허근 ω 보기

삼차방정식 $x ^{3} = 1$의 한 허근이 $\omega$이므로 $x ^{3} - 1 = \left( x - 1 \right) \left( x ^{2} + x + 1 \right) = 0$에서 $\omega ^{3} = 1$, $\omega ^{2} + \omega + 1 = 0$ $\omega$의 켤레복소수 $\displaystyle {\overline{\omega}}$는 $x ^{3} = 1$의 다른 한 허근이므로 $\displaystyle {\overline{\omega}} ^{3} = 1$, $\displaystyle {\overline{\omega}} ^{2} + {\overline{\omega}} + 1 = 0$, $\displaystyle \omega + {\overline{\omega}} = \mathit{-} 1$, $\displaystyle \omega \times {\overline{\omega}} = \mathit{1}$ ㄱ. $\displaystyle {\overline{\omega}} ^{3} = 1$ (참) ㄴ. $\displaystyle \frac{1}{\omega} + \left( \frac{1}{\omega} \right) ^{2} = \frac{\omega + 1}{\omega ^{2}} = \frac{- \omega ^{2}}{\omega ^{2}} = \mathit{-} 1$ $\displaystyle \frac{1}{{\overline{\omega}}} + \left( \frac{1}{{\overline{\omega}}} \right) ^{2} = \frac{{\overline{\omega}} + 1}{{\overline{\omega}} ^{2}} = \frac{- {\overline{\omega}} ^{2}}{{\overline{\omega}} ^{2}} = \mathit{-} 1$ $\therefore$ $\displaystyle \frac{1}{\omega} + \left( \frac{1}{\omega} \right) ^{2} = \frac{1}{{\overline{\omega}}} + \left( \frac{1}{{\overline{\omega}}} \right) ^{2}$ (참) ㄷ. $\left( - \omega - 1 \right) ^{n} = \left( \omega ^{2} \right) ^{n}$ $\displaystyle \left( \frac{{\overline{\omega}}}{\omega + {\overline{\omega}}} \right) ^{n}$$\displaystyle = \left( - {\overline{\omega}} \right) ^{n}$$\displaystyle = \left( - \frac{1}{\omega} \right) ^{n}$$\displaystyle = \left( - 1 \right) ^{n} \times \left( \frac{1}{\omega} \right) ^{n}$$= \left( - 1 \right) ^{n} \times \left( \omega ^{2} \right) ^{n}$ $\left( - \omega - 1 \right) ^{n} =$$\displaystyle \left( \frac{{\overline{\omega}}}{\omega + {\overline{\omega}}} \right) ^{n}$을 만족시키는 $n$은 $\left( \omega ^{2} \right) ^{n}$$= \left( - 1 \right) ^{n} \times \left( \omega ^{2} \right) ^{n}$, $1 = \left( - 1 \right) ^{n}$을 만족시키므로 $n$은 짝수이다. 그러므로 $100$ 이하의 짝수 $n$의 개수는 $50$ (참) 따라서 옳은 것은 ㄱ, ㄴ, ㄷ