Could some one please put up a solution for this question so i can check if i am right
Show that if the complex number w is a solution of z^n =1,then so is w^m,
is a solution of z^k =1.
where m and n are arbitrary integers
What i did:
let w = cosa + isina
mod w = 1 = mod w^m
arg w = a
arg w^m = ma
arg z= 2pi/k
therefore a = 2pi/k1 for a specific integer k
this ma = 2pi/k1 x m
if m<n than w^m must be a solution since it has the same mod and arg of z
please verify
Show that if the complex number w is a solution of z^n =1,then so is w^m,
is a solution of z^k =1.
where m and n are arbitrary integers
What i did:
let w = cosa + isina
mod w = 1 = mod w^m
arg w = a
arg w^m = ma
arg z= 2pi/k
therefore a = 2pi/k1 for a specific integer k
this ma = 2pi/k1 x m
if m<n than w^m must be a solution since it has the same mod and arg of z
please verify
