已知复数z满足z(1+根号3i)=根号3i,则z的共轭复数的虚部是多少
问题描述:
已知复数z满足z(1+根号3i)=根号3i,则z的共轭复数的虚部是多少
答
z=√3i/(1+√3i)
=√3i(1-√3i)/(1+√3i)(1-√3i)
=(√3i+3)/(1+3)
=3/4+√3i/4
所以z的共轭复数的虚部是-√3/4