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受欢迎的代数 >

0=z+i/z

解答

0 = z + \frac{i}{z}

解答

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

求解步骤

0 = z + \frac{i}{z}

替代

z = x + y i

0 = x + y i + \frac{i}{x + y i}

在两边乘以

x + y i

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

乘开

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

将

0

改写成标准复数形式:

0 + 0 i

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

复数仅在实部和虚部均相等时才相等改写为方程组:

[0 = x^{2} - y^{2} 0 = 1 + 2 x y]

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

z = x + y i

代回

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

流行的例子

15 - 28 = 3 l + (4 m) i

+4iz^2=2+2i,z^1=7

+ 4 i z^{2} = 2 + 2 i, z^{1} = 7

(2+ai)/(7+bi)=3+6i

\frac{2 + ai}{7 + bi} = 3 + 6 i

4x+i(5x-y)=2+i(-3)

4 x + i (5 x - y) = 2 + i (- 3)

solvefor x, x/(1-x)=1-5i

so l v e f or x, \frac{x}{1 - x} = 1 - 5 i