The value of `sum_(n=1)^(10) {sin(2npi)/11-icos(2npi)/11}`, is

Options

A. `-1`
B. 0
C. `-i`
D. `i`

Answer

D

Explanation

We have,
`sum_(n=1)^(10){sin(2npi)/11-icos(2npi)/11}`
`=-isum_(n=1)^(10){cos(2npi)/11+isin(2npi)/11}`
`=-isum_(n=1)^(10)e^(i2npi//11)`
`=-isum_(n=1)^(10)(e^(i2pi//11))^(n)`
`=-i(sum_(n=1)^(10)alpha^(n))`, where `alpha=e^(i2pi//11)`
`=-i(alpha+alpha^(2)+................+alpha^(10))`
`=-i{(alpha(1-alpha^(10)))/(1-alpha)}=-i{(alpha-alpha^(11))/(1-alpha)}`
`=-i{(alpha-1)/(1-alpha)}=i` `[therefore alpha^(11)=e^(i2pi)=cos2pi i+sin2pi=1]`