m/(1+i)+n=-m-1/i

解答

\frac{m}{1 + i} + n = - m - \frac{1}{i}

n = 3, m = - 2

求解步骤

\frac{m}{1 + i} + n = - m - \frac{1}{i}

找到

1 + i, i

的最小公倍数:

i (1 + i)

乘以最小公倍数=

i (1 + i)

\frac{m}{1 + i} i (1 + i) + ni (1 + i) = - mi (1 + i) - \frac{1}{i} i (1 + i)

乘开

- n + i (m + n) = (m - 1) + i (- 1 - m)

复数仅在实部和虚部均相等时才相等改写为方程组:

[- n = m - 1 m + n = - 1 - m]

[- n = m - 1 m + n = - 1 - m] : n = 3, m = - 2

n = 3, m = - 2

(a + 3 i) (1 + bi) = 25 - 39 i

x^{2} - 5 x + (7 + i) = 0

A = p (1 + \frac{i}{100})^{2}

9 x - 18 i = - 36 + 6

2 x - y + (2 x - 3 y - 2 - 9) i = 0