Complex Numbers in Polar Form raised to a power

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[1 + cos(π/6) + isin(π/6)]^6



For a question like this, the first thing I would think of using is the DeMoivre's Theorem, however, with the entire real part of it being (1+cos(π/6)), I'm not sure if the rule will apply.



How do I approach this?






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    up vote
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    down vote

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    [1 + cos(π/6) + isin(π/6)]^6



    For a question like this, the first thing I would think of using is the DeMoivre's Theorem, however, with the entire real part of it being (1+cos(π/6)), I'm not sure if the rule will apply.



    How do I approach this?






    complex-numbers







    share|cite|improve this question

























      up vote
      3
      down vote

      favorite
      1









      up vote
      3
      down vote

      favorite
      1






      1





      [1 + cos(π/6) + isin(π/6)]^6



      For a question like this, the first thing I would think of using is the DeMoivre's Theorem, however, with the entire real part of it being (1+cos(π/6)), I'm not sure if the rule will apply.



      How do I approach this?






      complex-numbers







      share|cite|improve this question















      [1 + cos(π/6) + isin(π/6)]^6



      For a question like this, the first thing I would think of using is the DeMoivre's Theorem, however, with the entire real part of it being (1+cos(π/6)), I'm not sure if the rule will apply.



      How do I approach this?






      complex-numbers




      complex-numbers






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      share|cite|improve this question













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      edited 1 hour ago

























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          Hint : remark that, for any $theta in mathbbR$, we have
          $$1+e^i theta= e^i 0+e^i theta=e^i theta/2(e^-i theta/2 +e^i theta/2)=2e^i theta/2 cos(theta/2).$$






          share|cite|improve this answer



























            up vote
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            Observe that $cosfracpi6+isinfracpi6$ is the unit vector of argument $pi/6$. It is easy to see that this vector when added to 1 gives a vector of argument $pi/12$ and length $sqrt1^2+1^2-2cosfrac5pi6=sqrt2+sqrt3$.



            Taking this to the sixth power multiplies the argument by six (hence it becomes $fracpi2$) and raises the length to the sixth power (hence $(2+sqrt3)^3=8+3cdot4sqrt3+3cdot6+3sqrt3=26+15sqrt3$). The final result is $(26+15sqrt3)e^ipi/2$ or $(26+15sqrt3)i$.






            share|cite|improve this answer



























              up vote
              0
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              Hint:



              To have lighter notations, set $;u=mathrm e^tfracipi6$, $bar u=mathrm e^-tfracipi6$, and rewrite the sum as
              beginalign
              (1&+u)^6+(1+bar u)^6= \
              2&+6(u+bar u)+15(u^2+bar u^2)+20(u^3+bar u^3)++15(u^4+bar u^4)+6(u^5+bar u^5)+u^6+bar u^6.
              endalign
              You can compute the sums of powers gradually. Note that $u+bar u=sqrt 3$, that $u^3=i$, $bar u^3=-i$, so $u^3+bar u^3=0$, and that $u,bar u=1$.



              Edit:



              We can use the same method (expanding the binomial) to the new formulation of the question:
              $$(1+u)^6=1+6u+15u^2+20u^3+15u^4+6u^5+u^6.$$
              Note in addition to the above that:




              • $u+u^5=2operatornameIm u=i$,


              • $u^2+u^4=2operatornameIm u^2=isqrt 3$,

              so $;(1+u)^6=1+(26+15sqrt 3)i.$






              share|cite|improve this answer






















              • That's not the stated sum!
                – Parcly Taxel
                42 mins ago










              • @ParclyTaxel: Initially, the O.P. posted a link to a screenshot which was the sum of what is the typedformula and its conjugate. I didn't check whether something had changed meanwhile…
                – Bernard
                36 mins ago










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              3 Answers
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              3 Answers
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              up vote
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              Hint : remark that, for any $theta in mathbbR$, we have
              $$1+e^i theta= e^i 0+e^i theta=e^i theta/2(e^-i theta/2 +e^i theta/2)=2e^i theta/2 cos(theta/2).$$






              share|cite|improve this answer
























                up vote
                3
                down vote













                Hint : remark that, for any $theta in mathbbR$, we have
                $$1+e^i theta= e^i 0+e^i theta=e^i theta/2(e^-i theta/2 +e^i theta/2)=2e^i theta/2 cos(theta/2).$$






                share|cite|improve this answer






















                  up vote
                  3
                  down vote










                  up vote
                  3
                  down vote









                  Hint : remark that, for any $theta in mathbbR$, we have
                  $$1+e^i theta= e^i 0+e^i theta=e^i theta/2(e^-i theta/2 +e^i theta/2)=2e^i theta/2 cos(theta/2).$$






                  share|cite|improve this answer












                  Hint : remark that, for any $theta in mathbbR$, we have
                  $$1+e^i theta= e^i 0+e^i theta=e^i theta/2(e^-i theta/2 +e^i theta/2)=2e^i theta/2 cos(theta/2).$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 51 mins ago









                  S. Cho

                  15610




                  15610




















                      up vote
                      3
                      down vote













                      Observe that $cosfracpi6+isinfracpi6$ is the unit vector of argument $pi/6$. It is easy to see that this vector when added to 1 gives a vector of argument $pi/12$ and length $sqrt1^2+1^2-2cosfrac5pi6=sqrt2+sqrt3$.



                      Taking this to the sixth power multiplies the argument by six (hence it becomes $fracpi2$) and raises the length to the sixth power (hence $(2+sqrt3)^3=8+3cdot4sqrt3+3cdot6+3sqrt3=26+15sqrt3$). The final result is $(26+15sqrt3)e^ipi/2$ or $(26+15sqrt3)i$.






                      share|cite|improve this answer
























                        up vote
                        3
                        down vote













                        Observe that $cosfracpi6+isinfracpi6$ is the unit vector of argument $pi/6$. It is easy to see that this vector when added to 1 gives a vector of argument $pi/12$ and length $sqrt1^2+1^2-2cosfrac5pi6=sqrt2+sqrt3$.



                        Taking this to the sixth power multiplies the argument by six (hence it becomes $fracpi2$) and raises the length to the sixth power (hence $(2+sqrt3)^3=8+3cdot4sqrt3+3cdot6+3sqrt3=26+15sqrt3$). The final result is $(26+15sqrt3)e^ipi/2$ or $(26+15sqrt3)i$.






                        share|cite|improve this answer






















                          up vote
                          3
                          down vote










                          up vote
                          3
                          down vote









                          Observe that $cosfracpi6+isinfracpi6$ is the unit vector of argument $pi/6$. It is easy to see that this vector when added to 1 gives a vector of argument $pi/12$ and length $sqrt1^2+1^2-2cosfrac5pi6=sqrt2+sqrt3$.



                          Taking this to the sixth power multiplies the argument by six (hence it becomes $fracpi2$) and raises the length to the sixth power (hence $(2+sqrt3)^3=8+3cdot4sqrt3+3cdot6+3sqrt3=26+15sqrt3$). The final result is $(26+15sqrt3)e^ipi/2$ or $(26+15sqrt3)i$.






                          share|cite|improve this answer












                          Observe that $cosfracpi6+isinfracpi6$ is the unit vector of argument $pi/6$. It is easy to see that this vector when added to 1 gives a vector of argument $pi/12$ and length $sqrt1^2+1^2-2cosfrac5pi6=sqrt2+sqrt3$.



                          Taking this to the sixth power multiplies the argument by six (hence it becomes $fracpi2$) and raises the length to the sixth power (hence $(2+sqrt3)^3=8+3cdot4sqrt3+3cdot6+3sqrt3=26+15sqrt3$). The final result is $(26+15sqrt3)e^ipi/2$ or $(26+15sqrt3)i$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 50 mins ago









                          Parcly Taxel

                          34.3k136890




                          34.3k136890




















                              up vote
                              0
                              down vote













                              Hint:



                              To have lighter notations, set $;u=mathrm e^tfracipi6$, $bar u=mathrm e^-tfracipi6$, and rewrite the sum as
                              beginalign
                              (1&+u)^6+(1+bar u)^6= \
                              2&+6(u+bar u)+15(u^2+bar u^2)+20(u^3+bar u^3)++15(u^4+bar u^4)+6(u^5+bar u^5)+u^6+bar u^6.
                              endalign
                              You can compute the sums of powers gradually. Note that $u+bar u=sqrt 3$, that $u^3=i$, $bar u^3=-i$, so $u^3+bar u^3=0$, and that $u,bar u=1$.



                              Edit:



                              We can use the same method (expanding the binomial) to the new formulation of the question:
                              $$(1+u)^6=1+6u+15u^2+20u^3+15u^4+6u^5+u^6.$$
                              Note in addition to the above that:




                              • $u+u^5=2operatornameIm u=i$,


                              • $u^2+u^4=2operatornameIm u^2=isqrt 3$,

                              so $;(1+u)^6=1+(26+15sqrt 3)i.$






                              share|cite|improve this answer






















                              • That's not the stated sum!
                                – Parcly Taxel
                                42 mins ago










                              • @ParclyTaxel: Initially, the O.P. posted a link to a screenshot which was the sum of what is the typedformula and its conjugate. I didn't check whether something had changed meanwhile…
                                – Bernard
                                36 mins ago














                              up vote
                              0
                              down vote













                              Hint:



                              To have lighter notations, set $;u=mathrm e^tfracipi6$, $bar u=mathrm e^-tfracipi6$, and rewrite the sum as
                              beginalign
                              (1&+u)^6+(1+bar u)^6= \
                              2&+6(u+bar u)+15(u^2+bar u^2)+20(u^3+bar u^3)++15(u^4+bar u^4)+6(u^5+bar u^5)+u^6+bar u^6.
                              endalign
                              You can compute the sums of powers gradually. Note that $u+bar u=sqrt 3$, that $u^3=i$, $bar u^3=-i$, so $u^3+bar u^3=0$, and that $u,bar u=1$.



                              Edit:



                              We can use the same method (expanding the binomial) to the new formulation of the question:
                              $$(1+u)^6=1+6u+15u^2+20u^3+15u^4+6u^5+u^6.$$
                              Note in addition to the above that:




                              • $u+u^5=2operatornameIm u=i$,


                              • $u^2+u^4=2operatornameIm u^2=isqrt 3$,

                              so $;(1+u)^6=1+(26+15sqrt 3)i.$






                              share|cite|improve this answer






















                              • That's not the stated sum!
                                – Parcly Taxel
                                42 mins ago










                              • @ParclyTaxel: Initially, the O.P. posted a link to a screenshot which was the sum of what is the typedformula and its conjugate. I didn't check whether something had changed meanwhile…
                                – Bernard
                                36 mins ago












                              up vote
                              0
                              down vote










                              up vote
                              0
                              down vote









                              Hint:



                              To have lighter notations, set $;u=mathrm e^tfracipi6$, $bar u=mathrm e^-tfracipi6$, and rewrite the sum as
                              beginalign
                              (1&+u)^6+(1+bar u)^6= \
                              2&+6(u+bar u)+15(u^2+bar u^2)+20(u^3+bar u^3)++15(u^4+bar u^4)+6(u^5+bar u^5)+u^6+bar u^6.
                              endalign
                              You can compute the sums of powers gradually. Note that $u+bar u=sqrt 3$, that $u^3=i$, $bar u^3=-i$, so $u^3+bar u^3=0$, and that $u,bar u=1$.



                              Edit:



                              We can use the same method (expanding the binomial) to the new formulation of the question:
                              $$(1+u)^6=1+6u+15u^2+20u^3+15u^4+6u^5+u^6.$$
                              Note in addition to the above that:




                              • $u+u^5=2operatornameIm u=i$,


                              • $u^2+u^4=2operatornameIm u^2=isqrt 3$,

                              so $;(1+u)^6=1+(26+15sqrt 3)i.$






                              share|cite|improve this answer














                              Hint:



                              To have lighter notations, set $;u=mathrm e^tfracipi6$, $bar u=mathrm e^-tfracipi6$, and rewrite the sum as
                              beginalign
                              (1&+u)^6+(1+bar u)^6= \
                              2&+6(u+bar u)+15(u^2+bar u^2)+20(u^3+bar u^3)++15(u^4+bar u^4)+6(u^5+bar u^5)+u^6+bar u^6.
                              endalign
                              You can compute the sums of powers gradually. Note that $u+bar u=sqrt 3$, that $u^3=i$, $bar u^3=-i$, so $u^3+bar u^3=0$, and that $u,bar u=1$.



                              Edit:



                              We can use the same method (expanding the binomial) to the new formulation of the question:
                              $$(1+u)^6=1+6u+15u^2+20u^3+15u^4+6u^5+u^6.$$
                              Note in addition to the above that:




                              • $u+u^5=2operatornameIm u=i$,


                              • $u^2+u^4=2operatornameIm u^2=isqrt 3$,

                              so $;(1+u)^6=1+(26+15sqrt 3)i.$







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited 18 mins ago

























                              answered 48 mins ago









                              Bernard

                              113k636104




                              113k636104











                              • That's not the stated sum!
                                – Parcly Taxel
                                42 mins ago










                              • @ParclyTaxel: Initially, the O.P. posted a link to a screenshot which was the sum of what is the typedformula and its conjugate. I didn't check whether something had changed meanwhile…
                                – Bernard
                                36 mins ago
















                              • That's not the stated sum!
                                – Parcly Taxel
                                42 mins ago










                              • @ParclyTaxel: Initially, the O.P. posted a link to a screenshot which was the sum of what is the typedformula and its conjugate. I didn't check whether something had changed meanwhile…
                                – Bernard
                                36 mins ago















                              That's not the stated sum!
                              – Parcly Taxel
                              42 mins ago




                              That's not the stated sum!
                              – Parcly Taxel
                              42 mins ago












                              @ParclyTaxel: Initially, the O.P. posted a link to a screenshot which was the sum of what is the typedformula and its conjugate. I didn't check whether something had changed meanwhile…
                              – Bernard
                              36 mins ago




                              @ParclyTaxel: Initially, the O.P. posted a link to a screenshot which was the sum of what is the typedformula and its conjugate. I didn't check whether something had changed meanwhile…
                              – Bernard
                              36 mins ago

















                               

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