simplification of complex numbers












2














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










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
















2














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




















  • 1




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










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














2












2








2


1





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















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






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share|cite|improve this question













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edited Dec 26 '18 at 18:21









Dr. Sonnhard Graubner

73.1k42865




73.1k42865










asked Dec 26 '18 at 18:15









Vicem0n

223




223








  • 1




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










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














  • 1




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










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








1




1




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




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












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




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










4 Answers
4






active

oldest

votes


















1














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





























    1














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






    share|cite|improve this answer





























      0














      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





























        0














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






        share|cite|improve this answer





















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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1














          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


























            1














            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
























              1












              1








              1






              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












              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









              Alex

              1777




              1777























                  1














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






                  share|cite|improve this answer


























                    1














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






                    share|cite|improve this answer
























                      1












                      1








                      1






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






                      share|cite|improve this answer












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







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 26 '18 at 21:23









                      robjohn♦

                      264k27303623




                      264k27303623























                          0














                          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


























                            0














                            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
























                              0












                              0








                              0






                              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












                              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 Graubner

                              73.1k42865




                              73.1k42865























                                  0














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






                                  share|cite|improve this answer


























                                    0














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






                                    share|cite|improve this answer
























                                      0












                                      0








                                      0






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






                                      share|cite|improve this answer












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







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Dec 26 '18 at 18:25









                                      lab bhattacharjee

                                      223k15156274




                                      223k15156274






























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