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







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














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






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

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





















          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3053170%2fsimplification-of-complex-numbers%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          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






























                                          draft saved

                                          draft discarded




















































                                          Thanks for contributing an answer to Mathematics Stack Exchange!


                                          • Please be sure to answer the question. Provide details and share your research!

                                          But avoid …



                                          • Asking for help, clarification, or responding to other answers.

                                          • Making statements based on opinion; back them up with references or personal experience.


                                          Use MathJax to format equations. MathJax reference.


                                          To learn more, see our tips on writing great answers.





                                          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                                          Please pay close attention to the following guidance:


                                          • Please be sure to answer the question. Provide details and share your research!

                                          But avoid …



                                          • Asking for help, clarification, or responding to other answers.

                                          • Making statements based on opinion; back them up with references or personal experience.


                                          To learn more, see our tips on writing great answers.




                                          draft saved


                                          draft discarded














                                          StackExchange.ready(
                                          function () {
                                          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3053170%2fsimplification-of-complex-numbers%23new-answer', 'question_page');
                                          }
                                          );

                                          Post as a guest















                                          Required, but never shown





















































                                          Required, but never shown














                                          Required, but never shown












                                          Required, but never shown







                                          Required, but never shown

































                                          Required, but never shown














                                          Required, but never shown












                                          Required, but never shown







                                          Required, but never shown







                                          -

                                          Popular posts from this blog

                                          Human spaceflight

                                          Image
                                          Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e
                                          Read more

                                          Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

                                          Image
                                          0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB]
                                          Read more

                                          File:DeusFollowingSea.jpg

                                          Image
                                          File:DeusFollowingSea.jpg From Wikipedia, the free encyclopedia Jump to navigation Jump to search File File history File usage No higher resolution available. DeusFollowingSea.jpg ‎ (238 × 211 pixels, file size: 12 KB, MIME type: image/jpeg ) Summary [ edit ] Media data and Non-free use rationale Description Cover art of the Deus CD Following Sea Author or copyright owner Deus Source (WP:NFCC#4) rocksucker.co.uk Date of publication 2012 Use in article (WP:NFCC#7) Following Sea Purpose of use in article (WP:NFCC#8) to serve as the primary means of visual identification at the top of the article dedicated to the work in question. Not replaceable with free media because (WP:NFCC#1) n.a. Minimal use (WP:NFCC#3) Low res image used in infobox of the article on the album Respect for commercial opportunities (WP:NFCC#2) n.a. Fair use Fair use of co
                                          Read more