Show that -1/2+ 1/2(i3 root) is a cube root of 1
Did you mean
show that -1/2 + (1/2)√3 i is a cube root of 1 ??
If so, then show that (-1/2 + (1/2)√3 i )^3 = 1
LS = [ (-1/2) ( 1 - √3i) ]^3
= (-1/2)^3 [ 1 + 3(-√3i) + 3(-√3i)^2 + (-√3i)^3 ]
= (-18) [ 1 - 3√3i + 9i^2 - 3√3 i^3 ]
= (-1/8)(1 - 3√3 - 9 + 3√3i ]
=(-1/8)[-8}
= 1
= RS
Yes, -1/2 + (1/2)√3 i is a cube root of 1