calculating complex numbers $ (1+isqrt{3})^{2017} + (1-isqrt{3})^{2017} $












0












$begingroup$


I have the following
$ (1+isqrt{3})^{2017} + (1-isqrt{3})^{2017} $
and I have to find the answser.



I think the answer might be $ 2^{2017} $ but I don't know how to find






complex-numbers







share|cite|improve this question









$endgroup$












  • $begingroup$
    Are you familiar with polar notation for complex numbers? i.e. converting from $x + iy$ to $r mbox{cis} theta$ (or $r e^{itheta}$)
    $endgroup$
    – ConMan
    Jan 7 at 22:28


















0












$begingroup$


I have the following
$ (1+isqrt{3})^{2017} + (1-isqrt{3})^{2017} $
and I have to find the answser.



I think the answer might be $ 2^{2017} $ but I don't know how to find






complex-numbers







share|cite|improve this question









$endgroup$












  • $begingroup$
    Are you familiar with polar notation for complex numbers? i.e. converting from $x + iy$ to $r mbox{cis} theta$ (or $r e^{itheta}$)
    $endgroup$
    – ConMan
    Jan 7 at 22:28
















0












0








0


1



$begingroup$


I have the following
$ (1+isqrt{3})^{2017} + (1-isqrt{3})^{2017} $
and I have to find the answser.



I think the answer might be $ 2^{2017} $ but I don't know how to find






complex-numbers







share|cite|improve this question









$endgroup$




I have the following
$ (1+isqrt{3})^{2017} + (1-isqrt{3})^{2017} $
and I have to find the answser.



I think the answer might be $ 2^{2017} $ but I don't know how to find






complex-numbers




complex-numbers






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 7 at 22:25









Student123Student123

536




536












  • $begingroup$
    Are you familiar with polar notation for complex numbers? i.e. converting from $x + iy$ to $r mbox{cis} theta$ (or $r e^{itheta}$)
    $endgroup$
    – ConMan
    Jan 7 at 22:28




















  • $begingroup$
    Are you familiar with polar notation for complex numbers? i.e. converting from $x + iy$ to $r mbox{cis} theta$ (or $r e^{itheta}$)
    $endgroup$
    – ConMan
    Jan 7 at 22:28


















$begingroup$
Are you familiar with polar notation for complex numbers? i.e. converting from $x + iy$ to $r mbox{cis} theta$ (or $r e^{itheta}$)
$endgroup$
– ConMan
Jan 7 at 22:28






$begingroup$
Are you familiar with polar notation for complex numbers? i.e. converting from $x + iy$ to $r mbox{cis} theta$ (or $r e^{itheta}$)
$endgroup$
– ConMan
Jan 7 at 22:28












2 Answers
2






active

oldest

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4












$begingroup$

Use $$(1+sqrt3i)^3=(1-sqrt3i)^3=-8$$ and $2017equiv1(mod3).$



I got $$2cdot(-8)^{672}=2^{2017}.$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    First, convert these to polar form, so $1+isqrt{3}=2e^{ifrac{pi}{3}},$ and $1-isqrt{3}=2e^{-ifrac{pi}{3}},$
    then raise these exponential terms to the desired power, and proceed with the computation, using the identity
    $$e^{it}=cos(t)+isin(t).$$






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      using $e^{i2pi n}=1$ might help too
      $endgroup$
      – Henry
      Jan 7 at 22:36












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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    Use $$(1+sqrt3i)^3=(1-sqrt3i)^3=-8$$ and $2017equiv1(mod3).$



    I got $$2cdot(-8)^{672}=2^{2017}.$$






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      Use $$(1+sqrt3i)^3=(1-sqrt3i)^3=-8$$ and $2017equiv1(mod3).$



      I got $$2cdot(-8)^{672}=2^{2017}.$$






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        Use $$(1+sqrt3i)^3=(1-sqrt3i)^3=-8$$ and $2017equiv1(mod3).$



        I got $$2cdot(-8)^{672}=2^{2017}.$$






        share|cite|improve this answer









        $endgroup$



        Use $$(1+sqrt3i)^3=(1-sqrt3i)^3=-8$$ and $2017equiv1(mod3).$



        I got $$2cdot(-8)^{672}=2^{2017}.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 7 at 22:38









        Michael RozenbergMichael Rozenberg

        110k1896201




        110k1896201























            1












            $begingroup$

            First, convert these to polar form, so $1+isqrt{3}=2e^{ifrac{pi}{3}},$ and $1-isqrt{3}=2e^{-ifrac{pi}{3}},$
            then raise these exponential terms to the desired power, and proceed with the computation, using the identity
            $$e^{it}=cos(t)+isin(t).$$






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              using $e^{i2pi n}=1$ might help too
              $endgroup$
              – Henry
              Jan 7 at 22:36
















            1












            $begingroup$

            First, convert these to polar form, so $1+isqrt{3}=2e^{ifrac{pi}{3}},$ and $1-isqrt{3}=2e^{-ifrac{pi}{3}},$
            then raise these exponential terms to the desired power, and proceed with the computation, using the identity
            $$e^{it}=cos(t)+isin(t).$$






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              using $e^{i2pi n}=1$ might help too
              $endgroup$
              – Henry
              Jan 7 at 22:36














            1












            1








            1





            $begingroup$

            First, convert these to polar form, so $1+isqrt{3}=2e^{ifrac{pi}{3}},$ and $1-isqrt{3}=2e^{-ifrac{pi}{3}},$
            then raise these exponential terms to the desired power, and proceed with the computation, using the identity
            $$e^{it}=cos(t)+isin(t).$$






            share|cite|improve this answer









            $endgroup$



            First, convert these to polar form, so $1+isqrt{3}=2e^{ifrac{pi}{3}},$ and $1-isqrt{3}=2e^{-ifrac{pi}{3}},$
            then raise these exponential terms to the desired power, and proceed with the computation, using the identity
            $$e^{it}=cos(t)+isin(t).$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 7 at 22:31









            ChickenmancerChickenmancer

            3,284724




            3,284724








            • 1




              $begingroup$
              using $e^{i2pi n}=1$ might help too
              $endgroup$
              – Henry
              Jan 7 at 22:36














            • 1




              $begingroup$
              using $e^{i2pi n}=1$ might help too
              $endgroup$
              – Henry
              Jan 7 at 22:36








            1




            1




            $begingroup$
            using $e^{i2pi n}=1$ might help too
            $endgroup$
            – Henry
            Jan 7 at 22:36




            $begingroup$
            using $e^{i2pi n}=1$ might help too
            $endgroup$
            – Henry
            Jan 7 at 22:36


















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