If `z_(r):r = 1,2,3,.....50` are the roots of the equaiton `sum_(r=0)^(50) z^(r)= 0`, then find the value of ` sum_(r=1)^(50)1//(z_(r) - 1)`

Answer

`-25`

Explanation

`E = (1)/(z_(1)-1) + (1)/(z_(2)-1) +......+ (1)/(z_(50)-1)`
Where `z_(1),z_(2),…..,z_(50)` are the roots of the equation `z^(51) -1=0` other than 1 Let .
`1+z+z^(2)+……+z^(50)) = (z-z_(2)) …..(1-z_(50))`
v`rArr log (1+z+z^(2) +…..+z^(50)) = log [(z-z_(1))(z -z_(2)).....(1-z_(50))]`
Differentiating both sides w.r.t.s and putting z = 1
`(1+2z+3z^(2)+......+50_(z)^(49))/(1+z+z^(2) +......+z^(50))`
`=(1)/(z-z_(1)) +(1)/(z-z_(2))+.......+(1)/(z-z_(50))`
`rArr(50xx 51)/(2xx51)=-[(1)/(z_(1)-1)+(1)/(z_(2)-1)+......+(1)/(z_(50) -1)]`
`therefore sum(1)/(z_(r)-1) = -25`