(1+i)z^2+(2-i)z+(1+3i)=0

解答

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

z = \frac{- 3.77205 \dots + 16.91271 \dots}{2} - 0.88602 \dots i, z = \frac{2.77205 \dots - 16.91271 \dots}{2} + 2.38602 \dots i

求解步骤

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

替代

z = x + y i

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

展开

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

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

将

0

改写成标准复数形式:

0 + 0 i

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

[x^{2} - y^{2} - 2 x y + 2 x + y + 1 = 0 3 - x + x^{2} - y^{2} + 2 y + 2 x y = 0] : (x = \frac{- 3.77205 \dots + 16.91271 \dots}{2}, x = \frac{2.77205 \dots - 16.91271 \dots}{2}, y = - 0.88602 \dots y = 2.38602 \dots)

z = x + y i

代回

z = \frac{- 3.77205 \dots + 16.91271 \dots}{2} - 0.88602 \dots i, z = \frac{2.77205 \dots - 16.91271 \dots}{2} + 2.38602 \dots i

z^{2} = 49 + i

z^{2} + (2 i - 3) z + 5 - 1 = 0

s i f (x) = \frac{5 x ^{2}}{x}, f (0)

z^{2} + 4 z i - 4 = 0

\frac{( a + bi ) - 3}{3 - ( a + bi )} = 3