Finite Set of Points in Complex Plane: $\{e^{n r \pi i}\}$

[Somefantastik]

Homework Statement

\left\{ e^{n r \pi i}: n \in \textbf{Z} \right\} , r \in \textbf{Q}

I'm trying to show that this set is finite.

Homework Equations

The Attempt at a Solution

Other than the fact that these points lie on the unit circle in the complex plane, I'm not sure where to start. Any direction would be helpful. clearly there's a way to choose r or n and use periodicity to show a finite set of points for this sets. But I'm not sure how r and n could be chosen.

 

[CompuChip]

Since r is a fraction, you could write it as p/q.

Knowing that e^{2 \pi i} = 1, can you show that the set has size {} \le q?

 

[Somefantastik]

exp(2*pi*i)? from where did the 2 come?

I know that exp(n*pi*i) = 1 for n integer. if r= p/q, then exp(n*r*pi*i) = exp(2*p*pi*i) if n = 2*q... but no, I don't know why the set has size \leq q.

 

[Mark44]

exp(2*pi*i)? from where did the 2 come?

I know that exp(n*pi*i) = 1 for n integer.

No, that's not true. exp(n*pi*i) alternates between 1 and -1, depending on whether n is even or odd, respectively.

if r= p/q, then exp(n*r*pi*i) = exp(2*p*pi*i) if n = 2*q... but no, I don't know why the set has size \leq q.

 

[Somefantastik]

No, that's not true. exp(n*pi*i) alternates between 1 and -1, depending on whether n is even or odd, respectively.

well in this case n = 2q which is even right? And exp(2*p*pi*i) = 1 since 2*p is even as well since p is an integer. Later I can set n = 2q+1 to handle the odd cases, but I'm still trying to figure out why the set is finite. so I still need some help please.

 

[Mark44]

Pick a value for r = p/q, then look at values of exp(n*r*pi*i) for n = 1, 2, 3, and so on. What is it that eventually happens at some value of n and thereafter?

 

[JSuarez]

Just a hint: start with r = 1/q, and plot e^{nr\pi i} in the unit circle. What happens? How could you reduce the the cases where p\neq 1 to this one?

 

[Somefantastik]

got it, thanks everybody.