+5iz^2=-2+3i,z^1=4

解答

+ 5 i z^{2} = - 2 + 3 i, z^{1} = 4

z = \frac{3 + 13}{10} + \frac{2}{5 3 + 13} i, z = - \frac{3 + 13}{10} - \frac{2}{5 3 + 13} i

求解步骤

5 i z^{2} = - 2 + 3 i

替代

z = x + y i

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

展开

5 i (x + y i)^{2} : - 10 x y + 5 i (x^{2} - y^{2})

- 10 x y + 5 i (x^{2} - y^{2}) = - 2 + 3 i

复数仅在实部和虚部均相等时才相等改写为方程组:

[- 10 x y = - 2 5 x^{2} - 5 y^{2} = 3]

[- 10 x y = - 2 5 x^{2} - 5 y^{2} = 3] : x = \frac{3 + 13}{10}, x = - \frac{3 + 13}{10}, y = \frac{2}{5 3 + 13} y = - \frac{2}{5 3 + 13}

z = x + y i

代回

z = \frac{3 + 13}{10} + \frac{2}{5 3 + 13} i, z = - \frac{3 + 13}{10} - \frac{2}{5 3 + 13} i

z^{1} + z^{2} = (4 + 3) + (2 + 5) i

z^{2} - (1 + 4 i) z - (6 - 12 i) = 0

(2 + i) \cdot a + (i - 3) \cdot b = 4

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

\frac{x + i y}{1 + ( x - i y )} = 3 + 4 i