Why the alternate form of $frac{sqrt3}2+frac i 2$ is $sqrt[6]{-1}$?












0












$begingroup$


The alternate form of:
$$frac{sqrt3}2+frac i 2$$
is
$$sqrt[6]{-1}$$
(I know that thanks to WolframAlpha.)



What are the arithmetic actions that gets us from the former to the latter?






algebra-precalculus complex-numbers arithmetic







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Well you can simply raise $sqrt{3}/2+i/2$ to the sixth power and find out that you get $-1$.
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 22:35










  • $begingroup$
    The two forms are not equivalent. $sqrt[6]{-1}$ is an ambiguous presentation of 6 different complex numbers.
    $endgroup$
    – user
    Dec 13 '18 at 22:44












  • $begingroup$
    Related: "Principal $n$th root of a complex number".
    $endgroup$
    – Blue
    Dec 13 '18 at 22:46
















0












$begingroup$


The alternate form of:
$$frac{sqrt3}2+frac i 2$$
is
$$sqrt[6]{-1}$$
(I know that thanks to WolframAlpha.)



What are the arithmetic actions that gets us from the former to the latter?






algebra-precalculus complex-numbers arithmetic







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Well you can simply raise $sqrt{3}/2+i/2$ to the sixth power and find out that you get $-1$.
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 22:35










  • $begingroup$
    The two forms are not equivalent. $sqrt[6]{-1}$ is an ambiguous presentation of 6 different complex numbers.
    $endgroup$
    – user
    Dec 13 '18 at 22:44












  • $begingroup$
    Related: "Principal $n$th root of a complex number".
    $endgroup$
    – Blue
    Dec 13 '18 at 22:46














0












0








0





$begingroup$


The alternate form of:
$$frac{sqrt3}2+frac i 2$$
is
$$sqrt[6]{-1}$$
(I know that thanks to WolframAlpha.)



What are the arithmetic actions that gets us from the former to the latter?






algebra-precalculus complex-numbers arithmetic







share|cite|improve this question











$endgroup$




The alternate form of:
$$frac{sqrt3}2+frac i 2$$
is
$$sqrt[6]{-1}$$
(I know that thanks to WolframAlpha.)



What are the arithmetic actions that gets us from the former to the latter?






algebra-precalculus complex-numbers arithmetic




algebra-precalculus complex-numbers arithmetic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 22:51









gimusi

92.9k84494




92.9k84494










asked Dec 13 '18 at 22:33









HeyJudeHeyJude

1687




1687








  • 2




    $begingroup$
    Well you can simply raise $sqrt{3}/2+i/2$ to the sixth power and find out that you get $-1$.
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 22:35










  • $begingroup$
    The two forms are not equivalent. $sqrt[6]{-1}$ is an ambiguous presentation of 6 different complex numbers.
    $endgroup$
    – user
    Dec 13 '18 at 22:44












  • $begingroup$
    Related: "Principal $n$th root of a complex number".
    $endgroup$
    – Blue
    Dec 13 '18 at 22:46














  • 2




    $begingroup$
    Well you can simply raise $sqrt{3}/2+i/2$ to the sixth power and find out that you get $-1$.
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 22:35










  • $begingroup$
    The two forms are not equivalent. $sqrt[6]{-1}$ is an ambiguous presentation of 6 different complex numbers.
    $endgroup$
    – user
    Dec 13 '18 at 22:44












  • $begingroup$
    Related: "Principal $n$th root of a complex number".
    $endgroup$
    – Blue
    Dec 13 '18 at 22:46








2




2




$begingroup$
Well you can simply raise $sqrt{3}/2+i/2$ to the sixth power and find out that you get $-1$.
$endgroup$
– SmileyCraft
Dec 13 '18 at 22:35




$begingroup$
Well you can simply raise $sqrt{3}/2+i/2$ to the sixth power and find out that you get $-1$.
$endgroup$
– SmileyCraft
Dec 13 '18 at 22:35












$begingroup$
The two forms are not equivalent. $sqrt[6]{-1}$ is an ambiguous presentation of 6 different complex numbers.
$endgroup$
– user
Dec 13 '18 at 22:44






$begingroup$
The two forms are not equivalent. $sqrt[6]{-1}$ is an ambiguous presentation of 6 different complex numbers.
$endgroup$
– user
Dec 13 '18 at 22:44














$begingroup$
Related: "Principal $n$th root of a complex number".
$endgroup$
– Blue
Dec 13 '18 at 22:46




$begingroup$
Related: "Principal $n$th root of a complex number".
$endgroup$
– Blue
Dec 13 '18 at 22:46










3 Answers
3






active

oldest

votes


















3












$begingroup$

That's a terrible notation by Wolfram. There are six sixth roots of $-1$, so you cannot tell which one it is when you write $sqrt[6]{-1}$.



The notation makes sense when we write $sqrt{2}$, say, because we take the usual convention that it is the positive root. But such choice is meaningless for arbitrary roots.



The sixth roots of $-1$ are, from De Moivre's formula (and writing $-1=cospi+isin pi$),
$$
omega_k=cosleft(tfrac{pi+2kpi}{6}right)+isinleft(tfrac{pi+2kpi}{6}right), k=0,ldots,5.
$$
Your root is $omega_0$ above. It is what we call a primitive root, in the sense that 6 is the smallest positive integer $r$ such that $omega_0^r=-1$.






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    $$sqrt{3}/2+i/2=cos(pi/6)+isin(pi/6)=e^{ipi/6}"="(e^{ipi})^{1/6}=(-1)^{1/6}.$$



    The second equality is Euler's theorem. The third equals sign is in quotes is because technically there are $6$ values of $(e^{ipi})^{1/6}$, and $e^{ipi/6}$ happens to be one of them.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      We simply have that



      $$left(frac{sqrt3}2+frac i 2right)^6=(e^{ipi/6})^6=e^{ipi}=-1$$



      and since $frac{sqrt3}2+frac i 2$ is the principal root of $z^6+1=0$ for that reason it is designed as $sqrt[6]{-1}$.



      Refer also to:




      • Principal $n$th root of a complex number






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        This demonstrates that $sqrt{3}/2+i/2$ is a sixth-root of $-1$, but it doesn't explain why the value deserves the designation "$sqrt[6]{-1}$".
        $endgroup$
        – Blue
        Dec 13 '18 at 22:44












      • $begingroup$
        That's the principal root! Ok I add something more on that.
        $endgroup$
        – gimusi
        Dec 13 '18 at 22:46











      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      That's a terrible notation by Wolfram. There are six sixth roots of $-1$, so you cannot tell which one it is when you write $sqrt[6]{-1}$.



      The notation makes sense when we write $sqrt{2}$, say, because we take the usual convention that it is the positive root. But such choice is meaningless for arbitrary roots.



      The sixth roots of $-1$ are, from De Moivre's formula (and writing $-1=cospi+isin pi$),
      $$
      omega_k=cosleft(tfrac{pi+2kpi}{6}right)+isinleft(tfrac{pi+2kpi}{6}right), k=0,ldots,5.
      $$
      Your root is $omega_0$ above. It is what we call a primitive root, in the sense that 6 is the smallest positive integer $r$ such that $omega_0^r=-1$.






      share|cite|improve this answer











      $endgroup$


















        3












        $begingroup$

        That's a terrible notation by Wolfram. There are six sixth roots of $-1$, so you cannot tell which one it is when you write $sqrt[6]{-1}$.



        The notation makes sense when we write $sqrt{2}$, say, because we take the usual convention that it is the positive root. But such choice is meaningless for arbitrary roots.



        The sixth roots of $-1$ are, from De Moivre's formula (and writing $-1=cospi+isin pi$),
        $$
        omega_k=cosleft(tfrac{pi+2kpi}{6}right)+isinleft(tfrac{pi+2kpi}{6}right), k=0,ldots,5.
        $$
        Your root is $omega_0$ above. It is what we call a primitive root, in the sense that 6 is the smallest positive integer $r$ such that $omega_0^r=-1$.






        share|cite|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$

          That's a terrible notation by Wolfram. There are six sixth roots of $-1$, so you cannot tell which one it is when you write $sqrt[6]{-1}$.



          The notation makes sense when we write $sqrt{2}$, say, because we take the usual convention that it is the positive root. But such choice is meaningless for arbitrary roots.



          The sixth roots of $-1$ are, from De Moivre's formula (and writing $-1=cospi+isin pi$),
          $$
          omega_k=cosleft(tfrac{pi+2kpi}{6}right)+isinleft(tfrac{pi+2kpi}{6}right), k=0,ldots,5.
          $$
          Your root is $omega_0$ above. It is what we call a primitive root, in the sense that 6 is the smallest positive integer $r$ such that $omega_0^r=-1$.






          share|cite|improve this answer











          $endgroup$



          That's a terrible notation by Wolfram. There are six sixth roots of $-1$, so you cannot tell which one it is when you write $sqrt[6]{-1}$.



          The notation makes sense when we write $sqrt{2}$, say, because we take the usual convention that it is the positive root. But such choice is meaningless for arbitrary roots.



          The sixth roots of $-1$ are, from De Moivre's formula (and writing $-1=cospi+isin pi$),
          $$
          omega_k=cosleft(tfrac{pi+2kpi}{6}right)+isinleft(tfrac{pi+2kpi}{6}right), k=0,ldots,5.
          $$
          Your root is $omega_0$ above. It is what we call a primitive root, in the sense that 6 is the smallest positive integer $r$ such that $omega_0^r=-1$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 14 '18 at 12:06









          Scientifica

          6,79641335




          6,79641335










          answered Dec 13 '18 at 22:41









          Martin ArgeramiMartin Argerami

          128k1182183




          128k1182183























              2












              $begingroup$

              $$sqrt{3}/2+i/2=cos(pi/6)+isin(pi/6)=e^{ipi/6}"="(e^{ipi})^{1/6}=(-1)^{1/6}.$$



              The second equality is Euler's theorem. The third equals sign is in quotes is because technically there are $6$ values of $(e^{ipi})^{1/6}$, and $e^{ipi/6}$ happens to be one of them.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                $$sqrt{3}/2+i/2=cos(pi/6)+isin(pi/6)=e^{ipi/6}"="(e^{ipi})^{1/6}=(-1)^{1/6}.$$



                The second equality is Euler's theorem. The third equals sign is in quotes is because technically there are $6$ values of $(e^{ipi})^{1/6}$, and $e^{ipi/6}$ happens to be one of them.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  $$sqrt{3}/2+i/2=cos(pi/6)+isin(pi/6)=e^{ipi/6}"="(e^{ipi})^{1/6}=(-1)^{1/6}.$$



                  The second equality is Euler's theorem. The third equals sign is in quotes is because technically there are $6$ values of $(e^{ipi})^{1/6}$, and $e^{ipi/6}$ happens to be one of them.






                  share|cite|improve this answer









                  $endgroup$



                  $$sqrt{3}/2+i/2=cos(pi/6)+isin(pi/6)=e^{ipi/6}"="(e^{ipi})^{1/6}=(-1)^{1/6}.$$



                  The second equality is Euler's theorem. The third equals sign is in quotes is because technically there are $6$ values of $(e^{ipi})^{1/6}$, and $e^{ipi/6}$ happens to be one of them.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 13 '18 at 22:39









                  Mike EarnestMike Earnest

                  23.9k12051




                  23.9k12051























                      0












                      $begingroup$

                      We simply have that



                      $$left(frac{sqrt3}2+frac i 2right)^6=(e^{ipi/6})^6=e^{ipi}=-1$$



                      and since $frac{sqrt3}2+frac i 2$ is the principal root of $z^6+1=0$ for that reason it is designed as $sqrt[6]{-1}$.



                      Refer also to:




                      • Principal $n$th root of a complex number






                      share|cite|improve this answer











                      $endgroup$









                      • 1




                        $begingroup$
                        This demonstrates that $sqrt{3}/2+i/2$ is a sixth-root of $-1$, but it doesn't explain why the value deserves the designation "$sqrt[6]{-1}$".
                        $endgroup$
                        – Blue
                        Dec 13 '18 at 22:44












                      • $begingroup$
                        That's the principal root! Ok I add something more on that.
                        $endgroup$
                        – gimusi
                        Dec 13 '18 at 22:46
















                      0












                      $begingroup$

                      We simply have that



                      $$left(frac{sqrt3}2+frac i 2right)^6=(e^{ipi/6})^6=e^{ipi}=-1$$



                      and since $frac{sqrt3}2+frac i 2$ is the principal root of $z^6+1=0$ for that reason it is designed as $sqrt[6]{-1}$.



                      Refer also to:




                      • Principal $n$th root of a complex number






                      share|cite|improve this answer











                      $endgroup$









                      • 1




                        $begingroup$
                        This demonstrates that $sqrt{3}/2+i/2$ is a sixth-root of $-1$, but it doesn't explain why the value deserves the designation "$sqrt[6]{-1}$".
                        $endgroup$
                        – Blue
                        Dec 13 '18 at 22:44












                      • $begingroup$
                        That's the principal root! Ok I add something more on that.
                        $endgroup$
                        – gimusi
                        Dec 13 '18 at 22:46














                      0












                      0








                      0





                      $begingroup$

                      We simply have that



                      $$left(frac{sqrt3}2+frac i 2right)^6=(e^{ipi/6})^6=e^{ipi}=-1$$



                      and since $frac{sqrt3}2+frac i 2$ is the principal root of $z^6+1=0$ for that reason it is designed as $sqrt[6]{-1}$.



                      Refer also to:




                      • Principal $n$th root of a complex number






                      share|cite|improve this answer











                      $endgroup$



                      We simply have that



                      $$left(frac{sqrt3}2+frac i 2right)^6=(e^{ipi/6})^6=e^{ipi}=-1$$



                      and since $frac{sqrt3}2+frac i 2$ is the principal root of $z^6+1=0$ for that reason it is designed as $sqrt[6]{-1}$.



                      Refer also to:




                      • Principal $n$th root of a complex number







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 13 '18 at 22:49

























                      answered Dec 13 '18 at 22:37









                      gimusigimusi

                      92.9k84494




                      92.9k84494








                      • 1




                        $begingroup$
                        This demonstrates that $sqrt{3}/2+i/2$ is a sixth-root of $-1$, but it doesn't explain why the value deserves the designation "$sqrt[6]{-1}$".
                        $endgroup$
                        – Blue
                        Dec 13 '18 at 22:44












                      • $begingroup$
                        That's the principal root! Ok I add something more on that.
                        $endgroup$
                        – gimusi
                        Dec 13 '18 at 22:46














                      • 1




                        $begingroup$
                        This demonstrates that $sqrt{3}/2+i/2$ is a sixth-root of $-1$, but it doesn't explain why the value deserves the designation "$sqrt[6]{-1}$".
                        $endgroup$
                        – Blue
                        Dec 13 '18 at 22:44












                      • $begingroup$
                        That's the principal root! Ok I add something more on that.
                        $endgroup$
                        – gimusi
                        Dec 13 '18 at 22:46








                      1




                      1




                      $begingroup$
                      This demonstrates that $sqrt{3}/2+i/2$ is a sixth-root of $-1$, but it doesn't explain why the value deserves the designation "$sqrt[6]{-1}$".
                      $endgroup$
                      – Blue
                      Dec 13 '18 at 22:44






                      $begingroup$
                      This demonstrates that $sqrt{3}/2+i/2$ is a sixth-root of $-1$, but it doesn't explain why the value deserves the designation "$sqrt[6]{-1}$".
                      $endgroup$
                      – Blue
                      Dec 13 '18 at 22:44














                      $begingroup$
                      That's the principal root! Ok I add something more on that.
                      $endgroup$
                      – gimusi
                      Dec 13 '18 at 22:46




                      $begingroup$
                      That's the principal root! Ok I add something more on that.
                      $endgroup$
                      – gimusi
                      Dec 13 '18 at 22:46


















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