x^2+2yi+2xyi+2x-y^2+1+i=0

解答

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

(x = \frac{- 2 - 1}{2}, x = - \frac{2 - 1}{2}, y = \frac{1}{2} y = - \frac{1}{2})

求解步骤

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

将

x^{2} + 2 y i + 2 x y i + 2 x - y^{2} + 1 + i

改写成标准复数形式:

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

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

将

0

改写成标准复数形式:

0 + 0 i

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

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

(x = \frac{- 2 - 1}{2}, x = - \frac{2 - 1}{2}, y = \frac{1}{2} y = - \frac{1}{2})

2 i x^{2} + 2 x + 9 i = 0

\frac{x - 3}{x + 1} = i

\frac{2 - 5 i}{4 - 7 i} = \frac{a}{b}

3 i z^{2} + (8 - 6 i) z - (8 - 6 i) = 0

b (a + bi) + (2 - i) = 3 + i