i(x)=5x-x^2

解答

i (x) = 5 x - x^{2}

x = 0, x = 5 - i

求解步骤

i (x) = 5 x - x^{2}

替代

x = a + bi

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

乘开

- b + ia = (5 a - a^{2} + b^{2}) + i (5 b - 2 ab)

复数仅在实部和虚部均相等时才相等改写为方程组:

[- b = 5 a - a^{2} + b^{2} a = 5 b - 2 ab]

[- b = 5 a - a^{2} + b^{2} a = 5 b - 2 ab] : (a = 0, a = 5, b = 0 b = - 1)

x = a + bi

代回

x = 0, x = 5 - i

z^{2} - (5 + i) z + (8 + i) = 0

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

x^{2} + 6 (3 - i) - 26 = 0

i z^{2} + 4 z + 3 = 0

20 - 12 i = 5 l + (4 m) i