Complex number help (1 Viewer)

manscux

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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
 

math man

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better way is:
we know w is a solution to

therefore

Now sub into

we get

and that's it.
 

Nws m8

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Do you think this type of question would pop up in around question 11-12?
I reckon that's going to be Q16! BRO THIS QUESTION IS SO EFFIN EASY, might be a MC or a 1 marker
 

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