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Homework Help: 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

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    Homework Helper

    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
     
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