복소수 조건식 값

[출제의도] 복소수의 연산에 관한 성질을 이해하여 문제해결하기 (나)에서 $\displaystyle \frac{1}{\alpha} + \frac{1}{\beta} = - \frac{1}{\gamma}$, $\displaystyle \frac{\alpha + \beta}{\alpha \beta} = - \frac{1}{\gamma}$ (가)에서 $\alpha + \beta = - \gamma$이므로 $\displaystyle \frac{- \gamma}{\alpha \beta} = - \frac{1}{\gamma}$ $\alpha \beta = \gamma ^{2}$ $\cdots \cdots \text{㉠}$ 같은 방법으로 $\beta \gamma = \alpha ^{2}$ $\cdots \cdots \text{㉡}$ $\gamma \alpha = \beta ^{2}$ $\cdots \cdots \text{㉢}$ (나)에서 $\displaystyle \frac{\alpha \beta + \beta \gamma + \gamma \alpha}{\alpha \beta \gamma} = 0$이므로 $\alpha \beta + \beta \gamma + \gamma \alpha = 0$ $\text{㉠}$, $\text{㉡}$, $\text{㉢}$에서 $\alpha ^{2} + \beta ^{2} + \gamma ^{2} = 0$ $\alpha ^{2} + \alpha \gamma + \gamma ^{2} = 0$ $\cdots \cdots \text{㉣}$ $\text{㉣}$에서 양변을 $\alpha ^{2}$으로 나누면 $\displaystyle \left( \frac{\gamma}{\alpha} \right) ^{2} + \frac{\gamma}{\alpha} + 1 = 0$ $\therefore$ $\displaystyle \frac{\gamma}{\alpha} = \frac{- 1 + \sqrt{3} i}{2}$ 또는 $\displaystyle \frac{\gamma}{\alpha} = \frac{- 1 - \sqrt{3} i}{2}$ $\text{㉡}$에서 $\displaystyle \frac{\alpha}{\beta} = \frac{\gamma}{\alpha}$이므로 $\displaystyle \frac{\gamma}{\alpha} + \overline{\left( \frac{\alpha}{\beta} \right)} = \frac{\gamma}{\alpha} + \overline{\left( \frac{\gamma}{\alpha} \right)}$이다. 따라서 $\displaystyle \frac{\gamma}{\alpha} + \overline{\left( \frac{\alpha}{\beta} \right)} = - 1$