On the Koebe Quarter Theorem for certain polynomials

Abstract

We study problems similar to the Koebe Quarter Theorem for close-to-convex polynomials with all zeros of derivative in ${\mathbb {T}}:=\{z\in {\mathbb {C}}:|z|=1\}$. We found minimal disc containing all images of ${\mathbb {D}}:=\{z\in {\mathbb {C}}: |z|<1\}$ and maximal disc contained in all images of ${\mathbb {D}}$ through polynomials of degree 3 and 4. Moreover we determine the extremal functions for both problems.

1 Introduction

Let ${\mathcal {S}}$ be the class of all analytic functions $f:{\mathbb {D}}\rightarrow {\mathbb {C}}$ normalized by $f(0)=f'(0)-1=0$ which are univalent. Let ${\mathcal {H}}_d$ be the class of all analytic functions $f:{\mathbb {D}}\rightarrow {\mathbb {C}}$ normalized by $f(0)=1$ and such that $f\ne 0$ in ${\mathbb {D}}$. For $\alpha , \beta \ge 0$ the Kaplan class $K(\alpha ,\beta )$ is the set of all functions $f\in {\mathcal {H}}_d$ satisfying the condition

$$\begin{aligned} \arg f(r\mathrm {e}^{\mathrm {i}\theta _2})-\arg f(r\mathrm {e}^{\mathrm {i}\theta _1})\le \beta \pi -\frac{1}{2}(\alpha -\beta )(\theta _1-\theta _2) \end{aligned}$$

(0.1)

for $0<r<1$ and $\theta _1<\theta _2<\theta _1+2\pi$ (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 $K(\alpha ,\beta )$ is called the Kaplan class because using the Kaplan method [8], one can show that a normalized function f analytic in ${\mathbb {D}}$ is close-to-convex of order $\alpha \ge 0$ if and only if $f'\in K(\alpha ,\alpha +2)$. Let ${\mathcal {C}}$ be the class of functions in ${\mathcal {S}}$ that are close-to-convex (see [4]). In particular $f\in {\mathcal {C}}$ if and only if $f'\in K(1,3)$. 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 ${\mathbb {T}}$ in the context of Kaplan classes).

Theorem A

(Sheil-Small) Every polynomial of degree n with all zeros on ${\mathbb {T}}$ belongs to $K(1,2\pi /\lambda -n+1)$, where $\lambda$ is the minimal arclength between each pair of zeros.

Define ${\mathbb {D}}(r):=\{z\in {\mathbb {C}}:|z|<r\}$ for $r>0$,

$$\begin{aligned} \mathrm r({\mathcal {F}}):=\sup \left\{ r>0:{\mathbb {D}}(r)\subset \bigcap \limits _{f\in {\mathcal {F}}} f({\mathbb {D}}) \right\} \end{aligned}$$

and

$$\begin{aligned} \mathrm R({\mathcal {F}}):=\inf \left\{ r>0: \bigcup \limits _{f\in {\mathcal {F}}} f({\mathbb {D}})\subset {\mathbb {D}}(r) \right\} \end{aligned}$$

for any ${\mathcal {F}}\subset {\mathcal {S}}$. Bieberbach in 1916 proved the following theorem (see [1]).

Theorem B

(Koebe Quarter Theorem) For every $f\in {\mathcal {S}}$, ${\mathbb {D}}(1/4)\subset f({\mathbb {D}})$, i.e. $\mathrm r({\mathcal {S}})=1/4$. Moreover the function

$$\begin{aligned} {\mathbb {D}}\ni z\mapsto \frac{z}{(1-z)^2} \end{aligned}$$

is extremal for this inclusion.

The problem of determining $r({\mathcal {F}})$ or $R({\mathcal {F}})$ for any class ${\mathcal {F}}\subset {\mathcal {S}}$ 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 ${\mathcal {F}}$ is similar to the concept of $r({\mathcal {F}})$ (usually they are equivalent). Directly from Theorem B we conclude that $r({\mathcal {F}})=1/4$ for ${\mathcal {F}}$ 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 ${\mathbb {D}}(1/4)$.

Let $n\in {\mathbb {N}}$. Denote by ${\mathcal {U}}_n$ the class of all polynomials of degree n belonging to ${\mathcal {S}}$. Additionally ${\mathcal {U}}_{n,{\mathbb {R}}}$ be the subclass of ${\mathcal {U}}_n$ containing only the polynomials with real coefficients. An interesting problem of determining $\mathrm r({\mathcal {U}}_n)$ and $\mathrm r({\mathcal {U}}_{n,{\mathbb {R}}})$ was stated in [2]. It is easy to show that $\mathrm r({\mathcal {U}}_{1,{\mathbb {R}}})=\mathrm r({\mathcal {U}}_1)=1$ and $\mathrm r({\mathcal {U}}_{2,{\mathbb {R}}})=\mathrm r({\mathcal {U}}_2)=1/2$, but cases for $n\ge 3$ are far from trivial. Dmitrishin et al. [2] proved that

$$\begin{aligned} \mathrm r({\mathcal {U}}_{3,{\mathbb {R}}})=\frac{3-\sqrt{5}}{2}\approx 0,382 \end{aligned}$$

with the extremal function given by the formula

$$\begin{aligned} {\mathbb {D}}\ni z\mapsto z+\frac{2\sqrt{5}}{5}z^2+\frac{5-\sqrt{5}}{10}z^3\ . \end{aligned}$$

Moreover they proposed a class of functions which can be extremal at the problem of determining $\mathrm r({\mathcal {U}}_{n,{\mathbb {R}}})$ for $n\ge 4$.

Let $n\in {\mathbb {N}}$. Denote by ${\mathcal {C}}_n(X)$ the class of all polynomials of degree n belonging to ${\mathcal {C}}$ with all zeros of derivative on $X\subset {\mathbb {C}}\setminus {\mathbb {D}}$. Inspired by [2] and [3] we state the problem of determining $\mathrm r({\mathcal {C}}_n({\mathbb {T}}))$ and $\mathrm R({\mathcal {C}}_n({\mathbb {T}}))$. The classes ${\mathcal {C}}_n({\mathbb {T}})$ are fairly large, ${\mathcal {C}}_n({\mathbb {T}})\subset {\mathcal {U}}_n$ and ${\mathcal {C}}_n({\mathbb {T}})\not \subset {\mathcal {U}}_{n,{\mathbb {R}}}$. Let us notice that cases $n=1$ and $n=2$ are trivial. It is easy to show that $\mathrm r({\mathcal {C}}_1({\mathbb {T}}))=\mathrm R({\mathcal {C}}_1({\mathbb {T}}))=1$, $\mathrm r({\mathcal {C}}_2({\mathbb {T}}))=1/2$ and $\mathrm R({\mathcal {C}}_2({\mathbb {T}}))=3/2$ with the extremal functions the same as in [2] for the problem of determining $\mathrm r({\mathcal {U}}_1)$ and $\mathrm r({\mathcal {U}}_2)$. In this article we determine $\mathrm r({\mathcal {C}}_n({\mathbb {T}}))$ and $\mathrm R({\mathcal {C}}_n({\mathbb {T}}))$ for $n= 3$ and $n=4$.

2 Main results

In this section we solve the problem of determinig $\mathrm r({\mathcal {C}}_n({\mathbb {T}}))$ and $\mathrm R({\mathcal {C}}_n({\mathbb {T}}))$ for $n=3$ and $n=4$.

Theorem 1.1

The following equalities hold:

$$\begin{aligned} \mathrm r\left( {\mathcal {C}}_3({\mathbb {T}})\right) =\frac{\sqrt{10}}{6} \end{aligned}$$

(1.1)

and

$$\begin{aligned} \mathrm R\left( {\mathcal {C}}_3({\mathbb {T}})\right) =\frac{8+3\sqrt{2}}{6}\ . \end{aligned}$$

(1.2)

Moreover the function

$$\begin{aligned} {\mathbb {D}}\ni z\mapsto z+\frac{-1+\mathrm {i}}{2}z^2-\frac{\mathrm {i}}{3}z^3 \end{aligned}$$

is extremal for (1.1) and (1.2).

Proof

Let $P\in {\mathcal {C}}_3({\mathbb {T}})$. Then by Theorem A,

$$\begin{aligned} {\mathbb {D}}\ni z\mapsto P(z):=\int \limits _0^z \left( 1-u\mathrm {e}^{-\mathrm {i}t_1}\right) \left( 1-u\mathrm {e}^{-\mathrm {i}t_2}\right) \mathrm {d}u \end{aligned}$$

for certain $t_1,t_2\in {\mathbb {R}}$ such that $\pi /2\le |t_1-t_2|\le \pi$. Without loss of generality we can assume that $t_1:=0$ and $\pi /2\le t_2\le \pi$. Therefore

$$\begin{aligned} {\mathbb {D}}\ni z\mapsto P(z):=z-\frac{1}{2}\left( 1+\mathrm {e}^{-\mathrm {i}t_2}\right) z^2+\frac{1}{3}\mathrm {e}^{-\mathrm {i}t_2}z^3\ . \end{aligned}$$

Since P is a polynomial, so there exists the holomorphic extension of P on ${\mathbb {T}}$. Setting $z:=\mathrm {e}^{\mathrm {i}\theta }$ for $\theta \in [0,2\pi )$ we get

$$\begin{aligned} g(\theta )&:=\left| P\left( \mathrm {e}^{\mathrm {i}\theta }\right) \right| ^2=\left| 1-\frac{1}{2}\mathrm {e}^{\mathrm {i}\theta }-\frac{1}{2}\mathrm {e}^{\mathrm {i}(\theta -t_2)}+\frac{1}{3}\mathrm {e}^{\mathrm {i}(2\theta -t_2)}\right| ^2\\&=\frac{29}{18}-\frac{4}{3}\cos (\theta )+\frac{1}{2}\cos (t_2)-\frac{4}{3}\cos (\theta -t_2)+\frac{2}{3}\cos (2\theta -t_2)\ . \end{aligned}$$

Therefore

$$\begin{aligned} g'(\theta )&=\frac{4}{3}(\sin (\theta )+\sin (\theta -t_2)-\sin (2\theta -t_2))\\&= \frac{16}{3}\sin \left( \frac{\theta }{2}\right) \sin \left( \frac{\theta -t_2}{2}\right) \sin \left( \frac{2\theta -t_2}{2}\right) \end{aligned}$$

and as a consequence

$$\begin{aligned} g'(\theta )=0 \iff \theta \in \left\{ 0,\frac{t_2}{2},t_2,\frac{t_2+2\pi }{2}\right\} \ . \end{aligned}$$

Function g has the minima at 0 and $t_2$ and the maxima at $t_2/2$ and $t_2/2+\pi$. Since

$$\begin{aligned} g(0)=g(t_2)=\frac{5}{18}-\frac{1}{6}\cos (t_2)\in \left[ \frac{5}{18},\frac{4}{9}\right] \end{aligned}$$

and

$$\begin{aligned} g\left( \frac{t_2}{2}\right)&=\left( \frac{4}{3}-\cos \left( \frac{t_2}{2}\right) \right) ^2\le \left( \frac{4}{3}+\cos \left( \frac{t_2}{2}\right) \right) ^2\\&=g\left( \frac{t_2}{2}+\pi \right) \in \left[ \frac{16}{9},\left( \frac{8+3\sqrt{2}}{6}\right) ^2\right] \ , \end{aligned}$$

so for every $z\in {\mathbb {T}}$,

$$\begin{aligned} \frac{\sqrt{10}}{6}\le \left| P(z)\right| \le \frac{8+3\sqrt{2}}{6}\ , \end{aligned}$$

which leads to (1.1) and (1.2).

Now consider the function

$$\begin{aligned} {\mathbb {C}}\ni z\mapsto f(z):=z+\frac{-1+\mathrm {i}}{2}z^2-\frac{\mathrm {i}}{3}z^3\ . \end{aligned}$$

Then $f_{|{\mathbb {D}}}\in {\mathcal {C}}_3({\mathbb {T}})$,

$$\begin{aligned} |f(1)|=\frac{\sqrt{10}}{6} \end{aligned}$$

and

$$\begin{aligned} \left| f\left( \frac{-1-\mathrm {i}}{\sqrt{2}}\right) \right| = \frac{8+3\sqrt{2}}{6}\ . \end{aligned}$$

$\square$

Fix $t_2\in [2/5\pi ,2/3\pi ]$ and $t_3\in [2t_2,\pi +t_2/2]$. Define

$$\begin{aligned}&\Omega _1:= \left[ \max \left\{ 0,-\frac{2}{3}\arccos \left( -\frac{6}{11}\right) +\frac{t_2+t_3}{3}\right\} ,-\frac{2}{3}\arccos \left( \frac{6}{11}\right) +\frac{t_2+t_3}{3}\right] \ ,\\&\Omega _2:=\left[ \frac{2}{3}\arccos \left( \frac{6}{11}\right) +\frac{t_2+t_3}{3},\frac{2}{3}\arccos \left( -\frac{6}{11}\right) +\frac{t_2+t_3}{3}\right] \ ,\\&\Omega _3:=\left[ \frac{4}{3}\pi -\frac{2}{3}\arccos \left( -\frac{6}{11}\right) +\frac{t_2+t_3}{3},\frac{4}{3}\pi -\frac{2}{3}\arccos \left( \frac{6}{11}\right) +\frac{t_2+t_3}{3}\right] \ ,\\&\Omega _4:=\left[ \frac{4}{3}\pi +\frac{2}{3}\arccos \left( \frac{6}{11}\right) +\frac{t_2+t_3}{3},2\pi \right) \ ,\\&[0,2\pi )\ni \theta \mapsto h_1(\theta ):=\cos \left( \frac{3\theta -t_2-t_3}{2}\right) \ ,\\&[0,2\pi )\ni \theta \mapsto h_2(\theta ):=\cos \left( \frac{\theta }{2}\right) \cos \left( \frac{\theta -t_2}{2}\right) \cos \left( \frac{\theta -t_3}{2}\right) \ ,\\&[0,2\pi )\ni \theta \mapsto h(\theta ):=\frac{11}{8}h_1(\theta )-h_2(\theta )\ . \end{aligned}$$

Now we prove the following lemma.

Lemma 1.2

The function h has exactly three zeros, one in each interval $(0,t_2)$, $(t_2,t_3)$ and $(t_3,2\pi )$.

Proof

Since

$$\begin{aligned} \begin{aligned}&\left| \cos \left( \frac{\theta }{2}\right) \cos \left( \frac{\theta -t_2}{2}\right) \cos \left( \frac{\theta -t_3}{2}\right) \right| \le \left| \cos \left( \frac{\theta }{2}\right) \cos \left( \frac{\theta -t_3}{2}\right) \right| \\&\quad =\frac{1}{2}\left| \cos \left( \frac{2\theta -t_3}{2}\right) +\cos \left( \frac{t_3}{2}\right) \right| \le \frac{1}{2}\left( 1+\left| \cos \left( \frac{t_3}{2}\right) \right| \right) \le \frac{3}{4} \end{aligned} \end{aligned}$$

(1.3)

for $\theta \in [0,2\pi )$, so $h(\theta )\ne 0$ if

$$\begin{aligned} \left| \cos \left( \frac{3\theta -t_2-t_3}{2}\right) \right| > \frac{6}{11}\ . \end{aligned}$$

Therefore $h(\theta )\ne 0$ if $\theta \in [0,2\pi )\setminus (\Omega _1\cup \Omega _2\cup \Omega _3\cup \Omega _4)$. For $\theta \in [0,2\pi )$ we obtain

$$\begin{aligned} h'(\theta )= -\frac{25}{16}\sin \left( \frac{3\theta -t_2-t_3}{2}\right) +\frac{1}{2}\sin \left( \frac{\theta }{2}\right) \sin \left( \frac{\theta -t_2}{2}\right) \sin \left( \frac{\theta -t_3}{2}\right) \ . \end{aligned}$$

Let us notice that

$$\begin{aligned} \frac{3}{10}\pi<\arccos \left( \frac{6}{11}\right) <\frac{3}{8}\pi \end{aligned}$$

(1.4)

and

$$\begin{aligned} \frac{3}{5}\pi<\arccos \left( -\frac{6}{11}\right) <\frac{3}{4}\pi \ . \end{aligned}$$

(1.5)

Assume that $\theta \in \Omega _1$. Since $t_2\in [2/5\pi ,2/3\pi ]$ and $t_3\in [2t_2,\pi +t_2/2]$, so by using (1.4) and (1.5) we get

$$\begin{aligned}&\frac{\theta }{2}\in \left[ 0,\frac{7}{30}\pi \right] \ ,\\&\frac{\theta -t_2}{2}\in \left[ -\frac{1}{4}\pi ,-\frac{1}{30}\pi \right] \ ,\\&\frac{\theta -t_3}{2}\in \left[ -\frac{7}{12}\pi ,-\frac{3}{10}\pi \right] \ ,\\&\frac{3\theta -t_2-t_3}{2}\in \left[ -\frac{3}{4}\pi ,-\frac{3}{10}\pi \right] \ . \end{aligned}$$

Therefore

$$\begin{aligned}&\sin \left( \frac{\theta -t_2}{2}\right) ,\ \sin \left( \frac{\theta -t_3}{2}\right) ,\ \sin \left( \frac{3\theta -t_2-t_3}{2}\right) <0\ ,\\&\quad \sin \left( \frac{\theta }{2}\right) \ge 0\ . \end{aligned}$$

Hence $h'(\theta )>0$ for $\theta \in \Omega _1$.

Assume that $\theta \in \Omega _2$. Since $t_2\in [2/5\pi ,2/3\pi ]$ and $t_3\in [2t_2,\pi +t_2/2]$, so by using (1.4) and (1.5) we get

$$\begin{aligned}&\frac{\theta }{2}\in \left[ \frac{3}{10}\pi ,\frac{7}{12}\pi \right] \ ,\\&\frac{\theta -t_2}{2}\in \left[ \frac{1}{10}\pi ,\frac{19}{60}\pi \right] \ ,\\&\frac{\theta -t_3}{2}\in \left[ -\frac{7}{30}\pi ,\frac{1}{20}\pi \right] \ ,\\&\frac{3\theta -t_2-t_3}{2}\in \left[ \frac{3}{10}\pi ,\frac{3}{4}\pi \right] \ . \end{aligned}$$

Therefore

$$\begin{aligned}&\sin \left( \frac{3\theta -t_2-t_3}{2}\right) \in \left[ \frac{\sqrt{2}}{2},1 \right] \ ,\\&\qquad \sin \left( \frac{\theta }{2}\right) \sin \left( \frac{\theta -t_3}{2}\right) \in \left[ \frac{1}{2}\left( \cos \left( \frac{2}{3}\pi \right) -1\right) ,\frac{1}{2}\left( \cos \left( \frac{2}{5}\pi \right) +1\right) \right] \\&\quad =\left[ -\frac{3}{4},\frac{3+\sqrt{5}}{8}\right] \ ,\\&\qquad \sin \left( \frac{\theta -t_2}{2}\right) \in \left( 0,\sin \left( \frac{19}{60}\pi \right) \right] \ . \end{aligned}$$

Hence $h'(\theta )<0$ for $\theta \in \Omega _2$.

Assume that $\theta \in \Omega _3$. Since $t_2\in [2/5\pi ,2/3\pi ]$ and $t_3\in [2t_2,\pi +t_2/2]$, so by using (1.4) and (1.5) we get

$$\begin{aligned}&\frac{\theta }{2}\in \left[ \frac{37}{60}\pi ,\frac{9}{10}\pi \right] \ ,\\&\frac{\theta -t_2}{2}\in \left[ \frac{5}{12}\pi ,\frac{19}{30}\pi \right] \ ,\\&\frac{\theta -t_3}{2}\in \left[ \frac{1}{12}\pi ,\frac{11}{30}\pi \right] \ ,\\&\frac{3\theta -t_2-t_3}{2}\in \left[ \frac{5}{4}\pi ,\frac{17}{10}\pi \right] \ . \end{aligned}$$

Therefore

$$\begin{aligned}&\sin \left( \frac{3\theta -t_2-t_3}{2}\right) <0\ ,\\&\quad \sin \left( \frac{\theta }{2}\right) ,\ \sin \left( \frac{\theta -t_2}{2}\right) ,\ \sin \left( \frac{\theta -t_3}{2}\right) >0\ . \end{aligned}$$

Hence $h'(\theta )>0$ for $\theta \in \Omega _3$.

Assume that $\theta \in \Omega _4$. Since $t_2\in [2/5\pi ,2/3\pi ]$ and $t_3\in [2t_2,\pi +t_2/2]$, so by using (1.4) and (1.5) we get

$$\begin{aligned}&\frac{\theta }{2}\in \left[ \frac{29}{30}\pi ,\pi \right] \ ,\\&\frac{\theta -t_2}{2}\in \left[ \frac{23}{30}\pi ,\frac{4}{5}\pi \right] \ ,\\&\frac{\theta -t_3}{2}\in \left[ \frac{13}{30}\pi ,\frac{3}{5}\pi \right] \ ,\\&\frac{3\theta -t_2-t_3}{2}\in \left[ \frac{23}{10}\pi ,\frac{12}{5}\pi \right] \ . \end{aligned}$$

Therefore

$$\begin{aligned}&\sin \left( \frac{3\theta -t_2-t_3}{2}\right) \in \left[ \frac{1+\sqrt{5}}{4},\sqrt{\frac{5+\sqrt{5}}{8}} \right] \ ,\\&\sin \left( \frac{\theta }{2}\right) \sin \left( \frac{\theta -t_3}{2}\right) \in \left[ \frac{1}{2}\left( \cos \left( \frac{2}{3}\pi \right) -1\right) ,\frac{1}{2}\left( \cos \left( \frac{2}{5}\pi \right) +1\right) \right] \\&\quad =\left[ -\frac{3}{4},\frac{3+\sqrt{5}}{8}\right] \ ,\\&\qquad \sin \left( \frac{\theta -t_2}{2}\right) \in \left[ \sqrt{\frac{5-\sqrt{5}}{8}},\sin \left( \frac{23}{30}\pi \right) \right] \ . \end{aligned}$$

Hence $h'(\theta )<0$ for $\theta \in \Omega _4$.

Let us notice that

$$\begin{aligned}&h_1\left( -\frac{2}{3}\arccos \left( -\frac{6}{11}\right) +\frac{t_2+t_3}{3}\right) =h_1\left( \frac{2}{3}\arccos \left( -\frac{6}{11}\right) +\frac{t_2+t_3}{3}\right) \\&\quad =h_1\left( \frac{4}{3}\pi -\frac{2}{3}\arccos \left( -\frac{6}{11}\right) +\frac{t_2+t_3}{3}\right) =-\frac{6}{11}\ ,\\&h_1\left( -\frac{2}{3}\arccos \left( \frac{6}{11}\right) +\frac{t_2+t_3}{3}\right) =h_1\left( \frac{2}{3}\arccos \left( \frac{6}{11}\right) +\frac{t_2+t_3}{3}\right) \\&\quad =h_1\left( \frac{4}{3}\pi -\frac{2}{3}\arccos \left( \frac{6}{11}\right) +\frac{t_2+t_3}{3}\right) \\&\quad =h_1\left( \frac{4}{3}\pi +\frac{2}{3}\arccos \left( \frac{6}{11}\right) +\frac{t_2+t_3}{3}\right) =\frac{6}{11}\ , \end{aligned}$$

Since $t_2\in [2/5\pi ,2/3\pi ]$ and $t_3\in [2t_2,\pi +t_2/2]$, so $\Omega _1\subset [0,t_2)$, $\Omega _3\cup \Omega _4\subset (t_3,2\pi )$ and $t_2<\min (\Omega _2)<t_3$. Since $t_2\notin \Omega _1\cup \Omega _2\cup \Omega _3\cup \Omega _4$, so by (1.3) we obtain $|h_2(\theta )|<3/4$ for $\theta \in \Omega _1\cup \Omega _2\cup \Omega _3\cup \Omega _4$. Therefore

$$\begin{aligned} h(\min (\Omega _1)),\ h(\max (\Omega _2)),\ h(\min (\Omega _3))<0 \end{aligned}$$

and

$$\begin{aligned} h(\max (\Omega _1)),\ h(\min (\Omega _2)),\ h(\max (\Omega _3)),\ h(\min (\Omega _4))>0\ . \end{aligned}$$

Moreover

$$\begin{aligned} h(0)&=\frac{7}{8}\cos \left( \frac{t_2+t_3}{2}\right) -\frac{1}{2}\cos \left( \frac{t_3-t_2}{2}\right)<0\ ,\\ h(t_3)&=\frac{7}{8}\cos \left( \frac{2t_3-t_2}{2}\right) -\frac{1}{2}\cos \left( \frac{t_2}{2}\right) <0\ ,\\ h(2\pi )&=-\frac{7}{8}\cos \left( \frac{t_2+t_3}{2}\right) +\frac{1}{2}\cos \left( \frac{t_3-t_2}{2}\right) >0\ . \end{aligned}$$

Hence and by monotonicity of h on intervals $\Omega _1$, $\Omega _2$, $\Omega _3$ and $\Omega _4$ we conclude that h has exactly three zeros, one in each interval $(0,t_2)$, $(t_2,t_3)$ and $(t_3,2\pi )$. $\square$

Using Lemma 1.2 we prove the following theorem.

Theorem 1.3

The following equalities hold:

$$\begin{aligned} \mathrm r({\mathcal {C}}_4({\mathbb {T}}))=\frac{8-\sqrt{5}}{12} \end{aligned}$$

(1.6)

and

$$\begin{aligned} \mathrm R({\mathcal {C}}_4({\mathbb {T}}))=\frac{20+5\sqrt{5}}{12}\ . \end{aligned}$$

(1.7)

Moreover the function

$$\begin{aligned}&{\mathbb {D}}\ni z\mapsto z-\frac{1-\mathrm {i}\sqrt{5+2\sqrt{5}}}{4}z^2-\frac{3+\sqrt{5}+\mathrm {i}\sqrt{10+2\sqrt{5}}}{12}z^3\\&\quad +\frac{1+\sqrt{5}-\mathrm {i}\sqrt{10-2\sqrt{5}}}{16}z^4 \end{aligned}$$

is extremal for (1.6) and (1.7).

Proof

Let $P\in {\mathcal {C}}_4({\mathbb {T}})$. Then by Theorem A,

$$\begin{aligned} {\mathbb {D}}\ni z\mapsto P(z):=\int \limits _0^z \left( 1-u\mathrm {e}^{-\mathrm {i}t_1}\right) \left( 1-u\mathrm {e}^{-\mathrm {i}t_2}\right) \left( 1-u\mathrm {e}^{-\mathrm {i}t_3}\right) \mathrm {d}u \end{aligned}$$

for certain $t_1,t_2,t_3\in {\mathbb {R}}$ such that $2\pi /5\le |t_1-t_2|\le |t_2-t_3|\le 4\pi /5$. Without loss of generality we can assume that $t_1:=0$, $2\pi /5\le t_2\le 2\pi /3$ and $2t_2\le t_3\le t_2/2+\pi$. Therefore P is given by the formula

$$\begin{aligned} P(z):= & {} z-\frac{1}{2}\left( 1+\mathrm {e}^{-\mathrm {i}t_2}+\mathrm {e}^{-\mathrm {i}t_3}\right) z^2+\frac{1}{3}\left( \mathrm {e}^{-\mathrm {i}t_2}+\mathrm {e}^{-\mathrm {i}t_3}\right. \\&\left. +\mathrm {e}^{-\mathrm {i}(t_2+t_3)}\right) z^3-\frac{1}{4}\mathrm {e}^{-\mathrm {i}(t_2+t_3)}z^4\ . \end{aligned}$$

Since P is a polynomial, so there exists the holomorphic extension of P on ${\mathbb {T}}$. Setting $z:=\mathrm {e}^{\mathrm {i}\theta }$ for $\theta \in [0,2\pi )$ we get

$$\begin{aligned} g(\theta )&:=\left| P\left( \mathrm {e}^{\mathrm {i}\theta }\right) \right| ^2=\frac{103}{48}+\frac{13}{18}\cos (t_2)+\frac{13}{18}\cos (t_3)+\frac{13}{18}\cos (t_3-t_2)\\&\quad -\frac{11}{6}\cos (\theta )-\frac{11}{6}\cos (\theta -t_2)-\frac{11}{6}\cos (\theta -t_3)-\frac{1}{3}\cos (\theta -t_2-t_3)\\&\quad -\frac{1}{3}\cos (\theta +t_2-t_3)-\frac{1}{3}\cos (\theta -t_2+t_3)+\frac{11}{12}\cos (2\theta -t_2)\\&\quad +\frac{11}{12}\cos (2\theta -t_3)\\&\quad +\frac{11}{12}\cos (2\theta -t_2-t_3)-\frac{1}{2}\cos (3\theta -t_2-t_3) \end{aligned}$$

Therefore

$$\begin{aligned} g'(\theta )=-\frac{32}{3}h(\theta )\sin \left( \frac{\theta }{2}\right) \sin \left( \frac{\theta -t_2}{2}\right) \sin \left( \frac{\theta -t_3}{2}\right) \ . \end{aligned}$$

The function

$$\begin{aligned}{}[0,2\pi )\ni \theta \mapsto \sin \left( \frac{\theta }{2}\right) \sin \left( \frac{\theta -t_2}{2}\right) \sin \left( \frac{\theta -t_3}{2}\right) \end{aligned}$$

has exactly three zeros at points 0, $t_2$ and $t_3$. Hence by Lemma 1.2 we conclude that the function $g'$ has exactly six zeros. Therefore g has three minima at points 0, $t_2$ and $t_3$ and three maxima one in each interval $(0,t_2)$, $(t_2,t_3)$ and $(t_3,2\pi )$. Moreover

$$\begin{aligned} g(0)&=\frac{5}{16}-\frac{7}{36}\cos (t_2)-\frac{7}{36}\cos (t_3)+\frac{1}{18}\cos (t_3-t_2)+\frac{1}{12}\cos (t_2+t_3)\\&\ge \frac{5}{16}-\frac{7}{36}\cos \left( \frac{2}{5}\pi \right) -\frac{7}{36}\cos \left( \frac{4}{5}\pi \right) +\frac{1}{18}\cos \left( \frac{4}{5}\pi \right) \\&\quad +\frac{1}{12}\cos \left( \frac{6}{5}\pi \right) =\frac{54-5\sqrt{5}}{144}\ , \end{aligned}$$

$$\begin{aligned} g(t_3)&=\frac{5}{16}+\frac{1}{18}\cos (t_2)-\frac{7}{36}\cos (t_3)-\frac{7}{36}\cos (t_3-t_2)+\frac{1}{12}\cos (2t_3-t_2)\\&\ge \frac{5}{16}+\frac{1}{18}\cos \left( \frac{4}{5}\pi \right) -\frac{7}{36}\cos \left( \frac{4}{5}\pi \right) -\frac{7}{36}\cos \left( \frac{2}{5}\pi \right) \\&\qquad +\frac{1}{12}\cos \left( \frac{6}{5}\pi \right) =\frac{54-5\sqrt{5}}{144}\ , \end{aligned}$$

and

$$\begin{aligned} g(t_2)=\frac{5}{16}-\frac{7}{36}\cos (t_2)+\frac{1}{18}\cos (t_3)-\frac{7}{36}\cos (t_3-t_2)+\frac{1}{12}\cos (t_3-2t_2)\ . \end{aligned}$$

Since $g(t_2)$ is increasing with respect to $t_2$ and $t_3$, so

$$\begin{aligned} g(t_2)&\ge \frac{5}{16}-\frac{7}{36}\cos \left( \frac{2}{5}\pi \right) +\frac{1}{18}\cos \left( \frac{4}{5}\pi \right) -\frac{7}{36}\cos \left( \frac{2}{5}\pi \right) +\frac{1}{12}\\&=\left( \frac{8-\sqrt{5}}{12}\right) ^2 \end{aligned}$$

and the equality holds if and only if $t_2=2/5\pi$ and $t_3=4/5\pi$. This ends the proof of (1.6).

Now we prove (1.7). Let us notice that for $z\in {\mathbb {D}}\cup {\mathbb {T}}$,

$$\begin{aligned} |P(z)|&\le \frac{5}{4}+\frac{1}{2}\left| 1+\mathrm {e}^{-\mathrm {i}t_2}+\mathrm {e}^{-\mathrm {i}t_3}\right| +\frac{1}{3}\left| \mathrm {e}^{-\mathrm {i}t_2}+\mathrm {e}^{-\mathrm {i}t_3}+\mathrm {e}^{-\mathrm {i}(t_2+t_3)}\right| \\&=\frac{5}{4}+\frac{5}{6}\left| 1+\mathrm {e}^{\mathrm {i}t_2}+\mathrm {e}^{\mathrm {i}t_3}\right| =\frac{5}{4}+\frac{5}{6}\sqrt{3+2\cos (t_2)+2\cos (t_3)+2\cos (t_3-t_2)}\\&=\frac{5}{4}+\frac{5}{6}\sqrt{1+4\cos ^2\left( \frac{t_2}{2}\right) +4\cos \left( \frac{t_2}{2}\right) \cos \left( \frac{2t_3-t_2}{2}\right) }\\&\le \frac{5}{4}+\frac{5}{6}\sqrt{1+4\cos ^2\left( \frac{t_2}{2}\right) +4\cos \left( \frac{t_2}{2}\right) \cos \left( \frac{3t_2}{2}\right) }=\frac{5}{4}+\frac{5}{6}\left( 1+2\cos (t_2)\right) \\&\le \frac{5}{4}+\frac{5}{6}\left( 1+\frac{\sqrt{5}-1}{2}\right) =\frac{20+5\sqrt{5}}{12}. \end{aligned}$$

and the equality holds if and only if $z=\mathrm {e}^{\frac{7}{5}\pi \mathrm {i}}$, $t_2=2/5\pi$ and $t_3=4/5\pi$. This ends the proof of (1.7).

Moreover from the above considerations we conclude that the extremal function for (1.6) and (1.7) is

$$\begin{aligned}&{\mathbb {D}}\ni z\mapsto z-\frac{1+\mathrm {e}^{-\frac{2}{5}\pi \mathrm {i}}+\mathrm {e}^{-\frac{4}{5}\pi \mathrm {i}}}{2}z^2+\frac{\mathrm {e}^{-\frac{2}{5}\pi \mathrm {i}}+\mathrm {e}^{-\frac{4}{5}\pi \mathrm {i}}+\mathrm {e}^{-\frac{6}{5}\pi \mathrm {i}}}{3}z^3-\frac{\mathrm {e}^{-\frac{6}{5}\pi \mathrm {i}}}{4}z^4\\&\quad =z-\frac{1-\mathrm {i}\sqrt{5+2\sqrt{5}}}{4}z^2-\frac{3+\sqrt{5}+\mathrm {i}\sqrt{10+2\sqrt{5}}}{12}z^3\\&\qquad +\frac{1+\sqrt{5}-\mathrm {i}\sqrt{10-2\sqrt{5}}}{16}z^4\ . \end{aligned}$$

$\square$