设z=-(1-i)/根号2,则z的100次方+z的50次方+1的值为?

题目

设z=-(1-i)/根号2,则z的100次方+z的50次方+1的值为?

答案

z=-(1-i)/根号2 = (i -1)/√2
求 z^100 + z^50 + 1
z = (i -1)/√2
所以
z^2 = (i^2 - 2i + 1)/2 = (-1 -2i + 1) = -2i/2 = -i
z^100 = (z^2)^50 = (-i)^50 = i^(50) = (i^2)^25 = (-1)^25 = -1
z^50 = (z^2)^25 = (-i)^25 = -(i)^25 = -i * (i)^24 = -i * (i^2)^12 = -i * (-1)^12 = -i
因此
z^100 + z^50 + 1 = -1 - i + 1 = -i