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)dot`

Explanation

Given that , `|z_(1)|=|z_(2)|=* * *=|z_(n)|=1`
`rArr |z_(1)|^(2)=|z_(2)|^(2)=* * *=|z_(n)|^(2)=1`
`rArr z_(1)barz_(1) =z_(2)barz_(2)=z_(3)barz_(3)=* * *=z_(n)barz_(n)=1`
`rArr z_(1)=(1)/(z_(1)),z_(2)=(1)/(z_(2))=* * *=z_(n)=(1)/(barz_(n))`
Now, `|z_(1)+z_(2)+z_(3)+z_(4)+* * * + z_(n)|`
`(z_(1)barz_(1))/(barz_(1)) +( z_(2)barz_(2))/(barz_(1))+( z_(3)barz_(3))/(barz_(1))+* * *+(z_(n)barz_(n))/barz_(n)|" " [:.z_(1)+z_(2)=1, "where" z_(1)=(1)/(z), z_(1)=(barz)/(barz-barz),z_(1)=barz]`
`|(|z_(1)|^(2))/(barz_(1))+ (|z_(2)|^(2))/(barz_(2))+(|z_(3)|^(2))/(barz_(3))+ * * *+(|z_(n)|^(2))/(barz_(n))|`
`|(1)/(barz_(1))+ 1/(barz_(2))+1/(barz_(3))+ * * *+1/(barz_(n))| = sqrt((1)/(barz_(1))+(1)/(barz_(2))+(1)/(barz_(3)))`
`|(1)/(z_(1))+(1)/(z_(2))+* * *+(1)/(z_(3))|` " "Hence proved.