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Let $f,g:\mathbb C \to \mathbb C$ be two analytic functions such that $f(z)(g(z)+z^2)=0$ for all $z$ .Then prove that either $f(z)=0$ or $g(z)=-z^2$.

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    The $z^2$ is only a distraction. $h(z)=g(z)+z^2$ is analytic if $g$ is. (I assume that you forgot to mention that $f$ and $g$ are analytic.) What do you know about the zeros of analytic functions?2011-02-04
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    yes ,f and g both are analytic functions .2011-02-04
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    Please tell me whether i am correct or not:2011-02-04
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    Is Uniqueness theorem applicable to f and this g2011-02-04
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    @N Jana: Yes, that's the result you want to apply. Do you see how to apply it?2011-02-04
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    Suppose that f(a) not zero for some a(C.Then ,by the continuity of f,there exists a disk D(a;d) such that f(z)not 0 in D(a;d).But then h(z)=0 for all z(D(a;d).By the uniqueness theorem ,f(z)=0 in C.2011-02-04
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    @N Jana: Good. (You meant $h$ the last time you wrote $f$.) If you want, now that you have answered your own question, you could post it as an official answer.2011-02-04

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I thought it might be a good idea to attempt to get this off of the Unanswered list by posting what is essentially the solution N Jana gave in the comments.

If there exists $a\in\mathbb C$ such that $f(a)\neq 0$, then by continuity there is an open disk at $a$ where $f$ is nonzero. Then $g(z)+z^2$ is zero on this disk, and by the identity theorem for analytic functions, $g(z)+z^2=0$ for all $z\in\mathbb{C}$.

The $z^2$ here adds nothing essential. The obvious generalization is that if a product of analytic functions on a connected open set is identically $0$, then one of the functions is identically zero. (And this means that the ring of analytic functions on a connected open set is an integral domain with pointwise operations.)

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If $f=0$ you have done, so can assume this is not the case. Can assume there is a point where both $f$ and $g-z^2$ are both different from zero otherwise you have done. Up to traslation can assume this point is the origin. So there exists integers $h,k$ such that $f=z^hf'$ and $g=z^kg'$ with $f',g'$ different from zero at the origin. then you have $f'(z^{h+k}g'-z^{2+k})=0$. Since $f'$ not zero at the origin this means that locally $z^{h+k}g'-z^{2+k}=0$ from which the thesis follows.