Complex Question - The Kings (1 Viewer)

[medhat tawfik]

Question 8a(ii) of this paper. I'd type it up on Latex but I'm not much of a professional with it. Thanks in advance.

 

[gurmies]

No idea...

 

[medhat tawfik]

Well I did the second part. But it's the last part with highest common factor that I can't get.

 

[Affinity]

It's not quite a complex number question... roughly:
you can use a counting argument
1. Notice that z^n =1 has n distinct roots and that n is a finite number.
2. also, z_k, z_k^2 , ... , z_k^n are all roots of z_k^n = 1 (trivial)
3. We just need to show that no two in the above list are equal.

To do (3), suppose z_k^i = z_k^j where 1<= i,j <= n and i different to j. without loss of generality, we can assume that j > i
then we know that z_k^(j-i) = 1
that is cos(2pi*(j-i)*k/n) = 1
and sin(2pi*(j-i)*k/n) = 0
That forces 2pi*(j-i)*k/n to be a multiple of 2pi which means that (j-i)*k/n = N
where N is some integer so
(j-i)k = nN
now here's the tricky part: the above implies that n divides k(j-i), but n and k share no common prime factors, which means that n much divide (j-i). but j - i < n, which is a contradiction.. so we can safely conclude that the assumption we made at the beginning of part 3 is false. so all n powers are distinct. and we are done.

There are probably cleaner ways of presenting this argument...

 

[gurmies]

It's not quite a complex number question... roughly:
you can use a counting argument
1. Notice that z^n =1 has n distinct roots and that n is a finite number.
2. also, z_k, z_k^2 , ... , z_k^n are all roots of z_k^n = 1 (trivial)
3. We just need to show that no two in the above list are equal.

To do (3), suppose z_k^i = z_k^j where 1<= i,j <= n and i different to j. without loss of generality, we can assume that j > i
then we know that z_k^(j-i) = 1
that is cos(2pi*(j-i)*k/n) = 1
and sin(2pi*(j-i)*k/n) = 0
That forces 2pi*(j-i)*k/n to be a multiple of 2pi which means that (j-i)*k/n = N
where N is some integer so
(j-i)k = nN
now here's the tricky part: the above implies that n divides k(j-i), but n and k share no common prime factors, which means that n much divide (j-i). but j - i < n, which is a contradiction.. so we can safely conclude that the assumption we made at the beginning of part 3 is false. so all n powers are distinct. and we are done.

There are probably cleaner ways of presenting this argument...

Nicely done. I had a feeling it could be done using a proof by contradiction, but I couldn't find it xD

 

[medhat tawfik]

It's not quite a complex number question... roughly:
you can use a counting argument
1. Notice that z^n =1 has n distinct roots and that n is a finite number.
2. also, z_k, z_k^2 , ... , z_k^n are all roots of z_k^n = 1 (trivial)
3. We just need to show that no two in the above list are equal.

To do (3), suppose z_k^i = z_k^j where 1<= i,j <= n and i different to j. without loss of generality, we can assume that j > i
then we know that z_k^(j-i) = 1
that is cos(2pi*(j-i)*k/n) = 1
and sin(2pi*(j-i)*k/n) = 0
That forces 2pi*(j-i)*k/n to be a multiple of 2pi which means that (j-i)*k/n = N
where N is some integer so
(j-i)k = nN
now here's the tricky part: the above implies that n divides k(j-i), but n and k share no common prime factors, which means that n much divide (j-i). but j - i < n, which is a contradiction.. so we can safely conclude that the assumption we made at the beginning of part 3 is false. so all n powers are distinct. and we are done.

There are probably cleaner ways of presenting this argument...

Got it. Thanks much. Reaaalllllyyyy appreciate it.