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受欢迎的代数 >

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

解答

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

解答

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

替代

z = x + y i

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

使用分式交叉相乘: 若

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

则

a \cdot d = b \cdot c

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

乘开

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

复数仅在实部和虚部均相等时才相等改写为方程组:

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

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

流行的例子

z^{2} + i z - 1 = 0

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

∣ z + 2 + 4 i ∣ = 2

z*z+3(z-z)=4-3i

z \cdot z + 3 (z - z) = 4 - 3 i

x+yi=(3+i)(2-3i)

x + y i = (3 + i) (2 - 3 i)