m=p[(1+i)t-1]i

解答

m = p [(1 + i) t - 1] i

p = - m, t = 1, m \neq = 0

求解步骤

m = p [(1 + i) t - 1] i

化简

p ((1 + i) t - 1) i : - pt + i p (t - 1)

m = - pt + i p (t - 1)

将

m

改写成标准复数形式:

m + 0 i

m + 0 i = - pt + i p (t - 1)

复数仅在实部和虚部均相等时才相等改写为方程组:

[m = - pt 0 = pt - p]

[m = - pt 0 = pt - p] : p = - m, t = 1, m \neq = 0

p = - m, t = 1, m \neq = 0

a - bi = - b + ai

(3 - 4 i) (a + bi) = 25

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

a + bi = - 9 + 4 i

a + ib = 1 - (a - ib) + 3 i