已知z1=1-3i，z2=6-8i．若1z+1z1=1z2，求z的值．

Question

已知z₁=1-3i，z₂=6-8i．若(1)/(z)+(1)/(z1)=(1)/(z2)，求z的值．

Answer 1

∵z₁=1-3i，z₂=6-8i，(1)/(z)+(1)/(z1)=(1)/(z2)，
∴z=(z1•z2)/(z1−z2)=((1−3i)(6−8i))/(1−3i−(6−8i))=(−18−26i)/(−5+5i)=((−18−26i)(−5−5i))/((−5+5i)(−5−5i))=(−40+220i)/(25+25)=-(4)/(5)+(22)/(5)i．