tailieunhanh - On a new subclass of bi-univalent functions defined by using Salagean operator
In this manuscript, by using the Salagean operator, new subclasses of bi-univalent functions in the open unit disk are defined. Moreover, for functions belonging to these new subclasses, upper bounds for the second and third coefficients are found. | Turk J Math (2018) 42: 2891 – 2896 © TÜBİTAK doi: Turkish Journal of Mathematics Research Article On a new subclass of bi-univalent functions defined by using Salagean operator Bilal ŞEKER∗, Department of Mathematics, Faculty of Science, Dicle University, Diyarbakır, Turkey Received: • Accepted/Published Online: • Final Version: Abstract: In this manuscript, by using the Salagean operator, new subclasses of bi-univalent functions in the open unit disk are defined. Moreover, for functions belonging to these new subclasses, upper bounds for the second and third coefficients are found. Key words: Univalent functions, bi-univalent functions, coefficient bounds and coefficient estimates, Salagean operator 1. Introduction Let A denote the class of functions of the form f (z) = z + ∞ ∑ an z n , () n=2 which are analytic in the open unit disk U = {z ∈ C : |z| 0, p(z) = 1 + c1 z + c2 z 2 + . for z ∈ U. 2. Coefficient bounds for the function class HΣm,n (α) By introducing the function class HΣm,n (α) , we start by means of the following definition. Definition A function f (z) given by () is said to be in the class HΣm,n (α) (0 n) if the following conditions are satisfied: ( m ) D f (z) απ f ∈ Σ and arg n) . Then 2α |a2 | ≤ √ m n m 2α(3 − 3 ) + (2 − 2n )2 − α(22m − 22n ) () and |a3 | ≤ 2892 4α2 2α + . 3m − 3n (2m − 2n )2 () ŞEKER/Turk J Math Proof It can be written that the inequalities () and () are equivalent to Dm f (z) α = [p(z)] Dn f (z) () Dm g(w) α = [q(w)] Dn g(w) () and where p(z) and q(w) are in P and have the forms p(z) = 1 + p1 z + p2 z 2 + p3 z 3 + · · · () q(w) = 1 + q1 w + q2 w2 + q3 w3 + · · · . () and Now, equating the coefficients in () and (), we obtain (2m − 2n )a2 = αp1 () (3m − 3n )a3 − 2n (2m − 2n )a22 = αp2 + α(α − 1) 2 p1 2 () −(2m − 2n )a2 = αq1 () and (3m −
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