-1+2i=(2i)/((z-1))

解答

- 1 + 2 i = \frac{2 i}{( z - 1 )}

z = \frac{9}{5} - \frac{2}{5} i

求解步骤

- 1 + 2 i = \frac{2 i}{( z - 1 )}

替代

z = x + y i

- 1 + 2 i = \frac{2 i}{x + y i - 1}

在两边乘以

x - 1 + i y

- 1 \cdot (x - 1 + i y) + 2 i (x - 1 + i y) = \frac{2 i}{x + y i - 1} (x - 1 + i y)

乘开

(- x + 1 - 2 y) + i (- 2 - y + 2 x) = 2 i

将

2 i

改写成标准复数形式:

0 + 2 i

(- x + 1 - 2 y) + i (- 2 - y + 2 x) = 0 + 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 + 1 - 2 y = 0 - 2 - y + 2 x = 2]

[- x + 1 - 2 y = 0 - 2 - y + 2 x = 2] : x = \frac{9}{5}, y = - \frac{2}{5}

z = x + y i

代回

z = \frac{9}{5} - \frac{2}{5} i

0 = z^{2} + i z - 6

(7 + m) + (4 l - 10) i = 3 - 6 i

(- 2 - 6 i) - p (4 - 2 i) = q i

(x - y i) + (y + x i) = (5 + i)

so l v e f or x, g (3 - i) = x^{2} - 6 x