Simplify expression $e^ipi/3(sqrt2e^ipi /4 -1)$

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Problem



Simplify expression $e^ipi/3(sqrt2e^ipi /4 -1)$



Attempt to solve



$$= sqrt2e^i 7pi/12-e^i pi / 3 = e^ifrac56pi$$



I can't seem to find how this expression is $e^ifrac56pi$ ?



I would like the final result be in polar form $re^itheta$










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    Problem



    Simplify expression $e^ipi/3(sqrt2e^ipi /4 -1)$



    Attempt to solve



    $$= sqrt2e^i 7pi/12-e^i pi / 3 = e^ifrac56pi$$



    I can't seem to find how this expression is $e^ifrac56pi$ ?



    I would like the final result be in polar form $re^itheta$










    share|cite|improve this question























      up vote
      0
      down vote

      favorite
      1









      up vote
      0
      down vote

      favorite
      1






      1





      Problem



      Simplify expression $e^ipi/3(sqrt2e^ipi /4 -1)$



      Attempt to solve



      $$= sqrt2e^i 7pi/12-e^i pi / 3 = e^ifrac56pi$$



      I can't seem to find how this expression is $e^ifrac56pi$ ?



      I would like the final result be in polar form $re^itheta$










      share|cite|improve this question













      Problem



      Simplify expression $e^ipi/3(sqrt2e^ipi /4 -1)$



      Attempt to solve



      $$= sqrt2e^i 7pi/12-e^i pi / 3 = e^ifrac56pi$$



      I can't seem to find how this expression is $e^ifrac56pi$ ?



      I would like the final result be in polar form $re^itheta$







      complex-numbers






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      asked Sep 2 at 9:55









      Tuki

      637316




      637316




















          2 Answers
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          It might be slightly simpler to notice that
          $$ sqrt2e^ipi/4-1 = sqrt2left(frac1sqrt2+fracisqrt2right)-1=i=e^ipi/2, $$
          so
          $$ e^ipi/3(sqrt2e^ipi/4-1)=e^ipi/3e^ipi/2=e^i5pi/6. $$






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            One may write trigonometric form
            beginalign
            e^ifracpi3(sqrt2e^ifracpi4 -1)
            &= (cosdfracpi3+isindfracpi3)left(sqrt2(cosdfracpi4+isindfracpi4)-1right) \
            &= left(dfrac12+idfracsqrt32right)left(1+i-1right) \
            &= dfrac-sqrt32+idfrac12 \
            &= e^frac56pi i
            endalign






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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              3
              down vote



              accepted










              It might be slightly simpler to notice that
              $$ sqrt2e^ipi/4-1 = sqrt2left(frac1sqrt2+fracisqrt2right)-1=i=e^ipi/2, $$
              so
              $$ e^ipi/3(sqrt2e^ipi/4-1)=e^ipi/3e^ipi/2=e^i5pi/6. $$






              share|cite|improve this answer
























                up vote
                3
                down vote



                accepted










                It might be slightly simpler to notice that
                $$ sqrt2e^ipi/4-1 = sqrt2left(frac1sqrt2+fracisqrt2right)-1=i=e^ipi/2, $$
                so
                $$ e^ipi/3(sqrt2e^ipi/4-1)=e^ipi/3e^ipi/2=e^i5pi/6. $$






                share|cite|improve this answer






















                  up vote
                  3
                  down vote



                  accepted







                  up vote
                  3
                  down vote



                  accepted






                  It might be slightly simpler to notice that
                  $$ sqrt2e^ipi/4-1 = sqrt2left(frac1sqrt2+fracisqrt2right)-1=i=e^ipi/2, $$
                  so
                  $$ e^ipi/3(sqrt2e^ipi/4-1)=e^ipi/3e^ipi/2=e^i5pi/6. $$






                  share|cite|improve this answer












                  It might be slightly simpler to notice that
                  $$ sqrt2e^ipi/4-1 = sqrt2left(frac1sqrt2+fracisqrt2right)-1=i=e^ipi/2, $$
                  so
                  $$ e^ipi/3(sqrt2e^ipi/4-1)=e^ipi/3e^ipi/2=e^i5pi/6. $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Sep 2 at 9:59









                  Sobi

                  2,843417




                  2,843417




















                      up vote
                      1
                      down vote













                      One may write trigonometric form
                      beginalign
                      e^ifracpi3(sqrt2e^ifracpi4 -1)
                      &= (cosdfracpi3+isindfracpi3)left(sqrt2(cosdfracpi4+isindfracpi4)-1right) \
                      &= left(dfrac12+idfracsqrt32right)left(1+i-1right) \
                      &= dfrac-sqrt32+idfrac12 \
                      &= e^frac56pi i
                      endalign






                      share|cite|improve this answer
























                        up vote
                        1
                        down vote













                        One may write trigonometric form
                        beginalign
                        e^ifracpi3(sqrt2e^ifracpi4 -1)
                        &= (cosdfracpi3+isindfracpi3)left(sqrt2(cosdfracpi4+isindfracpi4)-1right) \
                        &= left(dfrac12+idfracsqrt32right)left(1+i-1right) \
                        &= dfrac-sqrt32+idfrac12 \
                        &= e^frac56pi i
                        endalign






                        share|cite|improve this answer






















                          up vote
                          1
                          down vote










                          up vote
                          1
                          down vote









                          One may write trigonometric form
                          beginalign
                          e^ifracpi3(sqrt2e^ifracpi4 -1)
                          &= (cosdfracpi3+isindfracpi3)left(sqrt2(cosdfracpi4+isindfracpi4)-1right) \
                          &= left(dfrac12+idfracsqrt32right)left(1+i-1right) \
                          &= dfrac-sqrt32+idfrac12 \
                          &= e^frac56pi i
                          endalign






                          share|cite|improve this answer












                          One may write trigonometric form
                          beginalign
                          e^ifracpi3(sqrt2e^ifracpi4 -1)
                          &= (cosdfracpi3+isindfracpi3)left(sqrt2(cosdfracpi4+isindfracpi4)-1right) \
                          &= left(dfrac12+idfracsqrt32right)left(1+i-1right) \
                          &= dfrac-sqrt32+idfrac12 \
                          &= e^frac56pi i
                          endalign







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Sep 2 at 10:15









                          Nosrati

                          22.1k61747




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