Results & Lemmas (11)
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Lemma 1.
Lemma 1. [12] If f(z) = z + ∞ P n=2 anzn belongs to the class SLk, then we have |an| ≤|τk|n−1 Fk,n, (7) where τk = (k − √ k2 + 4)/2.…
Lemma 1. [12] If f(z) = z + ∞ P n=2 anzn belongs to the class SLk, then we have |an| ≤|τk|n−1 Fk,n , (7) where τk = (k − √ k2 + 4)/2. Equality holds in (7) for the function gk(z) = z 1 −kτkz −τ 2 kz2 =
Theorem 2.
Theorem 2. Let Fk,n be the sequence of k-Fibonacci numbers defined in. If epk(z) = 1 + τ 2 kz2 1 −kτkz −τ 2 kz2 = 1 + ∞ X n=1 pnzn, (9)…
Theorem 2. Let {Fk,n} be the sequence of k-Fibonacci numbers defined in . If epk(z) = 1 + τ 2 kz2 1 −kτkz −τ 2 kz2 = 1 + ∞ X n=1 pnzn, (9) where τk = k− √ k2+4 2
Lemma 3.
Lemma 3. [13] Let p ∈P with p(z) = 1 + c1z + c2z2 + · · ·, then |cn| ≤2, for n ≥1. (11) If |c1| = 2, then p(z) ≡p1(z) ≡(1 + xz)/(1 −xz)…
Lemma 3. [13] Let p ∈P with p(z) = 1 + c1z + c2z2 + · · · , then |cn| ≤2, for n ≥1. (11) If |c1| = 2, then p(z) ≡p1(z) ≡(1 + xz)/(1 −xz) with x = c1 2 . Conversely, if p(z) ≡p1(z) for some |x| = 1, then c1 = 2x. Furthermore, we have c2 −c2 1 2 ≤2 −|c1|2 2 . (12) If |c1| < 2, and
Lemma 4.
Lemma 4. ([9]) Let p ∈P with coefficients cn as above, then |c3 −2c1c2 + c3 1| ≤2. (13) In 1976, Noonan and Thomas [10] stated the sth Hankel…
Lemma 4. ([9]) Let p ∈P with coefficients cn as above, then |c3 −2c1c2 + c3 1| ≤2. (13) In 1976, Noonan and Thomas [10] stated the sth Hankel determinant for s ≥1 and q ≥1 as Hs(q) =
Theorem 5.
Theorem 5. If f(z) = z + ∞ P n=2 anzn belongs to the class KSLk, then we have |an| ≤|τk|n−1 Fk,n n, (15) where τk = (k − √ k2 + 4)/2.…
Theorem 5. If f(z) = z + ∞ P n=2 anzn belongs to the class KSLk, then we have |an| ≤|τk|n−1 Fk,n n , (15) where τk = (k − √ k2 + 4)/2. Equality holds in (7) for the function fk(z) = 1 1 + τ 2
Theorem 6.
Theorem 6. If p(z) = 1 + p1z + p2z2 + · · · and p(z) ≺epk(z) = 1 + τ 2 kz2 1 −kτkz −τ 2 kz2, τk = k − √ k2 + 4 2, z ∈D, then we have |p1| ≤…
Theorem 6. If p(z) = 1 + p1z + p2z2 + · · · and p(z) ≺epk(z) = 1 + τ 2 kz2 1 −kτkz −τ 2 kz2 , τk = k − √ k2 + 4 2 , z ∈D, then we have |p1| ≤ √ k2 + 4 −k
Theorem 7.
Theorem 7. If p(z) = 1 + p1z + p2z2 + · · · and p(z) ≺epk(z) = 1 + τ 2 kz2 1 −kτkz −τ 2 kz2, τk = k − √ k2 + 4 2, z ∈D, then we have |p3|…
Theorem 7. If p(z) = 1 + p1z + p2z2 + · · · and p(z) ≺epk(z) = 1 + τ 2 kz2 1 −kτkz −τ 2 kz2 , τk = k − √ k2 + 4 2 , z ∈D, then we have |p3| ≤(k3 + 3k) (√ k2 + 4 −k 2
Theorem 8.
Theorem 8. If f(z) = z + a2z2 +... belongs to SLk, then |a2a4 −a2 3| ≤2k4 + 6k2 + 3 3 (√ k2 + 4 −k 2 )4. (31)
Theorem 8. If f(z) = z + a2z2 + . . . belongs to SLk, then |a2a4 −a2 3| ≤2k4 + 6k2 + 3 3 (√ k2 + 4 −k 2 )4 . (31)
Theorem 9.
Theorem 9. If f(z) = z + a2z2 +... belongs to KSLk, then |a2a4 −a2 3| ≤3k4 + 9k2 + 4 36 (√ k2 + 4 −k 2 )4. 171
Theorem 9. If f(z) = z + a2z2 + . . . belongs to KSLk, then |a2a4 −a2 3| ≤3k4 + 9k2 + 4 36 (√ k2 + 4 −k 2 )4 . 171
Corollary 10.
Corollary 10. If f(z) = z + a2z2 +... belongs to SL, then |a2a4 −a2 3| ≤11 3 (√ 5 −1 2 )4. (34)
Corollary 10. If f(z) = z + a2z2 + . . . belongs to SL, then |a2a4 −a2 3| ≤11 3 (√ 5 −1 2 )4 . (34)
Corollary 11.
Corollary 11. If f(z) = z + a2z2 +... belongs to KSL, then |a2a4 −a2 3| ≤4 9 (√ 5 −1 2 )4. (35) Acknowledgement This research has been…
Corollary 11. If f(z) = z + a2z2 + . . . belongs to KSL, then |a2a4 −a2 3| ≤4 9 (√ 5 −1 2 )4 . (35) Acknowledgement This research has been supported with grant number FEN.17.026 by DUBAP (Dicle University Coordination Committee of Scientific Research Projects). The authors would like to thank DUBAP for their supporting and the referees for the helpful suggestions. References