+8iz^2=5-3i,z^1=10

解

+ 8 i z^{2} = 5 - 3 i, z^{1} = 10

z = \frac{- 3 + 34}{4} - \frac{5}{4 34 - 3} i, z = - \frac{- 3 + 34}{4} + \frac{5}{4 34 - 3} i

解答ステップ

8 i z^{2} = 5 - 3 i

代用

z = x + y i

8 i (x + y i)^{2} = 5 - 3 i

拡張

8 i (x + y i)^{2} : - 16 x y + 8 i (x^{2} - y^{2})

- 16 x y + 8 i (x^{2} - y^{2}) = 5 - 3 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:

[- 16 x y = 5 8 x^{2} - 8 y^{2} = - 3]

[- 16 x y = 5 8 x^{2} - 8 y^{2} = - 3] : x = \frac{- 3 + 34}{4}, x = - \frac{- 3 + 34}{4}, y = - \frac{5}{4 34 - 3} y = \frac{5}{4 34 - 3}

代用を戻す

z = x + y i

z = \frac{- 3 + 34}{4} - \frac{5}{4 34 - 3} i, z = - \frac{- 3 + 34}{4} + \frac{5}{4 34 - 3} i