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受欢迎的代数 >

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

解答

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

解答

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

求解步骤

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

替代

z = x + y i

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

展开

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

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

将

0

改写成标准复数形式:

0 + 0 i

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

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

∣ x + y i ∣ = y - i x

((z-2i))/((z+2))=4i

\frac{( z - 2 i )}{( z + 2 )} = 4 i

- y + x i = x - y i

(2+ai)/(4+bi)=2+5i

\frac{2 + ai}{4 + bi} = 2 + 5 i

w=(p+4i)/(1+ip)

w = \frac{p + 4 i}{1 + i p}