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I don't know how to solve the following problem. Any help will be appreciated.

Let $f:\mathbb{C}\setminus\{0\}\to\mathbb{C}$ be an analytic function. Suppose $0$ is accumulation point of the zeros of $f$, i.e. an accumulation point of $f^{-1}(0)$. Prove that either $f$ is identically zero, or $0$ is an essential singularity for $f$.

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    $\begingroup$ Complex analysis was a while ago, but I think this works. Zero is either a pole (possibly of order zero) or an essential singularity. If it is a pole then for sufficiently large $n$ $x^nf(x)$ is entire. An entire function whose zeroes have an accumulation point is identically zero and therefore $f$ must be identically zero. $\endgroup$ Dec 20, 2012 at 19:54

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Hint: the other possibilities are a removable singularity or a pole. A pole is easy to rule out. For a removable singularity, $f(z)$ has a Maclaurin series ...

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  • $\begingroup$ Yes, as you said a pole is easy to rule out. How about the removable singularity? $\endgroup$
    – albmiz-mth
    Dec 20, 2012 at 20:11
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    $\begingroup$ Consider the Maclaurin series; if $f$ is not identically zero, some coefficient in this series is nonzero. $f(z)/z^n$, for the first $n$ where the coefficient is nonzero, extends to a continuous function in a neighbourhood of $0$ which is nonzero at $0$. Do you see a contradiction arising from $0$ being an accumulation point of zeros of $f$? $\endgroup$ Dec 20, 2012 at 20:36
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    $\begingroup$ Using the Maclaurin series is overkill. All you need is continuity of $f$: take a sequence in $f^{-1}(0)$ going to zero and then take the limit under $f$. $\endgroup$
    – JSchlather
    Dec 21, 2012 at 0:28
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    $\begingroup$ @JacobSchlather: So? That just tells you $f(0)=0$. $\endgroup$ Dec 21, 2012 at 7:11

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