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Why is 1^(1/6) equal to cis(n*pi/3)?

  1. May 11, 2010 #1
    Hi, I have a doubt concerning the following simple problem. This is actually a textbook problem, so the answer is correct. I just want to understand why this is.

    1^(1/6) = exp ((1/6)(2n*[tex]\pi[/tex]i)) = cos (n*[tex]\pi[/tex]/3) + i sin (n*[tex]\pi[/tex]/3)

    where n = 0,1,2,3,...

    Ok, it is comprehensible that for n=0 we get cos(0) = 1, which is equal to 1^(1/6) = 1
    However, I can't grasp why with n=1, for example, cis (n[tex]\pi[/tex]/3) is still equal to 1. In that case we would have cos([tex]\pi[/tex]/3) + i sin([tex]\pi[/tex]/3). How can that be equal to 1?! It has an imaginary part! Not even the real parts are equal.
    The pi's appear as exponentials for some reason. Anyway, you get what I'm saying.
     
  2. jcsd
  3. May 11, 2010 #2
    Check out this http://en.wikipedia.org/wiki/Root_of_unity" [Broken] while I look for the recent thread that discusses this...
     
    Last edited by a moderator: May 4, 2017
  4. May 11, 2010 #3

    Mark44

    Staff: Mentor

    It just happens that cos(0) = 1. More to the point here is that [cos(0)]^6 = 1
    No, it's not equal to 1, but [cis (1[itex]\pi[/itex]/3)]^6 does equal 1.
    You can minimize this by using [ itex] tags for inline LaTeX.

    The numbers cis (n[itex]\pi[/itex]/3) for n = 0, 1, 2, 3, 4, and 5 are the 6th roots of 1. Most of them are complex, but if you raise any of the to the 6th power, you get 1.
     
  5. May 11, 2010 #4
    Yeah I get it. It's just very weird... you can conclude that 1^(1/6) is not equal to 1... because cis (pi/3) is not 1, but cis ((pi/3))^6 is.
     
  6. May 11, 2010 #5

    Mark44

    Staff: Mentor

    Another way to look at this is that you are solving z^6 - 1 = 0. There are 6 values of z that are solutions, two of which are 1 and - 1. The other four are complex.
     
  7. May 11, 2010 #6
    Aren't we getting into the "principle" root thing again?
     
  8. May 11, 2010 #7

    Mark44

    Staff: Mentor

    Which is why I'm trying to steer it in the direction of finding the solutions of z^6 - 1 = 0.
     
  9. May 11, 2010 #8
    May I ask what the principle root thing is?
     
  10. May 11, 2010 #9
    No, you have to have at least ten posts to ask 2 questions in one thread.

    Juuuuust kiddin.
    When we say "the square root of 9", there is an implied understanding that we mean the POSITIVE square root. It can be ambiguous because there are two numbers that equal 9 when squared.

    Extend this to the SIXTH root (or n-th root). There are 6 (or n) roots of 1 , but when we say the "sixth root of 1" (or "the n-th root"), we mean the "principle"/positive/simplest one.
    There has to be a better way to determine it in general, as I'm sure (and hoping!) Mark44 will point out.
     
  11. May 12, 2010 #10

    Mark44

    Staff: Mentor

    OK, I'll give it a shot.

    The problem as posed at the beginning of this thread is essentially "find all of the sixth roots of 1." That is equivalent to solving the equation z6 - 1 = 0 in the complex numbers. The principal sixth root of 1 is 1, but there are five others, one of which is -1. The other four are complex.

    They can be found using the Theorem of De Moivre.

    z6 - 1 = 0 ==> z6 = 1 = cos(0 + n*2pi) + isin(0 + n*2pi), n [itex]\in[/itex] I
    ==> z = (cos(n*2pi) + isin(n*2pi))1/6, n [itex]\in[/itex] I
    ==> z = cos(n*2pi/6) + isin(n*2pi/6), n [itex]\in[/itex] I
    In the last equation, when n = 6, we have the same complex number as when n = 0, and when n = 7, we have the same complex number as when n = 1, so we keep repeating the same 6 complex numbers. For that reason, it suffices to say that n = 0, 1, 2, 3, 4, or 5.

    So we have 6 distinct complex values, all equally spaced around the unit circle. Each of them is "a" sixth root of 1, but the principal sixth root of 1 corresponds to n = 0, and is cos(0) + isin(0) = 1 + i0 = 1.
     
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