How do the roots of unity relate to De Moivre's Theorem?












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How do the roots of unity relate to De Moivre's Theorem? Others always pair the two up, but I do not understand why.






algebra-precalculus







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    0












    $begingroup$


    How do the roots of unity relate to De Moivre's Theorem? Others always pair the two up, but I do not understand why.






    algebra-precalculus







    share|cite|improve this question











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      0












      0








      0





      $begingroup$


      How do the roots of unity relate to De Moivre's Theorem? Others always pair the two up, but I do not understand why.






      algebra-precalculus







      share|cite|improve this question











      $endgroup$




      How do the roots of unity relate to De Moivre's Theorem? Others always pair the two up, but I do not understand why.






      algebra-precalculus




      algebra-precalculus






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      share|cite|improve this question













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      edited Jan 4 at 23:51









      Blue

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      asked Jan 4 at 23:49









      M. C.M. C.

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          2 Answers
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          $begingroup$

          If we let the $n$th root of unity be $$z=cos(theta)+icdot sin(theta)$$
          then by De Moivre's Theorem$$z^n=(cos(theta)+icdot sin(theta))^n$$$$=cos(ncdot theta)+icdot sin(ncdot theta)$$where $ninmathbb{N}$. But:$$z^n=1=cos(2picdot m)+icdot sin(pi+2picdot m)$$ Where $minmathbb{Z}$. So, equating real parts,$$cos(ncdot theta)=cos(2picdot m)$$$$thereforetheta=frac{2picdot m}{n}$$
          i.e. the $n$th roots of unity are $$z=cos(frac{2picdot m}{n})+icdot sin(frac{2picdot m}{n})$$Where $0le m le n-1$ as all other values of $m$ will produce the same values of $z$.






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          • $begingroup$
            In the last line I would prefer that you said " an $n$th root" rather than " the $n$ root".
            $endgroup$
            – DanielWainfleet
            Jan 5 at 0:27



















          1












          $begingroup$

          The roots of unity are all in the form $cos(x)+isin(x)$. Nth roots of unity, raised to the n power, will be 1. Thus, De Moivre's Theorem, which states that $cis(x)^n=cis(nx)$ also tells us that if $cis(x)$ is a nth root of unity, then $cis(x)^n=cis(nx)=1$. This process can also be reversed to find nth roots of unity, as, substituting in a n value, we then have a trig equation we can solve to find the values of x, and the nth roots of unity, $cis(x)$.






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            2 Answers
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            2 Answers
            2






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            active

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            3












            $begingroup$

            If we let the $n$th root of unity be $$z=cos(theta)+icdot sin(theta)$$
            then by De Moivre's Theorem$$z^n=(cos(theta)+icdot sin(theta))^n$$$$=cos(ncdot theta)+icdot sin(ncdot theta)$$where $ninmathbb{N}$. But:$$z^n=1=cos(2picdot m)+icdot sin(pi+2picdot m)$$ Where $minmathbb{Z}$. So, equating real parts,$$cos(ncdot theta)=cos(2picdot m)$$$$thereforetheta=frac{2picdot m}{n}$$
            i.e. the $n$th roots of unity are $$z=cos(frac{2picdot m}{n})+icdot sin(frac{2picdot m}{n})$$Where $0le m le n-1$ as all other values of $m$ will produce the same values of $z$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              In the last line I would prefer that you said " an $n$th root" rather than " the $n$ root".
              $endgroup$
              – DanielWainfleet
              Jan 5 at 0:27
















            3












            $begingroup$

            If we let the $n$th root of unity be $$z=cos(theta)+icdot sin(theta)$$
            then by De Moivre's Theorem$$z^n=(cos(theta)+icdot sin(theta))^n$$$$=cos(ncdot theta)+icdot sin(ncdot theta)$$where $ninmathbb{N}$. But:$$z^n=1=cos(2picdot m)+icdot sin(pi+2picdot m)$$ Where $minmathbb{Z}$. So, equating real parts,$$cos(ncdot theta)=cos(2picdot m)$$$$thereforetheta=frac{2picdot m}{n}$$
            i.e. the $n$th roots of unity are $$z=cos(frac{2picdot m}{n})+icdot sin(frac{2picdot m}{n})$$Where $0le m le n-1$ as all other values of $m$ will produce the same values of $z$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              In the last line I would prefer that you said " an $n$th root" rather than " the $n$ root".
              $endgroup$
              – DanielWainfleet
              Jan 5 at 0:27














            3












            3








            3





            $begingroup$

            If we let the $n$th root of unity be $$z=cos(theta)+icdot sin(theta)$$
            then by De Moivre's Theorem$$z^n=(cos(theta)+icdot sin(theta))^n$$$$=cos(ncdot theta)+icdot sin(ncdot theta)$$where $ninmathbb{N}$. But:$$z^n=1=cos(2picdot m)+icdot sin(pi+2picdot m)$$ Where $minmathbb{Z}$. So, equating real parts,$$cos(ncdot theta)=cos(2picdot m)$$$$thereforetheta=frac{2picdot m}{n}$$
            i.e. the $n$th roots of unity are $$z=cos(frac{2picdot m}{n})+icdot sin(frac{2picdot m}{n})$$Where $0le m le n-1$ as all other values of $m$ will produce the same values of $z$.






            share|cite|improve this answer











            $endgroup$



            If we let the $n$th root of unity be $$z=cos(theta)+icdot sin(theta)$$
            then by De Moivre's Theorem$$z^n=(cos(theta)+icdot sin(theta))^n$$$$=cos(ncdot theta)+icdot sin(ncdot theta)$$where $ninmathbb{N}$. But:$$z^n=1=cos(2picdot m)+icdot sin(pi+2picdot m)$$ Where $minmathbb{Z}$. So, equating real parts,$$cos(ncdot theta)=cos(2picdot m)$$$$thereforetheta=frac{2picdot m}{n}$$
            i.e. the $n$th roots of unity are $$z=cos(frac{2picdot m}{n})+icdot sin(frac{2picdot m}{n})$$Where $0le m le n-1$ as all other values of $m$ will produce the same values of $z$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 5 at 0:35

























            answered Jan 5 at 0:04









            Peter ForemanPeter Foreman

            4177




            4177












            • $begingroup$
              In the last line I would prefer that you said " an $n$th root" rather than " the $n$ root".
              $endgroup$
              – DanielWainfleet
              Jan 5 at 0:27


















            • $begingroup$
              In the last line I would prefer that you said " an $n$th root" rather than " the $n$ root".
              $endgroup$
              – DanielWainfleet
              Jan 5 at 0:27
















            $begingroup$
            In the last line I would prefer that you said " an $n$th root" rather than " the $n$ root".
            $endgroup$
            – DanielWainfleet
            Jan 5 at 0:27




            $begingroup$
            In the last line I would prefer that you said " an $n$th root" rather than " the $n$ root".
            $endgroup$
            – DanielWainfleet
            Jan 5 at 0:27











            1












            $begingroup$

            The roots of unity are all in the form $cos(x)+isin(x)$. Nth roots of unity, raised to the n power, will be 1. Thus, De Moivre's Theorem, which states that $cis(x)^n=cis(nx)$ also tells us that if $cis(x)$ is a nth root of unity, then $cis(x)^n=cis(nx)=1$. This process can also be reversed to find nth roots of unity, as, substituting in a n value, we then have a trig equation we can solve to find the values of x, and the nth roots of unity, $cis(x)$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              The roots of unity are all in the form $cos(x)+isin(x)$. Nth roots of unity, raised to the n power, will be 1. Thus, De Moivre's Theorem, which states that $cis(x)^n=cis(nx)$ also tells us that if $cis(x)$ is a nth root of unity, then $cis(x)^n=cis(nx)=1$. This process can also be reversed to find nth roots of unity, as, substituting in a n value, we then have a trig equation we can solve to find the values of x, and the nth roots of unity, $cis(x)$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                The roots of unity are all in the form $cos(x)+isin(x)$. Nth roots of unity, raised to the n power, will be 1. Thus, De Moivre's Theorem, which states that $cis(x)^n=cis(nx)$ also tells us that if $cis(x)$ is a nth root of unity, then $cis(x)^n=cis(nx)=1$. This process can also be reversed to find nth roots of unity, as, substituting in a n value, we then have a trig equation we can solve to find the values of x, and the nth roots of unity, $cis(x)$.






                share|cite|improve this answer









                $endgroup$



                The roots of unity are all in the form $cos(x)+isin(x)$. Nth roots of unity, raised to the n power, will be 1. Thus, De Moivre's Theorem, which states that $cis(x)^n=cis(nx)$ also tells us that if $cis(x)$ is a nth root of unity, then $cis(x)^n=cis(nx)=1$. This process can also be reversed to find nth roots of unity, as, substituting in a n value, we then have a trig equation we can solve to find the values of x, and the nth roots of unity, $cis(x)$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 5 at 0:04









                H HuangH Huang

                7210




                7210






























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