For the polynomials $F(z)={\sum }_{j=1}^N{a}_j{z}^j$ with real coefficients and normalization $a_1=1$ we solve the extremal problem$\underset{a_2,\dots ,a_N}{\sup }(\underset{z\in \mathbb{D}}{\inf }\{\mathrm{Re}(F(z)):\mathrm{Im}(F(z))=0\}).$ We show that the solution is $-\frac{1}{4}{\sec }^2\frac{\pi }{N+2}$, and the extremal polynomial$\frac{1}{U_N^{\prime }(\cos \frac{\pi }{N+2})}\sum _{j=1}^N{U}_{N-j+1}^{\prime }(\cos \frac{\pi }{N+2})U_{j-1}(\cos \frac{\pi }{N+2})z^j$ is unique and univalent, where the $U_j(x)$ are the Chebyshev polynomials of the second kind, $j=1,\dots ,N$. As an application, we obtain the estimate of the Koebe radius for the univalent polynomials in $\mathbb{D}$ and formulate several conjectures.

中文翻译：

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多项式的极值问题

对于多项式 $F(z)={\sum }_{j=1}^N{\mathrm{一种}}_j{z}^j$ 具有实系数和归一化 ${\mathrm{一种}}_1=1$ 我们解决了极值问题$\underset{{\mathrm{一种}}_2,\dots ,{\mathrm{一种}}_N}{\mathrm{超}}(\underset{z\in \mathbb{D}}{\mathrm{信息}}\{\mathrm{关于}(F(z))：\mathrm{我是}(F(z))=0\}).$ 我们证明解决方案是 $-\frac{1}{4}{\mathrm{秒}}^2\frac{\pi }{N+2}$，以及极值多项式$\frac{1}{{你}_N^{\prime }(\cos \frac{\pi }{N+2})}\sum _{j=1}^N{你}_{N-j+1}^{\prime }(\cos \frac{\pi }{N+2}){你}_{j-1}(\cos \frac{\pi }{N+2})z^j$ 是唯一的和单价的，其中 ${你}_j(X)$ 是第二类切比雪夫多项式， $j=1,\dots ,N$. 作为一个应用，我们获得了单价多项式的 Koebe 半径的估计$\mathbb{D}$ 并提出几个猜想。