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How to find all complex polymials $p(z)$ and $q(z)$ such that $$p(z) \cos^2 (z) + q(z) \sin^2(z) =1$$ (where $z$ is complex number)? Clearly , $p(z)=q(z)=1$ is a solution. I think it is the only solution but I cannot prove that.

3 Answers 3

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At all points $z = k \pi $, $k \in \Bbb Z$, $p(z) = 1$ must hold, i.e. the polynomial $p(z) - 1$ has infinitely many zeros. It follows that $p(z) \equiv 1$.

$q(z) \equiv 1$ follows similarly.

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    Will you please explain it little more ..how is p(z)=1 for z= k*pi – 2017-01-12
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    @mathislove: What is $\sin(k \pi)$ and $\cos(k \pi)$? – 2017-01-12
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    Cos^2 (k* pi) =1 and sin^(k*pi)= 0 I am getting this – 2017-01-12
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    @mathislove: Now substitute that in your equation. – 2017-01-12
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    Yeah ,i got it..but how to prove q(z)=1? – 2017-01-12
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    I have q(z) sin^(z)= sin^2(z).. – 2017-01-12
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    Hoe does it follow similarly? – 2017-01-12
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    @mathislove: Can you find points $z$ where $\cos^2(z) = 0$ and $\sin^2(z)=1$ ? – 2017-01-12
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    Is that k*pi/2? – 2017-01-12
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    @mathislove: Almost. $z = (2k+1) \pi /2$. Now substitute that in your equation and you see that $q(z)-1$ has infinitely many zeros as well. – 2017-01-12
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    Sir please check my solution. – 2017-01-20
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Hint: For each integer $n$, we have $$p(\pi n)\cos^2(\pi n) + q(\pi n)\sin^2(\pi n) = 1$$ $$p(\pi n) = 1$$

Hence, $z = \ldots,-2\pi,-\pi,0,\pi,2\pi,\ldots$ are all roots of $p(z)-1$.

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    Sir please check my solution – 2017-01-20
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Given, p(z) * (cos z)^2 + q(z) * (sin z)^2 =1 .........(1)

=> (e^ 2iz + e^ --2iz) * (p(z) -- q(z)) + 2 * (p(z) + q(z)) =4

=>Cos (2z) *( p(z) -- q(z)) + (p(z) + q(z)) =2

Since,(1) is true for all z€C . So for z€ C such that Cos (2z)=1 yields , p(z)=1 ..........(2) And ,Since,(1) is true for all z€C . So for z€ C such that Cos (2z)=0 yields, p(z)+ q(z)=1 . So from (2) ,it yields q(z)=1. Thus p(z) =1 and q(z)=1 are the only polynomials.

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    People are requested to check this .I think my solution is ok.still I request others to check this.Thanks in advance. – 2017-01-20
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    Your answer is difficult to read. [This page](http://meta.math.stackexchange.com/q/5020/8348) should give you a start at learning how to typeset mathematics. – 2017-01-20
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    Actually sir I don't know how to do all this. – 2017-01-20
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    You have posted some questions and answers now, and always have *other* people edited those to typeset them properly. You really should start learning to to it yourself, the linked-to reference is full of examples. You can also consult any TeX tutorial because that's what it is, essentially. – 2017-01-20