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本帖最后由 hbghlyj 于 2024-10-30 10:50 编辑 对于 $k\in\{1,\dots,n\}$，
$$z_k^n + a_{n-1}z_k^{n-1} + \dots + a_1z_k + a_0 = 0$$即，$$z_k^n = - (a_{n-1}z_k^{n-1} + \dots + a_1z_k + a_0)$$取两边的模$$\left|z_k\right |^n = \left| a_{n-1}z_k^{n-1} + \dots + a_1z_k + a_0 \right|$$
两边平方$$\left|z_k\right |^{2n} = \left|a_{n-1}z_k^{n-1} + \dots + a_1z_k + a_0\right|^2$$
根据柯西不等式
\begin{align*}\left|z_k\right|^{2n}&= \left|a_{n-1}z_k^{n-1} + \dots + a_1z_k + a_0\right|^2 \\
&\le \left(1 + \left|z_k\right|^2 + \dots + \left|z_k\right| ^ {2n-2}\right) \left(\left|a_0\right|^2 + \dots + \left|a_{n-1}\right|^2\right)\\
&=\frac{\left|z_k\right|^{2n} - 1}{\left|z_k\right|^2 - 1}\left(\left|a_0\right|^2 + \dots + \left|a_{n-1}\right|^2\right),\end{align*}
若 $|z_k| > 1$，两边乘以$\displaystyle\frac{\left|z_k\right|^2-1}{|z_k|^{2n}}$，$$\left|z_k\right|^2-1 \le \frac{\left|z_k\right|^{2n}-1}{\left|z_k\right|^{2n}}\left(\left|a_0\right|^2 + \dots + \left|a_{n-1}\right|^2\right)$$
由于 $\displaystyle \frac{|z_k|^{2n}-1}{|z_k|^{2n}}<1$，得到$$|z_k|^2 < 1 + |a_0|^2 + \dots + |a_{n-1}|^2$$即，
$$|z_k| < \sqrt{1 + |a_0|^2 + \dots + |a_{n-1}|^2}$$ |
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