# An extremal problem for odd univalent polynomials

Dmitriy Dmitrishin  Odessa National Polytechnic University, 1 Shevchenko Ave.,
Odesa 65044, Ukraine; email<EMAIL_ADDRESS>,  Daniel Gray  Georgia
Southern University, Statesboro GA, 30460;
email<EMAIL_ADDRESS>,  Alex Stokolos  Georgia Southern
University, Statesboro GA, 30460;
email<EMAIL_ADDRESS>and  Iryna Tarasenko  Odessa National
Academy of Communications, 22/1 Staroportofrankivska Str., Odesa 65000,
Ukraine

###### Abstract.

For the univalent polynomials $F(z)=\sum\limits_{j=1}^{N}a_{j}z^{2j-1}$ with
real coefficients and normalization $a_{1}=1$ we solve the extremal problem

$\min_{a_{j}:\,a_{1}=1}\left(-iF(i)\right)=\min_{a_{j}:\,a_{1}=1}\sum\limits_{j=1}^{N}{(-1)^{j+1}a_{j}}.$

We show that the solution is $\frac{1}{2}\sec^{2}{\frac{\pi}{2N+2}},$ and the
extremal polynomial

$\sum_{j=1}^{N}\frac{U^{\prime}_{2(N-j+1)}\left(\cos\left(\frac{\pi}{2N+2}\right)\right)}{U^{\prime}_{2N}\left(\cos\left(\frac{\pi}{2N+2}\right)\right)}z^{2j-1}$

is unique and univalent, where the $U_{j}(x)$ are the Chebyshev polynomials of
the second kind and $U^{\prime}_{j}(x)$ denotes the derivative. As an
application, we obtain the estimate of the Koebe radius for the odd univalent
polynomials in $\mathbb{D}$ and formulate several conjectures.

Keywords. Chebyshev polynomials, odd univalent polynomials, Koebe one-quarter
theorem.

##  0\. Motivation

This work is motivated by problems of classical geometric complex analysis.
Let us recall the main milestones in the development of the theory.

The paper by Ludwig Bieberbach of 1916, in which he proved the exact estimate
$\left\lvert a_{2}\right\rvert\leq 2$ for the second coefficient of the
univalent in the unit disc functions $f(z)$ with the normalization $f(0)=0$,
$f^{\prime}(0)=1,$ can be considered the fundamental work. Such functions are
called schlicht functions. This estimate immediately implies the Koebe 1/4
theorem, i.e. the statement that the image of any schlicht function contains a
disc of radius 1/4. The assertion was put forward as a hypothesis by Paul
Koebe, and the function
$K(z)=\frac{z}{(1-z)^{2}}=z+2z^{2}+\ldots+jz^{j}+\ldots,$ now called the Koebe
function, is a schlicht function and has the property that
$K(-1)=-\frac{1}{4},$ thus showing sharpness of the constant.

In that same work, Bieberbach stated in a footnote the remark that, in general
$\left\lvert a_{j}\right\rvert\leq j$ is valid. This innocent-looking
statement became known as the Bieberbach conjecture, which, since then, has
been the engine of the development of geometric complex analysis for hundred
years.

In 1984, the Bieberbach conjecture was proved by Louis de Branges, who also
showed that the extremal function, up to rotation, is the Koebe function.

The odd Koebe function

$K^{(2)}(z)=(K(z^{2}))^{\frac{1}{2}}=\frac{z}{1-z^{2}}=z+z^{3}+z^{5}+\ldots+z^{2j+1}+\ldots$

has generated its own range of tasks, even more intriguing and unexpected than
the general case. Thus, J. Littlewood and R.E.A.C. Paley conjectured [1] that
all the coefficients of an odd, i.e. 2-symmetric, schlicht function are
bounded in absolute value by 1. The conjecture turned out to be false: M.
Fekete and G. Szegö proved the exact inequality $\left\lvert
a_{5}\right\rvert\leq 1/2+e^{-2/3}=1.013\ldots$ [2]. However, W. Hayman [3]
showed that for any odd schlicht function, except for the Koebe function
$F^{(2)}(z)$, the limit of the absolute value of the coefficients is less than
one. This implies that the Littlewood-Paley conjecture is valid for all but a
finite number of coefficients.

Many of the above-mentioned problems have the form of extremal problems and
can be formulated for the univalent polynomials. Note that the coefficients of
univalent functions can be sufficiently large, while the leading coefficient
of a schlicht polynomial of degree $N$ is at most $1/N.$ As a result, partial
sums of univalent functions, usually, are not univalent polynomials. And in
general, the number of interesting examples of univalent polynomials is quite
limited.

For instance, the value of the Koebe function $K(-1)$ is extreme for the
problem of covering a disc by the image of the unit disc. M. Brandt [4, (486)]
found the solution to the problem for the polynomials
$F(z)=z+a_{2}z^{2}+...+a_{N}z^{N}$ that are univalent in the disc
$\mathbb{D}$.

(1)
$\sup_{a_{j}:\,a_{1}=1}\left(F(-1)\right)=-\frac{1}{4}\sec^{2}{\frac{\pi}{N+2}}.$

Moreover, he wrote out the extremal polynomial [4, (390)-(392)]. In [5], a
more general formulation is considered; the extremal polynomial has the form

(2)
$\sum_{j=1}^{N}\frac{U^{\prime}_{N-j+1}\left(\cos\left(\frac{\pi}{N+2}\right)\right)}{U^{\prime}_{N}\left(\cos\left(\frac{\pi}{N+2}\right)\right)}U_{j-1}\left(\cos\left(\frac{\pi}{N+2}\right)\right)z^{j},$

which is more convenient for further generalizations, and it is shown that the
extremal polynomial is unique. This problem can be considered as the problem
of covering a segment on the real axis by the image of the unit disc under the
polynomial schlicht mapping. It is interesting to note that although the
problems of covering a disc and a segment are related, neither of them is a
consequence of the other. To the curious reader, we recommend to read [6].

##  1\. Introduction

Let us now consider an analogous problem for the univalent odd polynomials.

Optimization problem: on the class of odd polynomials of degree $2N-1$

$F(z)=\sum\limits_{j=1}^{N}{a_{j}z^{2j-1}}$

that are univalent in $\mathbb{D}=\left\\{z:\,\left\lvert
z\right\rvert<1\right\\}$ and have real coefficients, find

(3)
$J_{N}=\min_{a_{j}:\,a_{1}=1}\left(-iF(i)\right)=\min_{a_{j}:\,a_{1}=1}\left(\sum\limits_{j=1}^{N}{(-1)^{j+1}a_{j}}\right),$

as well as the polynomial (extremizer) that solves the optimization problem.
Above $i^{2}=-1.$

Note that the possibility of writing “minimum” instead of “infimum” is
guaranteed by DeBranges theorem, i.e., we can consider the infimum only on a
compact set since the coefficients of schlicht functions do not escape to
infinity.

##  2\. Main results

Consider this optimization problem for a wider class of polynomials than
univalent ones, namely for those polynomials $F(z)$ that map the semidisc
$\hat{D}=\left\\{z\in\mathbb{C}:\,\left\lvert
z\right\rvert<1,\,\text{Re}\\{z\\}>0\right\\}$ to the right half-plane
$\hat{C}=\left\\{z\in\mathbb{C}:\,\text{Re}\\{z\\}>0\right\\}.$

Note that the odd algebraic polynomial $F(z)$ satisfies the previous condition
if and only if the polynomial $iF(-iz)$ is typically real111 A polynomial
$f(z)$ is typically real if it is real in the interval $(-1,1)$ and in all
other points of the disc $\mathbb{D}$ we have $\Im\\{f(z)\\}\Im\\{z\\}>0$.
Every univalent polynomial with real coefficients is also typically real.

######  Theorem 1.

If the polynomial $iF(-iz)$ is typically real, then the equality

$J_{N}=\frac{1}{2}\sec^{2}{\frac{\pi}{2N+2}}=-iF^{(0)}(i)\leq-iF(i)$

holds. In this case, the extremizer is unique and is given by the formula

(4)
$F^{(0)}(z)=\sum\limits_{j=1}^{N}{\frac{U^{\prime}_{2(N-j+1)}\left(\cos\left(\frac{\pi}{2N+2}\right)\right)}{U^{\prime}_{2N}\left(\cos\left(\frac{\pi}{2N+2}\right)\right)}z^{2j-1}}.$

###### Proof.

Since $iF(-iz)$ is typically real, this means $F(z)$ maps the semidisc
$\hat{D}$ to the half-plane $\hat{C}$. It follows that
$\displaystyle\text{Re}(F(z))=\sum\limits_{j=1}^{N}{a_{j}\cos\left((2j-1)t\right)}\geq
0$. Thus, we have

$J_{N}=\min_{a_{j}:\,a_{1}=1}\left(\left\lvert F(i)\right\rvert\right)=$
$=\inf_{a_{j}}\left\\{1-a_{2}+a_{3}-\ldots:\sum\limits_{j=1}^{N}{a_{j}\cos\left((2j-1)t\right)}\geq
0,\,t\in\left[0,\frac{\pi}{2}\right]\right\\}.$

The equality

(5)
$\sum\limits_{j=1}^{N}{a_{j}\cos\left((2j-1)t\right)}=\gamma_{1}\cos\left(t\right)\left(1+2\sum\limits_{j=2}^{N}{\frac{\gamma_{j}}{\gamma_{1}}\cos\left(2(j-1)t\right)}\right)$

holds, where the coefficients $a_{1},\ldots,a_{N}$ and
$\gamma_{1},\ldots,\gamma_{N}$ are related by the bijective relationship

(6)
$\left\\{\begin{array}[]{ll}\gamma_{s}=\sum\limits_{j=s}^{N}{(-1)^{s+j}a_{j}},\\\
s=1,\ldots,N,\end{array}\right.$

which is equivalent to

(7) $\left\\{\begin{array}[]{ll}a_{s}=\gamma_{s}+\gamma_{s+1},\\\
s=1,\ldots,N.\end{array}\right.$

In (7), for convenience, it is assumed that $\gamma_{N+1}=0.$ It follows from
(6) or (7) that $\gamma_{1}=\sum\limits_{j=1}^{N}{(-1)^{s+1}a_{j}}$ and
$a_{1}=\gamma_{1}+\gamma_{2}=1.$ Then

$J_{N}=\min_{\gamma_{j}:\,\gamma_{1}+\gamma_{2}=1}\left\\{\gamma_{1}:\,1+2\sum\limits_{j=1}^{N-1}{\frac{\gamma_{j+1}}{\gamma_{1}}\cos\left(2tj\right)}\geq
0\right\\}.$

According to Problem 52 in [7, p. 79] or (1.3) in [8], a nonnegative
trigonometric polynomial of the form

$1+\sum_{j=1}^{N-1}\left(\lambda_{j}\cos\left(2tj\right)+\mu_{j}\sin\left(2tj\right)\right)$

must satisfy the inequality $\sqrt{\lambda_{1}^{2}+\mu_{1}^{2}}\leq
2\cos\left(\frac{\pi}{N+1}\right)$. Since our nonnegative trigonometric
polynomial

$1+2\sum\limits_{j=1}^{N-1}{\frac{\gamma_{j+1}}{\gamma_{1}}\cos\left(2tj\right)}$

has no sine terms, the inequality simplifies, yielding

$\sqrt{\left(2\cdot\frac{\gamma_{2}}{\gamma_{1}}\right)^{2}}\leq
2\cos\left(\frac{\pi}{N+1}\right)\implies\left\lvert\frac{\gamma_{2}}{\gamma_{1}}\right\rvert=\left\lvert\frac{1-\gamma_{1}}{\gamma_{1}}\right\rvert\leq\cos\left(\frac{\pi}{N+1}\right).$

From this, taking into account that $0<\gamma_{1}<1,$ we obtain

$\gamma_{1}\geq\frac{1}{1+\cos\left(\frac{\pi}{N+1}\right)}=\frac{1}{2\cos^{2}\left(\frac{\pi}{2N+2}\right)}.$

Therefore,
$\gamma_{1}=\frac{1}{1+\cos\left(\pi/(N+1)\right)}=\frac{1}{2\cos^{2}\left(\pi/(2N+2)\right)}.$
The remaining coefficients are uniquely determined [9] from the condition

$\displaystyle\left\lvert\frac{\gamma_{j+1}}{\gamma_{1}}\right\rvert=\frac{1}{2(N+1)\sin\left(\frac{\pi}{N+1}\right)}\left[(N-j+2)\sin\left(\frac{\pi(j+1)}{N+1}\right)-\right.$
$\displaystyle\left.-(N-j)\sin\left(\frac{\pi(j-1)}{N+1}\right)\right],$
$\displaystyle j=2,\ldots,N-1.$

Determine from (7) the coefficients of the extremizer (these are unique):

$\displaystyle
a^{0}_{j}=\frac{1}{2(N+1)\sin\left(\frac{\pi}{N+1}\right)\left(1+\cos\left(\frac{\pi}{N+1}\right)\right)}\left((N-j+3)\sin\left(\frac{\pi
j}{N+1}\right)-\right.$
$\displaystyle-(N-j+1)\sin\left(\frac{\pi(j-2)}{N+1}\right)+(N-j+2)\sin\left(\frac{\pi(j+1)}{N+1}\right)-$
$\displaystyle\left.-(N-j)\sin\left(\frac{\pi(j-1)}{N+1}\right)\right),$
$\displaystyle j=1,\ldots,N-1,$ $\displaystyle
a^{(0)}_{N}=\frac{1}{2(N+1)\sin\left(\frac{\pi}{N+1}\right)\left(1+\cos\left(\frac{\pi}{N+1}\right)\right)}\left(3\sin\left(\frac{\pi}{N+1}\right)-\right.$
$\displaystyle-\left.\sin\left(\frac{3\pi}{N+1}\right)\right)=$
$\displaystyle=\frac{2}{N+1}\left(1-\cos\left(\frac{\pi}{N+1}\right)\right)=\frac{4}{N+1}\sin^{2}\left(\frac{\pi}{2N+2}\right).$

The formulas for the extremizer coefficients can be simplified using an easily
verifiable identity:

$\displaystyle(N-j+3)\sin\left(\frac{\pi
j}{N+1}\right)-(N-j+1)\sin\left(\frac{\pi(j-2)}{N+1}\right)+$
$\displaystyle+(N-j+2)\sin\left(\frac{\pi(j+1)}{N+1}\right)-(N-j)\sin\left(\frac{\pi(j-1)}{N+1}\right)=$
$\displaystyle=2\left(1+\cos\left(\frac{\pi}{N+1}\right)\right)\left((N-j+2)\sin\left(\frac{\pi
j}{N+1}\right)-\right.$
$\displaystyle\left.-(N-j+1)\sin\left(\frac{\pi(j-1)}{N+1}\right)\right).$

Then

$\displaystyle
a^{0}_{j}=\frac{1}{(N+1)\sin\left(\frac{\pi}{N+1}\right)}\left((N-j+2)\sin\left(\frac{\pi
j}{N+1}\right)-\right.$
$\displaystyle\left.-(N-j+1)\sin\left(\frac{\pi(j-1)}{N+1}\right)\right),$
$\displaystyle j=1,\ldots,N.$

Applying the formula [5, Lemma 2]

$U^{\prime}_{k}(x)=\frac{1}{2(1-x^{2})}\left((k+2)U_{k-1}(x)-kU_{k+1}(x)\right),$

where
$U_{j}(x)=U_{j}(\cos\left(t\right))=\frac{\sin\left((j+1)t\right)}{\sin\left(t\right)}=2^{j}x^{j}+\ldots$
is the family of Chebyshev polynomials of the second kind, we can express the
coefficients $a_{j}^{0}$ in terms of the derivatives of Chebyshev polynomials
of the second kind:

$a_{j}^{0}=\frac{U^{\prime}_{2(N-j+1)}\left(\cos\left(\frac{\pi}{2N+2}\right)\right)}{U^{\prime}_{2N}\left(\cos\left(\frac{\pi}{2N+2}\right)\right)},\;j=1,\ldots,N.$

The theorem is fully proved. ∎

The extremizer allows a closed-form representation.

######  Theorem 2.

The following representation holds:

(8)
$F^{(0)}(z)=z\frac{4z^{2}\left(z^{2N+2}(1+z^{2})+N(1-z^{2})+2\right)\sin^{2}\left(\frac{\pi}{2N+2}\right)+(N+1)(1-z^{2})^{3}}{(N+1)\left(z^{4}-2z^{2}\cos\left(\frac{\pi}{N+1}\right)+1\right)^{2}}=$
$-\frac{z(z^{2}-1)}{z^{4}-2z^{2}\cos\left(\frac{\pi}{N+1}\right)+1}+4\sin^{2}\frac{\pi}{2N+2}\cdot\frac{z^{3}(z^{2}+1)(z^{2N+2}+1)}{(N+1)(z^{4}-2z^{2}\cos\left(\frac{\pi}{N+1}\right)+1)^{2}}.$

The proof follows from the formulas for the sum of a geometric series and for
the derivative of a geometric series.

Note that changing the schlicht normalization $F(0)=0,\,F^{\prime}(0)=1$ to
the normalization $F(0)=0,\,F(1)=1$ leads to other extremal polynomials,
namely, to the Suffridge and odd Fejér polynomials, which is interesting in
itself. Details can be found in [10].

##  3\. Extremizer univalence

Generally, the task of proving the univalence of functions, and polynomials in
particular, is far from simple. The number of popular examples of such
polynomials is quite limited, especially for odd polynomials. Nevertheless, we
managed to prove the univalence of the extremal polynomial in our case, which
can be considered an undoubted success.

######  Theorem 3.

The polynomial $F^{(0)}(z)$ is univalent in $\mathbb{D}.$

Denote

$u(t):=\text{Re}\left\\{F^{(0)}(e^{it})\right\\}=\frac{4\sin^{2}\left(\frac{\pi}{2N+2}\right)}{N+1}\frac{\cos\left(t\right)\cos^{2}\left((N+1)t\right)}{\left(\cos\left(2t\right)-\cos\left(\frac{\pi}{N+1}\right)\right)^{2}},$
$v(t):=\text{Im}\left\\{F^{(0)}(e^{it})\right\\}=$
$=\frac{\sin\left((2N+2)t\right)\cos\left(t\right)\left(1-\cos\left(\frac{\pi}{N+1}\right)\right)-(N+1)\sin\left(t\right)\left(\cos\left(2t\right)-\cos\left(\frac{\pi}{N+1}\right)\right)}{(N+1)\left(\cos\left(2t\right)-\cos\left(\frac{\pi}{N+1}\right)\right)^{2}}.$

We want to show that the set
$\Gamma=\left\\{F^{(0)}(e^{it}),\,t\in\left[0,\frac{\pi}{2}\right]\right\\}$
defines a simple curve (without self-crossing) on the complex plane, which
also does not intersect the real axis. That is, first it is to be proved that
there holds

Statement A. $v(t)\geq 0,\;t\in\left[0,\frac{\pi}{2}\right].$

Then the curve
$\hat{\Gamma}=\left\\{F^{(0)}(e^{it}),\,t\in\left[0,2\pi\right]\right\\}$
turns out to be a simple curve too. This follows from the fact that the curve
$\hat{\Gamma}$ is symmetric with respect to the imaginary and real axes. I.e.,
the image of the boundary of the disc $D$ is a simple curve on the complex
plane; therefore, the mapping will be univalent.

The curve $\Gamma$ will be simple if the system of equations

$\left\\{\begin{array}[]{ll}u(t_{1})=u(t_{2}),\\\
v(t_{1})=v(t_{2}),\end{array}\right.$

has no solutions in the region $t_{1}\in\left[0,\frac{\pi}{2}\right],$
$t_{2}\in\left[0,\frac{\pi}{2}\right],$ $t_{1}\neq t_{2}.$ This property will
certainly be true if the following statements are fulfilled:

Statement B. The function $u(t)$ decreases when
$t\in\left[0,\frac{\pi}{N+1}\right].$

Statement C. For $t\in\left(\frac{\pi}{N+1},\frac{\pi}{2}\right]$ the
inequality holds

$u(t)<u\left(\frac{\pi}{N+1}\right).$

Statement D. The function $v(t)$ decreases when
$t\in\left(\frac{\pi}{N+1},\frac{\pi}{2}\right].$

The proofs of Statements A and B will be given in Section 3.1, and the proof
of Statements C and D will be given in Section 3.2. In proving Statement B, we
will prove the following stronger statement: the function $u(t)$ is positive
and decreasing for $t\in\left(0,\frac{3\pi}{2N+2}\right).$

### 3.1 Proofs of Statements A, B

Proof of Statement A. Note that the function $v(t)$ is continuous, except for
the point $t_{0}=\frac{\pi}{2N+2}$ where the function has a removable
discontinuity. Let us use the formula 1.391.1 from [11, p.40]

$\sin\left((2N+2)t\right)=2(N+1)\sin\left(t\right)\cos\left(t\right)\prod\limits_{k=1}^{N}\left(1-\frac{\sin^{2}\left(t\right)}{\sin^{2}\left(\frac{k\pi}{2N+2}\right)}\right)$

and well-known trigonometric formulas. Then

$v(t)=\frac{\sin\left(t\right)}{\sin^{2}\left(t\right)-\sin^{2}\left(\frac{\pi}{2N+2}\right)}\left(\frac{1}{2}-\cos^{2}\left(t\right)\prod\limits_{k=2}^{N}\left(1-\frac{\sin^{2}\left(t\right)}{\sin^{2}\left(\frac{k\pi}{2N+2}\right)}\right)\right).$

Each expression in parentheses in the product
$\prod\limits_{k=2}^{N}\left(1-\frac{\sin^{2}\left(t\right)}{\sin^{2}\left(\frac{k\pi}{2N+2}\right)}\right)$
is positive and decreases when $t\in\left(0,\frac{\pi}{N+1}\right).$ Hence,
the function

$\frac{1}{2}-\cos^{2}\left(t\right)\prod\limits_{k=2}^{N}\left(1-\frac{\sin^{2}\left(t\right)}{\sin^{2}\left(\frac{k\pi}{2N+2}\right)}\right)$

increases on the interval $\left(0,\frac{\pi}{N+1}\right)$; moreover, it
increases from $-\frac{1}{2}$ to zero when
$t\in\left(0,\frac{\pi}{2N+2}\right)$ and from zero to $\frac{1}{2}$ when
$t\in\left(\frac{\pi}{2N+2},\frac{\pi}{N+1}\right).$ Thus, the function $v(t)$
is positive on $\left(0,\frac{\pi}{N+1}\right).$

Let $t\in\left(\frac{\pi}{N+1},\frac{\pi}{2}\right).$ We have

$\sin\left((2N+2)t\right)\cos\left(t\right)\left(1-\cos\left(\frac{\pi}{N+1}\right)\right)-(N+1)\sin\left(t\right)\left(\cos\left(2t\right)-\cos\left(\frac{\pi}{N+1}\right)\right)>$
$>-2\sin^{2}\left(\frac{\pi}{2N+2}\right)+2(N+1)\sin\left(t\right)\left(\sin^{2}\left(t\right)-\sin^{2}\left(\frac{\pi}{2N+2}\right)\right)\geq$
$\geq-2\sin^{2}\left(\frac{\pi}{2N+2}\right)+2(N+1)\sin\left(\frac{\pi}{N+1}\right)\left(\sin^{2}\left(\frac{\pi}{N+1}\right)-\sin^{2}\left(\frac{\pi}{2N+2}\right)\right),$

because the function
$-2\sin^{2}\left(\frac{\pi}{2N+2}\right)+2(N+1)\sin\left(t\right)\left(\sin^{2}\left(t\right)-\sin^{2}\left(\frac{\pi}{2N+2}\right)\right)$
increases on $\left(\frac{\pi}{N+1},\frac{\pi}{2}\right).$ The inequality

$-2\sin^{2}\left(\frac{\pi}{2N+2}\right)+2(N+1)\sin\left(\frac{\pi}{N+1}\right)\left(\sin^{2}\left(\frac{\pi}{N+1}\right)-\sin^{2}\left(\frac{\pi}{2N+2}\right)\right)>0$

is equivalent to the inequality

$(N+1)\sin\left(\frac{\pi}{N+1}\right)\left(1+2\cos\left(\frac{\pi}{N+1}\right)\right)>1.$

The last inequality is obvious, since
$(N+1)\sin\left(\frac{\pi}{N+1}\right)>1.$

Proof of Statement B. The function $u(t)$ is continuous, except for the point
$t_{0}=\frac{\pi}{2N+2}$ where the function has a removable discontinuity. The
formulas 1.391.2 and 1.391.4 from [11, p.40] say

$\cos\left((N+1)t\right)=\prod\limits_{k=1}^{\frac{N}{2}}{\left(1-\frac{\sin^{2}\left(t\right)}{\sin^{2}\left(\frac{(2k-1)\pi}{2N+2}\right)}\right)}\;\text{if
}N\text{ is even},$
$\cos\left((N+1)t\right)=\cos\left(t\right)\prod\limits_{k=1}^{\frac{N+1}{2}}{\left(1-\frac{\sin^{2}\left(t\right)}{\sin^{2}\left(\frac{(2k-1)\pi}{2N+2}\right)}\right)}\;\text{if
}N\text{ is odd}.$

Then

$\frac{\cos\left((N+1)t\right)}{\cos\left(2t\right)-\cos\left(\frac{\pi}{N+1}\right)}=\frac{1}{2\sin^{2}\left(\frac{\pi}{2N+2}\right)}\prod\limits_{k=2}^{\frac{N}{2}}{\left(1-\frac{\sin^{2}\left(t\right)}{\sin^{2}\left(\frac{(2k-1)\pi}{2N+2}\right)}\right)}\;\text{if
}N\text{ is even},$
$\frac{\cos\left((N+1)t\right)}{\cos\left(2t\right)-\cos\left(\frac{\pi}{N+1}\right)}=\frac{\cos\left(t\right)}{2\sin^{2}\left(\frac{\pi}{2N+2}\right)}\prod\limits_{k=2}^{\frac{N+1}{2}}{\left(1-\frac{\sin^{2}\left(t\right)}{\sin^{2}\left(\frac{(2k-1)\pi}{2N+2}\right)}\right)}\;\text{if
}N\text{ is odd}.$

When $t\in\left(0,\frac{3\pi}{2N+2}\right),$ each expression in parentheses in
the product is positive and decreases in $t.$ Hence, the entire product
decreases, and then the function $u(t)$ also decreases as the product of
positive decreasing functions. Because
$\frac{3\pi}{2N+2}=\frac{3}{2}\cdot\frac{\pi}{N+1}>\frac{\pi}{N+1}$, Statement
B is confirmed.

### 3.2 Proof of Statements C and D

####  3.0.1. Proof of Statement C

Let us note that function $\phi(t)=\frac{x}{(x^{2}-a^{2})^{2}}$ is increasing
for $x\in(0,\left\lvert a\right\rvert).$ Therefore,

$\frac{\cos\left(t\right)\cos^{2}\left((N+1)t\right)}{\left(\cos\left(2t\right)-\cos\left(\frac{\pi}{N+1}\right)\right)^{2}}\leq\frac{\cos\left(t\right)}{\left(\cos\left(2t\right)-\cos\left(\frac{\pi}{N+1}\right)\right)^{2}}=\frac{\cos\left(t\right)}{4\left(\cos^{2}\left(t\right)-\cos^{2}\left(\frac{\pi}{2(N+1)}\right)\right)^{2}}<$
$\frac{\cos\left(\frac{\pi}{N+1}\right)}{4\left(\cos^{2}\left(\frac{\pi}{N+1}\right)-\cos^{2}\left(\frac{\pi}{2(N+1)}\right)\right)^{2}}=\frac{\cos\left(\frac{\pi}{N+1}\right)}{\left(\cos\left(\frac{2\pi}{N+1}\right)-\cos\left(\frac{\pi}{N+1}\right)\right)^{2}},$

which implies the Statement C.

####  3.0.2. Proof of Statement D

We will begin the proof of Statement D with calculating the derivative of
$v(t)$:

$v^{\prime}(t)=\frac{1}{\sin^{2}\left(t\right)-\sin^{2}\left(\frac{\pi}{2N+2}\right)}\left(-\frac{\cos\left(t\right)}{2}\left(\sin^{2}\left(t\right)+\sin^{2}\left(\frac{\pi}{2N+2}\right)\right)+\right.$
$+\sin^{2}\left(\frac{\pi}{2N+2}\right)\cos\left(t\right)\cos\left((2N+2)t\right)-$
$\left.-\frac{\sin^{2}\left(\frac{\pi}{2N+2}\right)}{2N+2}\sin\left((2N+2)t\right)\sin\left(t\right)\left(1+\frac{4\cos^{2}\left(t\right)}{\sin^{2}\left(t\right)-\sin^{2}\left(\frac{\pi}{2N+2}\right)}\right)\right).$

We need to check that

(9)
$\begin{split}-\frac{\sin^{2}\left(t\right)+\sin^{2}\left(\frac{\pi}{2N+2}\right)}{2\sin^{2}\left(\frac{\pi}{2N+2}\right)}+\cos\left((2N+2)t\right)-\\\
-\frac{\sin\left((2N+2)t\right)}{2N+2}\left(\tan\left(t\right)+\frac{4\cos\left(t\right)\sin\left(t\right)}{\sin^{2}\left(t\right)-\sin^{2}\left(\frac{\pi}{2N+2}\right)}\right)<0\end{split}$

when $t\in\left(\frac{\pi}{N+1},\frac{\pi}{2}\right),$ $N=2,3,4,\ldots.$

Since

$\frac{\sin^{2}\left(t\right)+\sin^{2}\left(\frac{\pi}{2N+2}\right)}{2\sin^{2}\left(\frac{\pi}{2N+2}\right)}>\frac{\sin^{2}\left(\frac{\pi}{N+1}\right)+\sin^{2}\left(\frac{\pi}{2N+2}\right)}{2\sin^{2}\left(\frac{\pi}{2N+2}\right)}=2\cos^{2}\left(\frac{\pi}{2N+2}\right)+1\geq
2,$ $\left\lvert\cos\left((2N+2)t\right)\right\rvert\leq 1,$

then it suffices to prove the inequality

(10)
$\frac{\sin\left((2N+2)t\right)}{2N+2}\left(\tan\left(t\right)+\frac{4\cos\left(t\right)\sin\left(t\right)}{\sin^{2}\left(t\right)-\sin^{2}\left(\frac{\pi}{2N+2}\right)}\right)\geq-1.$

Let us split the interval $\left(\frac{\pi}{N+1},\frac{\pi}{2}\right)$ into
two intervals:

$\left(\frac{\pi}{N+1},\frac{\pi}{2}\right)=\left(\frac{\pi}{N+1},\frac{\pi}{2}-\frac{\pi}{2N+2}\right)\cup\left[\frac{\pi}{2}-\frac{\pi}{2N+2},\frac{\pi}{2}\right).$

We will consider the behavior of the function

$\Psi(t)=\frac{\sin\left((2N+2)t\right)}{2N+2}\left(\tan\left(t\right)+\frac{4\cos\left(t\right)\sin\left(t\right)}{\sin^{2}\left(t\right)-\sin^{2}\left(\frac{\pi}{2N+2}\right)}\right)$

on each of the specified intervals separately.

_Case 1:_ $t\in
I_{2}=\left[\frac{\pi}{2}-\frac{\pi}{2N+2},\frac{\pi}{2}\right).$

When $N$ is even, the function $\Psi(t)$ is not negative on the interval
$I_{2}$. Therefore, inequality (10) is satisfied. Let $N$ be odd. Then

$\left\lvert\frac{\sin\left((N+1)t\right)}{(N+1)\cos\left(t\right)}\right\rvert\leq
1,$
$\frac{1}{\sin^{2}\left(t\right)-\sin^{2}\left(\frac{\pi}{2N+2}\right)}\leq\frac{1}{\cos^{2}\left(\frac{\pi}{2N+2}\right)-\sin^{2}\left(\frac{\pi}{2N+2}\right)}=\frac{1}{\cos\left(\frac{\pi}{N+1}\right)}\leq\sqrt{2},$

hence,

$\left\lvert\Psi(t)\right\rvert\leq\left\lvert\cos\left((N+1)t\right)\right\rvert\left(1+4\sqrt{2}\cos^{2}\left(t\right)\right).$

It is convenient to make the substitution $\tau=\frac{\pi}{2}-t,$
$\tau\in\left(0,\frac{\pi}{2N+2}\right].$ Then

$\left\lvert\cos\left((N+1)\tau\right)\right\rvert\left(1+4\sqrt{2}\sin^{2}\left(\tau\right)\right)\leq\left(1-\frac{(N+1)^{2}\tau^{2}}{2}\right)\left(1+4\sqrt{2}\tau^{2}\right)\leq
1.$

The last inequality is true, since $\frac{2}{(N+1)^{2}}<\frac{1}{4\sqrt{2}}.$
Thus, $\Psi(t)\geq-1,$
$t\in\left[\frac{\pi}{2}-\frac{\pi}{2N+2},\frac{\pi}{2}\right).$

_Case 2:_ $t\in
I_{1}=\left(\frac{\pi}{N+1},\frac{\pi}{2}-\frac{\pi}{2N+2}\right).$

Note that for $N=2,$ $I_{1}=\varnothing;$ and for $N=3,$ $\Psi(t)\geq 0$ when
$t\in I_{1}.$ Therefore, inequality (10) is valid when $N=2$ or $N=3$.

Let $N\geq 4.$ Estimate the quantity
$\displaystyle\frac{\sin\left(t\right)}{\sin^{2}\left(t\right)-\sin^{2}\left(\pi/(2N+2)\right)}.$
This function can be represented as a sum of two functions which are both
decreasing on $I_{1}$,

$\frac{\sin\left(t\right)}{\sin^{2}\left(t\right)-\sin^{2}\left(\frac{\pi}{2N+2}\right)}=\frac{1}{\sin\left(t\right)+\sin\left(\frac{\pi}{2N+2}\right)}+\frac{\sin\left(\frac{\pi}{2N+2}\right)}{\sin^{2}\left(t\right)-\sin^{2}\left(\frac{\pi}{2N+2}\right)}.$

Therefore, it decreases on $I_{1}.$ Consider the function
$\frac{N+1}{2}\left(\cos\left(t\right)+\frac{1}{2}\frac{\sin^{2}\left(t\right)}{\cos\left(t\right)}\right).$
It increases on $I_{1},$ since
$\left(\cos\left(t\right)+\frac{1}{2}\frac{\sin^{2}\left(t\right)}{\cos\left(t\right)}\right)^{\prime}=\frac{\sin^{3}\left(t\right)}{2\cos^{2}\left(t\right)}.$
Let us show that for $t=\frac{\pi}{N+1}$, the inequality

(11)
$\frac{\sin\left(t\right)}{\sin^{2}\left(t\right)-\sin^{2}\left(\frac{\pi}{2N+2}\right)}<\frac{N+1}{2}\left(\cos\left(t\right)+\frac{1}{2}\frac{\sin^{2}\left(t\right)}{\cos\left(t\right)}\right)$

holds, which will also imply the validity of inequality (11) for all $t\in
I_{1}$.

Let us verify that inequality (11) is satisfied for $N=4$ and $N=5$, directly
calculating the expressions in both sides of this inequality with
$t=\frac{\pi}{N+1}.$ We will get $2.35\ldots<2.55\ldots$ and
$2.73\ldots<3.03\ldots$ for $N=4$ and $N=5$, respectively.

Let us show the validity of inequality (11) for the rest $N\geq 6.$ Substitute
$x=\frac{\pi}{N+1},$ then inequality (11) for $t=\frac{\pi}{N+1}$ becomes the
inequality

(12)
$\frac{\sin\left(x\right)}{\sin^{2}\left(x\right)-\sin^{2}\left(\frac{x}{2}\right)}<\frac{\pi}{2x}\left(\cos\left(x\right)+\frac{1}{2}\frac{\sin^{2}\left(x\right)}{\cos\left(x\right)}\right),\;x\in\left(0,\frac{\pi}{7}\right].$

Use the identity

$\sin^{2}\left(x\right)-\sin^{2}\left(\frac{x}{2}\right)=\frac{1}{2}\left(1+\cos\left(x\right)-2\cos^{2}\left(x\right)\right)=(1-\cos\left(x\right))\left(\frac{1}{2}+\cos\left(x\right)\right)$

and rewrite inequality (12) as

(13)
$\frac{4}{\pi}x\sin\left(x\right)\cos\left(x\right)<\left(1+\cos^{2}\left(x\right)\right)(1-\cos\left(x\right))\left(\frac{1}{2}+\cos\left(x\right)\right).$

Apply the inequality

(14)
$\frac{4}{\pi}x\sin\left(x\right)<1-\cos^{3}\left(x\right)=\left(1-\cos\left(x\right)\right)\left(1+\cos\left(x\right)+\cos^{2}\left(x\right)\right),\;x\in\left(0,\frac{\pi}{7}\right]$

to inequality (13), thus obtaining

(15)
$\displaystyle\cos\left(x\right)\left(1-\cos\left(x\right)\right)\left(1+\cos\left(x\right)+\cos^{2}\left(x\right)\right)$
$\displaystyle<\left(1-\cos\left(x\right)\right)\left(1+\cos^{2}\left(x\right)\right)\left(\frac{1}{2}+\cos\left(x\right)\right)$

We will give the proof of inequality (14) at the end of section 3.2, after
verifying (15).

Verifying (15) is equivalent to checking the inequality

$\left(1+\cos^{2}\left(x\right)\right)\left(\frac{1}{2}+\cos\left(x\right)\right)-\cos\left(x\right)\left(1+\cos\left(x\right)+\cos^{2}\left(x\right)\right)>0.$

We have

$\left(1+\cos^{2}\left(x\right)\right)\left(\frac{1}{2}+\cos\left(x\right)\right)-\cos\left(x\right)\left(1+\cos\left(x\right)+\cos^{2}\left(x\right)\right)=\frac{1}{2}\sin^{2}\left(x\right)>0$

when $x\in\left(0,\frac{\pi}{7}\right].$

Applying (11), we finally estimate the function $\Psi(t)$.

$\left\lvert\Psi(t)\right\rvert\leq\frac{1}{2N+2}\left(\tan\left(t\right)+\frac{4\cos\left(t\right)\sin\left(t\right)}{\sin^{2}\left(t\right)-\sin^{2}\left(\frac{\pi}{2N+2}\right)}\right)<$
$<\frac{1}{2N+2}\tan\left(t\right)+\cos^{2}\left(t\right)+\frac{1}{2}\sin^{2}\left(t\right)=1+\frac{1}{2N+2}\tan\left(t\right)-\frac{1}{2}\sin^{2}\left(t\right)=$
$=1+\frac{1}{4}\tan\left(t\right)\left(\frac{2}{N+1}-\sin\left(2t\right)\right).$

Since $t\in I_{1}$, this means that
$2t\in\left(\frac{2\pi}{N+1},\pi-\frac{\pi}{N+1}\right)$. Recall that the sine
wave increases on this interval for $2t<\frac{\pi}{2}$ and decreases for
$2t>\frac{\pi}{2}$. Thus, for $N\geq 6$, using the symmetry of the sine wave
we have

$\sin\left(2\left(\frac{\pi}{2}-\frac{\pi}{2N+2}\right)\right)=\sin\left(\pi-\frac{\pi}{N+1}\right)=\sin\left(\frac{\pi}{N+1}\right)<\sin\left(\frac{2\pi}{N+1}\right).$

Therefore,

$1+\frac{1}{4}\tan\left(t\right)\left(\frac{2}{N+1}-\sin\left(2t\right)\right)\leq$
$=1+\frac{1}{4}\tan\left(t\right)\left(\frac{2}{N+1}-\sin\left(\frac{\pi}{N+1}\right)\right)<1,$

since $\sin\left(\frac{\pi}{N+1}\right)>\frac{2}{N+1}$ when $N\geq 3.$ Whence,
Statement D is proved.

In conclusion, let us prove inequality (14). Rewrite it in the form

$\cos^{3}\left(x\right)<1-\frac{4}{\pi}x\sin\left(x\right).$

We have

$\cos^{3}\left(x\right)<\left(1-\frac{1}{2}x^{2}+\frac{1}{24}x^{4}\right)^{3}=$
$=1-\frac{3}{2}x^{2}+\frac{7}{8}x^{4}-\frac{1}{4}x^{6}+\frac{7}{192}x^{8}-\frac{1}{384}x^{10}+\frac{3}{13824}x^{12},$
$1-\frac{4}{\pi}x\sin\left(x\right)>1-\frac{4}{\pi}x^{2}.$

Calculate

$1-\frac{4}{\pi}x^{2}-\left(1-\frac{1}{2}x^{2}+\frac{1}{24}x^{4}\right)^{3}=$
$=x^{2}\left(\frac{3\pi-8}{2\pi}-\frac{7}{8}x^{2}\right)+x^{6}\left(\frac{1}{4}-\frac{7}{192}x^{2}\right)+x^{10}\left(\frac{1}{384}-\frac{1}{13824}x^{2}\right).$

This difference is obviously positive when
$0<x^{2}\leq\frac{4(3\pi-8)}{7\pi},$ i.e. when
$0<x\leq\sqrt{\frac{4(3\pi-8)}{7\pi}}=0.509\ldots.$ But
$\frac{\pi}{7}=0.44\ldots<0.509,$ therefore, inequality (14) holds at least
for $x\in\left(0,\frac{\pi}{7}\right].$

The assertion of Theorem 3 follows from Statements A, B, C, D, which we have
proved. Thus, Theorem 3 is also completely proved.

##  4\. Discussion of the results

The statement of the problems (1) and (3) considered above can be generalized
as follows.

Optimization problem: on the class of the univalent in
$\mathbb{D}=\left\\{z:\,\left\lvert z\right\rvert<1\right\\}$ polynomials with
the $T$-fold symmetry of degree $(N-1)T+1$ with real coefficients

$F(z)=\sum\limits_{j=1}^{N}{a_{j}z^{T(j-1)+1}}$

find

(16) $J^{(T)}_{N}=\min_{a_{j}:\,a_{1}=1}\left(\left\lvert
F\left(e^{\frac{i\pi}{T}}\right)\right\rvert\right),$

as well as the polynomial (extremizer) that solves the optimization problem.

Comparative analysis of polynomials (2) and (4) shows a remarkable feature in
the first factor. Namely, in the odd case it is the same multiplier, only with
odd numbers. Hence, it is logical to assume that in the $T$-symmetric case
there will be the same factor, only with the numbers forming a $T$-arithmetic
sequence.

The middle factor from (2) is absent in (4), so that we can assume it to equal
one. It is logical to assume that in the $T$-symmetric case this will be the
factor that turns into the middle one from formula (4) in the case $T=1,$ and
into one in the case $T=2.$ Such a factor has been proposed in [12] in the
form of the product that in the limit as $N\to\infty$ gives the corresponding
coefficient of the $T$-symmetric Koebe function. These arguments lead to the
following hypothesis.

######  Conjecture 1.

The extremizer for the problem (16) is as follows

$F^{(0)}(z)=z+\sum\limits_{j=2}^{N}{a^{(0)}_{j}z^{T(j-1)+1}},$
$a_{j}^{(0)}=\frac{U^{\prime}_{T(N-j+1)}\left(\cos\left(\frac{\pi}{TN+2}\right)\right)}{U^{\prime}_{TN}\left(\cos\left(\frac{\pi}{TN+2}\right)\right)}\prod\limits_{k=1}^{j-1}\frac{\sin\left(\frac{\pi(2+T(k-1))}{TN+2}\right)}{\sin\left(\frac{\pi
Tk}{TN+2}\right)},\;j=2,\ldots,N.$

Moreover, it is univalent.

In its turn, Conjecture 1 leads to the following problem.

######  Problem 1.

Determine the asymptotics of the coefficients of $F^{(0)}(z)$ and the quantity
$\left\lvert F^{(0)}\left(e^{\frac{i\pi}{T}}\right)\right\rvert$.

Figure 1. The image $F^{(0)}(\mathbb{D})$ at $N=7,$ $T=4.$

Figure 1 shows the image of the unit disc in the special case $N=7,$ $T=4.$
This picture argues in favor of the assumption that the polynomials are
univalent, which is confirmed in many other pictures plotted by the authors,
but not included in this work.

Note that Problem 1 is a polynomial version of the $T$-symmetric Koebe
theorem, which states that the image of the unit disc under the $T$-symmetric
schlicht mapping contains a disc of radius $4^{-1/T}.$ The sharpness of the
constant is given by the value $\left\lvert
K^{(T)}\left(e^{\frac{i\pi}{T}}\right)\right\rvert,$ where $\displaystyle
K^{(T)}=\frac{z}{(1-z^{T})^{2/T}}$ is $T$-symmetric Koebe function. This leads
to the next conjecture.

######  Conjecture 2.

The image of the unit disc under the $T$-symmetric schlicht polynomial degree
$N$ contains a disc of radius $\left\lvert
F^{(0)}\left(e^{\frac{i\pi}{T}}\right)\right\rvert.$ The value is sharp.

Note that in case $T=1$ the proposed value is
$\displaystyle\frac{1}{4}\sec^{2}{\frac{\pi}{N+2}}$ while in case $T=2$ it is
$\displaystyle\frac{1}{2}\sec^{2}{\frac{\pi}{2N+2}}.$

##  5\. Acknowledgment

The authors would like to thank Larie Ward for her help in preparation of the
manuscript.

## References

  * [1] J.E. Littlewood, R.E.A.C. Paley.  A proof that an odd schlicht function has bounded coefficients.  J. London Math. Soc., s1-7 (3):167–169, 1932.
  * [2] G. Szegö M. Fekete.  Eine Bemerkung über ungerade schlichte Funktionen.  J. London Math. Soc., 8:85–89, 1933.
  * [3] W.K. Hayman.  The asymptotic behaviour of p-valent functions.  Proceedings of the London Mathematical Society, Third Series, vol. 5:257–284, 1955.
  * [4] M. Brandt.  Variationsmethoden für in der Einheitskreisscheibe schlichte Polynome.  PhD thesis, Humboldt-Univ., Berlin, 1987.
  * [5] D. Dmitrishin, A. Smorodin, and A. Stokolos.  An extremal problem for polynomials.  Appl. Comput. Harmon. Anal., 56:283–305, 2022.
  * [6] Stokolos A. Dmitrishin D., Dyakonov K.  Univalent polynomials and koebe’s one-quarter theorem.  Anal. Math. Phys., 9:991–1004, 2019.
  * [7] Pólya G., Szegö G.  Problems and theorems in analysis. II. Theory of functions, zeros, polynomials, determinants, number theory, geometry.  Springer-Verlag, Berlin, 1998.  392 p.
  * [8] Roberto Andreani and Dimitar Dimitrov.  An extremal nonnegative sine polynomial.  Rocky Mountain Journal of Mathematics, 33, 07 2001.
  * [9] E.V. Egerváry, O. Szász.  Einige Extremalprobleme in Bereiche der trigonometrischen Polynomen.  Math. Z., 27:641–652, 1928.
  * [10] Dmitrishin D., Hagelstein P., Khamitova A., Korenovskyi A., Stokolos A.  Fejér polynomials and control of nonlinear discrete systems.  Constr Approx, 51:383–412, 2020.
  * [11] I.S. Gradshteyn, I.M. Ryzhik.  Table of Integrals, Series, and Products.  Academic Press, Amsterdam, Eighth edition, 2014.
  * [12] Dmitrishin D., Smorodin A., Stokolos A., Tohaneanu M.  Symmetrization of Suffridge polynomials and approximation of $T$-symmetric Koebe functions.  J. Math. Anal. Appl., 503(2), 2020.