ON CERTAIN SUBCLASSES OF UNIVALENT FUNCTIONS OF COMPLEX ORDER ASSOCIATED WITH PASCAL DISTRIBUTION SERIES

In this study, by establishing a connection between normalized univalent functions in the unit disc and Pascal distribution series, we have obtained the necessary and su¢ cient conditions for these functions to belong to some subclasses of univalent functions of complex-order. We also determined some conditions by considering the integral operator for these functions. 1. Introduction Let A stand for the standard class of analytic functions of the form f (z) = z + 1 X k=2 akz ; z 2 U = fz 2 C : jzj < 1g : (1) Moreover, let S be the class of functions in A, which are univalent in U (see [5]). The necessary and su¢ cient condition for a function f 2 A to be called starlike of complex order ( 2 C = C n f0g) is f(z) z 6= 0; z 2 U, and Re 1 + 1 zf 0(z) f(z) 1 > 0; (z 2 U): (2) We denote the class of these functions with S ( ). The class S ( ) introduced by Nasr and Aouf [10]. The necessary and su¢ cient condition for a function f 2 A to be called convex function of order ( 2 C ), that is f 2 C( ) is f 0(z) 6= 0 in U and Re 1 + 1 zf 00(z) f 0(z) > 0; (z 2 U): (3) 2020 Mathematics Subject Classication. Primary 30C45, 30C50, 30C55. Keywords and phrases. Univalent functions, complex order, Pascal distribution, coe¢ cient bounds, coe¢ cient estimates. bilal.seker@dicle.edu.tr; sevtaps@dicle.edu.tr-Corresponding author 0000-0003-1777-8145; 0000-0002-2573-0726. c 2021 Ankara University Communications Facu lty of Sciences University of Ankara-Series A1 Mathematics and Statistics 849 850 B. ŞEKER, S. SÜMER EKER The class C( ) was introduced by Wiatrowski [15]. It follows from (2) and (3) that for a function f 2 A we have the equivalence f 2 C( ), zf 0 2 S ( ): For a function f 2 A, we say that it is close-to-convex function of order ( 2 C ), that is f 2 R( ), if and only if Re 1 + 1 (f 0(z) 1) > 0; (z 2 U): The class R( ) was studied by Halim [6] and Owa [11]. Let T A represent the functions of the form f (z) = z 1 X k=2 akz ; (ak 0): (4) Many important results for the class T have been given by Silverman [14]. A lot of consequences have obtained by researchers about the functions in the class T . Using the functions of the form f (z) = z P1 k=n+1 akz , Alt¬ntaş et al. [2] dened following subclasses of A(n), which generalizes the results of Nasr et al. and Wiatrowski [10, 15], and obtained several results for this class. It is clear that for n = 1, we obtain the class T . Denition 1. [2] Let Sn( ; ; ) denote the subclass of T consisting of functions f which satisfy the inequality 1 zf 02f 00(z) zf 0(z) + (1 )f(z) 1 < ; (z 2 U; 2 C ; 0 < 1; 0 1): Also let Rn( ; ; ) denote the subclass of T consisting of functions f which satisfy the inequality 1 (f 0(z) + zf 00(z) 1) < ; (z 2 U; 2 C ; 0 < 1; 0 1): We note that Sn( ; 0; 1) S n( ) and Rn( ; 0; 1) Rn( ): Recently, it has been established a power series that its coe¢ cients were probabilities of the elementary distributions such as Poisson, Pascal, Binomial, etc. Many researchers have obtained several results about some subclasses of univalent functions using these series. (see, for example [1, 3, 7, 8, 9, 12,13] ) ON CERTAIN SUBCLASSES OF UNIVALENT FUNCTIONS 851 A variable x is said to have the Pascal distribution if it takes on the values 0; 1; 2; 3; ::: with the probabilities (1 q), qr(1 q) r 1! , qr(r+1)(1 q) 2! , qr(r+1)(r+2)(1 q) 3! ,..., respectively, where q and r are parameters. Hence P (X = k) = k + r 1 r 1 q(1 q); k 2 f0; 1; 2; :::g: Recently, El-Deeb et al. [4] introduced the following power series whose coe¢ cients are probabilities of the Pascal distribution and stated some su¢ cient conditions for the Pascal distribution series and other related series to be in some subclasses of analytic functions. Kq(z) := z + 1 X k=2 k + r 2 r 1 q (1 q)z (5) (z 2 U; r 1; 0 q 1): Now let us introduce the following new power series whose coe¢ cients are probabilities of the Pascal distribution. q(z) := 2z Kq(z) = z 1 X k=2 k + r 2 r 1 q (1 q)z (6) (z 2 U; r 1; 0 q 1): It is clear that q(z) is in the class T . Note that, by using ratio test we deduce that the radius of convergence of the power series Kq(z) and r q(z) are innity. We will need the following Lemmas from Alt¬ntaş et al. [2] to prove our main results. Lemma 2. [2] Let the function f 2 A(n), then f is in the class Sn( ; ; ) if and only if 1 X k=n+1 [ (k 1) + 1] (k + j j 1) ak j j: (7) Lemma 3. [2] Let the function f 2 A(n), then f is in the class Rn( ; ; ) if and only if 1 X k=n+1 k [ (k 1) + 1] ak j j: (8) Throughout this paper, we suppose that n = 1 for the functions in the classes Sn( ; ; ) andRn( ; ; ) and we will write S1( ; ; ) = S( ; ; ) andR1( ; ; ) = R( ; ; ) for briey. In the present paper, we established necessary and su¢ cient conditions for the functions that coe¢ cients consist of Pascal distribution series to be in S( ; ; ) and R( ; ; ). Also, we studied similar properties for integral transforms related to these series. 852 B. ŞEKER, S. SÜMER EKER 2. Main Results Theorem 4. q(z) given by (6) is in the class S( ; ; ) if and only if qr(r + 1) (1 q)2 + qr( j j+ + 1) 1 q j j(1 q) : (9) Proof. To prove that q 2 S( ; ; ), according to Lemma 2, it is su¢ cient to show that 1 X k=2 [ (k 1) + 1] (k + j j 1) k + r 2 r 1 q (1 q) j j: (10) We will use the following very known relation 1 X k=0 k + r 1 r 1 q = 1 (1 q)r ; 0 q 1: and the corresponding ones obtained by replacing the value of r with r 1,r + 1 and r + 2 in our proofs. By making calculations on the left hand side of the inequality (10) we obtain, 1 X k=2 [ (k 1) + 1] (k + j j 1) k + r 2 r 1 q (1 q) = (1 q) " 1 X k=2 k + r 2 r 1 q 1 (k 1)(k 2) + 1 X k=2 k + r 2 r 1 q 1 j j


Introduction
Let A stand for the standard class of analytic functions of the form Moreover, let S be the class of functions in A, which are univalent in U (see [5]). The necessary and su¢ cient condition for a function f 2 A to be called starlike of complex order ( 2 C = C n f0g) is f (z) z 6 = 0; z 2 U, and We denote the class of these functions with S ( ). The class S ( ) introduced by Nasr and Aouf [10].
The necessary and su¢ cient condition for a function f 2 A to be called convex function of order ( 2 C ), that is f 2 C( ) is f 0 (z) 6 = 0 in U and The class C( ) was introduced by Wiatrowski [15]. It follows from (2) and (3) that for a function f 2 A we have the equivalence For a function f 2 A, we say that it is close-to-convex function of order ( 2 C ), that is f 2 R( ), if and only if The class R( ) was studied by Halim [6] and Owa [11]. Let T A represent the functions of the form Many important results for the class T have been given by Silverman [14]. A lot of consequences have obtained by researchers about the functions in the class T . Using the functions of the form f (z) = z P 1 k=n+1 a k z k , Alt¬ntaş et al. [2] de…ned following subclasses of A(n), which generalizes the results of Nasr et al. and Wiatrowski [10,15], and obtained several results for this class. It is clear that for n = 1, we obtain the class T . zf 02 f 00 (z) zf 0 (z) + (1 )f (z) 1 < ; (z 2 U; 2 C ; 0 < 1; 0 1): Also let R n ( ; ; ) denote the subclass of T consisting of functions f which satisfy the inequality 1 (f 0 (z) + zf 00 (z) 1) < ; (z 2 U; 2 C ; 0 < 1; 0 1): We note that S n ( ; 0; 1) S n ( ) and R n ( ; 0; 1) R n ( ): Recently, it has been established a power series that its coe¢ cients were probabilities of the elementary distributions such as Poisson, Pascal, Binomial, etc. Many researchers have obtained several results about some subclasses of univalent functions using these series. (see, for example [1,3,7,8,9,12,13 ,..., respectively, where q and r are parameters. Hence k 2 f0; 1; 2; :::g: Recently, El-Deeb et al. [4] introduced the following power series whose coe¢cients are probabilities of the Pascal distribution and stated some su¢ cient conditions for the Pascal distribution series and other related series to be in some subclasses of analytic functions. (z 2 U; r 1; 0 q 1): Now let us introduce the following new power series whose coe¢ cients are probabilities of the Pascal distribution.
is in the class T . Note that, by using ratio test we deduce that the radius of convergence of the power series K r q (z) and r q (z) are in…nity. We will need the following Lemmas from Alt¬ntaş et al. [2] to prove our main results.
In the present paper, we established necessary and su¢ cient conditions for the functions that coe¢ cients consist of Pascal distribution series to be in S( ; ; ) and R( ; ; ). Also, we studied similar properties for integral transforms related to these series.

Main Results
Theorem 4. r q (z) given by (6) is in the class S( ; ; ) if and only if q 2 r(r + 1) (1 q) 2 + qr( j j + + 1) 1 q j j(1 q) r : Proof. To prove that r q 2 S( ; ; ), according to Lemma 2, it is su¢ cient to show that We will use the following very known relation and the corresponding ones obtained by replacing the value of r with r 1,r + 1 and r + 2 in our proofs.
By making calculations on the left hand side of the inequality (10) we obtain, Therefore the inequality (10) holds if and only if which is equivalent to (9). This completes the proof.
Corollary 6. r q (z) given by (6) is in the class S(1; 0; 1) S if and only if qr (1 q) r+1 1: Theorem 7. r q (z) given by (6) is in the class R( ; ; ) if and only if q 2 r(r + 1) Proof. To prove that r q 2 R( ; ; ), according to Lemma 3, it is su¢ cient to show that Now,using the same method as in the proof of Theorem 4, we obtain Therefore the inequality (12) holds if and only if This completes the proof.
As a special case of Theorem 7, if we put = 0 and = 1, we arrive at the following result. Taking = 0 and = = 1, we obtain the following corollary.

Integral Operators
In this section, we will give analog results for the integral operators de…ned as follows: where r q (t) is given by (6).
Proof. From (13), we can write According to Lemma 2, it is enough to show that Using the assumption (14), a simple computation shows that From (14), we conclude that H r q (z) 2 S( ; ; ). This completes the proof.
Theorem 11. H r q (z) given by (13) is in the class R( ; ; ) if and only if qr (1 q) + 1 (1 q) r j j: Proof. Since according to Lemma 3, it is enough to show that 1 X k=2 k [ (k 1) + 1] k k + r 2 r 1 q k 1 (1 q) r j j: Using the assumption (17), some simple computations shows that From (17), we conclude that H r q (z) 2 R( ; ; ). This completes the proof.
Author Contribution Statements All authors contributed equally to the planning, execution, and analysis of this research paper.
Declaration of Competing Interests No potential con ‡ict of interest and there is no funding was reported by the authors.