All the complex polynomials $p$ and $q$ for which $p(z)sin^2(z)+q(z)cos^2(z)=1$ holds.
$begingroup$
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
edited Jan 6 at 20:35
rtybase
10.7k21533
asked Jan 6 at 17:29
abcdmathabcdmath
317110
$endgroup$
add a comment |
$begingroup$
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
edited Jan 6 at 20:35
rtybase
10.7k21533
asked Jan 6 at 17:29
abcdmathabcdmath
317110
$endgroup$
add a comment |
$begingroup$
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
edited Jan 6 at 20:35
rtybase
10.7k21533
asked Jan 6 at 17:29
abcdmathabcdmath
317110
$endgroup$
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
edited Jan 6 at 20:35
rtybase
10.7k21533
asked Jan 6 at 17:29
abcdmathabcdmath
317110
edited Jan 6 at 20:35
rtybase
10.7k21533
asked Jan 6 at 17:29
abcdmathabcdmath
317110
edited Jan 6 at 20:35
rtybase
10.7k21533
edited Jan 6 at 20:35
rtybase
10.7k21533
edited Jan 6 at 20:35
rtybase
10.7k21533
10.7k21533
asked Jan 6 at 17:29
abcdmathabcdmath
317110
asked Jan 6 at 17:29
abcdmathabcdmath
317110
asked Jan 6 at 17:29
abcdmathabcdmath
317110
317110
add a comment |
add a comment |
1 Answer
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 $p=q=1$.
answered Jan 6 at 17:38
MindlackMindlack
2,84217
$endgroup$
add a comment |
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1 Answer
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1 Answer
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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 $p=q=1$.
answered Jan 6 at 17:38
MindlackMindlack
2,84217
$endgroup$
add a comment |
$begingroup$
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 $p=q=1$.
answered Jan 6 at 17:38
MindlackMindlack
2,84217
$endgroup$
add a comment |
$begingroup$
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 $p=q=1$.
answered Jan 6 at 17:38
MindlackMindlack
2,84217
$endgroup$
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 $p=q=1$.
answered Jan 6 at 17:38
MindlackMindlack
2,84217
answered Jan 6 at 17:38
MindlackMindlack
2,84217
answered Jan 6 at 17:38
MindlackMindlack
2,84217
answered Jan 6 at 17:38
MindlackMindlack
2,84217
2,84217
add a comment |
add a comment |
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