NOTE ON THE ZEROS OF FUNCTIONS WITH UNIVALENT DERIVATIVES

Let E denote the class of functions f(z) analytic in the unit disc D, normalized so that f(0) 0 f'(0) -I, such that each f(k)(z), k>0 is univalent in D. In this paper we establish conditions for some functions to belong to class E.

f(z) zeBZ(l Z/Zl).Then f(z) and all of its derivatives are close-to-convex in D. In particular fEE.
In this paper we prove the following: THEOREM I. Let f(z) be defined by (I.I), suppose that (1.2) holds and B z < I. Then; f'(z) is univalent in Izl 0 (0 < O < I) if and only if

2+B202-4pB
Zl -< B (2-Bp) (2.1) Let F be the class of functions which are derivatives of univalent functions of the from (I.I).For a fixed B, the radius of univalence of F, O F is equal to 2 (B) + (B)2+8 (1. --Cl B 6 as in f"(z) (1.2), iS convex in D. If we can show that Ret, 0 for Izl E 0 then f'(z) will be close-to-convex in Izl E 0 and consequently univalent there (see [3]).

I
By the maximum principle it suffices to prove that O(x) 0 for x in [-1,13.For simplicity we write 0(x) ax2+bx+c Observe that b2-4ac O. Thus 0(x) has two real roots, and we will be done if we can show that -b-/-4ac 2a (3.1) (The larger root of (x) is 2a See fgure I).
. (x) - From this, noting that 60-2 then f"(z) has a root in Izl < 0, which means that f'(z) is not univalent there.The  .Og, the radius of univalence of definition, O F is the largest number such that g(OFZ) is univalent for all geF in D.
Let geF.Then g f' for some f of the form (I.I).In [2] it is shown that f is This proves the first part of the theorem.To prove the second part note that by g, is non zero because f"(0) # 0. Therefore, by the first part of the theorem, the condition 2+82g 4OgB is the necessary and sufficient condition for f(z) and g(OgZ) to be univalent in D. The case of equality in (3.7) corresponds to the case where both inequalities in Q.6) 2+8 are equalities.That is the radius of univalence of the g for which z T is pre- cisely the expression on the right of (3.7).and z in D. This wiiI show that f(z) and f(n)(z), n > 2 are ciose-to-convex in D. In 2+8 [2] it was shown that, if (1.2) hoids and ]-< z < 2, then Re{ f'(z)} 0 in D. Thus eSZ we need oniy show that Re{ f(n)(z)} < 0 for n 3 in D.

eSZ f(n)(z)
If we denote the reai part of 8z on the unit circie for n 3 by Qn(X) where e x Rez it wiii be sufficient to show that Qn(X) 0 for x in [-I,I].
Henceforth we assume that n J and note that n-I) 8 n 2n) x 28n x 2 n(X =nn-2(_ ++n-(_ -7 The quadratic Qn(X) wiII be nonpositive for aii x e[-I,I] if its discriminant is non- positive.
(We may note that the case when n(X) has two reaI roots is not of interest).
PROOF OF THEOREM 3 Note that 3-0.4226 is the smaiIer zero of 2-68 + 382 Thus n(X) Re{ f(n)(z)} on the unit circle where x Rez.We will prove the theorem by eSZ showing that l(X) O, 2(x) 0 and n(X) 0 for n 3 and x in [-I,I].

+I
We will have i(-I) eOand i(I) e 0 if e a and e a, respectively.
But both inequalities are true; this follows from (2.4) and the fact that, for 8 < 0.4,
This inequality will hold for n e 3 if it holds fo n .2314,cslculations show that (2.$) implies that 3.7964 Zl 3.9798.