(x^2+6)+(2x)i=15+6i

解答

(x^{2} + 6) + (2 x) i = 15 + 6 i

x = 3, x = - 3 - 2 i

求解步骤

(x^{2} + 6) + (2 x) i = 15 + 6 i

替代

x = a + bi

(a + bi)^{2} + 6 + 2 (a + bi) i = 15 + 6 i

展开

(a + bi)^{2} + 6 + 2 (a + bi) i : (a^{2} - b^{2} + 6 - 2 b) + i (2 a + 2 ab)

(a^{2} - b^{2} + 6 - 2 b) + i (2 a + 2 ab) = 15 + 6 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} + 6 - 2 b = 15 2 a + 2 ab = 6]

[a^{2} - b^{2} + 6 - 2 b = 15 2 a + 2 ab = 6] : (a = 3, a = - 3, b = 0 b = - 2)

x = a + bi

代回

x = 3, x = - 3 - 2 i

z^{2} - (1 + 6 i) z - 10 = 0

\frac{48 - 12 i}{x} = 16 - 4 i

so l v e f or y, (2 x - y) - 12 i = 12 + 6

x + 3 y + (2 x - 3 y - 9) i = 0

a + bi + \frac{1}{a + bi} = 1