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

解答

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

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

求解步骤

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

替代

z = x + y i

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

在两边乘以

x + y i

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

乘开

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

复数仅在实部和虚部均相等时才相等改写为方程组:

[1 = - 2 y - 2 = 2 x]

[1 = - 2 y - 2 = 2 x] : x = - 1, y = - \frac{1}{2}

z = x + y i

代回

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

\frac{a + 2 i}{3 + bi} = 3

(a + bi) (x + y i) = 2 + 2 i

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

(a + ib)^{2} = - 5 + 12 i

3 x - 5 s i x = - 5