(z-i)/(z-2)=sqrt(2)

解答

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

z = (4 + 22) + (- 1 - 2) i

求解步骤

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

替代

z = x + y i

\frac{x + y i - i}{x + y i - 2} = 2

在两边乘以

x - 2 + i y

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

乘开

x + i (- 1 + y) = (2 x - 22) + 2 i y

复数仅在实部和虚部均相等时才相等改写为方程组:

[x = 2 x - 22 - 1 + y = 2 y]

[x = 2 x - 22 - 1 + y = 2 y] : x = 4 + 22, y = - 1 - 2

z = x + y i

代回

z = (4 + 22) + (- 1 - 2) i

(x + 3) - 3 i = \frac{5 + 3 y i}{2}

z + z ba r = 5 i + 6

a + 2 bi + 3 = 5 - 4 i

\frac{z ^{2} - i}{z - 2} = 2

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