i(x)=e(2x^2-1)

解答

i (x) = e (2 x^{2} - 1)

x = \frac{8 e ^{2} - 1}{4 e} + \frac{1}{4 e} i, x = - \frac{8 e ^{2} - 1}{4 e} + \frac{1}{4 e} i

求解步骤

i (x) = e (2 x^{2} - 1)

替代

x = a + bi

i (a + bi) = e (2 (a + bi)^{2} - 1)

乘开

- b + ia = (2 e a^{2} - 2 e b^{2} - e) + 4 e iab

复数仅在实部和虚部均相等时才相等改写为方程组:

[- b = 2 e a^{2} - 2 e b^{2} - e a = 4 e ab]

[- b = 2 e a^{2} - 2 e b^{2} - e a = 4 e ab] : (a = \frac{8 e ^{2} - 1}{4 e}, a = - \frac{8 e ^{2} - 1}{4 e}, b = \frac{1}{4 e} b = \frac{1}{4 e})

x = a + bi

代回

x = \frac{8 e ^{2} - 1}{4 e} + \frac{1}{4 e} i, x = - \frac{8 e ^{2} - 1}{4 e} + \frac{1}{4 e} i

z \cdot z + z = 3 - i

a + bi = - i

x^{2} + i x 3 + i = 0

3 x^{2} + (1 - i) x + 3 i = 0

x^{2} + (4 - 2 i) x - (1 + i) = 0