How to show that...












0












$begingroup$


Wolfram Alpha gives me that
$$begin{align}
sqrt{1+i}&=sqrt[4]{2}cosleft(frac{1}{2}tan^{-1}(1)right)+isqrt[4]{2}sinleft(frac{1}{2}tan^{-1}(1)right)\
&=sqrt[4]{2}e^{frac{1}{2}itan^{-1}(1)}\
sqrt{1+2i}&=sqrt[4]{5}cosleft(frac{1}{2}tan^{-1}(2)right)+isqrt[4]{5}sinleft(frac{1}{2}tan^{-1}(2)right)\
&=sqrt[4]{5}e^{frac{1}{2}itan^{-1}(2)}\
sqrt{1+3i}&=sqrt[4]{10}cosleft(frac{1}{2}tan^{-1}(3)right)+isqrt[4]{10}sinleft(frac{1}{2}tan^{-1}(3)right)\
&=sqrt[4]{10}e^{frac{1}{2}itan^{-1}(3)}
end{align}$$
In general, how can we show that
$$begin{align}
sqrt{1+ni}&=sqrt[4]{n^2+1}cosleft(frac{1}{2}tan^{-1}(n)right)+isqrt[4]{n^2+1}sinleft(frac{1}{2}tan^{-1}(n)right)\
&=sqrt[4]{n^2+1}e^{frac{1}{2}itan^{-1}(n)}
end{align}$$



I tried to draw a complex plane, but it did not help significantly. Perhaps I forgot some fundamental concepts. Any hints?






complex-analysis radicals







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    Hint: write $1+i n$ in polar form, that is as $r exp(iphi)$. What is $r$, what is $phi$?
    $endgroup$
    – Fabian
    Dec 2 '18 at 16:31










  • $begingroup$
    Thanks Fabian, I think I have figured it out.
    $endgroup$
    – Larry
    Dec 2 '18 at 16:37
















0












$begingroup$


Wolfram Alpha gives me that
$$begin{align}
sqrt{1+i}&=sqrt[4]{2}cosleft(frac{1}{2}tan^{-1}(1)right)+isqrt[4]{2}sinleft(frac{1}{2}tan^{-1}(1)right)\
&=sqrt[4]{2}e^{frac{1}{2}itan^{-1}(1)}\
sqrt{1+2i}&=sqrt[4]{5}cosleft(frac{1}{2}tan^{-1}(2)right)+isqrt[4]{5}sinleft(frac{1}{2}tan^{-1}(2)right)\
&=sqrt[4]{5}e^{frac{1}{2}itan^{-1}(2)}\
sqrt{1+3i}&=sqrt[4]{10}cosleft(frac{1}{2}tan^{-1}(3)right)+isqrt[4]{10}sinleft(frac{1}{2}tan^{-1}(3)right)\
&=sqrt[4]{10}e^{frac{1}{2}itan^{-1}(3)}
end{align}$$
In general, how can we show that
$$begin{align}
sqrt{1+ni}&=sqrt[4]{n^2+1}cosleft(frac{1}{2}tan^{-1}(n)right)+isqrt[4]{n^2+1}sinleft(frac{1}{2}tan^{-1}(n)right)\
&=sqrt[4]{n^2+1}e^{frac{1}{2}itan^{-1}(n)}
end{align}$$



I tried to draw a complex plane, but it did not help significantly. Perhaps I forgot some fundamental concepts. Any hints?






complex-analysis radicals







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    Hint: write $1+i n$ in polar form, that is as $r exp(iphi)$. What is $r$, what is $phi$?
    $endgroup$
    – Fabian
    Dec 2 '18 at 16:31










  • $begingroup$
    Thanks Fabian, I think I have figured it out.
    $endgroup$
    – Larry
    Dec 2 '18 at 16:37














0












0








0





$begingroup$


Wolfram Alpha gives me that
$$begin{align}
sqrt{1+i}&=sqrt[4]{2}cosleft(frac{1}{2}tan^{-1}(1)right)+isqrt[4]{2}sinleft(frac{1}{2}tan^{-1}(1)right)\
&=sqrt[4]{2}e^{frac{1}{2}itan^{-1}(1)}\
sqrt{1+2i}&=sqrt[4]{5}cosleft(frac{1}{2}tan^{-1}(2)right)+isqrt[4]{5}sinleft(frac{1}{2}tan^{-1}(2)right)\
&=sqrt[4]{5}e^{frac{1}{2}itan^{-1}(2)}\
sqrt{1+3i}&=sqrt[4]{10}cosleft(frac{1}{2}tan^{-1}(3)right)+isqrt[4]{10}sinleft(frac{1}{2}tan^{-1}(3)right)\
&=sqrt[4]{10}e^{frac{1}{2}itan^{-1}(3)}
end{align}$$
In general, how can we show that
$$begin{align}
sqrt{1+ni}&=sqrt[4]{n^2+1}cosleft(frac{1}{2}tan^{-1}(n)right)+isqrt[4]{n^2+1}sinleft(frac{1}{2}tan^{-1}(n)right)\
&=sqrt[4]{n^2+1}e^{frac{1}{2}itan^{-1}(n)}
end{align}$$



I tried to draw a complex plane, but it did not help significantly. Perhaps I forgot some fundamental concepts. Any hints?






complex-analysis radicals







share|cite|improve this question









$endgroup$




Wolfram Alpha gives me that
$$begin{align}
sqrt{1+i}&=sqrt[4]{2}cosleft(frac{1}{2}tan^{-1}(1)right)+isqrt[4]{2}sinleft(frac{1}{2}tan^{-1}(1)right)\
&=sqrt[4]{2}e^{frac{1}{2}itan^{-1}(1)}\
sqrt{1+2i}&=sqrt[4]{5}cosleft(frac{1}{2}tan^{-1}(2)right)+isqrt[4]{5}sinleft(frac{1}{2}tan^{-1}(2)right)\
&=sqrt[4]{5}e^{frac{1}{2}itan^{-1}(2)}\
sqrt{1+3i}&=sqrt[4]{10}cosleft(frac{1}{2}tan^{-1}(3)right)+isqrt[4]{10}sinleft(frac{1}{2}tan^{-1}(3)right)\
&=sqrt[4]{10}e^{frac{1}{2}itan^{-1}(3)}
end{align}$$
In general, how can we show that
$$begin{align}
sqrt{1+ni}&=sqrt[4]{n^2+1}cosleft(frac{1}{2}tan^{-1}(n)right)+isqrt[4]{n^2+1}sinleft(frac{1}{2}tan^{-1}(n)right)\
&=sqrt[4]{n^2+1}e^{frac{1}{2}itan^{-1}(n)}
end{align}$$



I tried to draw a complex plane, but it did not help significantly. Perhaps I forgot some fundamental concepts. Any hints?






complex-analysis radicals




complex-analysis radicals






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 2 '18 at 16:26









LarryLarry

2,40131129




2,40131129








  • 4




    $begingroup$
    Hint: write $1+i n$ in polar form, that is as $r exp(iphi)$. What is $r$, what is $phi$?
    $endgroup$
    – Fabian
    Dec 2 '18 at 16:31










  • $begingroup$
    Thanks Fabian, I think I have figured it out.
    $endgroup$
    – Larry
    Dec 2 '18 at 16:37














  • 4




    $begingroup$
    Hint: write $1+i n$ in polar form, that is as $r exp(iphi)$. What is $r$, what is $phi$?
    $endgroup$
    – Fabian
    Dec 2 '18 at 16:31










  • $begingroup$
    Thanks Fabian, I think I have figured it out.
    $endgroup$
    – Larry
    Dec 2 '18 at 16:37








4




4




$begingroup$
Hint: write $1+i n$ in polar form, that is as $r exp(iphi)$. What is $r$, what is $phi$?
$endgroup$
– Fabian
Dec 2 '18 at 16:31




$begingroup$
Hint: write $1+i n$ in polar form, that is as $r exp(iphi)$. What is $r$, what is $phi$?
$endgroup$
– Fabian
Dec 2 '18 at 16:31












$begingroup$
Thanks Fabian, I think I have figured it out.
$endgroup$
– Larry
Dec 2 '18 at 16:37




$begingroup$
Thanks Fabian, I think I have figured it out.
$endgroup$
– Larry
Dec 2 '18 at 16:37










2 Answers
2






active

oldest

votes


















1












$begingroup$

From the polar form $$z=sqrt{n^2+1},e^{iarctan n}$$ you draw



$$sqrt z=pmsqrt[4]{n^2+1},e^{iarctan n,/2}.$$



Then back to Cartesian coordinates,



$$pmsqrt[4]{n^2+1}left(cosfrac{arctan n}2+isinfrac{arctan n}2right).$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    $$begin{align}
    z&=1+ni,~r=sqrt{n^2+1}\
    1+ni &=sqrt{n^2+1}cos(tan^{-1}(n))+isqrt{n^2+1}sin(tan^{-1}(n))\
    &= sqrt{n^2+1}e^{itan^{-1}(n)}\
    sqrt{1+ni}&=sqrt[4]{n^2+1}e^{frac{1}{2}tan^{-1}(n)}\
    &=sqrt[4]{n^2+1}cosleft(frac{1}{2}tan^{-1}(n)right)+isqrt[4]{n^2+1}sinleft(frac{1}{2}tan^{-1}(n)right)
    end{align}$$






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      From the polar form $$z=sqrt{n^2+1},e^{iarctan n}$$ you draw



      $$sqrt z=pmsqrt[4]{n^2+1},e^{iarctan n,/2}.$$



      Then back to Cartesian coordinates,



      $$pmsqrt[4]{n^2+1}left(cosfrac{arctan n}2+isinfrac{arctan n}2right).$$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        From the polar form $$z=sqrt{n^2+1},e^{iarctan n}$$ you draw



        $$sqrt z=pmsqrt[4]{n^2+1},e^{iarctan n,/2}.$$



        Then back to Cartesian coordinates,



        $$pmsqrt[4]{n^2+1}left(cosfrac{arctan n}2+isinfrac{arctan n}2right).$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          From the polar form $$z=sqrt{n^2+1},e^{iarctan n}$$ you draw



          $$sqrt z=pmsqrt[4]{n^2+1},e^{iarctan n,/2}.$$



          Then back to Cartesian coordinates,



          $$pmsqrt[4]{n^2+1}left(cosfrac{arctan n}2+isinfrac{arctan n}2right).$$






          share|cite|improve this answer









          $endgroup$



          From the polar form $$z=sqrt{n^2+1},e^{iarctan n}$$ you draw



          $$sqrt z=pmsqrt[4]{n^2+1},e^{iarctan n,/2}.$$



          Then back to Cartesian coordinates,



          $$pmsqrt[4]{n^2+1}left(cosfrac{arctan n}2+isinfrac{arctan n}2right).$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 2 '18 at 16:55









          Yves DaoustYves Daoust

          128k675227




          128k675227























              0












              $begingroup$

              $$begin{align}
              z&=1+ni,~r=sqrt{n^2+1}\
              1+ni &=sqrt{n^2+1}cos(tan^{-1}(n))+isqrt{n^2+1}sin(tan^{-1}(n))\
              &= sqrt{n^2+1}e^{itan^{-1}(n)}\
              sqrt{1+ni}&=sqrt[4]{n^2+1}e^{frac{1}{2}tan^{-1}(n)}\
              &=sqrt[4]{n^2+1}cosleft(frac{1}{2}tan^{-1}(n)right)+isqrt[4]{n^2+1}sinleft(frac{1}{2}tan^{-1}(n)right)
              end{align}$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                $$begin{align}
                z&=1+ni,~r=sqrt{n^2+1}\
                1+ni &=sqrt{n^2+1}cos(tan^{-1}(n))+isqrt{n^2+1}sin(tan^{-1}(n))\
                &= sqrt{n^2+1}e^{itan^{-1}(n)}\
                sqrt{1+ni}&=sqrt[4]{n^2+1}e^{frac{1}{2}tan^{-1}(n)}\
                &=sqrt[4]{n^2+1}cosleft(frac{1}{2}tan^{-1}(n)right)+isqrt[4]{n^2+1}sinleft(frac{1}{2}tan^{-1}(n)right)
                end{align}$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $$begin{align}
                  z&=1+ni,~r=sqrt{n^2+1}\
                  1+ni &=sqrt{n^2+1}cos(tan^{-1}(n))+isqrt{n^2+1}sin(tan^{-1}(n))\
                  &= sqrt{n^2+1}e^{itan^{-1}(n)}\
                  sqrt{1+ni}&=sqrt[4]{n^2+1}e^{frac{1}{2}tan^{-1}(n)}\
                  &=sqrt[4]{n^2+1}cosleft(frac{1}{2}tan^{-1}(n)right)+isqrt[4]{n^2+1}sinleft(frac{1}{2}tan^{-1}(n)right)
                  end{align}$$






                  share|cite|improve this answer









                  $endgroup$



                  $$begin{align}
                  z&=1+ni,~r=sqrt{n^2+1}\
                  1+ni &=sqrt{n^2+1}cos(tan^{-1}(n))+isqrt{n^2+1}sin(tan^{-1}(n))\
                  &= sqrt{n^2+1}e^{itan^{-1}(n)}\
                  sqrt{1+ni}&=sqrt[4]{n^2+1}e^{frac{1}{2}tan^{-1}(n)}\
                  &=sqrt[4]{n^2+1}cosleft(frac{1}{2}tan^{-1}(n)right)+isqrt[4]{n^2+1}sinleft(frac{1}{2}tan^{-1}(n)right)
                  end{align}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 2 '18 at 16:48









                  LarryLarry

                  2,40131129




                  2,40131129






























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