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受欢迎的代数 >

(z+1)^2-i=0

解答

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

解答

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

求解步骤

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

替代

z = x + y i

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

展开

(x + y i + 1)^{2} - i : (x^{2} + 2 x - y^{2} + 1) + i (- 1 + 2 y + 2 x y)

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

将

0

改写成标准复数形式:

0 + 0 i

(x^{2} + 2 x - y^{2} + 1) + i (- 1 + 2 y + 2 x y) = 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:

[x^{2} + 2 x - y^{2} + 1 = 0 - 1 + 2 y + 2 x y = 0]

[x^{2} + 2 x - y^{2} + 1 = 0 - 1 + 2 y + 2 x y = 0] : (x = - \frac{- 2 + 2}{2}, x = - \frac{2 + 2}{2}, y = \frac{1}{2} y = - \frac{1}{2})

z = x + y i

代回

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

流行的例子

∣ z + 2 + 4 i ∣ = 2

z*z+3(z-z)=4-3i

z \cdot z + 3 (z - z) = 4 - 3 i

x+yi=(3+i)(2-3i)

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

z^{2} - 4 i z - 6 = 0

z^2-(3-2i)z+(5-i)=0

z^{2} - (3 - 2 i) z + (5 - i) = 0