f ′(z) 1 + f (z) zf ′(z) +…
Corollary 2.4 Let f (z) = z + a2z2 + · · · be analytic in D. Then ℜ ⎧ ⎪⎪⎨ ⎪⎪⎩ 2 f (z) zf ′(z) 1 + zf ′′(z) f ′(z) 1 + f (z) zf ′(z) + zf ′′(z) f ′(z)
Theorem 2.5
Theorem 2.5 Let p(z) = 1 + a1z + a2z2 + · · · be analytic in D with p(z) ̸≡1, and let 0 < M < 1 3. Then
Theorem 2.5 Let p(z) = 1 + a1z + a2z2 + · · · be analytic in D with p(z) ̸≡1, and let 0 < M < 1 3. Then
Corollary 2.5
Corollary 2.5 Let f (z) = z +a2z2 +· · · be analytic in D, and let 0 < M < 1. Then
Corollary 2.5 Let f (z) = z +a2z2 +· · · be analytic in D, and let 0 < M < 1. Then
Corollary 2.6
Corollary 2.6 Let f (z) = z +a2z2 +· · · be analytic in D, and let 0 < M < 1. Then
Corollary 2.6 Let f (z) = z +a2z2 +· · · be analytic in D, and let 0 < M < 1. Then
Corollary 2.7
Corollary 2.7 Let f (z) = z +a2z2 +· · · be analytic in D, and let 0 < M < 1. Then
Corollary 2.7 Let f (z) = z +a2z2 +· · · be analytic in D, and let 0 < M < 1. Then
Theorem 2.6
Theorem 2.6 Let p(z) = 1 + a1z + · · · be analytic in D with p(z) ̸≡1 and let γ ∈(0, 1]. Then arg 2p(z)
Theorem 2.6 Let p(z) = 1 + a1z + · · · be analytic in D with p(z) ̸≡1 and let γ ∈(0, 1]. Then arg 2p(z)
Theorem 2.1
Theorem 2.1 is complete. ⊓⊔ Setting p(z) = f (z) z we obtain the following corollary:
Theorem 2.1 is complete. ⊓⊔ Setting p(z) = f (z) z we obtain the following corollary:
Corollary 2.8
Corollary 2.8 Let f (z) = z + a2z2 + · · · be analytic in D. Then arg 2 f (z) f ′(z) f (z) + zf ′(z) < γ π 2 ⇒ arg f (z) z < γ π 2.
Corollary 2.8 Let f (z) = z + a2z2 + · · · be analytic in D. Then arg 2 f (z) f ′(z) f (z) + zf ′(z) < γ π 2 ⇒ arg f (z) z < γ π 2 .
Theorem 2.7
Theorem 2.7 Let p(z) = 1 + a1z + · · · be analytic in D with p(z) ̸≡1 and let γ ∈(0, 1]. Then
Theorem 2.7 Let p(z) = 1 + a1z + · · · be analytic in D with p(z) ̸≡1 and let γ ∈(0, 1]. Then
Theorem 2.6
Theorem 2.6, there exist a point z0 ∈D such that the equalities (2.7) and (2.8) holds. We have 123
Theorem 2.6, there exist a point z0 ∈D such that the equalities (2.7) and (2.8) holds. We have 123
Corollary 2.9
Corollary 2.9 Let f (z) = z + a2z2 + · · · be analytic in D. Then
Corollary 2.9 Let f (z) = z + a2z2 + · · · be analytic in D. Then