x^2-ix-0.5=0

解答

x^{2} - i x - 0.5 = 0

x = \frac{1}{2} + \frac{1}{2} i, x = - \frac{1}{2} + \frac{1}{2} i

求解步骤

x^{2} - i x - 0.5 = 0

替代

x = a + bi

(a + bi)^{2} - i (a + bi) - 0.5 = 0

展开

(a + bi)^{2} - i (a + bi) - 0.5 : (a^{2} - b^{2} + b - 0.5) + i (- a + 2 ab)

(a^{2} - b^{2} + b - 0.5) + i (- a + 2 ab) = 0

将

0

改写成标准复数形式:

0 + 0 i

(a^{2} - b^{2} + b - 0.5) + i (- a + 2 ab) = 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:

[a^{2} - b^{2} + b - 0.5 = 0 - a + 2 ab = 0]

[a^{2} - b^{2} + b - 0.5 = 0 - a + 2 ab = 0] : (a = \frac{1}{2}, a = - \frac{1}{2}, b = \frac{1}{2} b = \frac{1}{2})

x = a + bi

代回

x = \frac{1}{2} + \frac{1}{2} i, x = - \frac{1}{2} + \frac{1}{2} i

x + i y - 4 i = 3 y - 2 i x + 9

(x + y) - 16 i = 12 + (x - y) i

x + y i = 3 i - 4 + 2 - 6 i + \frac{1}{2} i

x (2 - 4 i) - 5 = x - (7 + y i)

so l v e f or l, y = \frac{x}{l in x}