已知 ${\mathrm{e}^{\theta \mathrm{i}}}=\cos \theta +\mathrm{i}\sin \theta $，求值 $\left| 2+2{\mathrm{e}^{\frac{2}{5}\pi\mathrm{i}}}+{\mathrm{e}^{\frac{6}{5}\pi\mathrm{i}}} \right|$．
【难度】
【出处】
2009年清华大学自主招生试题
【标注】
【答案】
$\sqrt{5}$
【解析】
题中代数式可进行如下变形\[\begin{split}\left| 2+2{\mathrm{e}^{\frac{2}{5}\pi\mathrm{i}}}+{\mathrm{e}^{\frac{6}{5}\pi\mathrm{i}}} \right|&=\left| 2+2\cos \frac{2\pi}{5}+2\mathrm{i}\sin \frac{2\pi}{5}+\cos \frac{6\pi}{5}+\mathrm{i}\sin \frac{6\pi}{5} \right|\\&=\sqrt{{{\left( 2+2\cos \frac{2\pi}{5}+\cos \frac{6\pi}{5} \right)}^{2}}+{{\left( 2\sin \frac{2\pi}{5}+\sin \frac{6\pi}{5} \right)}^{2}}}\\&=\sqrt{9+8\cos \frac{2\pi}{5}+4\cos \frac{6\pi}{5}+4\cos \frac{2\pi}{5}\cos \frac{6\pi}{5}+4\sin \frac{2\pi}{5}\sin \frac{6\pi}{5}}\\&=\sqrt{9+8\cos \frac{2\pi}{5}+4\cos \frac{6\pi}{5}+4\cos \frac{4\pi}{5}}\\&=\sqrt{9+8\left( \cos \frac{2\pi}{5}+\cos \frac{4\pi}{5} \right)}\end{split},\]又因为$$\cos 0+\cos \dfrac{2\pi}{5}+\cos \dfrac{4\pi}{5}+\cos \dfrac{6\pi}{5}+\cos \dfrac{8\pi}{5}=0,$$所以$$\cos \dfrac{2\pi}{5}+\cos \dfrac{4\pi}{5}=-\dfrac{1}{2}.$$因此原式 $=\sqrt{5}$．
答案
解析
备注
