(1-i)z+w=3+2i

解答

(1 - i) z + w = 3 + 2 i

z = (- 2 + \frac{- w + 5}{2}) + \frac{- w + 5}{2} i

求解步骤

(1 - i) z + w = 3 + 2 i

替代

z = x + y i

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

展开

(1 - i) (x + y i) + w : (x + y + w) + i (- x + y)

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

复数仅在实部和虚部均相等时才相等改写为方程组:

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

[x + y + w = 3 - x + y = 2] : x = - 2 + \frac{- w + 5}{2}, y = \frac{- w + 5}{2}

z = x + y i

代回

z = (- 2 + \frac{- w + 5}{2}) + \frac{- w + 5}{2} i

i = \frac{e}{r}

(3 x - 4 y) + (6 x + 2 y) i = 5 i

z^{2} = 1 - 3 i

(1 + 3 i)^{11} = a + ib

∣ z - i ∣ = 5