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受欢迎的代数 >

-1/(z-i)-2i-z=0

解答

- \frac{1}{z - i} - 2 i - z = 0

解答

z = \frac{- 1 + 13}{2} i, z = \frac{- 1 - 13}{2} i

求解步骤

- \frac{1}{z - i} - 2 i - z = 0

替代

z = x + y i

- \frac{1}{x + y i - i} - 2 i - (x + y i) = 0

在两边乘以

x + i (- 1 + y)

- \frac{1}{x + y i - i} (x + i (- 1 + y)) - 2 i (x + i (- 1 + y)) - (x + y i) (x + i (- 1 + y)) = 0 \cdot (x + i (- 1 + y))

乘开

(- x^{2} + y^{2} + y - 3) + i (- x - 2 x y) = 0

将

0

改写成标准复数形式:

0 + 0 i

(- x^{2} + y^{2} + y - 3) + i (- x - 2 x y) = 0 + 0 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} + y^{2} + y - 3 = 0 - x - 2 x y = 0]

[- x^{2} + y^{2} + y - 3 = 0 - x - 2 x y = 0] : (x = 0, x = 0, y = \frac{- 1 + 13}{2} y = \frac{- 1 - 13}{2})

z = x + y i

代回

z = \frac{- 1 + 13}{2} i, z = \frac{- 1 - 13}{2} i

流行的例子

solvefor x,g[-6]ifg[x]=10x+9

so l v e f or x, g [- 6] i f g [x] = 10 x + 9

((z-2))/((z+1))=3i

\frac{( z - 2 )}{( z + 1 )} = 3 i

z^{2} = 8 i

x^2-(2+3i)x-1+3i=0

x^{2} - (2 + 3 i) x - 1 + 3 i = 0

(m+ni)(2+5i)=36+3i

(m + ni) (2 + 5 i) = 36 + 3 i