Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

De moivre formula Question

  1. Nov 22, 2005 #1
    Hi
    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

    Bob
     
  2. jcsd
  3. Nov 22, 2005 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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: 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,

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

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?