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受欢迎的代数 >

z^2-(5+4i)z+(3-5i)=0

解答

z^{2} - (5 + 4 i) z + (3 - 5 i) = 0

解答

z = - 0.17102 \dots - 0.80791 \dots i, z = 5.17102 \dots + 4.80791 \dots i

求解步骤

z^{2} - (5 + 4 i) z + (3 - 5 i) = 0

替代

z = x + y i

(x + y i)^{2} - (5 + 4 i) (x + y i) + 3 - 5 i = 0

展开

(x + y i)^{2} - (5 + 4 i) (x + y i) + 3 - 5 i : (x^{2} - y^{2} - 5 x + 4 y + 3) + i (- 5 - 4 x - 5 y + 2 x y)

(x^{2} - y^{2} - 5 x + 4 y + 3) + i (- 5 - 4 x - 5 y + 2 x y) = 0

将

0

改写成标准复数形式:

0 + 0 i

(x^{2} - y^{2} - 5 x + 4 y + 3) + i (- 5 - 4 x - 5 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} - 5 x + 4 y + 3 = 0 - 5 - 4 x - 5 y + 2 x y = 0]

[x^{2} - y^{2} - 5 x + 4 y + 3 = 0 - 5 - 4 x - 5 y + 2 x y = 0] : (x = - 0.17102 \dots, x = 5.17102 \dots, y = - 0.80791 \dots y = 4.80791 \dots)

z = x + y i

代回

z = - 0.17102 \dots - 0.80791 \dots i, z = 5.17102 \dots + 4.80791 \dots i

流行的例子

x^{2} + 3 x + 2 - i = 0

i x = x^{2} - 2 x + 2

(2 + i) 2 = c + d i

v = \frac{2 r}{i}

(x-yi)^2=9(x+yi)

(x - y i)^{2} = 9 (x + y i)