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

解答

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

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

求解步骤

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

替代

z = x + y i

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

使用分式交叉相乘: 若

\frac{a}{b} = \frac{c}{d}

则

a \cdot d = b \cdot c

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

乘开

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

复数仅在实部和虚部均相等时才相等改写为方程组:

[4 x y = - 1 - 2 x^{2} + 2 y^{2} = 1]

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

z = x + y i

代回

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

x + i y = 1

2 x^{2} + 2 x i - y^{2} - y i - 2 i = 2

a + bi = - 12 + 7 i

z^{2} - (2 + i) z - 1 + 7 i = 0

(a + bi)^{2} = - 16 - 30 i