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受欢迎的代数 >

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

解答

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

解答

z = 0.92505 \dots - 0.67543 \dots i, z = - 1.92505 \dots - 0.32456 \dots i

求解步骤

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

替代

z = x + y i

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

展开

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

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

[x^{2} - y^{2} + x - y - 2 = 0 1 + x + y + 2 x y = 0] : (x = 0.92505 \dots, x = - 1.92505 \dots, y = - 0.67543 \dots y = - 0.32456 \dots)

z = x + y i

代回

z = 0.92505 \dots - 0.67543 \dots i, z = - 1.92505 \dots - 0.32456 \dots i

流行的例子

(5i)/(4+9i)= a/b

\frac{5 i}{4 + 9 i} = \frac{a}{b}

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

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

5 x - 2 y i = - 8 x + 9

5 + (x + 3 y) i = x + 2 i

4 + 3 i = x + y i