Why $|z^5|-2^5=0$ has $infty$ solutions in $mathbb{C}$?












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Given $z=x+iy$ a complex number, I can't understand why $|z^5|-2^5=0$ has infinite solutions in $mathbb{C}$.






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  • $begingroup$
    For any $thetain[0, 2pi/5),$ the complex number $z=2(costheta+isintheta)$ satisfies your equation.
    $endgroup$
    – Bumblebee
    Jan 16 at 18:18


















0












$begingroup$


Given $z=x+iy$ a complex number, I can't understand why $|z^5|-2^5=0$ has infinite solutions in $mathbb{C}$.






complex-numbers







share|cite|improve this question









$endgroup$












  • $begingroup$
    For any $thetain[0, 2pi/5),$ the complex number $z=2(costheta+isintheta)$ satisfies your equation.
    $endgroup$
    – Bumblebee
    Jan 16 at 18:18
















0












0








0





$begingroup$


Given $z=x+iy$ a complex number, I can't understand why $|z^5|-2^5=0$ has infinite solutions in $mathbb{C}$.






complex-numbers







share|cite|improve this question









$endgroup$




Given $z=x+iy$ a complex number, I can't understand why $|z^5|-2^5=0$ has infinite solutions in $mathbb{C}$.






complex-numbers




complex-numbers






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asked Jan 16 at 18:15









KevinKevin

16211




16211












  • $begingroup$
    For any $thetain[0, 2pi/5),$ the complex number $z=2(costheta+isintheta)$ satisfies your equation.
    $endgroup$
    – Bumblebee
    Jan 16 at 18:18




















  • $begingroup$
    For any $thetain[0, 2pi/5),$ the complex number $z=2(costheta+isintheta)$ satisfies your equation.
    $endgroup$
    – Bumblebee
    Jan 16 at 18:18


















$begingroup$
For any $thetain[0, 2pi/5),$ the complex number $z=2(costheta+isintheta)$ satisfies your equation.
$endgroup$
– Bumblebee
Jan 16 at 18:18






$begingroup$
For any $thetain[0, 2pi/5),$ the complex number $z=2(costheta+isintheta)$ satisfies your equation.
$endgroup$
– Bumblebee
Jan 16 at 18:18












3 Answers
3






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oldest

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2












$begingroup$

Let's start by showing that $|z|-2=0$ has infinitely many solutions. Rewrite this as $|z|=2$. This equation is solved if $z$ has magnitude $2$; in other words, if it lies on the circle $x^{2}+y^{2}=2^{2}$. Clearly there are infinitely many points on this circle, so there are infinitely many such complex numbers.



Now, we have shown there are infinitely many complex numbers with magnitude $2$. Since $|z^{5}|=|z|^{5}$, each of the above numbers satisfies $|z^{5}|=|z|^{5}=2^{5}$






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  • $begingroup$
    This is quite clear.
    $endgroup$
    – Randall
    Jan 16 at 19:28



















0












$begingroup$

First off, note that $|z^5| = |z|^5$, so we can just drop the two $5$s, and ask why $|z| = 2$ has infinitely many roots in $mathbb{C}$ (because it does, and all of these are necessarily solutions to your equation). But this is precisely what $z mapsto |z|$ does: it takes each circle centred on the origin, and maps all points of that circle to the unique point on the positive real half-line. Thus, the infinitely many solutions that you speak of are exactly the infinitely many points on the circle of radius $2$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    The stated equation is equivalant to $|z|=2$. Now there are infinitely many $z$ that fullfill this:



    $$z=2e^{iphi}, 0le phi < 2pi$$



    would be one way to describe them, or



    $$z=xpm isqrt{4-x^2}, -2 le x le 2$$



    would be another. Using the usual visualization of the Gaussian plane, those are the complex $z$ that lie on the circle of radious $2$ around the origin.






    share|cite|improve this answer









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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Let's start by showing that $|z|-2=0$ has infinitely many solutions. Rewrite this as $|z|=2$. This equation is solved if $z$ has magnitude $2$; in other words, if it lies on the circle $x^{2}+y^{2}=2^{2}$. Clearly there are infinitely many points on this circle, so there are infinitely many such complex numbers.



      Now, we have shown there are infinitely many complex numbers with magnitude $2$. Since $|z^{5}|=|z|^{5}$, each of the above numbers satisfies $|z^{5}|=|z|^{5}=2^{5}$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        This is quite clear.
        $endgroup$
        – Randall
        Jan 16 at 19:28
















      2












      $begingroup$

      Let's start by showing that $|z|-2=0$ has infinitely many solutions. Rewrite this as $|z|=2$. This equation is solved if $z$ has magnitude $2$; in other words, if it lies on the circle $x^{2}+y^{2}=2^{2}$. Clearly there are infinitely many points on this circle, so there are infinitely many such complex numbers.



      Now, we have shown there are infinitely many complex numbers with magnitude $2$. Since $|z^{5}|=|z|^{5}$, each of the above numbers satisfies $|z^{5}|=|z|^{5}=2^{5}$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        This is quite clear.
        $endgroup$
        – Randall
        Jan 16 at 19:28














      2












      2








      2





      $begingroup$

      Let's start by showing that $|z|-2=0$ has infinitely many solutions. Rewrite this as $|z|=2$. This equation is solved if $z$ has magnitude $2$; in other words, if it lies on the circle $x^{2}+y^{2}=2^{2}$. Clearly there are infinitely many points on this circle, so there are infinitely many such complex numbers.



      Now, we have shown there are infinitely many complex numbers with magnitude $2$. Since $|z^{5}|=|z|^{5}$, each of the above numbers satisfies $|z^{5}|=|z|^{5}=2^{5}$






      share|cite|improve this answer









      $endgroup$



      Let's start by showing that $|z|-2=0$ has infinitely many solutions. Rewrite this as $|z|=2$. This equation is solved if $z$ has magnitude $2$; in other words, if it lies on the circle $x^{2}+y^{2}=2^{2}$. Clearly there are infinitely many points on this circle, so there are infinitely many such complex numbers.



      Now, we have shown there are infinitely many complex numbers with magnitude $2$. Since $|z^{5}|=|z|^{5}$, each of the above numbers satisfies $|z^{5}|=|z|^{5}=2^{5}$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 16 at 18:20









      pwerthpwerth

      3,340417




      3,340417












      • $begingroup$
        This is quite clear.
        $endgroup$
        – Randall
        Jan 16 at 19:28


















      • $begingroup$
        This is quite clear.
        $endgroup$
        – Randall
        Jan 16 at 19:28
















      $begingroup$
      This is quite clear.
      $endgroup$
      – Randall
      Jan 16 at 19:28




      $begingroup$
      This is quite clear.
      $endgroup$
      – Randall
      Jan 16 at 19:28











      0












      $begingroup$

      First off, note that $|z^5| = |z|^5$, so we can just drop the two $5$s, and ask why $|z| = 2$ has infinitely many roots in $mathbb{C}$ (because it does, and all of these are necessarily solutions to your equation). But this is precisely what $z mapsto |z|$ does: it takes each circle centred on the origin, and maps all points of that circle to the unique point on the positive real half-line. Thus, the infinitely many solutions that you speak of are exactly the infinitely many points on the circle of radius $2$.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        First off, note that $|z^5| = |z|^5$, so we can just drop the two $5$s, and ask why $|z| = 2$ has infinitely many roots in $mathbb{C}$ (because it does, and all of these are necessarily solutions to your equation). But this is precisely what $z mapsto |z|$ does: it takes each circle centred on the origin, and maps all points of that circle to the unique point on the positive real half-line. Thus, the infinitely many solutions that you speak of are exactly the infinitely many points on the circle of radius $2$.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          First off, note that $|z^5| = |z|^5$, so we can just drop the two $5$s, and ask why $|z| = 2$ has infinitely many roots in $mathbb{C}$ (because it does, and all of these are necessarily solutions to your equation). But this is precisely what $z mapsto |z|$ does: it takes each circle centred on the origin, and maps all points of that circle to the unique point on the positive real half-line. Thus, the infinitely many solutions that you speak of are exactly the infinitely many points on the circle of radius $2$.






          share|cite|improve this answer









          $endgroup$



          First off, note that $|z^5| = |z|^5$, so we can just drop the two $5$s, and ask why $|z| = 2$ has infinitely many roots in $mathbb{C}$ (because it does, and all of these are necessarily solutions to your equation). But this is precisely what $z mapsto |z|$ does: it takes each circle centred on the origin, and maps all points of that circle to the unique point on the positive real half-line. Thus, the infinitely many solutions that you speak of are exactly the infinitely many points on the circle of radius $2$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 16 at 18:19









          user3482749user3482749

          4,3291119




          4,3291119























              0












              $begingroup$

              The stated equation is equivalant to $|z|=2$. Now there are infinitely many $z$ that fullfill this:



              $$z=2e^{iphi}, 0le phi < 2pi$$



              would be one way to describe them, or



              $$z=xpm isqrt{4-x^2}, -2 le x le 2$$



              would be another. Using the usual visualization of the Gaussian plane, those are the complex $z$ that lie on the circle of radious $2$ around the origin.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                The stated equation is equivalant to $|z|=2$. Now there are infinitely many $z$ that fullfill this:



                $$z=2e^{iphi}, 0le phi < 2pi$$



                would be one way to describe them, or



                $$z=xpm isqrt{4-x^2}, -2 le x le 2$$



                would be another. Using the usual visualization of the Gaussian plane, those are the complex $z$ that lie on the circle of radious $2$ around the origin.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  The stated equation is equivalant to $|z|=2$. Now there are infinitely many $z$ that fullfill this:



                  $$z=2e^{iphi}, 0le phi < 2pi$$



                  would be one way to describe them, or



                  $$z=xpm isqrt{4-x^2}, -2 le x le 2$$



                  would be another. Using the usual visualization of the Gaussian plane, those are the complex $z$ that lie on the circle of radious $2$ around the origin.






                  share|cite|improve this answer









                  $endgroup$



                  The stated equation is equivalant to $|z|=2$. Now there are infinitely many $z$ that fullfill this:



                  $$z=2e^{iphi}, 0le phi < 2pi$$



                  would be one way to describe them, or



                  $$z=xpm isqrt{4-x^2}, -2 le x le 2$$



                  would be another. Using the usual visualization of the Gaussian plane, those are the complex $z$ that lie on the circle of radious $2$ around the origin.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 16 at 18:22









                  IngixIngix

                  5,192159




                  5,192159






























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