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题:试证$\,z\underset{\,}{\,}$距$\odot(O,r)$的内接正$n$边形各顶点的平方和是$\,n(r^2+|z|^2).$W8< 证:设$n$个顶点是$\,r\exp(i\varphi_k)\;(\varphi_k=\varphi_0+\frac{2k\pi}{n},\,k=\overline{0,n-1})\underset{\,}{\;}$:r $\because\quad|z-re^{i\varphi}|^2=(z-re^{i\varphi})(\bar{z}-re^{-i\varphi})=r^2+|z|^2-r(ze^{-i\varphi}+\bar{z}e^{i\varphi}),$ibSgT $\qquad$等比数列和$\;\displaystyle{\small\sum_{k=0}^{n-1}}e^{i\varphi_k}=e^{i\varphi_0}{\small\sum_{k=0}^{n-1}}\big(e^{i\frac{2\pi}{n}}\big)^k=0$}EXi $\therefore\quad\displaystyle{\small\sum_{k=0}^{n-1}}|z-re^{i\varphi_k}|^2=n(r^2+|z|^2)-r{\small\sum_{k=0}^{n-1}}(ze^{-i\varphi_k}+\bar{z}e^{i\varphi_k})=n(r^2+|z|^2).$,aUJA>
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