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受欢迎的代数 >

z^2+2iz-1-i=0

解答

z^{2} + 2 \cdot i \cdot z - 1 - i = 0

解答

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

求解步骤

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

替代

z = x + y i

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

展开

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

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

将

0

改写成标准复数形式:

0 + 0 i

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

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

z = x + y i

代回

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

流行的例子

x^2-2(1+i)x+(2i-1)=0

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

(x + i y) (3 + i) = 5

(1+i)(x+2y)-(3-2i)(x-y)=8+3i

(1 + i) (x + 2 y) - (3 - 2 i) (x - y) = 8 + 3 i

x + y i = 2

(x+yi)(1+2i)=-1+8i

(x + y i) (1 + 2 i) = - 1 + 8 i