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Abstract
The main motive of this paper is to find an upper bound of the fourth Hankel
determinant H4,1
f
for a subclass S, with hyperbolic domain.
Results & Lemmas (15)
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.
Lemma 1.
Lemma 1. [18] If p (z) ∈P and of the form p(z) = z + P∞ n=2 an zn, z ∈A, then c2 – νc2 1 ≤ –4ν + 2 (ν ≤0), 2 (0 ≤ν ≤1) 4ν – 2 (ν…
Lemma 1. [18] If p (z) ∈P and of the form p(z) = z + P∞ n=2 an zn, z ∈A, then c2 – νc2 1 ≤ –4ν + 2 (ν ≤0), 2 (0 ≤ν ≤1) 4ν – 2 (ν ≥1).
Lemma 2.
Lemma 2. [22] If p (z) ∈P and of the form p(z) = z + P∞ n=2 an zn, z ∈A,, then, for all n, m ∈N |µcncm – cn+m| ≤ ( 2 0 ≤µ ≤1 2|2µ – 1|…
Lemma 2. [22] If p (z) ∈P and of the form p(z) = z + P∞ n=2 an zn, z ∈A,, then, for all n, m ∈N |µcncm – cn+m| ≤ ( 2 0 ≤µ ≤1 2|2µ – 1| otherwise
Lemma 3.
Lemma 3. [16] If p (z) ∈P and of the form p(z) = z + P∞ n=2 an zn, z ∈A,, then, for all n, m ∈N c3 1 – 2c1c2 + c3
Lemma 3. [16] If p (z) ∈P and of the form p(z) = z + P∞ n=2 an zn, z ∈A,, then, for all n, m ∈N c3 1 – 2c1c2 + c3
Theorem 1.
Theorem 1. Let f(z) ∈K – SL a, b where 0 < k < 1, and a, b are taken according to (4) and (5) and has the form (1). Then |an| ≤2T2 |1 –…
Theorem 1. Let f(z) ∈K – SL a, b where 0 < k < 1, and a, b are taken according to (4) and (5) and has the form (1). Then |an| ≤2T2 |1 – b| n 1 – k2 , for all n ≥2 (9) These inqualities is sharp.
Theorem 2.
Theorem 2. If f (z) ∈K – SL a, b where 0 < k < 1, and a, b are taken according to (4) and (5). Then for a real number µ, we have a3 –…
Theorem 2. If f (z) ∈K – SL a, b where 0 < k < 1, and a, b are taken according to (4) and (5) . Then for a real number µ, we have a3 – µa2 2 ≤T2 |1 – b| 3 1 – k2
Corollary 1.
Corollary 1. If f (z) ∈K – SL a, b where 0 < k < 1 and a, b are taken according to (4) and (5), then for µ = 1,we have a3 – a2 2 ≤2T2 |1…
Corollary 1. If f (z) ∈K – SL a, b where 0 < k < 1 and a, b are taken according to (4) and (5) , then for µ = 1,we have a3 – a2 2 ≤2T2 |1 – b| 3 1 – k2 . (19) 06
Theorem 3.
Theorem 3. If f (z) ∈K – SL a, b where 0 < k < 1 and a, b are taken according to (4) and (5), then |a4 – a2a3| ≤T2 |1 – b| 6 1 – k2.…
Theorem 3. If f (z) ∈K – SL a, b where 0 < k < 1 and a, b are taken according to (4) and (5) , then |a4 – a2a3| ≤T2 |1 – b| 6 1 – k2 . (20)
Corollary 2.
Corollary 2. If f (z) ∈K – SL a, b where 0 < k < 1, and a, b are taken according to (4) and (5), then |a4 – µa2a3| ≤T2 |1 – b| 6 1 –…
Corollary 2. If f (z) ∈K – SL a, b where 0 < k < 1, and a, b are taken according to (4) and (5) , then |a4 – µa2a3| ≤T2 |1 – b| 6 1 – k2 + (1 – µ) 2T4 1 – b 2 3 1 – k22 . (22)
Theorem 4.
Theorem 4. If f (z) ∈K – ST a, b where 0 < k < 1, and a, b are taken according to (4) and (5), then a2a4 – a2 3 ≤T4 1 – b 2 18 1 –…
Theorem 4. If f (z) ∈K – ST a, b where 0 < k < 1, and a, b are taken according to (4) and (5) , then a2a4 – a2 3 ≤T4 1 – b 2 18 1 – k22 . (23)
Corollary 3.
Corollary 3. If f (z) ∈K – SL a, b where 0 < k < 1, and a, b are taken according to (4) and (5), then a2a4 – µa2 3 ≤T4 1 – b 2 2 1 –…
Corollary 3. If f (z) ∈K – SL a, b where 0 < k < 1, and a, b are taken according to (4) and (5) , then a2a4 – µa2 3 ≤T4 1 – b 2 2 1 – k22 1 – 4 9µ .
Theorem 5.
Theorem 5. If f (z) ∈K – ST a, b where 0 < k < 1, and a, b are taken according to (4) and (5), then H3,1 f ≤T6 1 – b 3 3 1 – k23…
Theorem 5. If f (z) ∈K – ST a, b where 0 < k < 1, and a, b are taken according to (4) and (5) , then H3,1 f ≤T6 1 – b 3 3 1 – k23 + 17T4 1 – b 2 18 1 – k22 . (26)
Theorem 6.
Theorem 6. If f (z) ∈K – SL a, b where 0 < k < 1, and a, b are taken according to (4) and (5), then a5 – a2 3 ≤4T2 1 – b 45 1 – k2.…
Theorem 6. If f (z) ∈K – SL a, b where 0 < k < 1, and a, b are taken according to (4) and (5) , then a5 – a2 3 ≤4T2 1 – b 45 1 – k2 . (28)
Theorem 7.
Theorem 7. If f (z) ∈K – SL a, b where 0 < k < 1, and a, b are taken according to (4) and (5), then |a2a5 – a3a4| ≤2T4 1 – b 2 15 1 –…
Theorem 7. If f (z) ∈K – SL a, b where 0 < k < 1, and a, b are taken according to (4) and (5) , then |a2a5 – a3a4| ≤2T4 1 – b 2 15 1 – k22 . (29)
Theorem 8.
Theorem 8. If f (z) ∈K – SL a, b where 0 < k < 1, and a, b are taken according to (4) and (5), then a3a5 – a2 4 ≤T4 1 – b 2 15 1 –…
Theorem 8. If f (z) ∈K – SL a, b where 0 < k < 1, and a, b are taken according to (4) and (5) , then a3a5 – a2 4 ≤T4 1 – b 2 15 1 – k22 . (30)
Theorem 9.
Theorem 9. If f (z) ∈K – SL a, b where 0 < k < 1, and a, b are taken according to (4) and (5), then H4,1 f ≤44T8 1 – b 4 1575 1 –…
Theorem 9. If f (z) ∈K – SL a, b where 0 < k < 1, and a, b are taken according to (4) and (5) , then H4,1 f ≤44T8 1 – b 4 1575 1 – k24 + 91T6 1 – b 3 750 1 – k23 . (31)
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