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De moivre formula Question

  1. Nov 22, 2005 #1
    I have a question regarding the use of de moivre formula:
    I'm presented with a complex number [tex] z = (cos(v) + i sin(v))^n = 1[/tex]
    I'm suppose to show that [tex]z^n = 1[/tex] is a root of unity. Is there a procedure on how to show this? If n = 6 and [tex]v = \frac{4 \pi}{6}[/tex]

    Sincerely and Best Regards

  2. jcsd
  3. Nov 22, 2005 #2


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    Are you sure you've copied this correctly? You are saying that you are told that z= 1 ?? And you want to prove that 1 is a "root of unity"??

    I very much doubt that! Please copy the problem carefully.

    Perhaps you are given that z= cos v+ i sin(v) and want to show that (cos v+ i sin(v))n= zn= 1, thus showing that z is a "root of unity".

    Furthermore, you then say "if n= 6 and [itex]v= \frac{4\pi}{6}[/itex]". Is that a separate problem or is the original problem to show that
    [tex]\left(cos\left(\frac{4\pi}{6})+ i sin(\frac{4\pi}{6}\right)\right)^6= 1[/tex]?

    If it is the latter, since YOU titled this "DeMoivre's Formula Question", what does DeMoivre's formula tell you [itex]\left(\(cos\left(\frac{4\pi}{6})+ i sin(\frac{4\pi}{6}\right)\right)^6[/itex] is?
    Last edited by a moderator: Nov 23, 2005
  4. Nov 23, 2005 #3
    Hello I have been looking through my textbook which gives the following procedure on how to show if [tex](cos(\frac{4 \pi}{6}) + i sin(\frac{4 \pi}{6}))[/tex]

    Is the six root of unity for [tex]z^{6} = 1[/tex]

    Its know that [tex]1 = cos(2 k \pi) + i sin(2 k \pi)[/tex] where k = 1,2,3,.....m

    According to De Moivre's formula the n'th root unity can be expressed as

    [tex] (cos(\frac{2 k \pi}{n}) + i sin(\frac{2 k \pi}{n})) [/tex]

    My case

    k = 2 and n = 6

    If I insert these into De Moivre's formula I get [tex]z = (cos(\frac{4 \pi}{6}) + i sin(\frac{4 \pi}{6}))[/tex]

    I insert z into the initial equation and get

    [tex](cos(\frac{4 \pi}{6}) + i sin(\frac{4 \pi}{6}))^6 = (cos(\frac{24 \pi}{6}) + i sin(\frac{24 \pi}{6})) = 1[/tex]

    Therefore [tex](cos(\frac{4 \pi}{6}) + i sin(\frac{4 \pi}{6}))[/tex] is the 6'th root of unity for

    [tex]z^6 = 1[/tex]

    I have hand it in tomorrow so therefore am I on the right track?

    Sincerely and Best Regards,

    Last edited: Nov 23, 2005
  5. Nov 23, 2005 #4


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    Yes, you can do that but it would be easier, since you are already given [tex]z = (cos(\frac{4 \pi}{6}) + i sin(\frac{4 \pi}{6}))[/tex],
    to just take the 6 th power of that and show that it is 1.
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