设方程 $x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}=0$ 的系数都是实数，且适合条件 $0<a_{0} \leqslant a_{1} \leqslant \cdots \leqslant a_{n-1} \leqslant 1$ 已知 $\lambda$ 为此方程的复数根，且适合条件 $|\lambda| \geqslant 1$．试证：$\lambda^{n+1}=1$
【难度】
【出处】
1992第7届CMO试题
【标注】
【答案】
略
【解析】
由条件可知 $\lambda \neq 0,1$．考虑
$ 0=(\lambda-1)\left(\lambda^{n}+a_{n-1} \lambda^{n-1}+\cdots+a_{1} \lambda+a_{0}\right)= \lambda^{n+1}+\left(a_{n-1}-1\right) \lambda^{n}+\left(a_{n-2}-a_{n-1}\right) \lambda^{n-1}+\cdots+\left(a_{0}-a_{1}\right) \lambda-a_{0} $
于是 $ \lambda^{n+1}=\left(1-a_{n-1}\right) \lambda^{n}+\left(a_{n-1}-a_{n-2}\right) \lambda^{n-1}+\cdots+\left(a_{1}-a_{0}\right) \lambda+a_{0} $ 其中右式系数都是非负实数．因此
$|\lambda|^{n+1}= |\left(1-a_{n-1}\right) \lambda^{n}+\left(a_{n-1}-a_{n-2}\right) \lambda^{n-1} 1+\cdots+\left(a_{1}-a_{0}\right) \lambda+a_{0} |\\ \leqslant \left(1-a_{n-1}\right)|\lambda|^{n}+\left(a_{n-1}-a_{n-2}\right)|\lambda|^{n-1}+\cdots+\left(a_{1}-a_{0}\right)|\lambda|+a_{0} $
由假设 $|\lambda| \geqslant 1$，所以有
$\begin{aligned}|\lambda|^{n+1} \leqslant &\left|\left(1-a_{n-1}\right)\right|\left.\lambda\right|^{n}+\left(a_{n-1}-a_{n-2}\right)|\lambda|^{n}+\cdots+ \left(a_{1}-a_{0}\right)|\lambda|^{n}+a_{0}|\lambda|^{n} \end{aligned}$
这证明 $|\lambda| \leqslant 1$．而由假设 $|\lambda| \geqslant 1$，可知 $|\lambda| = 1$．因此，上面不等式变为等式．所以有如下辐角关系 $\begin{aligned} \arg \left(1-a_{n-1}\right) \lambda^{n}=& \arg \left(a_{n-1}-a_{n-2}\right) \lambda^{n-1}=\cdots= \arg \left(a_{1}-a_{0}\right) \lambda=\arg a_{0}=0 \end{aligned}$ 因此 $\left(1-a_{\mathrm{n}-1}\right) \lambda^{n} \geqslant 0\left(a_{n-1}-a_{n-2}\right) \lambda^{\mathfrak{n}-1} \geqslant 0, \cdots,\left(a_{1}-a_{0}\right) \lambda \geqslant 0$ 从而 $\lambda^{n+1}=\left(1-a_{n-1}\right) \lambda^{n}+\cdots+a_{0} \geqslant 0$ 所以 $\lambda^{n+1}=\left|\lambda^{n+1}\right|=|\lambda|^{n+1}=1$
$ 0=(\lambda-1)\left(\lambda^{n}+a_{n-1} \lambda^{n-1}+\cdots+a_{1} \lambda+a_{0}\right)= \lambda^{n+1}+\left(a_{n-1}-1\right) \lambda^{n}+\left(a_{n-2}-a_{n-1}\right) \lambda^{n-1}+\cdots+\left(a_{0}-a_{1}\right) \lambda-a_{0} $
于是 $ \lambda^{n+1}=\left(1-a_{n-1}\right) \lambda^{n}+\left(a_{n-1}-a_{n-2}\right) \lambda^{n-1}+\cdots+\left(a_{1}-a_{0}\right) \lambda+a_{0} $ 其中右式系数都是非负实数．因此
$|\lambda|^{n+1}= |\left(1-a_{n-1}\right) \lambda^{n}+\left(a_{n-1}-a_{n-2}\right) \lambda^{n-1} 1+\cdots+\left(a_{1}-a_{0}\right) \lambda+a_{0} |\\ \leqslant \left(1-a_{n-1}\right)|\lambda|^{n}+\left(a_{n-1}-a_{n-2}\right)|\lambda|^{n-1}+\cdots+\left(a_{1}-a_{0}\right)|\lambda|+a_{0} $
由假设 $|\lambda| \geqslant 1$，所以有
$\begin{aligned}|\lambda|^{n+1} \leqslant &\left|\left(1-a_{n-1}\right)\right|\left.\lambda\right|^{n}+\left(a_{n-1}-a_{n-2}\right)|\lambda|^{n}+\cdots+ \left(a_{1}-a_{0}\right)|\lambda|^{n}+a_{0}|\lambda|^{n} \end{aligned}$
这证明 $|\lambda| \leqslant 1$．而由假设 $|\lambda| \geqslant 1$，可知 $|\lambda| = 1$．因此，上面不等式变为等式．所以有如下辐角关系 $\begin{aligned} \arg \left(1-a_{n-1}\right) \lambda^{n}=& \arg \left(a_{n-1}-a_{n-2}\right) \lambda^{n-1}=\cdots= \arg \left(a_{1}-a_{0}\right) \lambda=\arg a_{0}=0 \end{aligned}$ 因此 $\left(1-a_{\mathrm{n}-1}\right) \lambda^{n} \geqslant 0\left(a_{n-1}-a_{n-2}\right) \lambda^{\mathfrak{n}-1} \geqslant 0, \cdots,\left(a_{1}-a_{0}\right) \lambda \geqslant 0$ 从而 $\lambda^{n+1}=\left(1-a_{n-1}\right) \lambda^{n}+\cdots+a_{0} \geqslant 0$ 所以 $\lambda^{n+1}=\left|\lambda^{n+1}\right|=|\lambda|^{n+1}=1$
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