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

解

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

a = - \frac{12}{13}, b = - \frac{5}{13}

解答ステップ

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

拡張

(3 + 2 i) (a + bi) + (5 + 2 i) : (3 a - 2 b + 5) + i (2 + 2 a + 3 b)

(3 a - 2 b + 5) + i (2 + 2 a + 3 b) = 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:

[3 a - 2 b + 5 = 3 2 + 2 a + 3 b = - 1]

[3 a - 2 b + 5 = 3 2 + 2 a + 3 b = - 1] : a = - \frac{12}{13}, b = - \frac{5}{13}

a = - \frac{12}{13}, b = - \frac{5}{13}