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Assume that f is a holomorphic function on $\mathbb{C}$ except at the origin. Also, suppose that $f(n)=(-1)^n$ for each positive integer n. Prove that $\inf_{z\not=0}\lvert f(z) \rvert =0$.

Now, we know that $z=0$ is the only singular point of $f(z)$. There are three possibilities. Firstly, zero is a removable singularity, so we can remove zero and the resulting function is entire, and thus the result follows, by Liouville's theorem. Secondly, if zero is an essential singularity, then by Casorati-Weierstrass theorem we have that the image of f is dense so the result follows. Now, if zero is a pole of order k then we must have $f(n)=\frac{g(n)}{n^k}=(-1)^n$ for each positive integer, where g is an entire function. Then I was trying to get a contradiction from this but since the integers do not have an accumulation point in $\mathbb{C}$ then we cannot use the identity theorem, so I got stuck at this point. Any help would be appreciated.

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2 Answers 2

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It is actually much simpler than you imagine. You don't even have to know what an essential singularity is.

If $\inf_{z \neq 0} |f(z)|=\epsilon >0$ then $|\frac 1 {f(z)}| \leq \frac 1 {\epsilon}$ for $z \neq 0$. This implies that $\frac 1 f$ has removable singularity at $0$ and we can apply Liouville's Theorem to conclude that $\frac 1 f$, hence also $f$, is a constant. This contradicts the fact that $f(n)=(-1)^{n}$.

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Consider the function $g(z):=f(1/z)$, which is holomorphic in $\mathbb{C}\setminus \{ 0\}$, and $g(1/n)=(-1)^n$. This means $g$ can't have a removable singularity or a pole at $z=0$ (in the first case $g$ would be continous, which can't happen; while in the second, we'd have $\lim_{z\to 0} |g(z)|=\infty$ which also doesn't happen). Therefore $g$ has an essential singularity at $z=0$ and the result follows from Cassorati-Weierstrass.

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