Miskolc Mathematical Notes

Theorem 1. Let the function f(z) be defined by (1.1). Also let the following in- equality holds true: ∞ ∑ k=2  (k −1)+|ζ(1+m)+m(k −1)|

Corollary 1. Let the function f(z) be defined by (1.1). Also let the following inequality holds true: ∞ ∑ k=2 {(k −1)+|ζ(1+m)+m(k −1)|}|ak|· k! (λ+1)k−1 m+1 m+k α ≦|ζ(1+m)|. (2.2) Then the function f(z) belongs to the class S λ,α m (ζ,M ).

Corollary 2. Let the function f(z) be defined by (1.1). Also let the following inequality holds true: ∞ ∑ k=2 {(k −1)+|ζ(1+m)+m(k −1)|}|ak| [k]q! [λ+1]q,k−1 ≦|ζ(1+m)|. (2.3) Then the function f(z) belongs to the class F λ q (ζ,M ).

Theorem 2. Let the function f(z) given by (1.1) be in the normalized complex- order analytic function class H λ,α m,q (ζ,M ). (a) If 2m(k −1)ℜ(ζ) > (k −1)2(1−m)−|ζ|2 (1+m), let G = " 2m(k −1)ℜ(ζ) (k −1)2(1−m)−|ζ|2 (1+m) # (k = 2,3,4,··· , j −1), where N = [G] (the Gaussian symbol) and [G] is the greatest integer not greater than G. Then aj

Corollary 3. Let the function f(z) defined by (1.1) be in the class S λ,α m (ζ,M ). (a) If 2m(k −1)ℜ(ζ) > (k −1)2(1−m)−|ζ|2 (1+m), let G = " 2m(k −1)ℜ(ζ) (k −1)2(1−m)−|ζ|2 (1+m) # (k = 2,3,4,··· , j −1), where N = [G](the Gaussian symbol) and [G] is the greatest integer not greater than G. Then aj ≦

Corollary 4. Let the function f(z) be defined by (1.1) be in the class F λ q (ζ,M ). (a) If 2m(k −1)ℜ(ζ) > (k −1)2(1−m)−|ζ|2 (1+m), let G = " 2m(k −1)ℜ(ζ) (k −1)2(1−m)−|ζ|2 (1+m) # (k = 2,3,4,··· , j −1), where N = [G](the Gaussian symbol) and [G] is the greatest integer not greater than G. Then aj ≦

Lemma 1 ([10]). Let w(z) = ∞ ∑ k=1 ckzk ∈Ω. If µ is any complex number, then c2 −µc2 1 ≦max{1,|µ|} (4.1) for any complex number µ. Equality in (4.1) may be attained with the functions w(z) = z2 and w(z) = z for |µ| < 1 and |µ| ≧1, respectively. We now state and prove our main result in this section.

Theorem 3. Let the function f(z) defined by (1.1) be in the normalized complex- order analytic function class H λ,α m,q (ζ,M ). Suppose also that µ is any complex num- ber. Then a3 −µa2 2 ≦ |ζ(1+m)| 2 [3]q! [λ+1]q,2 m+1 m+3 α max{1,|δ|}, (4.2)