(x+iy)^2+2(x-iy)=-1+6i

解答

(x + i y)^{2} + 2 (x - i y) = - 1 + 6 i

(x = - 2, x = 2, y = - 1 y = 3)

求解步骤

(x + i y)^{2} + 2 (x - i y) = - 1 + 6 i

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(x + i y)^{2} + 2 (x - i y) : (x^{2} - y^{2} + 2 x) + i (- 2 y + 2 x y)

(x^{2} - y^{2} + 2 x) + i (- 2 y + 2 x y) = - 1 + 6 i

复数仅在实部和虚部均相等时才相等改写为方程组:

[x^{2} - y^{2} + 2 x = - 1 - 2 y + 2 x y = 6]

[x^{2} - y^{2} + 2 x = - 1 - 2 y + 2 x y = 6] : (x = - 2, x = 2, y = - 1 y = 3)

(x = - 2, x = 2, y = - 1 y = 3)

z^{2} = - 2 - 23 i

z - \frac{1}{z} = 4 + 3 i

∣ i + z ∣^{2} - i - 2 = z

a^{2} = i

i x^{2} + (1 - 2 i) x + 3 i - 1 = 0