z^2=-4-2i

解答

z^{2} = - 4 - 2 i

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

求解步骤

z^{2} = - 4 - 2 i

替代

z = x + y i

(x + y i)^{2} = - 4 - 2 i

展开

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

(x^{2} - y^{2}) + 2 i x y = - 4 - 2 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} = - 4 2 x y = - 2]

[x^{2} - y^{2} = - 4 2 x y = - 2] : x = - 2 + 5, x = - - 2 + 5, y = - \frac{1}{- 2 + 5} y = \frac{1}{- 2 + 5}

z = x + y i

代回

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

\frac{( 2 + ai )}{1 + i} = re a l

2 x^{2} = 14 x - 30 i

s i f (x) = - 2 x + 5, E (s, x) = f (1) + f (- 1)

3 z^{2} + (2 - 3 i) z - 1 - 3 i = 0

r (t) = (2 t + 1) i + (t^{2} - 1) j