simplification of complex numbers












2












$begingroup$


How does $$2πi(1/e^{5πi/6} +1/e^{3πi/6}+1/e^{πi/6})$$ reduce to π/3.
i tried using exponential form but for some reason this does not give me the answer. what approaches could you take to simplifying this?






complex-numbers







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$endgroup$








  • 1




    $begingroup$
    You might start trying to simplify $e^{-5πi/6} +e^{-3πi/6}+e^{-πi/6}$
    $endgroup$
    – Henry
    Dec 26 '18 at 18:18










  • $begingroup$
    @Henry ... And think geometrically. Draw them if you have to
    $endgroup$
    – Arthur
    Dec 26 '18 at 19:21
















2












$begingroup$


How does $$2πi(1/e^{5πi/6} +1/e^{3πi/6}+1/e^{πi/6})$$ reduce to π/3.
i tried using exponential form but for some reason this does not give me the answer. what approaches could you take to simplifying this?






complex-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You might start trying to simplify $e^{-5πi/6} +e^{-3πi/6}+e^{-πi/6}$
    $endgroup$
    – Henry
    Dec 26 '18 at 18:18










  • $begingroup$
    @Henry ... And think geometrically. Draw them if you have to
    $endgroup$
    – Arthur
    Dec 26 '18 at 19:21














2












2








2


1



$begingroup$


How does $$2πi(1/e^{5πi/6} +1/e^{3πi/6}+1/e^{πi/6})$$ reduce to π/3.
i tried using exponential form but for some reason this does not give me the answer. what approaches could you take to simplifying this?






complex-numbers







share|cite|improve this question











$endgroup$




How does $$2πi(1/e^{5πi/6} +1/e^{3πi/6}+1/e^{πi/6})$$ reduce to π/3.
i tried using exponential form but for some reason this does not give me the answer. what approaches could you take to simplifying this?






complex-numbers




complex-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 26 '18 at 18:21









Dr. Sonnhard Graubner

77.7k42866




77.7k42866










asked Dec 26 '18 at 18:15









Vicem0nVicem0n

343




343








  • 1




    $begingroup$
    You might start trying to simplify $e^{-5πi/6} +e^{-3πi/6}+e^{-πi/6}$
    $endgroup$
    – Henry
    Dec 26 '18 at 18:18










  • $begingroup$
    @Henry ... And think geometrically. Draw them if you have to
    $endgroup$
    – Arthur
    Dec 26 '18 at 19:21














  • 1




    $begingroup$
    You might start trying to simplify $e^{-5πi/6} +e^{-3πi/6}+e^{-πi/6}$
    $endgroup$
    – Henry
    Dec 26 '18 at 18:18










  • $begingroup$
    @Henry ... And think geometrically. Draw them if you have to
    $endgroup$
    – Arthur
    Dec 26 '18 at 19:21








1




1




$begingroup$
You might start trying to simplify $e^{-5πi/6} +e^{-3πi/6}+e^{-πi/6}$
$endgroup$
– Henry
Dec 26 '18 at 18:18




$begingroup$
You might start trying to simplify $e^{-5πi/6} +e^{-3πi/6}+e^{-πi/6}$
$endgroup$
– Henry
Dec 26 '18 at 18:18












$begingroup$
@Henry ... And think geometrically. Draw them if you have to
$endgroup$
– Arthur
Dec 26 '18 at 19:21




$begingroup$
@Henry ... And think geometrically. Draw them if you have to
$endgroup$
– Arthur
Dec 26 '18 at 19:21










4 Answers
4






active

oldest

votes


















1












$begingroup$

When adding complex numbers, it is often easier to use rectangular rather than polar form. This is straightforward in this case since the angles we're dealing with are multiples of $pi/6$ and thus have exact rectangular forms. Going off of @Henry's comment, converting to negative exponents will greatly simplify the process. Recall that
$$
e^{itheta} = cos(theta) + isin(theta)
$$



So we have $e^{-5ipi/6} = -frac{sqrt{3}}{2}-ifrac{1}{2}$, $e^{-3ipi/6} = e^{-ipi/2} = -i$, and $e^{-ipi/6} = frac{sqrt{3}}{2}-ifrac{1}{2}$.



Then the expression is
$$
2pi i(-2i) = 4pi
$$



So in fact the expression does not simplify to $pi/3$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    $$
    begin{align}
    2πileft(1/e^{5πi/6}+1/e^{3πi/6}+1/e^{πi/6}right)
    &=2πi,color{#C00}{e^{-3πi/6}}left(color{#090}{e^{-2πi/6}}+1+color{#090}{e^{2πi/6}}right)\
    &=2πi,color{#C00}{frac1i}left(1+color{#090}{2cosleft(fracpi3right)}right)\
    &=4pi
    end{align}
    $$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Note that $$e^{-frac{5pi i}{6}}=-frac{sqrt{3}}{2}-frac{i}{2}$$
      $$e^{frac{3pi i}{6}}=-i$$ and $$e^{frac{pi i}{6}}=frac{sqrt{3}}{2}-frac{i}{2}$$






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        $$e^{-pi i/6}sum_{r=0}^2(e^{-pi i/3})^r=dfrac{-1-1}{e^{-pi i/6}-e^{pi i/6}}=dfrac2{2isindfracpi6}=?$$






        share|cite|improve this answer









        $endgroup$













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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          When adding complex numbers, it is often easier to use rectangular rather than polar form. This is straightforward in this case since the angles we're dealing with are multiples of $pi/6$ and thus have exact rectangular forms. Going off of @Henry's comment, converting to negative exponents will greatly simplify the process. Recall that
          $$
          e^{itheta} = cos(theta) + isin(theta)
          $$



          So we have $e^{-5ipi/6} = -frac{sqrt{3}}{2}-ifrac{1}{2}$, $e^{-3ipi/6} = e^{-ipi/2} = -i$, and $e^{-ipi/6} = frac{sqrt{3}}{2}-ifrac{1}{2}$.



          Then the expression is
          $$
          2pi i(-2i) = 4pi
          $$



          So in fact the expression does not simplify to $pi/3$.






          share|cite|improve this answer









          $endgroup$


















            1












            $begingroup$

            When adding complex numbers, it is often easier to use rectangular rather than polar form. This is straightforward in this case since the angles we're dealing with are multiples of $pi/6$ and thus have exact rectangular forms. Going off of @Henry's comment, converting to negative exponents will greatly simplify the process. Recall that
            $$
            e^{itheta} = cos(theta) + isin(theta)
            $$



            So we have $e^{-5ipi/6} = -frac{sqrt{3}}{2}-ifrac{1}{2}$, $e^{-3ipi/6} = e^{-ipi/2} = -i$, and $e^{-ipi/6} = frac{sqrt{3}}{2}-ifrac{1}{2}$.



            Then the expression is
            $$
            2pi i(-2i) = 4pi
            $$



            So in fact the expression does not simplify to $pi/3$.






            share|cite|improve this answer









            $endgroup$
















              1












              1








              1





              $begingroup$

              When adding complex numbers, it is often easier to use rectangular rather than polar form. This is straightforward in this case since the angles we're dealing with are multiples of $pi/6$ and thus have exact rectangular forms. Going off of @Henry's comment, converting to negative exponents will greatly simplify the process. Recall that
              $$
              e^{itheta} = cos(theta) + isin(theta)
              $$



              So we have $e^{-5ipi/6} = -frac{sqrt{3}}{2}-ifrac{1}{2}$, $e^{-3ipi/6} = e^{-ipi/2} = -i$, and $e^{-ipi/6} = frac{sqrt{3}}{2}-ifrac{1}{2}$.



              Then the expression is
              $$
              2pi i(-2i) = 4pi
              $$



              So in fact the expression does not simplify to $pi/3$.






              share|cite|improve this answer









              $endgroup$



              When adding complex numbers, it is often easier to use rectangular rather than polar form. This is straightforward in this case since the angles we're dealing with are multiples of $pi/6$ and thus have exact rectangular forms. Going off of @Henry's comment, converting to negative exponents will greatly simplify the process. Recall that
              $$
              e^{itheta} = cos(theta) + isin(theta)
              $$



              So we have $e^{-5ipi/6} = -frac{sqrt{3}}{2}-ifrac{1}{2}$, $e^{-3ipi/6} = e^{-ipi/2} = -i$, and $e^{-ipi/6} = frac{sqrt{3}}{2}-ifrac{1}{2}$.



              Then the expression is
              $$
              2pi i(-2i) = 4pi
              $$



              So in fact the expression does not simplify to $pi/3$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 26 '18 at 18:34









              AlexAlex

              52138




              52138























                  1












                  $begingroup$

                  $$
                  begin{align}
                  2πileft(1/e^{5πi/6}+1/e^{3πi/6}+1/e^{πi/6}right)
                  &=2πi,color{#C00}{e^{-3πi/6}}left(color{#090}{e^{-2πi/6}}+1+color{#090}{e^{2πi/6}}right)\
                  &=2πi,color{#C00}{frac1i}left(1+color{#090}{2cosleft(fracpi3right)}right)\
                  &=4pi
                  end{align}
                  $$






                  share|cite|improve this answer









                  $endgroup$


















                    1












                    $begingroup$

                    $$
                    begin{align}
                    2πileft(1/e^{5πi/6}+1/e^{3πi/6}+1/e^{πi/6}right)
                    &=2πi,color{#C00}{e^{-3πi/6}}left(color{#090}{e^{-2πi/6}}+1+color{#090}{e^{2πi/6}}right)\
                    &=2πi,color{#C00}{frac1i}left(1+color{#090}{2cosleft(fracpi3right)}right)\
                    &=4pi
                    end{align}
                    $$






                    share|cite|improve this answer









                    $endgroup$
















                      1












                      1








                      1





                      $begingroup$

                      $$
                      begin{align}
                      2πileft(1/e^{5πi/6}+1/e^{3πi/6}+1/e^{πi/6}right)
                      &=2πi,color{#C00}{e^{-3πi/6}}left(color{#090}{e^{-2πi/6}}+1+color{#090}{e^{2πi/6}}right)\
                      &=2πi,color{#C00}{frac1i}left(1+color{#090}{2cosleft(fracpi3right)}right)\
                      &=4pi
                      end{align}
                      $$






                      share|cite|improve this answer









                      $endgroup$



                      $$
                      begin{align}
                      2πileft(1/e^{5πi/6}+1/e^{3πi/6}+1/e^{πi/6}right)
                      &=2πi,color{#C00}{e^{-3πi/6}}left(color{#090}{e^{-2πi/6}}+1+color{#090}{e^{2πi/6}}right)\
                      &=2πi,color{#C00}{frac1i}left(1+color{#090}{2cosleft(fracpi3right)}right)\
                      &=4pi
                      end{align}
                      $$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 26 '18 at 21:23









                      robjohn♦robjohn

                      269k27311638




                      269k27311638























                          0












                          $begingroup$

                          Note that $$e^{-frac{5pi i}{6}}=-frac{sqrt{3}}{2}-frac{i}{2}$$
                          $$e^{frac{3pi i}{6}}=-i$$ and $$e^{frac{pi i}{6}}=frac{sqrt{3}}{2}-frac{i}{2}$$






                          share|cite|improve this answer









                          $endgroup$


















                            0












                            $begingroup$

                            Note that $$e^{-frac{5pi i}{6}}=-frac{sqrt{3}}{2}-frac{i}{2}$$
                            $$e^{frac{3pi i}{6}}=-i$$ and $$e^{frac{pi i}{6}}=frac{sqrt{3}}{2}-frac{i}{2}$$






                            share|cite|improve this answer









                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              Note that $$e^{-frac{5pi i}{6}}=-frac{sqrt{3}}{2}-frac{i}{2}$$
                              $$e^{frac{3pi i}{6}}=-i$$ and $$e^{frac{pi i}{6}}=frac{sqrt{3}}{2}-frac{i}{2}$$






                              share|cite|improve this answer









                              $endgroup$



                              Note that $$e^{-frac{5pi i}{6}}=-frac{sqrt{3}}{2}-frac{i}{2}$$
                              $$e^{frac{3pi i}{6}}=-i$$ and $$e^{frac{pi i}{6}}=frac{sqrt{3}}{2}-frac{i}{2}$$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Dec 26 '18 at 18:25









                              Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                              77.7k42866




                              77.7k42866























                                  0












                                  $begingroup$

                                  $$e^{-pi i/6}sum_{r=0}^2(e^{-pi i/3})^r=dfrac{-1-1}{e^{-pi i/6}-e^{pi i/6}}=dfrac2{2isindfracpi6}=?$$






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    $$e^{-pi i/6}sum_{r=0}^2(e^{-pi i/3})^r=dfrac{-1-1}{e^{-pi i/6}-e^{pi i/6}}=dfrac2{2isindfracpi6}=?$$






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      $$e^{-pi i/6}sum_{r=0}^2(e^{-pi i/3})^r=dfrac{-1-1}{e^{-pi i/6}-e^{pi i/6}}=dfrac2{2isindfracpi6}=?$$






                                      share|cite|improve this answer









                                      $endgroup$



                                      $$e^{-pi i/6}sum_{r=0}^2(e^{-pi i/3})^r=dfrac{-1-1}{e^{-pi i/6}-e^{pi i/6}}=dfrac2{2isindfracpi6}=?$$







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Dec 26 '18 at 18:25









                                      lab bhattacharjeelab bhattacharjee

                                      226k15158275




                                      226k15158275






























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