(3+bi)(3+di)=11-3i

解

(3 + bi) (3 + d i) = 11 - 3 i

(b = - 2, d = 1 b = 1, d = - 2)

解答ステップ

(3 + bi) (3 + d i) = 11 - 3 i

拡張

(3 + bi) (3 + d i) : (9 - b d) + i (3 b + 3 d)

(9 - b d) + i (3 b + 3 d) = 11 - 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:

[9 - b d = 11 3 b + 3 d = - 3]

[9 - b d = 11 3 b + 3 d = - 3] : (b = - 2, d = 1 b = 1, d = - 2)

(b = - 2, d = 1 b = 1, d = - 2)