(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

展开

(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

(a + ib)^{2} + 2 (a - ib) + 6 = 0

z^{2} - z (1 + 4 i) + 12 i - 6 = 0

∣ z ∣^{2} + i + i 2 i = 6 z

\frac{( 1 + 3 i )}{z + 5} = i

x^{2} - 9 i = 0