(1+z)/(1-z)=8i-5

解答

\frac{1 + z}{1 - z} = 8 i - 5

z = \frac{11}{10} + \frac{1}{5} i

求解步骤

\frac{1 + z}{1 - z} = 8 i - 5

替代

z = x + y i

\frac{1 + x + y i}{1 - ( x + y i )} = 8 i - 5

在两边乘以

1 - x - i y

\frac{1 + x + y i}{1 - ( x + y i )} (1 - x - i y) = 8 i (1 - x - i y) - 5 (1 - x - i y)

乘开

(1 + x) + i y = (8 y - 5 + 5 x) + i (8 + 5 y - 8 x)

复数仅在实部和虚部均相等时才相等改写为方程组:

[1 + x = 8 y - 5 + 5 x y = 8 + 5 y - 8 x]

[1 + x = 8 y - 5 + 5 x y = 8 + 5 y - 8 x] : x = \frac{11}{10}, y = \frac{1}{5}

z = x + y i

代回

z = \frac{11}{10} + \frac{1}{5} i

\frac{1 + 2 z}{3 + 2 z} = i

\frac{- 1}{b + i} = \frac{1}{- b + i}

(x + y) + (x - y) \cdot i = 2 + 5 \cdot i

(1 + i) (x + y i) = 1 + 3 i

4 z^{2} - 4 z + 3 - 23 i = 0