ON CERTAIN SUFFICIENT CONDITIONS FOR STARLIKENESS

We consider certain properties of f(z)f ′′(z)/f ′2(z) as a sufficient condition for starlikeness.


Introduction and preliminaries.
Let A denote the class of functions f (z) which are analytic in the unit disc U = {z : |z| < 1} with f (0) = f (0) − 1 = 0.
For a function f (z) ∈ A we say that it is starlike in the unit disc U if and only if for all z ∈ U .We denote by S * the class of all such functions.We denote by K the class of convex functions in the unit disc U , i.e., the class of univalent functions f (z) ∈ A for which for all z ∈ U.Both of the above mentioned classes are subclasses of univalent functions in U and more K ⊂ S * ( [1,2]).
Let f (z) and g(z) be analytic in the unit disc.Then we say that f (z) is subordinate to g(z), and we write f (z) ≺ g(z), if g(z) is univalent in U , f (0) = g(0) and f (U) ⊆ g(U).
In this paper, we use the method of differential subordinations.The general theory of differential subordinations introduced by Miler and Mocanu is given in [5].Namely, if φ : We say that the univalent function q(z) is dominant of the differential subordination (1.3) if p(z) ≺ q(z) for all p(z) satisfying (1.3).If q(z) is a dominant of (1.3) and q(z) ≺ q(z) for all dominants of (1.3), then we say that q(z) is the best dominant of the differential subordination (1.3).
In the following section, we need the following lemma of Miller and Mocanu [6].
Even more we need the following lemma, which in more general form is due to Hallenbeck and Ruscheweyh [3]. (1.5)

Main results and consequences.
In this part, we use Lemmas 1.1 and 1.2 to obtain some conditions for f (z)f (z)/f 2 (z) which lead to starlikeness. (2.1) is starlike in U , and for the function Therefore the conditions of Lemma 1.1 are satisfied and we obtain that if and it is enough to prove that h(U ) = D.After some transformations we obtain Re (2.9)
Example 2.4.The function smaller than 3/2 for all z ∈ U.So f (z) is starlike according to Corollary 2.3(ii).It have been more complicated to realize it from zf (z)/f (z) = z/(e z − 1).Now, using Lemma 1.2 we prove a theorem which we used to improve the results from Corollary 2.3(ii) and (iii) and to obtain some other results. (2.11) (2.12) Therefore the conditions of Lemma 1.2 are satisfied and for n = 0 we obtain (2.13) If we apply the definitions of F(z) and G(z) in the result above and use the following fact which is true because we obtain that In the following corollaries, we deliver some interesting results using Theorem 2.5. (2.16) Further, h(z) is convex, so the conditions from Theorem 2.5 are satisfied, and we obtain Remark 2.8.The result from Corollary 2.7 is the same as in [7] (Theorem 1, for a = 0 and b = −1) and it is better than the result from Corollary 2.3(iii).
Example 2.9.The same function as in Example 2.4, f (z) = 1 − e −z , can be used to illustrate Corollary 2.7: Proof.(i) From h(0) = 0 and h(z) is a convex function in the unit disc U , by Theorem 2.5 we get that where z = x + iy, it follows that g(U) is symmetric with respect to the x-axis.It is also convex (Remark 2.6) and so Now, from Re h e iθ = α and h(0) = 0 < α we get that h(z) maps the unit disc U into the half plane with real part less than α.So the condition from (i) is equivalent with (2.25) If we put α = 1/2(1 − ln 2) here, using (i) we obtain the statement of (ii).