計算第2題
設
\(\begin{align}
& {{z}_{1}}=\cos \alpha +i\sin \alpha \\
& {{z}_{2}}=\cos \beta +i\sin \beta \\
& {{z}_{3}}=\cos \gamma +i\sin \gamma \\
\end{align}\)
\(\begin{align}
& \cos \alpha +\cos \beta +\cos \gamma =0 \\
& \sin \alpha +\sin \beta +\sin \gamma =0 \\
& {{z}_{1}}+{{z}_{2}}+{{z}_{3}}=0 \\
& \overline{{{z}_{1}}}+\overline{{{z}_{2}}}+\overline{{{z}_{3}}}=0 \\
& {{z}_{1}}\overline{{{z}_{1}}}={{z}_{2}}\overline{{{z}_{2}}}={{z}_{3}}\overline{{{z}_{3}}}=1 \\
& \\
& {{z}_{1}}^{2}+{{z}_{2}}^{2}+{{z}_{3}}^{2} \\
& ={{\left( {{z}_{1}}+{{z}_{2}}+{{z}_{3}} \right)}^{2}}-2\left( {{z}_{1}}{{z}_{2}}+{{z}_{2}}{{z}_{3}}+{{z}_{3}}{{z}_{1}} \right) \\
& =-2{{z}_{1}}{{z}_{2}}{{z}_{3}}\left( \overline{{{z}_{1}}}+\overline{{{z}_{2}}}+\overline{{{z}_{3}}} \right) \\
& =0 \\
& \\
& \cos 2\alpha +\cos 2\beta +\cos 2\gamma =0 \\
& \sin 2\alpha +\sin 2\beta +\sin 2\gamma =0 \\
\end{align}\)
[ 本帖最後由 thepiano 於 2014-5-27 09:59 PM 編輯 ]