i(x)=x^2+1

解答

i (x) = x^{2} + 1

x = - \frac{- 1 + 5}{2} i, x = \frac{1 + 5}{2} i

求解步骤

i (x) = x^{2} + 1

替代

x = a + bi

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

乘开

- b + ia = (a^{2} - b^{2} + 1) + 2 iab

复数仅在实部和虚部均相等时才相等改写为方程组:

[- b = a^{2} - b^{2} + 1 a = 2 ab]

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

x = a + bi

代回

x = - \frac{- 1 + 5}{2} i, x = \frac{1 + 5}{2} i

z^{2} = 4 (- 1 - i 3)

i (t) = 4 t^{2} - 48 t + 154

i (t) = 8 t^{2} - 4 t A

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

∣ z + i ∣ = 3 + 0 i