z^2=sqrt(3)+3i

解答

z^{2} = 3 + 3 i

z = \frac{4 3 ^{3}}{2} + \frac{4 3}{2} i, z = - \frac{4 3 ^{3}}{2} - \frac{4 3}{2} i

求解步骤

z^{2} = 3 + 3 i

替代

z = x + y i

(x + y i)^{2} = 3 + 3 i

展开

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

(x^{2} - y^{2}) + 2 i x y = 3 + 3 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} = 3 2 x y = 3]

[x^{2} - y^{2} = 3 2 x y = 3] : (x = \frac{4 3 ^{3}}{2}, x = - \frac{4 3 ^{3}}{2}, y = \frac{4 3}{2} y = - \frac{4 3}{2})

z = x + y i

代回

z = \frac{4 3 ^{3}}{2} + \frac{4 3}{2} i, z = - \frac{4 3 ^{3}}{2} - \frac{4 3}{2} i

z^{2} = i (1 - 0. 5^{2})

x (1 + i) + y (1 - i) = 0

7 i - 3 j = a (i - j) + b (i + j)

x^{2} + 6 x i - 9 = (5 + 12 i) y

3 x + 8 i = 12 - 12