x+iy=(3x+4i)(1+x-iy)

解答

x + i y = (3 x + 4 i) (1 + x - i y)

(x = - 0.92203 \dots, y = - 0.17659 \dots)

求解步骤

x + i y = (3 x + 4 i) (1 + x - i y)

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(3 x + 4 i) (1 + x - i y) : (3 x + 3 x^{2} + 4 y) + i (4 + 4 x - 3 x y)

x + i y = (3 x + 3 x^{2} + 4 y) + i (4 + 4 x - 3 x y)

复数仅在实部和虚部均相等时才相等改写为方程组:

[x = 3 x + 3 x^{2} + 4 y y = 4 + 4 x - 3 x y]

[x = 3 x + 3 x^{2} + 4 y y = 4 + 4 x - 3 x y] : (x = - 0.92203 \dots, y = - 0.17659 \dots)

(x = - 0.92203 \dots, y = - 0.17659 \dots)

z^{2} = i (∣ z ∣^{2} - 4)

(2 - 3 i) x + (3 - 2 i) y = 14 + 8 i

y - in y = x^{2} + 1

(a + b + 3) + (8 + b) i = 2 a + (3 a - b) i

(x + 1) + 6 y i = 4 + 24 i