a+bi=(a-bi)^2

解答

a + bi = (a - bi)^{2}

a = 1, a = 0, a = - \frac{1}{2}, a = - \frac{1}{2}, b = 0 b = 0 b = \frac{3}{2} b = - \frac{3}{2}

求解步骤

a + bi = (a - bi)^{2}

展开

(a - bi)^{2} : (a^{2} - b^{2}) - 2 iab

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

复数仅在实部和虚部均相等时才相等改写为方程组:

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

[a = a^{2} - b^{2} b = - 2 ab] : a = 1, a = 0, a = - \frac{1}{2}, a = - \frac{1}{2}, b = 0 b = 0 b = \frac{3}{2} b = - \frac{3}{2}

a = 1, a = 0, a = - \frac{1}{2}, a = - \frac{1}{2}, b = 0 b = 0 b = \frac{3}{2} b = - \frac{3}{2}

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

(x + y \cdot i)^{2} = i

(x + y i) = - 8

so l v e f or x, S i z = \frac{x y}{x + y}

- 5 x + 3 i y = 10 - 9 i