J. Indones. Math. Soc.

Proposition 1.1. [3] Let q be univalent in D and let φ be analytic in a domain containing q(D) . Let zq′(z)φ[q(z)] be starlike. If p is analytic in D, p(0) = q(0) and satisfies zp′(z)φ[p(z)] ≺zq′(z)φ[q(z)] then p ≺q and q is the best dominant.

Lemma 1.2. Let β0 = 2 √ 2|D−E| (1−|E|) . If 1 + βzp′(z) ≺1+Dz 1+Ez (β ≥β0) then p(z) ≺ √1 + z .

Lemma 1.3. Let β0 = 4|D−E| (1−|E|). If 1+β zp′(z) p(z) ≺1+Dz 1+Ez (β ≥β0) then p(z) ≺√1 + z.

Lemma 1.4. Let β0 = 4 √ 2|D−E| (1−|E|) . If 1 + β zp′(z) p2(z) (β ≥β0) ≺1+Dz 1+Ez then p(z) ≺ √1 + z . With appropriate choices of φ in Proposition 1.1, and using similar approach as above Lemma 1.3 and Lemma 1.4 is easily verified. Details of proving these lemmas can be found in [4]. 2. Main Results

Theorem 2.1. Let β0 = 2 √ 2(|D−E|) 1−|E| , |E| < 1 , |D| ≤1 , D ̸= E , β ≥β0 and f ∈A. If f satisfies 1+β { (1 −α)zf ′(z) f(z) ( zf ′′(z) f ′(z) −zf ′(z)

Corollary 2.2. Let β0 = 2 √ 2(|D−E|) 1−|E| , |E| < 1 , |D| ≤1 , D ̸= E , β ≥β0 and f ∈A. If f satisfies 1 + β zf ′′(z) f ′(z) {z[f ′′(z)]′ f ′′(z) −zf ′′(z) f ′(z) + 1 } ≺1 + Dz

Theorem 2.3. Let β0 = 4|D−E| (1−|E|) and β ≥β0. Suppose f ∈A and 1+β        (1 −α)zf ′(z) [ 1 −zf ′(z) f(z) + zf ′′(z)

Corollary 2.4. Let β0 = 4|D−E| (1−|E|) , β ≥β0 and f ∈A. If 1 + β zf ′′(z) [f ′(z) + zf ′′(z)] {z[f ′′(z)]′ f ′′(z) −zf ′′(z) f ′(z) + 1 } ≺1 + Dz 1 + Ez then f ∈SLc.

Theorem 2.5. Let β0 = 4 √ 2|D−E| (1−|E|) , β ≥β0 and f ∈A. 1+βz        (1 −α)[f ′(z)]3f(z) [

Corollary 2.6. Let β0 = 4 √ 2|D−E| (1−|E|) , β ≥β0 and f ∈A. 1 + β zf ′(z)f ′′(z) [f ′(z) + zf ′′(z)]2 {z[f ′′(z)]′ f ′′(z) + 1 −zf ′′(z) f ′(z) } ≺1 + Dz 1 + Ez ⇒f ∈SLc.