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Homework Help: Finite set

  1. Apr 30, 2010 #1
    1. The problem statement, all variables and given/known data

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

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

    2. Relevant equations



    3. 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.
     
  2. jcsd
  3. Apr 30, 2010 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

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

    Knowing that [itex]e^{2 \pi i} = 1[/itex], can you show that the set has size [itex]{} \le q[/itex]?
     
  4. Apr 30, 2010 #3
    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 [tex] \leq q[/tex].
     
    Last edited: Apr 30, 2010
  5. Apr 30, 2010 #4

    Mark44

    Staff: Mentor

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

     
  6. Apr 30, 2010 #5
    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.
     
  7. Apr 30, 2010 #6

    Mark44

    Staff: Mentor

    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?
     
  8. Apr 30, 2010 #7
    Just a hint: start with r = 1/q, and plot [itex]e^{nr\pi i}[/itex] in the unit circle. What happens? How could you reduce the the cases where [itex]p\neq 1[/itex] to this one?
     
  9. May 4, 2010 #8
    got it, thanks everybody.
     
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