On a Length Problem for Univalent Functions

Let g be an analytic function with the normalization in the open unit disk. Let L(r) be the length of g({z : |z| = r}). In this paper we present a correspondence between g and L(r) for the case when g is not necessary univalent. Furthermore, some other results related to the length of analytic functions are also discussed.


Introduction
Let A be the family of functions of the form which are analytic in the open unit disk D = {z ∈ C : |z| < 1}.Let S denote the subfamily of A consisting of all univalent functions in D.
Let C(r) denote the image curve of the |z| = r < 1 under the function g ∈ A which bound the area A(r).Furthermore, let L(r) be the length of C(r) and M(r) = max |z|=r<1 |g(z)|.
If g ∈ A satisfies Re zg (z) g(z) > 0, z ∈ D, then g is said to be starlike with respect to the origin in D and we write g ∈ S * .It is known (for details, see [1,2]) that S * ⊂ S.
The aim of the present paper is to prove, using a modified methodology, that in the following implication where O denotes the Landau's symbol, the assumption that g is starlike univalent can be changed by a weaker one.Result (2) was proved by Keogh [3].Moreover, some other length problems for analytic functions are investigated.Several interesting developments related to length problems for univalent functions were considered in [4][5][6][7][8][9][10][11][12][13][14][15].

Main Results
Theorem 1.Let g be of the form (1) and suppose that zg (z) Then and O means Landau's symbol.
Applying [3] (Theorem 1) and the hypothesis of Theorem 1, we have Remark 1.If g satisfies the condition of Theorem 1, then g is not necessary univalent in D. It is well known that if g ∈ S, then it follows that for some h ∈ S * and some γ ∈ (0, ∞), then g is said to be a Bazilevic function of type γ [13].The class of Bazilevic functions of type γ is denoted by g ∈ B(γ) .We note that Theorem 1 improves the implication (2) by Keogh [3] and it is also related to Theorem 3 given by Thomas [13].
We will need the following Tsuji's result. Moreover, Theorem 2. Let g be of the form (1) and suppose that zg (z) and where where O means Landau's symbol.
Proof.From the hypotheses ( 6) and ( 7), it follows that Hence, we obtain Therefore, we complete the proof of Theorem 2.
Let us recall the following Fejér-Riesz's result.

Lemma 2 ([16]
).Let h be analytic in D and continuous on D. Then where p > 0.
Theorem 3. Let g be of the form (1) and suppose that Then and O means Landau's symbol.
Proof.From the assumption, we have In fact, if g(z) = 0 in D, it contradicts hypothesis (8).
Applying Fejér-Riesz's Lemma 2, we have While, we obtain Therefore, we complete the proof of Theorem 3.
From Theorem 3, we have the following result.
Corollary 1.Let g be of the form (1) and suppose that g is univalent in D. Then we have where m(r) and M(r) are given by (9), respectively.
Proof.From the hypothesis, we have which completes the proof.
Theorem 4. Let g be of the form (1) and suppose that zg (z) Then where 0 < |z| = r < 1 and O means Landau's symbol.
Proof.It follows that Applying (12), we have where 0 ≤ ρ ≤ r < t < 1.Then, applying Schwarz's lemma, we have M(r) ≤ max Putting 0 < r 1 < r and t = (1 + ρ 2 )/2, we have Then we have where C is a bounded positive constant.On the other hand, putting t → 1 − , we have Using (5), we have Therefore we complete the proof of (13).
Remark 2. In Theorem 5, we do not suppose that g is univalent in |z| < 1 and therefore, it improves the result by Pommerenke [2].