1.

If `|z_1|=|z_2|=dot=|z_n|=1,`prove that `|z_1+z_2+z_3+...+z_n|=1/(z_1)+1/(z_2)+1/(z_3)+...+1/(z_n).`

Answer» We have
`|z_(1)|=|z_(2)|=|z_(3)|=...=|z_(n)|=1`
`rArr" "|z_(1)|^(2)=|z_(2)|^(2)=|z_(3)|^(2)=...=|z_(n)|^(2)=1`
`rArr" "z_(1)bar(z)_(1)=1, z_(2)bar(z)_(2)=1, z_(3)bar(z)_(3)=1,...,z_(n)bar(z)_(n)=1`
`rArr" "(1)/(z_(1))=bar(z)_(1), (1)/(z_(2))=bar(z)_(2), (1)/(z_(3))=bar(z)_(3),...,(1)/(z_(n))=bar(z)_(n)`
`rArr" "|(1)/(z_(1))+(1)/(z_(2))+(1)/(z_(3))+...+(1)/(z_(n))|=|bar(z)_(1)+bar(z)_(2)+bar(z)_(3)+...+bar(z)_(n)|`
`" "=|bar(z_(1)+z_(2)+z_(3)+...+z_(n))|`
`" "=|z_(1)+z_(2)+z_(3)+...+z_(n)|" "because |bar(z)|=|z|]`
`therefore" "|(1)/(z_(1))+(1)/(z_(2))+(1)/(z_(3))+...+(1)/(z_(n))|=|z_(1)+z_(2)+z_(3)+...+z_(n)|`.


Discussion

No Comment Found