+4iz^2=2+2i,z^1=7

解答

+ 4 i z^{2} = 2 + 2 i, z^{1} = 7

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

求解步骤

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

替代

z = x + y i

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

展开

4 i (x + y i)^{2} : - 8 x y + 4 i (x^{2} - y^{2})

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

复数仅在实部和虚部均相等时才相等改写为方程组:

[- 8 x y = 2 4 x^{2} - 4 y^{2} = 2]

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

z = x + y i

代回

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

\frac{2 + ai}{7 + bi} = 3 + 6 i

4 x + i (5 x - y) = 2 + i (- 3)

so l v e f or x, \frac{x}{1 - x} = 1 - 5 i

(x - y i) + ∣ x + y i ∣ = 25 - 5 i

z^{2} = 5 - 6 i