(a+bi)(x-yi)=2+2i

解

(a + bi) (x - y i) = 2 + 2 i

x = \frac{2 ( a + b )}{a ^{2} + b ^{2}}, y = - \frac{2 a - 2 b}{a ^{2} + b ^{2}}, a \neq = - ib, a \neq = ib, a \neq = 0

解答ステップ

(a + bi) (x - y i) = 2 + 2 i

拡張

(a + bi) (x - y i) : (a x + b y) + i (- a y + b x)

(a x + b y) + i (- a y + b x) = 2 + 2 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 x + b y = 2 - a y + b x = 2]

[a x + b y = 2 - a y + b x = 2] : x = \frac{2 ( a + b )}{a ^{2} + b ^{2}}, y = - \frac{2 a - 2 b}{a ^{2} + b ^{2}}, a \neq = - ib, a \neq = ib, a \neq = 0

x = \frac{2 ( a + b )}{a ^{2} + b ^{2}}, y = - \frac{2 a - 2 b}{a ^{2} + b ^{2}}, a \neq = - ib, a \neq = ib, a \neq = 0