设z=-(1-i)/根号2,则z的100次方+z的50次方+1的值为?
问题描述：
设z=-(1-i)/根号2,则z的100次方+z的50次方+1的值为?
答
方法一：z=cos(0.75pi)+isin(0.75pi)
z^100+z^50+1=cos(75pi)+isin(75pi)+cos(37.5pi)+isin(37.5pi)+1
=-1+0-i+1=-i
利用欧拉公式很简便
方法二：利用一元二次方程化简
简单是简单，步骤多，不值得
答
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