Let $f$ be defined on $\mathbb{C}$ be holomorphic. Suppose there exists and $n \in \mathbb{Z}^+$ such that

$$\int_{\partial B_1(0)}\frac{f(z)}{(z-a)^n}=0$$

for all $a \in B_1(0)$.

prove $f$ is a polynomial.

Here is what I have, by corollary of Cauchy's theorem we have that

$$f^{(n-1)}(a)(2 \pi i)/(n-1)!=0$$

Which implies $f^{(n-1)}(a)=0$ and this implies $f$ is a polynomial since its derivative at some point eventually is zero, specially on some finite radius ball.

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    $\begingroup$ That's basically the idea! I don't understand what your question is $\endgroup$ Jun 29, 2022 at 3:47
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    $\begingroup$ Ah, yup! If you want to be really precise, you should show that $f^{(N)}(a) = 0$ for all $N \geq n-1$ to show that it's a polynomial, rather than just checking that $f^{(n-1)} = 0$. Though obviously the argument is basically the same ^_^ $\endgroup$ Jun 29, 2022 at 3:49
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    $\begingroup$ Thanks yeah that could be a small lemma to show that part! $\endgroup$
    – homosapien
    Jun 29, 2022 at 3:50
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    $\begingroup$ @Hossien Sahebjame Yes. I meant that you did not write that $f$ was supposed to be an entire function, since you did not write what is $f$. That is why I ask you to give the assumptions. $\endgroup$ Jun 29, 2022 at 19:35
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    $\begingroup$ You have shown that $f^{(n-1)}(a)=0$ for every $a$ inside the open unit disc. Therefore, by the identity theorem, $f^{(n-1)}=0$ on $\mathbb C$. Hence $f^{(n)},f^{(n+1)},\ldots$ etc. are also zero on $\mathbb C$. This means $f$ is a polynomial. $\endgroup$
    – user1551
    Jun 30, 2022 at 3:28


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