z^2+(i-1)z+i=0

解答

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

z = - 0.36602 \dots + 0.36602 \dots i, z = 1.36602 \dots - 1.36602 \dots i

求解步骤

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

替代

z = x + y i

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

展开

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

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

将

0

改写成标准复数形式:

0 + 0 i

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

[x^{2} - y^{2} - x - y = 0 1 + x - y + 2 x y = 0] : (x = - 0.36602 \dots, x = 1.36602 \dots, y = 0.36602 \dots y = - 1.36602 \dots)

z = x + y i

代回

z = - 0.36602 \dots + 0.36602 \dots i, z = 1.36602 \dots - 1.36602 \dots i

(1 + i) x + (- 2 + i) y = - 3 + 16 i

24 + 42 y i = - 0.5 x + 0.5 i

6 (4 + 7 i)^{2} + A (4 + 7 i) + B = 0

x + i y = 1, 1 x + y = 1 + i

3 x^{2} - 125 i x - 3500 = 121 i x + 1543