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Proving that (z^m)^(1/n)=(z^(1/n))^m

  1. Nov 16, 2013 #1
    1. The problem statement, all variables and given/known data

    I want to prove something that seems trivial at first:
    [itex] (z^m)^{1/n}=(z^{1/n})^m [/itex], where m and n dont have a common divisor.

    3. The attempt at a solution

    When I'm using [itex] z=re^{i\theta}[/itex], I arrive that above is true if and only if the following is true:
    [itex] e^{ \frac{2\pi \theta mk}{n}}=e^{\frac{2\pi \theta k}{n}} [/itex]
    and this confuses me.
     
  2. jcsd
  3. Nov 16, 2013 #2

    mfb

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    Then you are doing something wrong.

    The equality is true even for complex m,n, so you can consider the more general problem (z^m)^n =? (z^n)^m or (z^m)^n =? z^(mn) and save some fractions.
     
  4. Nov 16, 2013 #3
    This is what puzzlies me:

    [itex] (z^{\frac {1} {n}})^{m} = r^{\frac {m}{n}}e^{\frac {m\theta + 2m \pi k} {n}i} =
    r^{\frac {m}{n}}e^{\frac {m\theta} {n}i}e^{\frac {2m \pi k} {n}i}[/itex]

    [itex] (z^m)^{\frac {1} {n}} = r^{\frac {m}{n}}e^{\frac {m\theta + 2 \pi k} {n}i} =
    r^{\frac {m}{n}}e^{\frac {m\theta} {n}i}e^{\frac {2 \pi k} {n}i}[/itex]

    How is this possible?
     
  5. Nov 16, 2013 #4

    mfb

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    You are using n and m with two different meanings, and you don't need that +2pi k there.
     
  6. Nov 16, 2013 #5
    Sorry, but I'm not sure I understand.
    m and n are constants while k goes from 0 to n-1.

    I do need +2kpi, as I need to prove that both represent the same set of values.
     
  7. Nov 16, 2013 #6

    Dick

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    I think you are right to to be confused about this. z^(1/n) isn't even a function. It's multivalued. You need to define a branch. If you are naively using the polar form, the answer will depend on the argument you pick for z.
     
  8. Nov 16, 2013 #7
    So how can I prove that both expression represent the following:
    https://dl.dropboxusercontent.com/u/27412797/q1.png [Broken]
     
    Last edited by a moderator: May 6, 2017
  9. Nov 16, 2013 #8

    Dick

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    You don't do it by assuming that (z^(1/n))^m=(z^m)^(1/n). That's a complex numbers mistake. You can only safely take (z^a)^b=(z^b)^a to be true if a and b are integers. You know from deMoivre that all of those numbers are nth roots of z^m. What I think they want you to show is that that is all of them. For example putting k=n doesn't give you a new root, and that for no two different k in k=0,...n-1 are they the same root.
     
    Last edited by a moderator: May 6, 2017
  10. Nov 16, 2013 #9
    The question is formulated like this:
    Show that set of values represented by (z^(1/n))^m are the same set of values represented by (z^m)^(1/n).

    And both these equivalent to set: https://dl.dropboxusercontent.com/u/27412797/q1.png [Broken]
     
    Last edited by a moderator: May 6, 2017
  11. Nov 16, 2013 #10

    Dick

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    Alright. The full question has it make a little more sense. I'll give you an example of what can go wrong. Take z=1, m=2 and n=2. Then the values of (1^2)^(1/2) are 1^(1/2) which is +1 or -1. The values of (1^(1/2))^2 are (+1)^2 or (-1)^2 which are both 1. So the value sets aren't equal. Because 2 and 2 aren't relatively prime. That's the problem you are dealing with.
     
    Last edited by a moderator: May 6, 2017
  12. Nov 16, 2013 #11
    So I should show that
    [itex]e^{km\frac {2\pi i} {n}} = e^{k\frac {2\pi i} {n}} [/itex] because m mod(n) != 0?
     
  13. Nov 16, 2013 #12

    Dick

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    Even if that were true, why? I think you should think about it some more. Figure out why it works if n=2 and m=3, those have no common divisor. Choosing z=1 is perfectly general. You've already factored out |z|. You just need to think about the angles.
     
  14. Nov 16, 2013 #13
    e^(2pi*(m/n)k) represent same angle when m and n have common divisor, but when they do the whole expressions becomes equal to 1.


    But I'm still not sure how to translate this into proving what I need to prove. I'm not even sure why the above is true...
     
  15. Nov 16, 2013 #14
    I think I'm closer to the solution as I understand now that:

    [itex] e^{2pi \frac {mk}{n} i} = e^{2pi \frac {mk(mod(n))}{n} i} [/itex]
     
  16. Nov 16, 2013 #15

    Dick

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    That's a good start. So you know k=0,1,2,...n-1 are the only possible solutions. k=n would just repeat k=0, k=n+1 would just repeat k=1 etc. Now I think you just want to show that they are all different solutions if m and n have no common divisor. This is really turning into a number theory problem.
     
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