+4iz^2=2+2i,z^1=7

解

+ 4 i z^{2} = 2 + 2 i, z^{1} = 7

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

解答ステップ

4 i z^{2} = 2 + 2 i

代用

z = x + y i

4 i (x + y i)^{2} = 2 + 2 i

拡張

4 i (x + y i)^{2} : - 8 x y + 4 i (x^{2} - y^{2})

- 8 x y + 4 i (x^{2} - y^{2}) = 2 + 2 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:

[- 8 x y = 2 4 x^{2} - 4 y^{2} = 2]

[- 8 x y = 2 4 x^{2} - 4 y^{2} = 2] : x = \frac{1 + 2}{2}, x = - \frac{1 + 2}{2}, y = - \frac{1}{2 1 + 2} y = \frac{1}{2 1 + 2}

代用を戻す

z = x + y i

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