복소수 거듭제곱의 실수부·허수부 계산

$( 1 + i ) ^{2} = 2 i$, $\displaystyle ( 1 - \sqrt{3} i ) ^{3} = - 8$이므로 $\displaystyle \left( \frac{1 + i}{1 - \sqrt{3} i} \right) ^{13}$$\displaystyle = \left( \frac{1 + i}{1 - \sqrt{3} i} \right) ^{12} \times \left( \frac{1 + i}{1 - \sqrt{3} i} \right)$ $\displaystyle = \frac{( 1 + i ) ^{12}}{( 1 - \sqrt{3} ) ^{12}} \times \left( \frac{1 + i}{1 - \sqrt{3} i} \right)$ $\displaystyle = \frac{\left\{ ( 1 + i ) ^{2} \right\} ^{6}}{\left\{ ( 1 - \sqrt{3} ) ^{3} \right\} ^{4}} \times \left( \frac{1 + i}{1 - \sqrt{3} i} \right) \times \frac{1 + \sqrt{3} i}{1 + \sqrt{3} i}$ $\displaystyle = \frac{( 2 i ) ^{6}}{( - 8 ) ^{4}} \times \frac{1 - \sqrt{3} + ( 1 + \sqrt{3} ) i}{4}$ $\displaystyle = \frac{- 2 ^{6}}{2 ^{12}} \times \frac{1 - \sqrt{3} + ( 1 + \sqrt{3} ) i}{4}$ $\displaystyle = \frac{\sqrt{3} - 1}{2 ^{8}} - \frac{\sqrt{3} + 1}{2 ^{8}}$ ∴ $\displaystyle x = \frac{\sqrt{3} - 1}{2 ^{8}}$, $\displaystyle y = - \frac{\sqrt{3} + 1}{2 ^{8}}$ ∴ $\displaystyle | x | + | y | = \frac{\sqrt{3} - 1}{2 ^{8}} + \frac{\sqrt{3} + 1}{2 ^{8}} = \frac{\sqrt{3}}{2 ^{7}}$