Paragraph Roots of z^n-1=0 are 1, , ^2, ^3, . . ^ n-1 where = 2 n +i 2 n so z^n-1=(z-1)(z- ) ( z - ^ n -1 ) On the basis of above information, answer the following questions : Question _ r =1 ^ n -1 r n is equal to ( n I ⁺ )

Step-by-step solution

z^ n -1=( z -1)( z - ) ( z - ^2 ) . ( z - ^ n -1 ) z^n-1 z-1 =(z- ) (z- ^2 ) (z- ^ n-1 ) _ z 1 z^n-1 z-1 = _ z 1 (z- ) (z- ^2 ) . (z- ^ n-1 ) n = _ r =1 ^ n -1 (1- ^ r )= _ r =1 ^ n -1 (1- ^ 2 ri n )= _ r=1 ^ n-1 -e^ i n (e^ i n -e^ - i n ) = _ r=1 ^ n -1 -e^ i n 2 i ( r n ) |n|= _ r=1 ^ n -1 2 ( r n ) |n|=2^ n -1 _ r =1 ^ n -1 ( r n ) _ r =1 ^ n -1 ( r n )= n 2^ n -1 ( n 1) So _ r=1 ^ n-1 n r n = n ( _ r=1 ^ n-1 r n )= n n 2^ n -1 = nn -( n -1) n 2