TAMKANG JOURNAL OF MATHEMATICS

Theorem 2.1. Let f ∈Σα. If ∞ X n=1 h (n + α)(1 + B) + (A + 1) i |an| ≤(B −A) −α(1 −B), (z ∈U) (3) holds and B(1 + α) > A + α, then f ∈Sα(A, B).

Theorem 2.2. Let f ∈Σα. If ∞ X n=1 (n + α) h (n + α)(1 + B) + (A + 1) i |an| ≤(1 + α) h (B −A) −α(1 −B) i , (z ∈U) (5) holds and B(1 + α) > A + α, then f ∈Cα(A, B). Note that when α = 0, Theorem 2.1 and Theorem 2.2 reduce to Theorem 2.2 and

Theorem 2.1 in [5] respectively. Further, we note that these sufficient conditions are also necessary for functions of the form (1) when α = 0, A = 2µ −1, B = 1 with positive or negative coefficients ([1, 2, 3]). 3. Main results We consider in this section partial sums of functions in the classes Sα(A, B) and Cα(A, B) and obtain the sharp lower bounds for the ratio of real part of f(z) to fk(z) and f ′(z) to f ′ k(z). In the sequel, we will make use of the generalized result such that ℜ{(1 + wα(z))/

Theorem 3.1. Let f be given by (1) and satisfies (3) then ℜ n f(z) fk(z) o ≥ 2(1 + k + α + A) 2(k + α) + (2 + A + B), (z ∈U). (6) The result is sharp for every k with extremal function f(z) = 1 z1+α + (B −A) −α(1 −B) 2(k + α) + (2 + A + B)zk+1+α, k ≥0.

Corollary 3.1. Let f be given by (1) and satisfies (3) then ℜ n f(z) fk(z) o ≥ 2(1 + k + A) 2k + 2 + A + B , (z ∈U). (9)

Corollary 3.2. Let the assumptions of Theorem 3.1 hold. Then for f of the form (1) satisfies condition ∞ X n=1 (n + µ)|an| ≤1 −µ, (z ∈U), ℜ n f(z) fk(z) o ≥ k + 2µ k + 1 + µ, (z ∈U). (11) The result is sharp for every k with extremal function

Theorem 3.2. Let f ∈Σα and ∞ X n=1 (n + α) h (n + α)(1 + B) + (A + 1) i |an| ≤(1 + α) h (B −A) −α(1 −B) i , (z ∈U) holds, then ℜ

Corollary 3.3. Let f be given by (1) and satisfies (5) then ℜ n f(z) fk(z) o ≥ (k + 2)(2k + A + B) (k + 1)(2k + 2 + A + B), (z ∈U). (16) The result is sharp for every k with extremal function f(z) = 1 z + (B −A) (k + 1)(2k + 2 + A + B)zk+1, k ≥0. (17)

Corollary 3.4. Let the assumptions of Theorem 3.2 hold. Then for f(z) of the form (1) satisfies condition ∞ X n=1 n(n + µ)|an| ≤1 −µ, (z ∈U), ℜ n f(z) fk(z) o ≥ (k + 2)(k + µ) (k + 1)(k + 1 + µ), (z ∈U). (18) The result is sharp for every k with extremal function

Theorem 3.3. Let f ∈Σα such that ∞ X n=1 h (n + α)(1 + B) + (A + 1) i |an| ≤(B −A) −α(1 −B), (z ∈U) holds. Then ℜ nfk(z) f(z) o ≥2(k + 1) + A + B + α(1 + B) k + 2 + 2αB

Corollary 3.5. Let the assumptions of Corollary 3.1 hold. Then ℜ nfk(z) f(z) o ≥2(k + 1) + A + B k + 2 , (z ∈U). (22)

Corollary 3.6. Let the assumptions of Corollary 3.2 hold. Then ℜ nfk(z) f(z) o ≥k + 1 + µ k + 2 , (z ∈U). (23)

Theorem 3.4. Let f be given by (1) and satisfies (5) then ℜ nfk(z) f(z) o ≥(k + 1)2(2k + 2 + A + B) + αν 2(k + 1)(k + 2) −(B −A) + αω , (z ∈U). (24) where ν :=  (k + 1 + α)(1 + B) + (A + 1) + (k + 1)(B + 1)  and ω :=

Corollary 3.7. Let the assumptions of Corollary 3.3 hold. Then ℜ nfk(z) f(z) o ≥(k + 1)2(2k + 2 + A + B) 2(k + 1)(k + 2) −(B −A), (z ∈U). (25)

Corollary 3.8. Let the assumptions of Corollary 3.4 hold. Then ℜ nfk(z) f(z) o ≥ (k + 1)(k + 1 + µ) (k + 1)(k + 2) −k(1 −µ), (z ∈U). (26)

Theorem 3.5. Let f be given by (1) and satisfies (3) with A = −B. Then ℜ n f ′(z) f ′ k(z) o ≥0, (z ∈U), (27) ℜ nf ′ k(z) f ′(z) o ≥1 + 2α 2(1 + α), (z ∈U).

Theorem 3.6. Let f be given by (1) and satisfies (5). Then ℜ n f ′(z) f ′ k(z) o ≥2(k + A + B) + φ −α[(k + 1)(B −A −α(1 −B))] (2 + 2k + A + B) + φ , (z ∈U), (29) ℜ nf ′ k(z) f ′(z) o