Question

复数z=1–i/2+i(i为虚数单位)的虚部为多少

Answer 1

.复数z=1–i/2+i=(1-i)(2-i)/(2+i)(2-i)=(2-i-2i-1)/(4+1)=1/5-3/5i
虚部为-3/5
对于复数z=a+bi,实部是a,虚部是b
当虚数是纯虚数时就有实部a=0,虚部b不=0.