a+bi-5=i(10+4(a+bi))

解答

a + bi - 5 = i (10 + 4 (a + bi))

a = - \frac{35}{17}, b = \frac{30}{17}

求解步骤

a + bi - 5 = i (10 + 4 (a + bi))

化简

(a - 5) + ib = - 4 b + i (10 + 4 a)

复数仅在实部和虚部均相等时才相等改写为方程组:

[a - 5 = - 4 b b = 10 + 4 a]

[a - 5 = - 4 b b = 10 + 4 a] : a = - \frac{35}{17}, b = \frac{30}{17}

a = - \frac{35}{17}, b = \frac{30}{17}

z^{2} + 2 i z + 3 = 0

(1 + q) (2 + i) + p = ((p + q - 1) (3 + i) + 2)

a^{2} - 4 ia - 4 = 0

\frac{i \cdot z - 3}{z ^{2} - 1} = - 1

4 i = x^{2} - 3 x + 7