(a+i)^2=8+bi

解

(a + i)^{2} = 8 + bi

(a = - 3, a = 3, b = - 6 b = 6)

解答ステップ

(a + i)^{2} = 8 + bi

拡張

(a + i)^{2} : (a^{2} - 1) + 2 ia

(a^{2} - 1) + 2 ia = 8 + bi

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} - 1 = 8 2 a = b]

[a^{2} - 1 = 8 2 a = b] : (a = - 3, a = 3, b = - 6 b = 6)

(a = - 3, a = 3, b = - 6 b = 6)