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[不等式] 三个根均不同且模为1

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hbghlyj posted 2023-4-28 23:10 |Read mode

已知三次复系数多项式$p(x)=az^3+3bz^2+3\bar{b}z+\bar{a}$满足其三个根均不同且模为1，证明：$2\left|a\bar{b}-b^2\right|+\left|b\right|^2<\left|a\right|^2$

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等价转化下: 单位圆周在 $w\mapsto aw^3+3bw$ 下的像与实轴有 $6$ 个不同交点. 然后不知了. posted 2023-5-1 14:40

试了下, 令 $z=\frac{1+iw}{1-iw}$, 将 $p$ 转化作 $w$ 的实系数三次多项式, 再用判别式暴力求解可行. posted 2023-5-1 15:36

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original poster hbghlyj posted 2024-12-18 03:40

关于$w$的多项式有3个实根$\Rightarrow\Delta>0$然后呢

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