An extremal problem for odd univalent polynomials

Theorem 1. If the polynomial iF(-iz) is typically real, then the equality $$J_N = \frac{1}{2}\sec^2\frac{\pi}{2N+2} = -iF^{(0)}(i) \le -iF(i)$$ holds. In this case, the extremizer is unique and is given by the formula (4) $$F^{(0)}(z) = \sum_{j=1}^{N} \frac{U'_{2(N-j+1)}\left(\cos\left(\frac{\pi}{2N+2}\right)\right)}{U'_{2N}\left(\cos\left(\frac{\pi}{2N+2}\right)\right)} z^{2j-1}.$$

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'(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].

Theorem 3. The polynomial $F^{(0)}(z)$ is univalent in $\mathbb{D}$ . Denote $$u(t) := \operatorname{Re}\left\{F^{(0)}(e^{it})\right\} = \frac{4\sin^2\left(\frac{\pi}{2N+2}\right)}{N+1} \frac{\cos(t)\cos^2((N+1)t)}{\left(\cos(2t) - \cos\left(\frac{\pi}{N+1}\right)\right)^2},$$ $$v(t) := \operatorname{Im}\left\{F^{(0)}(e^{it})\right\} =$$ $$= \frac{\sin((2N+2)t)\cos(t)\left(1 - \cos\left(\frac{\pi}{N+1}\right)\right) - (N+1)\sin(t)\left(\cos(2t) - \cos\left(\frac{\pi}{N+1}\right)\right)}{(N+1)\left(\cos(2t) - \cos\left(\frac{\pi}{N+1}\right)\right)^2}.$$ We want to show that the set $\Gamma = \{F^{(0)}(e^{it}), t \in [0, \frac{\pi}{2}]\}$ 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 $\hat{\Gamma} = \{F^{(0)}(e^{it}), t \in [0, 2\pi]\}$ 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 $$\begin{cases} u(t_1) = u(t_2), \\ v(t_1) = v(t_2), \end{cases}$$ 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 $t \in \left(0, \frac{3\pi}{2N+2}\right)$ .