设z1=2-i，z2=1-3i，则虚数z=iz1+.z25的实部为．

Question

设z₁=2-i，z₂=1-3i，则虚数z=(i)/(z1)+(.z2)/(5)的实部为______．

Answer 1

z=(i)/(2−i)+(1+3i)/(5)=(i(2+i))/((2−i)(2+i))+(1+3i)/(5)=(2i−1)/(5)+(1+3i)/(5)=(5i)/(5)=i．
故其实部为0．
故答案为0．