(-1+2i)/(8i)=a+bi

解答

\frac{- 1 + 2 i}{8 i} = a + bi

a = \frac{1}{4}, b = \frac{1}{8}

求解步骤

\frac{- 1 + 2 i}{8 i} = a + bi

展开

\frac{- 1 + 2 i}{8 i} : \frac{1}{4} + \frac{i}{8}

\frac{1}{4} + \frac{i}{8} = a + bi

复数仅在实部和虚部均相等时才相等改写为方程组:

[\frac{1}{4} = a \frac{1}{8} = b]

[\frac{1}{4} = a \frac{1}{8} = b] : a = \frac{1}{4}, b = \frac{1}{8}

a = \frac{1}{4}, b = \frac{1}{8}

z^{2} - 8 z + 18 = 8 i

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

z^{2} + z = 5 + 5 i

so l v e f or z, z - \frac{1}{z} = 5 + 4 i

∣ z - i ∣ = 4