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

解答

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

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

求解步骤

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

替代

z = x + y i

\frac{1 + 2 ( x + y i )}{3 + 2 ( x + y i )} = i

在两边乘以

3 + 2 x + 2 i y

\frac{1 + 2 ( x + y i )}{3 + 2 ( x + y i )} (3 + 2 x + 2 i y) = i (3 + 2 x + 2 i y)

乘开

(1 + 2 x) + 2 i y = - 2 y + i (3 + 2 x)

复数仅在实部和虚部均相等时才相等改写为方程组:

[1 + 2 x = - 2 y 2 y = 3 + 2 x]

[1 + 2 x = - 2 y 2 y = 3 + 2 x] : x = - 1, y = \frac{1}{2}

z = x + y i

代回

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

(x + y i) (2 + 3 i) = 1 - 5 i