On a class of analytic functions associated to a complex domain concerning q-differential-difference operator

Lemma 2.1 Suppose the following data: a ∈C, a positive integer n and H[ϑ,n] =

Theorem 3.1 For υ ∈Λ if one of the following statements is given: • Sκ,k q υ(z) is of bounded boundary rotation; • υ satisfies the subordination structure Sκ,k q υ(z) ′ ≺ 1 + z 1 – z ♭ , ♭> 0,z ∈U; • υ fulfills the layout ℜ Sκ,k

Theorem 3.2 Consider υ ∈S∗ q(κ,k,h), where h(z) is convex univalent function in U. Then Sκ,k q υ(z) ≺zexp  z 0 h(ð(w)) – 1 w dw  , where ð(z) is analytic in U, with ð(0) = 0 and |ð(z)| < 1. Moreover, for |z| = χ, Sκ,k q υ(z) fulfills the formula exp

Corollary 3.1 ([8]) Let q −→1 in Theorem 3.2. Then Sκ,k 1 υ(z) ≺zexp  z 0 h(ð(w)) – 1 w dw  . Note that all the special cases of the class S∗ q(κ,k,h) can be considered as consequences of Theorem 3.2.

Theorem 3.3 If υ ∈Jκ,b q (A,B,k) then the odd function B(z) = 1 2  υ(z) – υ(–z)

Corollary 3.2 Let λ = 1 in Theorem 3.3. Then 1 + 1 b S0,k+1 q B(z) S0,k q B(z) – 1  ≺1 + Az 1 + Bz .

Corollary 3.3 Let κ = 0,k = 1 and q −→1 in Theorem 3.3. Then 1 + 1 b S0,2 q B(z) S0,1 q B(z) – 1  ≺1 + Az 1 + Bz .

Corollary 3.4 Let q −→1 in Theorem 3.3. Then 1 + 1 b Sκ,k+1 q B(z) Sκ,k q B(z) – 1  ≺1 + Az 1 + Bz .

Theorem 4.1 Consider Eq. (6) with β = 0 and ψ ∈Λ with non-negative coefficients. If h(z),z ∈U is univalent convex in U then there exists a solution satisfying the subordination (major solution) Sκ,k q υ(z) ≺zexp  z 0 h(ð(w)) – 1 w dw  , (7) where ð(z) is analytic in U, with ð(0) = 0 and |ð(z)| < 1.