(2-i)^2-(2-yi)=x+5i

解答

(2 - i)^{2} - (2 - y i) = x + 5 i

x = 1, y = 9

求解步骤

(2 - i)^{2} - (2 - y i) = x + 5 i

展开

(2 - i)^{2} - (2 - y i) : 1 + i (- 4 + y)

1 + i (- 4 + y) = x + 5 i

复数仅在实部和虚部均相等时才相等改写为方程组:

[1 = x - 4 + y = 5]

[1 = x - 4 + y = 5] : x = 1, y = 9

x = 1, y = 9

∣ z - (1 - i) ∣ = 2

a - bi = i (a + bi)

\frac{z + 1}{z - 1} = 2 + i^{25}

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

\frac{26 + 28 i}{2 + i} = a + bi