$[1+(frac1+i2)][1+(frac1+i2)^2][1+(frac1+i2)^2^2]…[1+(frac1+i2)^2^n]=(1-frac12^2^n)(1+i)$,where $nge 2$

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Show that $$!!!left[1+left(frac1+i2right)right]!!!left[1+left(frac1+i2right)^2right]!!!left[1+left(frac1+i2right)^2^2right]cdotsleft[1+left(frac1+i2right)^2^nright]!!!=left(1-frac12^2^nright)(1+i)$$ for $nge 2$.




I took $frac1+i2=frac1sqrt2e^ifracpi4$ and tried solving but i could not reach the RHS.Please help.




algebra-precalculus complex-numbers




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  • Thank you,all three answers are helpful.
    – learner_avid
    Jul 30 at 8:22














up vote
3
down vote

favorite
1













Show that $$!!!left[1+left(frac1+i2right)right]!!!left[1+left(frac1+i2right)^2right]!!!left[1+left(frac1+i2right)^2^2right]cdotsleft[1+left(frac1+i2right)^2^nright]!!!=left(1-frac12^2^nright)(1+i)$$ for $nge 2$.




I took $frac1+i2=frac1sqrt2e^ifracpi4$ and tried solving but i could not reach the RHS.Please help.




algebra-precalculus complex-numbers




share|cite|improve this question





















  • Thank you,all three answers are helpful.
    – learner_avid
    Jul 30 at 8:22












up vote
3
down vote

favorite
1









up vote
3
down vote

favorite
1






1






Show that $$!!!left[1+left(frac1+i2right)right]!!!left[1+left(frac1+i2right)^2right]!!!left[1+left(frac1+i2right)^2^2right]cdotsleft[1+left(frac1+i2right)^2^nright]!!!=left(1-frac12^2^nright)(1+i)$$ for $nge 2$.




I took $frac1+i2=frac1sqrt2e^ifracpi4$ and tried solving but i could not reach the RHS.Please help.




algebra-precalculus complex-numbers




share|cite|improve this question














Show that $$!!!left[1+left(frac1+i2right)right]!!!left[1+left(frac1+i2right)^2right]!!!left[1+left(frac1+i2right)^2^2right]cdotsleft[1+left(frac1+i2right)^2^nright]!!!=left(1-frac12^2^nright)(1+i)$$ for $nge 2$.




I took $frac1+i2=frac1sqrt2e^ifracpi4$ and tried solving but i could not reach the RHS.Please help.





algebra-precalculus complex-numbers





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 30 at 6:05









Math Lover

12.3k21232




12.3k21232









asked Jul 30 at 4:57









learner_avid

687213




687213











  • Thank you,all three answers are helpful.
    – learner_avid
    Jul 30 at 8:22
















  • Thank you,all three answers are helpful.
    – learner_avid
    Jul 30 at 8:22















Thank you,all three answers are helpful.
– learner_avid
Jul 30 at 8:22




Thank you,all three answers are helpful.
– learner_avid
Jul 30 at 8:22










4 Answers
4






active

oldest

votes

















up vote
2
down vote



accepted










Hints:



  • $(1-a) cdot (1+a) = 1 - a^2,$, $,(1-a) cdot (1+a)(1+a^2) = 1 - a^4,$, $;ldots$


  • $(1+i)^2 = 2i$


  • $1 - dfrac1+i2=dfrac1-i2 = dfrac11+i$






share|cite|improve this answer




























    up vote
    2
    down vote













    Notice that, for any $x neq 1,$
    beginalign*
    prod_k=0^n(1+x^2^k)&=(1+x)(1+x^2)(1+x^2^2)cdots(1+x^2^n)\&=frac(1-x)(1+x)(1+x^2)(1+x^2^2)cdots(1+x^2^n)1-x\&=frac(1-x^2)(1+x^2)(1+x^2^2)cdots(1+x^2^n)1-x\&vdots\&=frac1-x^2^n+11-x.
    endalign*



    Thus, beginalign*prod_k=0^nleft[1+left(frac1+i2right)^2^kright]&=frac1-left(dfrac1+i2right)^2^n+11-dfrac1+i2=frac1-left[left(dfrac1+i2right)^2right]^2^ndfrac11+i\&=left[1-left(fraci2right)^2^nright](1+i)\&=left(1-frac12^2^nright)(1+i).endalign*






    share|cite|improve this answer






























      up vote
      1
      down vote













      Hint: Use mathematical induction. Also, for $n > 2$, $$left(frac1+i2right)^2^n = frac12^2^n-1,$$
      and
      $$left(1-frac12^2^n-1right)left(1+frac12^2^n-1right)=1-frac12^2^n.$$






      share|cite|improve this answer




























        up vote
        1
        down vote













        Hint: Use the general case
        $$(1+x)(1+x^2)(1+x^4)cdots(1+x^2^n)=dfrac(1-x^2^n+1)1-x$$






        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








          up vote
          2
          down vote



          accepted










          Hints:



          • $(1-a) cdot (1+a) = 1 - a^2,$, $,(1-a) cdot (1+a)(1+a^2) = 1 - a^4,$, $;ldots$


          • $(1+i)^2 = 2i$


          • $1 - dfrac1+i2=dfrac1-i2 = dfrac11+i$






          share|cite|improve this answer

























            up vote
            2
            down vote



            accepted










            Hints:



            • $(1-a) cdot (1+a) = 1 - a^2,$, $,(1-a) cdot (1+a)(1+a^2) = 1 - a^4,$, $;ldots$


            • $(1+i)^2 = 2i$


            • $1 - dfrac1+i2=dfrac1-i2 = dfrac11+i$






            share|cite|improve this answer























              up vote
              2
              down vote



              accepted







              up vote
              2
              down vote



              accepted






              Hints:



              • $(1-a) cdot (1+a) = 1 - a^2,$, $,(1-a) cdot (1+a)(1+a^2) = 1 - a^4,$, $;ldots$


              • $(1+i)^2 = 2i$


              • $1 - dfrac1+i2=dfrac1-i2 = dfrac11+i$






              share|cite|improve this answer













              Hints:



              • $(1-a) cdot (1+a) = 1 - a^2,$, $,(1-a) cdot (1+a)(1+a^2) = 1 - a^4,$, $;ldots$


              • $(1+i)^2 = 2i$


              • $1 - dfrac1+i2=dfrac1-i2 = dfrac11+i$







              share|cite|improve this answer













              share|cite|improve this answer



              share|cite|improve this answer











              answered Jul 30 at 5:06









              dxiv

              53.8k64796




              53.8k64796




















                  up vote
                  2
                  down vote













                  Notice that, for any $x neq 1,$
                  beginalign*
                  prod_k=0^n(1+x^2^k)&=(1+x)(1+x^2)(1+x^2^2)cdots(1+x^2^n)\&=frac(1-x)(1+x)(1+x^2)(1+x^2^2)cdots(1+x^2^n)1-x\&=frac(1-x^2)(1+x^2)(1+x^2^2)cdots(1+x^2^n)1-x\&vdots\&=frac1-x^2^n+11-x.
                  endalign*



                  Thus, beginalign*prod_k=0^nleft[1+left(frac1+i2right)^2^kright]&=frac1-left(dfrac1+i2right)^2^n+11-dfrac1+i2=frac1-left[left(dfrac1+i2right)^2right]^2^ndfrac11+i\&=left[1-left(fraci2right)^2^nright](1+i)\&=left(1-frac12^2^nright)(1+i).endalign*






                  share|cite|improve this answer



























                    up vote
                    2
                    down vote













                    Notice that, for any $x neq 1,$
                    beginalign*
                    prod_k=0^n(1+x^2^k)&=(1+x)(1+x^2)(1+x^2^2)cdots(1+x^2^n)\&=frac(1-x)(1+x)(1+x^2)(1+x^2^2)cdots(1+x^2^n)1-x\&=frac(1-x^2)(1+x^2)(1+x^2^2)cdots(1+x^2^n)1-x\&vdots\&=frac1-x^2^n+11-x.
                    endalign*



                    Thus, beginalign*prod_k=0^nleft[1+left(frac1+i2right)^2^kright]&=frac1-left(dfrac1+i2right)^2^n+11-dfrac1+i2=frac1-left[left(dfrac1+i2right)^2right]^2^ndfrac11+i\&=left[1-left(fraci2right)^2^nright](1+i)\&=left(1-frac12^2^nright)(1+i).endalign*






                    share|cite|improve this answer

























                      up vote
                      2
                      down vote










                      up vote
                      2
                      down vote









                      Notice that, for any $x neq 1,$
                      beginalign*
                      prod_k=0^n(1+x^2^k)&=(1+x)(1+x^2)(1+x^2^2)cdots(1+x^2^n)\&=frac(1-x)(1+x)(1+x^2)(1+x^2^2)cdots(1+x^2^n)1-x\&=frac(1-x^2)(1+x^2)(1+x^2^2)cdots(1+x^2^n)1-x\&vdots\&=frac1-x^2^n+11-x.
                      endalign*



                      Thus, beginalign*prod_k=0^nleft[1+left(frac1+i2right)^2^kright]&=frac1-left(dfrac1+i2right)^2^n+11-dfrac1+i2=frac1-left[left(dfrac1+i2right)^2right]^2^ndfrac11+i\&=left[1-left(fraci2right)^2^nright](1+i)\&=left(1-frac12^2^nright)(1+i).endalign*






                      share|cite|improve this answer















                      Notice that, for any $x neq 1,$
                      beginalign*
                      prod_k=0^n(1+x^2^k)&=(1+x)(1+x^2)(1+x^2^2)cdots(1+x^2^n)\&=frac(1-x)(1+x)(1+x^2)(1+x^2^2)cdots(1+x^2^n)1-x\&=frac(1-x^2)(1+x^2)(1+x^2^2)cdots(1+x^2^n)1-x\&vdots\&=frac1-x^2^n+11-x.
                      endalign*



                      Thus, beginalign*prod_k=0^nleft[1+left(frac1+i2right)^2^kright]&=frac1-left(dfrac1+i2right)^2^n+11-dfrac1+i2=frac1-left[left(dfrac1+i2right)^2right]^2^ndfrac11+i\&=left[1-left(fraci2right)^2^nright](1+i)\&=left(1-frac12^2^nright)(1+i).endalign*







                      share|cite|improve this answer















                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Jul 30 at 11:30


























                      answered Jul 30 at 9:06









                      mengdie1982

                      2,827216




                      2,827216




















                          up vote
                          1
                          down vote













                          Hint: Use mathematical induction. Also, for $n > 2$, $$left(frac1+i2right)^2^n = frac12^2^n-1,$$
                          and
                          $$left(1-frac12^2^n-1right)left(1+frac12^2^n-1right)=1-frac12^2^n.$$






                          share|cite|improve this answer

























                            up vote
                            1
                            down vote













                            Hint: Use mathematical induction. Also, for $n > 2$, $$left(frac1+i2right)^2^n = frac12^2^n-1,$$
                            and
                            $$left(1-frac12^2^n-1right)left(1+frac12^2^n-1right)=1-frac12^2^n.$$






                            share|cite|improve this answer























                              up vote
                              1
                              down vote










                              up vote
                              1
                              down vote









                              Hint: Use mathematical induction. Also, for $n > 2$, $$left(frac1+i2right)^2^n = frac12^2^n-1,$$
                              and
                              $$left(1-frac12^2^n-1right)left(1+frac12^2^n-1right)=1-frac12^2^n.$$






                              share|cite|improve this answer













                              Hint: Use mathematical induction. Also, for $n > 2$, $$left(frac1+i2right)^2^n = frac12^2^n-1,$$
                              and
                              $$left(1-frac12^2^n-1right)left(1+frac12^2^n-1right)=1-frac12^2^n.$$







                              share|cite|improve this answer













                              share|cite|improve this answer



                              share|cite|improve this answer











                              answered Jul 30 at 5:12









                              Math Lover

                              12.3k21232




                              12.3k21232




















                                  up vote
                                  1
                                  down vote













                                  Hint: Use the general case
                                  $$(1+x)(1+x^2)(1+x^4)cdots(1+x^2^n)=dfrac(1-x^2^n+1)1-x$$






                                  share|cite|improve this answer



























                                    up vote
                                    1
                                    down vote













                                    Hint: Use the general case
                                    $$(1+x)(1+x^2)(1+x^4)cdots(1+x^2^n)=dfrac(1-x^2^n+1)1-x$$






                                    share|cite|improve this answer

























                                      up vote
                                      1
                                      down vote










                                      up vote
                                      1
                                      down vote









                                      Hint: Use the general case
                                      $$(1+x)(1+x^2)(1+x^4)cdots(1+x^2^n)=dfrac(1-x^2^n+1)1-x$$






                                      share|cite|improve this answer















                                      Hint: Use the general case
                                      $$(1+x)(1+x^2)(1+x^4)cdots(1+x^2^n)=dfrac(1-x^2^n+1)1-x$$







                                      share|cite|improve this answer















                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited Jul 30 at 15:53









                                      Math Lover

                                      12.3k21232




                                      12.3k21232











                                      answered Jul 30 at 5:08









                                      user 108128

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