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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? – Jonas Meyer Feb 4 '11 at 15:06
yes ,f and g both are analytic functions . – user6618 Feb 4 '11 at 15:40
Please tell me whether i am correct or not: – user6618 Feb 4 '11 at 15:44
Is Uniqueness theorem applicable to f and this g – user6618 Feb 4 '11 at 15:48
@N Jana: Yes, that's the result you want to apply. Do you see how to apply it? – Jonas Meyer Feb 4 '11 at 15:50

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.

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