Argument Properties for a Class of Analytic Functions Involving Libera Transform

1Mathematics Department, College of Science, King Saud University, Riyadh 11989, Saudi Arabia 2Faculty of Mathematics and Computer Science, Babeş-Bolyai University, 400084 Cluj-Napoca, Romania 3Department of Mathematics, Amity University Rajasthan, NH-11C, Jaipur 302002, India 4Department of Mathematics and Statistics, College of Science, Al-ImamMohammad Ibn Saud Islamic University (IMSIU), P.O. Box 65892, Riyadh 11566, Saudi Arabia


Introduction
Let A denote the class of functions of the form A function  ∈ A is said to be in the class K * () of convex functions of order  in U, if and only if it satisfies the condition Re (1 +   ()   () ) > ,  ∈ U, ( < 1) .
If  ∈ A satisfies the condition Re  ()   () > ,  ∈ U, (0 ≤  < 1) , then  is said to be star-like of reciprocal order , and we denote this class by NS * ().
The above definition was recently discussed by Nunokawa et al. [1] and Ravichandran and Sivaprasad Kumar [2].
Motivated by their works, we define a class of convex functions of reciprocal order  as follows.
then  is said to be convex function of reciprocal order , and we denote this class by NK().
It is well-known that (see, e.g., [3,4]) if  is univalent in U, then the following subordination property holds: it follows that a star-like (convex) function of reciprocal order  is a star-like (convex) function.Thus, and the equality holds in both cases if and only if  = 0.
(2) Let 0 <  < (3) We note that, for 0 ≤  < 1, we have In the following, we give some examples of functions belonging to the class of star-like functions of reciprocal order and the class of convex functions of reciprocal order.
Since   ∈ A and a simple computation shows that Re   () we conclude that   ∈ NS * ().
(2) Using the third part of the Remarks 1, we deduce that Since we proved that   ∈ NS * (), and from the above differential equation we obtain it follows that   ∈ NK().
This gives that Re   () where Since Re  ≥ 1/2, it follows that ℎ is analytic in U. Using the fact that ℎ is a convex (not necessary normalized) function in U and ℎ(U) is symmetric with respect to the real axis, we deduce that Re ℎ() >  for all  ∈ U; hence   ∈ NS * ().
(2) From the above result, using the same reasons like in the second part of Example 3, we obtain that We mention that several authors have investigated the strongly star-like functions and the strongly convex functions (see [5][6][7][8][9][10][11][12][13][14][15][16][17][18]).In the present investigation we give some argument properties of analytic functions belonging to A, such that the images of these functions by the Libera transform have bounded arguments.

Main Results
The following lemma will be used to prove our main results.
Lemma 5 (see [16]).Let  0 be the solution of for a suitable fixed  > 0, so that  : and let If  is analytic in U, (0) = 1 and then Remark that in the article [19] the author considered some special situations improving many results with a lot of applications.Thus, in [19, Lemma 2], the author proved the next result.
Let  be a function defined on U satisfying and let We emphasize that a special case of this lemma improves the conclusion of Lemma 1 from [19], and in the same article the author derived a number of interesting consequences of it.
Let L : A → A defined by be the well-known Libera transform [20].We first prove the following theorem, which is essential for proving our other results.Theorem 6.For  ∈ A suppose that the Libera transform  = L satisfies the condition (21), with where 0 <  ≤  0 and  0 is defined by (20) Suppose that there exists a number  0 ∈ U, such that   ( 0 ) = 0; since   (0) = 1, then  0 ∈ U\{0}.It follows that there exists a unique number  ∈ N and a unique function , analytic in U, with ( 0 ) ̸ = 0, such that From the above relation we deduce that which implies that  0 is a simple pole for the function   ()/  ().Consequently, from (29) we obtain that ( 0 ) = 0, and using the fact that  satisfies the inequality (21) for all  ∈ U it follows that which contradicts the assumption  > 0. Thus,   () ̸ = 0 for all  ∈ U, which implies that the function  is analytic in U, and (0) = 1.Now, a simple calculus shows that and our result follows immediately from Lemma 5.
Taking () =  in Theorem 6, for the function  defined by (29) we have () = 1/2, and we obtain the following result.
Corollary 7. Suppose that the parameters  and  satisfy the conditions (20) Proof.Let  be the function defined by Similarly like in the proof of Theorem 6, we will prove that the function  is analytic in U. Supposing that there exists a number  0 ∈ U, such that   ( 0 ) = 0, since   (0) = 1, then  0 ∈ U \ {0}.Hence, there exists a unique number  ∈ N and a unique function , analytic in U, with ( 0 ) ̸ = 0, such that From here we deduce that which implies that  0 is a simple pole for the function   ()/  ().Now, using the fact that  = L ∈ NK(), we obtain which contradicts the assumption 0 ≤ .Consequently,   () ̸ = 0 for all  ∈ U, and this implies that the function  is analytic in U, with (0) = 1.
From (45) we easily get where the function  is given by (29).
The assumption  = L ∈ NK() is equivalent to hence there exists a Schwarz function  such that By simple calculations, we get that is, Since the circular transform  maps the unit disk U onto the disk from the subordination property (6) we deduce that the subordination relation ( 53) is equivalent to With the above notation, the condition (21) becomes We will prove that if  satisfies the assumption (42), that is, 0 <  ≤ γ, then the subordination (53) implies that the inequality (56) holds, that is, where Since the sets Ω and Δ are symmetric with respect to the real axis, the above inclusion is equivalent to where The line () containing the half-line which is a part of the boundary the set Δ, has the equation Thus, the line () intersects the real axis in the point / and contains the point −/ (see Figure 1).The smallest set Δ that includes the domain Ω is obtained for the case when the line (), containing the point −/, is tangent to the upper boundary of Ω, which is the half-circle In this case, as it is shown in the figure, the line () becomes the tangent line (  ) to (C).
We will determine now the equation of the tangent line (  ).If we consider the family (  ) of all the lines containing the point −/, with nonnegative angular coefficient, that is, we will solve the system It follows that and the line is tangent to the half-circle if and only if the discriminant of the above quadratic form is zero, that is, The tangent (  ) intersects the real axis in the point 1/(m); hence we deduce that the inclusion (59) holds if and only if  Thus, if the parameter  satisfies the assumption (42), then the condition (21) of Lemma 5 is fulfilled, and our result follows from Theorem 6.
which are analytic in the open unit disk U = { ∈ C : || < 1}.A function  ∈ A is said to be in the class S * () of starlike functions of order  in U, if and only if it satisfies the condition Re   ()  () > ,  ∈ U, ( < 1) .