(1+i)(a+bi)+5=2-3i

解

(1 + i) (a + bi) + 5 = 2 - 3 i

a = - 3, b = 0

解答ステップ

(1 + i) (a + bi) + 5 = 2 - 3 i

拡張

(1 + i) (a + bi) + 5 : (a - b + 5) + i (a + b)

(a - b + 5) + i (a + b) = 2 - 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:

[a - b + 5 = 2 a + b = - 3]

[a - b + 5 = 2 a + b = - 3] : a = - 3, b = 0

a = - 3, b = 0