z^2=-5+i

解答

z^{2} = - 5 + i

z = \frac{- 5 + 26}{2} + \frac{1}{2 26 - 5} i, z = - \frac{- 5 + 26}{2} - \frac{1}{2 26 - 5} i

求解步骤

z^{2} = - 5 + i

替代

z = x + y i

(x + y i)^{2} = - 5 + i

展开

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

(x^{2} - y^{2}) + 2 i x y = - 5 + 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} = - 5 2 x y = 1]

[x^{2} - y^{2} = - 5 2 x y = 1] : x = \frac{- 5 + 26}{2}, x = - \frac{- 5 + 26}{2}, y = \frac{1}{2 26 - 5} y = - \frac{1}{2 26 - 5}

z = x + y i

代回

z = \frac{- 5 + 26}{2} + \frac{1}{2 26 - 5} i, z = - \frac{- 5 + 26}{2} - \frac{1}{2 26 - 5} i

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

\frac{3}{x - y i} = \frac{1 + i}{2 - i}

4 x i + 3 y - 3 = x + 5 i - 8

z^{2} + z i = 2 i^{2}

z^{2} + 16 = 30 i