+ix^2=1-i,x^1=1

解

+ i x^{2} = 1 - i, x^{1} = 1

x = \frac{- 1 + 2}{2} - \frac{1}{2 - 1 + 2} i, x = - \frac{- 1 + 2}{2} + \frac{1}{2 - 1 + 2} i

解答ステップ

i x^{2} = 1 - i

代用

x = a + bi

i (a + bi)^{2} = 1 - i

拡張

i (a + bi)^{2} : - 2 ab + i (a^{2} - b^{2})

- 2 ab + i (a^{2} - b^{2}) = 1 - i

複素数は, その実数部と虚数部が等しい場合にのみ等しくなるequation系として書き換える:

[- 2 ab = 1 a^{2} - b^{2} = - 1]

[- 2 ab = 1 a^{2} - b^{2} = - 1] : a = \frac{- 1 + 2}{2}, a = - \frac{- 1 + 2}{2}, b = - \frac{1}{2 - 1 + 2} b = \frac{1}{2 - 1 + 2}

代用を戻す

x = a + bi

x = \frac{- 1 + 2}{2} - \frac{1}{2 - 1 + 2} i, x = - \frac{- 1 + 2}{2} + \frac{1}{2 - 1 + 2} i