solvefor x,x^2-y^2ifx=-3

解

解く

x, x^{2} - y^{2} i f x = - 3

x = - \frac{- y ^{2} f + y ^{4} f ^{2} + 12}{2} i, x = - \frac{- y ^{2} f - y ^{4} f ^{2} + 12}{2} i, x = \frac{- 12 - y ^{4} f ^{2}}{2} + \frac{y ^{2} f}{2} i, x = - \frac{- 12 - y ^{4} f ^{2}}{2} + \frac{y ^{2} f}{2} i

解答ステップ

x^{2} - y^{2} i f x = - 3

代用

x = a + bi

(a + bi)^{2} - y^{2} i f (a + bi) = - 3

拡張

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

a^{2} - b^{2} + y^{2} f b + i (2 ab - y^{2} f a) = - 3

標準的な複素数形式で

- 3

を書き換える:

- 3 + 0 i

a^{2} - b^{2} + y^{2} f b + i (2 ab - y^{2} f a) = - 3 + 0 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:

[a^{2} - b^{2} + y^{2} f b = - 3 2 ab - y^{2} f a = 0]

[a^{2} - b^{2} + y^{2} f b = - 3 2 ab - y^{2} f a = 0] : a = 0, a = 0, a = \frac{- 12 - y ^{4} f ^{2}}{2}, a = - \frac{- 12 - y ^{4} f ^{2}}{2}, b = - \frac{- y ^{2} f + y ^{4} f ^{2} + 12}{2} b = - \frac{- y ^{2} f - y ^{4} f ^{2} + 12}{2} b = \frac{y ^{2} f}{2} b = \frac{y ^{2} f}{2}

代用を戻す

x = a + bi

x = - \frac{- y ^{2} f + y ^{4} f ^{2} + 12}{2} i, x = - \frac{- y ^{2} f - y ^{4} f ^{2} + 12}{2} i, x = \frac{- 12 - y ^{4} f ^{2}}{2} + \frac{y ^{2} f}{2} i, x = - \frac{- 12 - y ^{4} f ^{2}}{2} + \frac{y ^{2} f}{2} i