+2ic^2=-8+3i,c^1=5

解答

+ 2 i c^{2} = - 8 + 3 i, c^{1} = 5

c = \frac{3 + 73}{2} + \frac{4}{3 + 73} i, c = - \frac{3 + 73}{2} - \frac{4}{3 + 73} i

求解步骤

2 i c^{2} = - 8 + 3 i

替代

c = x + y i

2 i (x + y i)^{2} = - 8 + 3 i

展开

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

- 4 x y + 2 i (x^{2} - y^{2}) = - 8 + 3 i

复数仅在实部和虚部均相等时才相等改写为方程组:

[- 4 x y = - 8 2 x^{2} - 2 y^{2} = 3]

[- 4 x y = - 8 2 x^{2} - 2 y^{2} = 3] : x = \frac{3 + 73}{2}, x = - \frac{3 + 73}{2}, y = \frac{4}{3 + 73} y = - \frac{4}{3 + 73}

c = x + y i

代回

c = \frac{3 + 73}{2} + \frac{4}{3 + 73} i, c = - \frac{3 + 73}{2} - \frac{4}{3 + 73} i

x + i y = 3 i - i x

7 - 4 y i = 9 x + 3 i

so l v e f or t, \frac{L d i}{d t} + R i = E

15 - 3 y i = \frac{1}{2} x + 2 i

z^{2} = - 1 + 3 i