On the Koebe Quarter Theorem for certain polynomials

We study problems similar to the Koebe Quarter Theorem for close-to-convex polynomials with all zeros of derivative in T:={z∈C:|z|=1}\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathbb {T}}:=\{z\in {\mathbb {C}}:|z|=1\}$$\end{document}. We found minimal disc containing all images of D:={z∈C:|z|<1}\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathbb {D}}:=\{z\in {\mathbb {C}}: |z|<1\}$$\end{document} and maximal disc contained in all images of D\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathbb {D}}$$\end{document} through polynomials of degree 3 and 4. Moreover we determine the extremal functions for both problems.


Introduction
for 0 < r < 1 and θ 1 < θ 2 < θ 1 + 2π (see [9, pp. 32-33]). Kaplan clases were first defined by Sheil-Small in [10] and the restated version as in the condition (0.1) originally comes from [11]. The class K (α, β) is called the Kaplan class because using the Kaplan method [8], one can show that a normalized function f analytic in D is close-to-convex of order α ≥ 0 if and only if f ∈ K (α, α + 2). Let C be the class of functions in S that are close-to-convex (see [4]). In particular f ∈ C if and only if f ∈ K (1,3). The analytical properties and nice geometric interpretation make Kaplan classes find many applications to this day. In [7] Jahangiri and Ponnusamy use Kaplan classes both as a tool in assumptions as well as part of the result. In this article we will need the following theorem by Sheil-Small (see [12, p. 248], in [5] and [6] one can find additional information about polynomials with all zeros of derivative in T in the context of Kaplan classes).
Theorem A (Sheil-Small) Every polynomial of degree n with all zeros on T belongs to K (1, 2π/λ − n + 1), where λ is the minimal arclength between each pair of zeros.
The problem of determining r (F) or R(F) for any class F ⊂ S is well known in the literature. Goodman devoted an entire chapter (see [4, pp. 113-120]) for so called Koebe domains. The concept of the Koebe domain for a set F is similar to the concept of r (F) (usually they are equivalent). Directly from Theorem B we conclude that r (F) = 1/4 for F being the class of starlike functions as well as the class of close-to-convex functions since the Koebe function is starlike and in the consequences close-to-convex. Hence, the Koebe domain for starlike functions and close-to-convex functions is a disc D(1/4).
Let n ∈ N. Denote by U n the class of all polynomials of degree n belonging to S. Additionally U n,R be the subclass of U n containing only the polynomials with real coefficients. An interesting problem of determining r(U n ) and r(U n,R ) was stated in [2]. It is easy to show that r(U 1,R ) = r(U 1 ) = 1 and r(U 2,R ) = r(U 2 ) = 1/2, but cases for n ≥ 3 are far from trivial. Dmitrishin et al. [2] proved that with the extremal function given by the formula Moreover they proposed a class of functions which can be extremal at the problem of determining r(U n,R ) for n ≥ 4. Let n ∈ N. Denote by C n (X ) the class of all polynomials of degree n belonging to C with all zeros of derivative on X ⊂ C \ D. Inspired by [2] and [3] we state the problem of determining r(C n (T)) and R(C n (T)). The classes C n (T) are fairly large, C n (T) ⊂ U n and C n (T) ⊂ U n,R . Let us notice that cases n = 1 and n = 2 are trivial. It is easy to show that r(C 1 (T)) = R(C 1 (T)) = 1, r(C 2 (T)) = 1/2 and R(C 2 (T)) = 3/2 with the extremal functions the same as in [2] for the problem of determining r(U 1 ) and r(U 2 ). In this article we determine r(C n (T)) and R(C n (T)) for n = 3 and n = 4.

Main results
In this section we solve the problem of determinig r(C n (T)) and R(C n (T)) for n = 3 and n = 4.

Theorem 1.1
The following equalities hold: and .
Proof Let P ∈ C 3 (T). Then by Theorem A, Without loss of generality we can assume that t 1 := 0 and π/2 ≤ t 2 ≤ π . Therefore Since P is a polynomial, so there exists the holomorphic extension of P on T. Setting z := e iθ for θ ∈ [0, 2π) we get and as a consequence Function g has the minima at 0 and t 2 and the maxima at t 2 /2 and t 2 /2 + π . Since which leads to (1.1) and (1.2).

Now consider the function
.
Now we prove the following lemma.
Using Lemma 1.2 we prove the following theorem.

Compliance with ethical standards
Conflict of interest The authors have no relevant financial or non-financial interests to disclose. The authors have no conflicts of interest to declare that are relevant to the content of this article. All authors certify that they have no affiliations with or involvement in any organization or entity with any financial interest or non-financial interest in the subject matter or materials discussed in this manuscript. The authors have no financial or proprietary interests in any material discussed in this article.
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