3a^2-3b^2-i6ab=g(a+bi)

解答

3 a^{2} - 3 b^{2} - i 6 ab = g (a + bi)

a = \frac{g}{3}, a = 0, a = - \frac{g}{6}, a = - \frac{g}{6}, b = 0 b = 0 b = \frac{g}{2 3} b = - \frac{g}{2 3}

求解步骤

3 a^{2} - 3 b^{2} - i 6 ab = g (a + bi)

将

g (a + bi)

改写成标准复数形式:

g a + g bi

3 a^{2} - 3 b^{2} - i 6 ab = g a + g bi

复数仅在实部和虚部均相等时才相等改写为方程组:

[3 a^{2} - 3 b^{2} = g a - 6 ab = g b]

[3 a^{2} - 3 b^{2} = g a - 6 ab = g b] : a = \frac{g}{3}, a = 0, a = - \frac{g}{6}, a = - \frac{g}{6}, b = 0 b = 0 b = \frac{g}{2 3} b = - \frac{g}{2 3}

a = \frac{g}{3}, a = 0, a = - \frac{g}{6}, a = - \frac{g}{6}, b = 0 b = 0 b = \frac{g}{2 3} b = - \frac{g}{2 3}

a - bi = ai + b i^{2}

i^{2} + x^{2} = 1

(x + y i) = \frac{1 - ai}{1 + ai}

so l v e f or m, r = \frac{i - m}{1 + m}

- 9 x + 8 i = - 54 + (16 y) i