求复数 $2 + 2{{\rm{e}}^{\frac{2}{5}{\rm{\pi i}}}} + {{\rm{e}}^{\frac{6}{5}{\rm{\pi i}}}}$ 的模，其中 $r{{\rm{e}}^{{\rm{i}}\theta }} = r\left( {\cos \theta + {\rm{i}}\sin \theta } \right)$．
【难度】
【出处】
无
【标注】
【答案】
$\sqrt 5$
【解析】
由复数的指数形式的定义有\[\begin{split} \left|2+2{\rm e} ^{\frac 25\pi{\rm i}}+{\rm e}^{\frac 65\pi{\rm i}}\right|&=\left|2+2\cos\dfrac{2\pi}5+2{\rm i}\sin\dfrac{2\pi}5+\cos\dfrac{6\pi}5+{\rm i}\dfrac{6\pi}5\right|\\
&=\sqrt{\left(2+2\cos\dfrac{2\pi}5+\cos\dfrac{6\pi}5\right)^2+\left(2\sin\dfrac{2\pi}5+\sin\dfrac{6\pi}5\right)^2}\\
&=\sqrt{9+8\cos\dfrac{2\pi}5+4\cos\dfrac{6\pi}5+4\cos\dfrac{2\pi}5\cos\dfrac{6\pi}5+4\sin\dfrac{2\pi}5\sin\dfrac{6\pi}5}\\
&=\sqrt{9+8\cos\dfrac{2\pi}5+4\cos\dfrac{6\pi}5+4\cos\dfrac{4\pi}5}\\
&=\sqrt{5+4\cos 0+4\cos\dfrac{2\pi}5+4\cos\dfrac{4\pi}5+4\cos\dfrac{6\pi}5+4\cos\dfrac{8\pi}5}\\
&=\sqrt 5.\end{split} \]
&=\sqrt{\left(2+2\cos\dfrac{2\pi}5+\cos\dfrac{6\pi}5\right)^2+\left(2\sin\dfrac{2\pi}5+\sin\dfrac{6\pi}5\right)^2}\\
&=\sqrt{9+8\cos\dfrac{2\pi}5+4\cos\dfrac{6\pi}5+4\cos\dfrac{2\pi}5\cos\dfrac{6\pi}5+4\sin\dfrac{2\pi}5\sin\dfrac{6\pi}5}\\
&=\sqrt{9+8\cos\dfrac{2\pi}5+4\cos\dfrac{6\pi}5+4\cos\dfrac{4\pi}5}\\
&=\sqrt{5+4\cos 0+4\cos\dfrac{2\pi}5+4\cos\dfrac{4\pi}5+4\cos\dfrac{6\pi}5+4\cos\dfrac{8\pi}5}\\
&=\sqrt 5.\end{split} \]
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