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

解答

(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

求解步骤

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

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(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

复数仅在实部和虚部均相等时才相等改写为方程组:

[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

z^{2} + 2 (i - 2) z + (i - 2) = 0

(a + ib)^{2} = - 5 + 12 i

3 x - 5 s i x = - 5

(3 - 6 i) (2 - ai) = x

3 + 8 i = x + y i