All the complex polynomials $p$ and $q$ for which $p(z)sin^2(z)+q(z)cos^2(z)=1$ holds.












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Find all the complex polynomials $p$ and $q$ for which $$p(z)sin^2(z)+q(z)cos^2(z)=1$$ holds.

We know that the function $sin^2(z)+cos^2(z)-1$ is analytic on $mathbb{C}$ and bounded, so it is constant. And since this function in $0$ on real line, $$sin^2(z)+cos^2(z)-1=0 text{ on } mathbb{C}$$ But I have no idea for the polynomials $p$ and $q$ other than $1$. Please help me to find all such complex polynomials $p$ and $q$ such that the above holds.






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    0












    $begingroup$


    Find all the complex polynomials $p$ and $q$ for which $$p(z)sin^2(z)+q(z)cos^2(z)=1$$ holds.

    We know that the function $sin^2(z)+cos^2(z)-1$ is analytic on $mathbb{C}$ and bounded, so it is constant. And since this function in $0$ on real line, $$sin^2(z)+cos^2(z)-1=0 text{ on } mathbb{C}$$ But I have no idea for the polynomials $p$ and $q$ other than $1$. Please help me to find all such complex polynomials $p$ and $q$ such that the above holds.






    complex-analysis







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Find all the complex polynomials $p$ and $q$ for which $$p(z)sin^2(z)+q(z)cos^2(z)=1$$ holds.

      We know that the function $sin^2(z)+cos^2(z)-1$ is analytic on $mathbb{C}$ and bounded, so it is constant. And since this function in $0$ on real line, $$sin^2(z)+cos^2(z)-1=0 text{ on } mathbb{C}$$ But I have no idea for the polynomials $p$ and $q$ other than $1$. Please help me to find all such complex polynomials $p$ and $q$ such that the above holds.






      complex-analysis







      share|cite|improve this question











      $endgroup$




      Find all the complex polynomials $p$ and $q$ for which $$p(z)sin^2(z)+q(z)cos^2(z)=1$$ holds.

      We know that the function $sin^2(z)+cos^2(z)-1$ is analytic on $mathbb{C}$ and bounded, so it is constant. And since this function in $0$ on real line, $$sin^2(z)+cos^2(z)-1=0 text{ on } mathbb{C}$$ But I have no idea for the polynomials $p$ and $q$ other than $1$. Please help me to find all such complex polynomials $p$ and $q$ such that the above holds.






      complex-analysis




      complex-analysis






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      edited Jan 6 at 20:35









      rtybase

      11.5k31534




      11.5k31534










      asked Jan 6 at 17:29









      abcdmathabcdmath

      337110




      337110






















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          Let $p_1=p-1$, $q_1=q-1$, then $q_1(z)cos^2(z)+p_1(z)sin^2(z)=0$.



          Now take $z=kpi$, $k$ an integer, then $q_1(z)=0$. Thus $q_1$ has infinitely many roots so is zero. Similarly, $p_1=0$, hence $p=q=1$.






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            $begingroup$

            Let $p_1=p-1$, $q_1=q-1$, then $q_1(z)cos^2(z)+p_1(z)sin^2(z)=0$.



            Now take $z=kpi$, $k$ an integer, then $q_1(z)=0$. Thus $q_1$ has infinitely many roots so is zero. Similarly, $p_1=0$, hence $p=q=1$.






            share|cite|improve this answer









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              2












              $begingroup$

              Let $p_1=p-1$, $q_1=q-1$, then $q_1(z)cos^2(z)+p_1(z)sin^2(z)=0$.



              Now take $z=kpi$, $k$ an integer, then $q_1(z)=0$. Thus $q_1$ has infinitely many roots so is zero. Similarly, $p_1=0$, hence $p=q=1$.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Let $p_1=p-1$, $q_1=q-1$, then $q_1(z)cos^2(z)+p_1(z)sin^2(z)=0$.



                Now take $z=kpi$, $k$ an integer, then $q_1(z)=0$. Thus $q_1$ has infinitely many roots so is zero. Similarly, $p_1=0$, hence $p=q=1$.






                share|cite|improve this answer









                $endgroup$



                Let $p_1=p-1$, $q_1=q-1$, then $q_1(z)cos^2(z)+p_1(z)sin^2(z)=0$.



                Now take $z=kpi$, $k$ an integer, then $q_1(z)=0$. Thus $q_1$ has infinitely many roots so is zero. Similarly, $p_1=0$, hence $p=q=1$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 6 at 17:38









                MindlackMindlack

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