1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Roots of Unity - is this correct?

  1. Dec 2, 2011 #1
    I'm going over some things I didn't do too well on in my latest Algebra test.
    One question was: List all of the roots of [itex]x^{8}\:-1\:=0[/itex], and write them in the form [itex]a+bi[/itex].

    So I knew I had to list all the 8th roots of unity. In other places in the test they used the notation [itex]e^{iθ}[/itex] and this was a huge problem for me, I didn't know that [itex]r\ast e^{iθ}[/itex]= [itex]r*(cos(θ)\:+isin(θ))[/itex].

    What I'm looking for: [itex]e^{i*2\pi*\frac{k}{n}}[/itex], or in this case, [itex]e^{i*2\pi*\frac{k}{8}}[/itex], where k = 0, 1, 2, 3, 4, 5, 6, 7. (Could also write this as [itex]cos(\frac{2\pi*k}{8})\:+isin(\frac{2\pi*k}{8})[/itex] ).

    For the 1st root: [itex]cos(\frac{2\pi*0}{8})\:+isin(\frac{2\pi*0}{8})\:=cos(0)\:+isin(0)[/itex] = [itex]1+0i\:=1[/itex]

    For the 2nd root: [itex]cos(\frac{2\pi*1}{8})\:+isin(\frac{2\pi*1}{8})\:=cos(\frac{\pi}{4})\:+isin(\frac{\pi}{4})\:=\frac{ \sqrt{2}}{2}+\frac{ \sqrt{2}}{2}i[/itex]

    For the 3rd root: [itex]cos(\frac{2\pi*2}{8})\:+isin(\frac{2\pi*2}{8})\:=cos(\frac{\pi}{2})\:+isin(\frac{\pi}{2})\:=0+1i\:=i[/itex]

    For the 4th root: [itex]cos(\frac{2\pi*3}{8})\:+isin(\frac{2\pi*3}{8})\:=cos(\frac{3\pi}{4})\:+isin(\frac{3\pi}{4})\:=-\frac{ \sqrt{2}}{2}+\frac{ \sqrt{2}}{2}i[/itex]

    5th root: -1
    6th root: [itex]-\frac{ \sqrt{2}}{2}-\frac{ \sqrt{2}}{2}i[/itex]
    7th root: -i
    8th root:[itex]\frac{ \sqrt{2}}{2}-\frac{ \sqrt{2}}{2}i[/itex]


    So the correct answer to this question is [itex]±1,\:\:±i,\:\:and\:\:±\frac{ \sqrt{2}}{2}±\frac{ \sqrt{2}}{2}i[/itex]


    Is this right? And if I just apply the same method to other similar problems I should be fine?
     
  2. jcsd
  3. Dec 2, 2011 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Congratulation, you did a good job! Your derivation is rigth, and you can apply this method for all roots of unity.
    If you need to find the roots of an equation zn-r e=0 do the following:

    [tex]z=(r\exp(i\psi))^{1/n}(e^{i 2\pi k/n})=r^{1/n} e^{i(\psi /n+2\pi k/n)}=r^{1/n} \left( \cos(\psi /n+2\pi k/n)+i \sin(\psi /n+2\pi k/n)\right)[/tex]

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




Similar Discussions: Roots of Unity - is this correct?
  1. Roots of unity (Replies: 18)

Loading...