(3-i)(a+bi)=0

解答

(3 - i) (a + bi) = 0

a = 0, b = 0

求解步骤

(3 - i) (a + bi) = 0

展开

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

(3 a + b) + i (- a + 3 b) = 0

将

0

改写成标准复数形式:

0 + 0 i

(3 a + b) + i (- a + 3 b) = 0 + 0 i

复数仅在实部和虚部均相等时才相等改写为方程组:

[3 a + b = 0 - a + 3 b = 0]

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

a = 0, b = 0

x + i y = y + i x

(2 - 5 i) x + (1 + 3 i) y = 8 - 9 i

(z - i)^{2} = 3 + 4 i

x^{2} - 12 i x - 1 = 0

a - bi (a + bi + 1) = 11 + 3 i