IIT JEE , AIEEE Forum - Mathematics , Algebra

anant_c

x^n-1=0
x^n=cos2k(pi)+i sin2k(pi)
x = cos(2k(pi))/n+ i sin2k(pi)/n

now why is k=0,1,2,3.....n-1.....y nt a negative integer...or y are the roots in GP?

shashankparewar

you can also take negative values of k
but it will still remain the same.
for example if you take k= -1
then
arguement of x = -2(pi)/n
= 2(pi)+ {-2(pi)/n
= 2(pi) {n-1}/n
which is same as when k=n-1

anant_c

y r the roots in GP?

shashankparewar

if you see the coorect soln of the roots of roots of inity .you will find that the the roots lie in the circle of radius 1 and each soln differ from each other by an arguement of 2(pi)/n.thats why they are in gp.

Ask & Discuss Questions with Community & Experts