具有折叠对称性的三项式的Koebe定理
Koebe's theorem for trinomials with fold symmetry
摘要：
实系数单叶多项式的Koebe问题仅在三项式中得到完全解决，这意味着在这种情况下，Koebe半径和极值多项式（极值化子）已经找到。一般情况仍然开放，但已经制定了公式。对于实系数和$T$重旋转对称的单叶多项式，也假设了相应的结构.本文证实了这些假说的三项$z + az^{T + 1} + bz^{2T + 1}$。也就是说，Koebe半径是$r=4\cos^2 \frac{π(1+T)}{2+3T}$，而Koebe问题的唯一极值化子是三项式\begin{gather*} B^{(T)}(z)=z+\frac2{2+3T}\left(-T+(2+2T)\cos\frac{πT}{2+3T}\right)z^{1+T}+\\ +\frac1{2+3T}\left(2+T-2T\cos\frac{πT}{2+3T}\right)z^{1+2T}. \end{gather*}关键词和短语：Koebe四分之一定理，Koebe半径，单叶多项式，具有折叠对称性的三项式。
英文摘要：
The Koebe problem for univalent polynomials with real coefficients is fully solved only for trinomials, which means that in this case the Koebe radius and the extremal polynomial (extremizer) have been found. The general case remains open, but conjectures have been formulated. The corresponding conjectures have also been hypothesized for univalent polynomials with real coefficients and $T$-fold rotational symmetry. This paper provides confirmation of these hypotheses for trinomials $z + az^{T + 1} + bz^{2T + 1}$. Namely, the Koebe radius is $r=4\cos^2 \frac{π(1+T)}{2+3T}$, and the only extremizer of the Koebe problem is the trinomial \begin{gather*} B^{(T)}(z)=z+\frac2{2+3T}\left(-T+(2+2T)\cos\frac{πT}{2+3T}\right)z^{1+T}+\\ +\frac1{2+3T}\left(2+T-2T\cos\frac{πT}{2+3T}\right)z^{1+2T}. \end{gather*} Key words and phrases: Koebe one-quarter theorem, Koebe radius, univalent polynomial, trinomials with fold symmetry.
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