Rewriting this complex square root












2












$begingroup$


for some reason I can to figure out how to rewrite this square root.
I have:



$sqrt{2+i}$



And I need to rewrite it into:



$frac{sqrt{2(sqrt{5} + 2)} + sqrt{-2(sqrt{5} - 2)}}{2}$



Can anybody show me I to do this? I've been trying for an hour now..






complex-numbers







share|cite|improve this question









$endgroup$








  • 6




    $begingroup$
    Hint: $sqrt{cos theta + i sin theta} = cos(theta/2) + i sin(theta / 2)$. Note $2 + i = sqrt{5} cdot left(frac{2}{sqrt 5} + i frac{1}{sqrt 5}right)$
    $endgroup$
    – Omnomnomnom
    Jun 27 '14 at 19:44












  • $begingroup$
    Solve $(x+iy)^2=x^2-y^2+2ixy=2+i$.
    $endgroup$
    – Yves Daoust
    Jun 27 '14 at 19:54








  • 1




    $begingroup$
    @DaneBouchie Read that again. I say $$ 2 + i = sqrt{5} cdot left(frac{2}{sqrt 5} + i frac{1}{sqrt 5}right) $$
    $endgroup$
    – Omnomnomnom
    Jun 27 '14 at 20:18










  • $begingroup$
    Ah, my bad. I read $.$ as a period for some reason. Even then, lazy reading on my behalf.
    $endgroup$
    – Dane Bouchie
    Jun 27 '14 at 20:22


















2












$begingroup$


for some reason I can to figure out how to rewrite this square root.
I have:



$sqrt{2+i}$



And I need to rewrite it into:



$frac{sqrt{2(sqrt{5} + 2)} + sqrt{-2(sqrt{5} - 2)}}{2}$



Can anybody show me I to do this? I've been trying for an hour now..






complex-numbers







share|cite|improve this question









$endgroup$








  • 6




    $begingroup$
    Hint: $sqrt{cos theta + i sin theta} = cos(theta/2) + i sin(theta / 2)$. Note $2 + i = sqrt{5} cdot left(frac{2}{sqrt 5} + i frac{1}{sqrt 5}right)$
    $endgroup$
    – Omnomnomnom
    Jun 27 '14 at 19:44












  • $begingroup$
    Solve $(x+iy)^2=x^2-y^2+2ixy=2+i$.
    $endgroup$
    – Yves Daoust
    Jun 27 '14 at 19:54








  • 1




    $begingroup$
    @DaneBouchie Read that again. I say $$ 2 + i = sqrt{5} cdot left(frac{2}{sqrt 5} + i frac{1}{sqrt 5}right) $$
    $endgroup$
    – Omnomnomnom
    Jun 27 '14 at 20:18










  • $begingroup$
    Ah, my bad. I read $.$ as a period for some reason. Even then, lazy reading on my behalf.
    $endgroup$
    – Dane Bouchie
    Jun 27 '14 at 20:22
















2












2








2


0



$begingroup$


for some reason I can to figure out how to rewrite this square root.
I have:



$sqrt{2+i}$



And I need to rewrite it into:



$frac{sqrt{2(sqrt{5} + 2)} + sqrt{-2(sqrt{5} - 2)}}{2}$



Can anybody show me I to do this? I've been trying for an hour now..






complex-numbers







share|cite|improve this question









$endgroup$




for some reason I can to figure out how to rewrite this square root.
I have:



$sqrt{2+i}$



And I need to rewrite it into:



$frac{sqrt{2(sqrt{5} + 2)} + sqrt{-2(sqrt{5} - 2)}}{2}$



Can anybody show me I to do this? I've been trying for an hour now..






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 Jun 27 '14 at 19:40









user160513user160513

141




141








  • 6




    $begingroup$
    Hint: $sqrt{cos theta + i sin theta} = cos(theta/2) + i sin(theta / 2)$. Note $2 + i = sqrt{5} cdot left(frac{2}{sqrt 5} + i frac{1}{sqrt 5}right)$
    $endgroup$
    – Omnomnomnom
    Jun 27 '14 at 19:44












  • $begingroup$
    Solve $(x+iy)^2=x^2-y^2+2ixy=2+i$.
    $endgroup$
    – Yves Daoust
    Jun 27 '14 at 19:54








  • 1




    $begingroup$
    @DaneBouchie Read that again. I say $$ 2 + i = sqrt{5} cdot left(frac{2}{sqrt 5} + i frac{1}{sqrt 5}right) $$
    $endgroup$
    – Omnomnomnom
    Jun 27 '14 at 20:18










  • $begingroup$
    Ah, my bad. I read $.$ as a period for some reason. Even then, lazy reading on my behalf.
    $endgroup$
    – Dane Bouchie
    Jun 27 '14 at 20:22
















  • 6




    $begingroup$
    Hint: $sqrt{cos theta + i sin theta} = cos(theta/2) + i sin(theta / 2)$. Note $2 + i = sqrt{5} cdot left(frac{2}{sqrt 5} + i frac{1}{sqrt 5}right)$
    $endgroup$
    – Omnomnomnom
    Jun 27 '14 at 19:44












  • $begingroup$
    Solve $(x+iy)^2=x^2-y^2+2ixy=2+i$.
    $endgroup$
    – Yves Daoust
    Jun 27 '14 at 19:54








  • 1




    $begingroup$
    @DaneBouchie Read that again. I say $$ 2 + i = sqrt{5} cdot left(frac{2}{sqrt 5} + i frac{1}{sqrt 5}right) $$
    $endgroup$
    – Omnomnomnom
    Jun 27 '14 at 20:18










  • $begingroup$
    Ah, my bad. I read $.$ as a period for some reason. Even then, lazy reading on my behalf.
    $endgroup$
    – Dane Bouchie
    Jun 27 '14 at 20:22










6




6




$begingroup$
Hint: $sqrt{cos theta + i sin theta} = cos(theta/2) + i sin(theta / 2)$. Note $2 + i = sqrt{5} cdot left(frac{2}{sqrt 5} + i frac{1}{sqrt 5}right)$
$endgroup$
– Omnomnomnom
Jun 27 '14 at 19:44






$begingroup$
Hint: $sqrt{cos theta + i sin theta} = cos(theta/2) + i sin(theta / 2)$. Note $2 + i = sqrt{5} cdot left(frac{2}{sqrt 5} + i frac{1}{sqrt 5}right)$
$endgroup$
– Omnomnomnom
Jun 27 '14 at 19:44














$begingroup$
Solve $(x+iy)^2=x^2-y^2+2ixy=2+i$.
$endgroup$
– Yves Daoust
Jun 27 '14 at 19:54






$begingroup$
Solve $(x+iy)^2=x^2-y^2+2ixy=2+i$.
$endgroup$
– Yves Daoust
Jun 27 '14 at 19:54






1




1




$begingroup$
@DaneBouchie Read that again. I say $$ 2 + i = sqrt{5} cdot left(frac{2}{sqrt 5} + i frac{1}{sqrt 5}right) $$
$endgroup$
– Omnomnomnom
Jun 27 '14 at 20:18




$begingroup$
@DaneBouchie Read that again. I say $$ 2 + i = sqrt{5} cdot left(frac{2}{sqrt 5} + i frac{1}{sqrt 5}right) $$
$endgroup$
– Omnomnomnom
Jun 27 '14 at 20:18












$begingroup$
Ah, my bad. I read $.$ as a period for some reason. Even then, lazy reading on my behalf.
$endgroup$
– Dane Bouchie
Jun 27 '14 at 20:22






$begingroup$
Ah, my bad. I read $.$ as a period for some reason. Even then, lazy reading on my behalf.
$endgroup$
– Dane Bouchie
Jun 27 '14 at 20:22












1 Answer
1






active

oldest

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2












$begingroup$

One can use trigonometry for sure but I thing the fastest way to solve this is just to solve the quadratic.
$$(x+yi)^2=2+i$$ so



$$x^2-y^2=2$$
$$2xy=1$$



and $$(x^2+y^2)^2=(x^2-y^2)^2+4x^2y^2=4+1=5$$



So $$x^2+y^2=sqrt{5}$$
(positive since $x$ and $y$ real)
and
$$x^2=frac{sqrt{5}+2}{2}$$
$$y^2=frac{sqrt{5}-2}{2}$$



so



$$x=pmsqrt{frac{sqrt{5}+2}{2}}$$
$$y=pmsqrt{frac{sqrt{5}-2}{2}}$$



Both $x$ and $y$ have same sign since $2xy=1$.






share|cite|improve this answer











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    $begingroup$

    One can use trigonometry for sure but I thing the fastest way to solve this is just to solve the quadratic.
    $$(x+yi)^2=2+i$$ so



    $$x^2-y^2=2$$
    $$2xy=1$$



    and $$(x^2+y^2)^2=(x^2-y^2)^2+4x^2y^2=4+1=5$$



    So $$x^2+y^2=sqrt{5}$$
    (positive since $x$ and $y$ real)
    and
    $$x^2=frac{sqrt{5}+2}{2}$$
    $$y^2=frac{sqrt{5}-2}{2}$$



    so



    $$x=pmsqrt{frac{sqrt{5}+2}{2}}$$
    $$y=pmsqrt{frac{sqrt{5}-2}{2}}$$



    Both $x$ and $y$ have same sign since $2xy=1$.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      One can use trigonometry for sure but I thing the fastest way to solve this is just to solve the quadratic.
      $$(x+yi)^2=2+i$$ so



      $$x^2-y^2=2$$
      $$2xy=1$$



      and $$(x^2+y^2)^2=(x^2-y^2)^2+4x^2y^2=4+1=5$$



      So $$x^2+y^2=sqrt{5}$$
      (positive since $x$ and $y$ real)
      and
      $$x^2=frac{sqrt{5}+2}{2}$$
      $$y^2=frac{sqrt{5}-2}{2}$$



      so



      $$x=pmsqrt{frac{sqrt{5}+2}{2}}$$
      $$y=pmsqrt{frac{sqrt{5}-2}{2}}$$



      Both $x$ and $y$ have same sign since $2xy=1$.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        One can use trigonometry for sure but I thing the fastest way to solve this is just to solve the quadratic.
        $$(x+yi)^2=2+i$$ so



        $$x^2-y^2=2$$
        $$2xy=1$$



        and $$(x^2+y^2)^2=(x^2-y^2)^2+4x^2y^2=4+1=5$$



        So $$x^2+y^2=sqrt{5}$$
        (positive since $x$ and $y$ real)
        and
        $$x^2=frac{sqrt{5}+2}{2}$$
        $$y^2=frac{sqrt{5}-2}{2}$$



        so



        $$x=pmsqrt{frac{sqrt{5}+2}{2}}$$
        $$y=pmsqrt{frac{sqrt{5}-2}{2}}$$



        Both $x$ and $y$ have same sign since $2xy=1$.






        share|cite|improve this answer











        $endgroup$



        One can use trigonometry for sure but I thing the fastest way to solve this is just to solve the quadratic.
        $$(x+yi)^2=2+i$$ so



        $$x^2-y^2=2$$
        $$2xy=1$$



        and $$(x^2+y^2)^2=(x^2-y^2)^2+4x^2y^2=4+1=5$$



        So $$x^2+y^2=sqrt{5}$$
        (positive since $x$ and $y$ real)
        and
        $$x^2=frac{sqrt{5}+2}{2}$$
        $$y^2=frac{sqrt{5}-2}{2}$$



        so



        $$x=pmsqrt{frac{sqrt{5}+2}{2}}$$
        $$y=pmsqrt{frac{sqrt{5}-2}{2}}$$



        Both $x$ and $y$ have same sign since $2xy=1$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 16 '18 at 10:39









        Did

        248k23225463




        248k23225463










        answered Jun 27 '14 at 20:50









        Rene SchipperusRene Schipperus

        32.4k11960




        32.4k11960






























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