If `alpha` Is the fifth root of unity, then :

Options

A. `|1+alpha+alpha^(2)+alpha^(3)+alpha^(4)|=0`
B. `|1+alpha+alpha^(2)+alpha^(3)|=1`
C. `|1+alpha+alpha^(2)|=2cos"(pi)/(5)`
D. `|1+alpha|=2cos"(pi)/(10)`

Answer

A, B, C

Explanation

`(a,b,c)` We have `alpha=cos"(2pi)/(5)+isin"(2pi)/(5)`
and `1+alpha+alpha^(2)+alpha^(3)+alpha^(4)=0`
`:. |1+alpha+alpha^(2)+alpha^(3)|=|-alpha^(4)|=|alpha^(4)|=1`
Also, `|1+alpha+alpha^(2)|=|-alpha^(3)(1+alpha)|=|1+alpha|` ..........`(i)`
`=|1+cos'(2pi)/(5)+isin"(2pi)/(5)|`
`=|2cos'(pi)/(5)(cos'(pi)/(5)+isin"(pi)/(5))|`
`=2cos"(pi)/(5)`
Again from `(i)`, `|1+alpha|=2cos"(pi)/(5)`