Abstract
Let $ \mathcal{H} $ be the class of harmonic functions $ f=h+\bar{g} $ in the unit disk $\mathbb{D}:=\{z\in\mathbb{C} : |z|<1\}$, where $ h $ and $ g $ are analytic in $ \mathbb{D} $. Let
$$\mathcal{P}_{\mathcal{H}}^{0}(α)=\{f=h+\overline{g} \in \mathcal{H} : \real (h^{\prime}(z)-α)>|g^{\prime}(z)|\; \mbox{with}\; 0\leqα<1,\; g^{\prime}(0)=0,\; z \in \mathbb{D}\}
$$ be the class of close-to-convex mappings defined by Li and Ponnusamy \cite{Injectivity section}. In this paper, we obtain the s
Results & Lemmas (18)
Each result is auto-extracted from the paper via OCR - the math is shown verbatim and may carry occasional transcription artifacts. Click to expand the full statement.
Theorem 1.5.
Theorem 1.5. [27] Let f(z) = P∞ n=0 anzn be analytic in D and |f(z)| ≤1. Then (1.6) |f(z)| + ∞ X n=N |an||z|n ≤1 for |z| = r ≤RN, where RN…
Theorem 1.5. [27] Let f(z) = P∞ n=0 anzn be analytic in D and |f(z)| ≤1. Then (1.6) |f(z)| + ∞ X n=N |an||z|n ≤1 for |z| = r ≤RN, where RN is the positive root of the equation ψN(r) = 0, ψN(r) = 2(1 + r)rN −(1 −r)2. The radius RN is the best possible. Moreover, (1.7) |f(z)|2 + ∞ X n=N
Theorem 1.8.
Theorem 1.8. [27] Let f(z) = P∞ n=0 anzn be analytic in D, |f(z)| ≤1 and Sr denote the image of the subdisk |z| < r under mapping f. Then…
Theorem 1.8. [27] Let f(z) = P∞ n=0 anzn be analytic in D, |f(z)| ≤1 and Sr denote the image of the subdisk |z| < r under mapping f. Then (1.9) B1(r) := ∞ X n=0 |an|rn + 16 9 Sr π ≤1 for
Lemma 1.12.
Lemma 1.12. [34] Let f ∈P0 H(α) and be given by (1.11). Then for any n ≥2, (i) |an| + |bn| ≤2(1 −α) n; (ii) ||an| −|bn|| ≤2(1 −α) n;
Lemma 1.12. [34] Let f ∈P0 H(α) and be given by (1.11). Then for any n ≥2, (i) |an| + |bn| ≤2(1 −α) n ; (ii) ||an| −|bn|| ≤2(1 −α) n ;
Lemma 1.13.
Lemma 1.13. [13] Let f = h + g ∈P0 H(α) with 0 ≤α < 1. Then (1.14) |z| + ∞ X n=2 2(1 −α)(−1)n−1 n |z|n ≤|f(z)| ≤|z| + ∞ X n=2 2(1 −α) n
Lemma 1.13. [13] Let f = h + g ∈P0 H(α) with 0 ≤α < 1. Then (1.14) |z| + ∞ X n=2 2(1 −α)(−1)n−1 n |z|n ≤|f(z)| ≤|z| + ∞ X n=2 2(1 −α) n
Theorem 2.1.
Theorem 2.1. Let f ∈P0 H(α) be given by (1.11). Then, for N ≥2, (2.2) |f(z)| + ∞ X n=N (|an| + |bn|)|z|n ≤d (f(0), ∂f(D)) for |z| = r…
Theorem 2.1. Let f ∈P0 H(α) be given by (1.11). Then, for N ≥2, (2.2) |f(z)| + ∞ X n=N (|an| + |bn|)|z|n ≤d (f(0), ∂f(D)) for |z| = r ≤rN(α), where rN(α) is the smallest root of the equation (2.3) r −1 −2(1 −α)
Theorem 2.4.
Theorem 2.4. Let f ∈P0 H(α) be given by (1.11). Then, N ≥2, (2.5) |f(z)|2 + ∞ X n=N |an||z|n ≤d (f(0), ∂f(D)) for |z| = r ≤rN(α), where…
Theorem 2.4. Let f ∈P0 H(α) be given by (1.11). Then, N ≥2, (2.5) |f(z)|2 + ∞ X n=N |an||z|n ≤d (f(0), ∂f(D)) for |z| = r ≤rN(α), where rN(α) ∈(0, 1) is the smallest root of the equation (2.6) r−2(1−α)(r+ln (1−r)) 2 −2(1−α)
Theorem 2.7.
Theorem 2.7. Let f ∈P0 H(α) be given by (1.11). Then for a positive integer N ≥2, (2.8) |f(zm)| + ∞ X n=N |an||z|n ≤d (f(0), ∂f(D))
Theorem 2.7. Let f ∈P0 H(α) be given by (1.11). Then for a positive integer N ≥2, (2.8) |f(zm)| + ∞ X n=N |an||z|n ≤d (f(0), ∂f(D))
Theorem 2.10.
Theorem 2.10. Let f ∈P0 H(α) be given by (1.11). Then r + |h(r)|p + ∞ X n=2 (|an| + |bn|)rn ≤d (f(0), ∂f(D)), for r ≤rp(α), (2.11) where…
Theorem 2.10. Let f ∈P0 H(α) be given by (1.11). Then r + |h(r)|p + ∞ X n=2 (|an| + |bn|)rn ≤d (f(0), ∂f(D)) , for r ≤rp(α), (2.11) where rp(α) is the smallest root of the equation (2.12) rp + r −1 −2(1 −α) (r −1 + ln (2 −2r)) = 0 in (0, 1). The radius rp(α) is the best possible.
Theorem 3.1.
Theorem 3.1. [33] Let f(z) = P∞ n=0 anzn be analytic in D, |f(z)| ≤1 and Sr denotes the image of the subdisk |z| < r under mapping f. Then…
Theorem 3.1. [33] Let f(z) = P∞ n=0 anzn be analytic in D, |f(z)| ≤1 and Sr denotes the image of the subdisk |z| < r under mapping f. Then (3.2) B1(r) := ∞ X n=0 |an|rn + 16 9 Sr π ≤1 for
Theorem 3.4.
Theorem 3.4. [33] Let f(z) = P∞ n=0 anzn be analytic in D and |f(z)| ≤1. Then (3.5) |a0| + ∞ X n=1 |an| + 1 2|an|2 rn ≤1 for r ≤1 3
Theorem 3.4. [33] Let f(z) = P∞ n=0 anzn be analytic in D and |f(z)| ≤1. Then (3.5) |a0| + ∞ X n=1 |an| + 1 2|an|2 rn ≤1 for r ≤1 3
Theorem 3.6.
Theorem 3.6. Let f ∈P0 H(α) be given by (1.11). Then (3.7) r + ∞ X n=2 (|an| + |bn|)rn + P Sr π ≤d (f(0), ∂f(D)) for r ≤rN(α), where…
Theorem 3.6. Let f ∈P0 H(α) be given by (1.11). Then (3.7) r + ∞ X n=2 (|an| + |bn|)rn + P Sr π ≤d (f(0), ∂f(D)) for r ≤rN(α), where P(w) = wN + wN−1 + · · · + w, a polynomial in w of degree N −1, and rN(α) ∈(0, 1) is the smallest root of the equation (3.8) r−1−2(1−α) (r −1 + ln (2 −2r))+P
Corollary 3.9.
Corollary 3.9. Let f ∈P0 H(α) be given by (1.11). Then (3.10) r + ∞ X n=2 (|an| + |bn|)rn + Sr π ≤d (f(0), ∂f(D)) for r ≤rα, where rα…
Corollary 3.9. Let f ∈P0 H(α) be given by (1.11). Then (3.10) r + ∞ X n=2 (|an| + |bn|)rn + Sr π ≤d (f(0), ∂f(D)) for r ≤rα, where rα ∈(0, 1) is the smallest root of the equation (3.11) r2 + r −1 −2(1 −α)(3 −2α) (r + ln (1 −r)) −2(1 −α)(ln 2 −1) = 0.
Theorem 3.12.
Theorem 3.12. Let f ∈P0 H(α) be given by (1.11). Then (3.13) r + ∞ X n=2 |an| + |bn| + (|an| + |bn|)2 rn ≤d (f(0), ∂f(D)), for r ≤rα,…
Theorem 3.12. Let f ∈P0 H(α) be given by (1.11). Then (3.13) r + ∞ X n=2 |an| + |bn| + (|an| + |bn|)2 rn ≤d (f(0), ∂f(D)) , for r ≤rα, where rα ∈(0, 1) is the smallest root of the equation (3.14) r −1 −2(1 −α) (r −1 + ln (2 −2r)) + 4(1 −α2)(Li2(r) −r) = 0, where Li2(z) is a dilogarithm. The constant rα is best possible. 4. Refined Bohr Radius for the class P0 H(α)
Theorem 4.1.
Theorem 4.1. [38] Let f(z) = P∞ n=0 anzn be analytic in D and |f(z)| ≤1. Then ∞ X n=0 |an|rn + 1 1 + |a0| + r 1 −r ∞ X n=1
Theorem 4.1. [38] Let f(z) = P∞ n=0 anzn be analytic in D and |f(z)| ≤1. Then ∞ X n=0 |an|rn + 1 1 + |a0| + r 1 −r ∞ X n=1
Theorem 4.2.
Theorem 4.2. Let f ∈P0 H(α) be given by (1.11). Then r + ∞ X n=2 (|an| + |bn|)rn + 1 1 −rN ∞ X n=2 n(|an| + |bn|)2r2n (4.3) ≤d (f(0),…
Theorem 4.2. Let f ∈P0 H(α) be given by (1.11). Then r + ∞ X n=2 (|an| + |bn|)rn + 1 1 −rN ∞ X n=2 n(|an| + |bn|)2r2n (4.3) ≤d (f(0), ∂f(D)) for r ≤rN(α),
Theorem 4.5.
Theorem 4.5. Let f ∈P0 H(α) be given by (1.11). Then r + ∞ X n=2 (|an| + |bn|)rn + 1 1 + |a2| + |b2| + rm 1 −rm ∞ X
Theorem 4.5. Let f ∈P0 H(α) be given by (1.11). Then r + ∞ X n=2 (|an| + |bn|)rn + 1 1 + |a2| + |b2| + rm 1 −rm ∞ X
Theorem 4.8.
Theorem 4.8. Let f ∈P0 H(α) be given by (1.11). Then r + (1 −(1 + |a2| + |b2| −(|a2| + |b2|)2)) r 1 −(|a2| + |b2|)r + ∞ X n=3 (|an| +…
Theorem 4.8. Let f ∈P0 H(α) be given by (1.11). Then r + (1 −(1 + |a2| + |b2| −(|a2| + |b2|)2)) r 1 −(|a2| + |b2|)r + ∞ X n=3 (|an| + |bn|)rn (4.9) ≤d (f(0), ∂f(D)) for r ≤rα, where rα ∈(0, 1) is the smallest root of the equation (4.10) r −(1 −(|a2| + |b2|)2) r 1 −(|a2| + |b2|)r
Theorem 5.1.
Theorem 5.1. Let f ∈P0 H(α) be given by (1.11). Then |f(z)| + q |Jf(z)|r + ∞ X n=N (|an| + |bn|)rn ≤d (f(0), ∂f(D)) for r ≤rN(α), (5.2)…
Theorem 5.1. Let f ∈P0 H(α) be given by (1.11). Then |f(z)| + q |Jf(z)|r + ∞ X n=N (|an| + |bn|)rn ≤d (f(0), ∂f(D)) for r ≤rN(α), (5.2) where rN(α) ∈(0, 1) is the smallest root of the equation r −1 −2(1 −α) 2r −1 + r2 2 + · · · + rN−1
Function classes studied:
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