(x^2-9)+(2y-5)i=7i

解答

(x^{2} - 9) + (2 y - 5) i = 7 i

(x = - 3, x = 3, y = 6 y = 6)

求解步骤

(x^{2} - 9) + (2 y - 5) i = 7 i

将

7 i

改写成标准复数形式:

0 + 7 i

(x^{2} - 9) + (2 y - 5) i = 0 + 7 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} - 9 = 0 2 y - 5 = 7]

[x^{2} - 9 = 0 2 y - 5 = 7] : (x = - 3, x = 3, y = 6 y = 6)

(x = - 3, x = 3, y = 6 y = 6)

x + 3 i + 3 = 5 + y i

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

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

\frac{1}{z} = \frac{2 + 8 i}{8 i}

z^{2} = - 3 + 4 i