(3+3i)(a+bi)=3+18i

解

(3 + 3 i) (a + bi) = 3 + 18 i

a = \frac{7}{2}, b = \frac{5}{2}

解答ステップ

(3 + 3 i) (a + bi) = 3 + 18 i

拡張

(3 + 3 i) (a + bi) : (3 a - 3 b) + i (3 a + 3 b)

(3 a - 3 b) + i (3 a + 3 b) = 3 + 18 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:

[3 a - 3 b = 3 3 a + 3 b = 18]

[3 a - 3 b = 3 3 a + 3 b = 18] : a = \frac{7}{2}, b = \frac{5}{2}

a = \frac{7}{2}, b = \frac{5}{2}