复数z=sin(π/6)_icos(π/3)的模是多少啊?
问题描述:
复数z=sin(π/6)_icos(π/3)的模是多少啊?
答
|z|=根号[(sinπ/6)^2+(cosπ/3)^2]
=根号(1/4+1/4)=根号2/2