यदि `|z_(1)|=|z_(2)|= . . .=|z_(n)|=1`, तो सिद्ध कीजिए कि -
`|z_(1)+z_(2)+ . . . +z_(n)|=|(1)/(z_(1))+(1)/(z_(2))+ . . . +(1)/(z_(n))|`

Explanation

`|z_(1)|=|z_(2)|= . . .=|z_(n)|=1` . . . (i)
इसलिए `|{:(1)/(z_(1))+(1)/(z_(2))+ . . . +(1)/(n_(n)):}|=|{:(bar(z_(1)))/(z_(1)bar(z_(1)))+(bar(z_(2)))/(z_(2)bar(z_(2)))+ . . . +(bar(z_(n)))/(z_(n)bar(z_(n))):}|`
`|{:(bar(z_(1)))/(|z_(1)|^(2))+(bar(z_(2)))/(|z_(1)|^(2))+ . . . +(bar(z_(n)))/(|z_(n)|^(2)):}|`
`=|bar(z_(1))+bar(z_(2))+ . . . +bar(z_(n))|" "` [समीकरण (i) से]
`bar(|z_(1)+z_(2)+ . . .+z_(n)|)`
`=|z_(1)+z_(2)+ . . . +z_(n)|" "(because|bar(z)|=|z|)`
`" "` यही सिद्ध करना था ।