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I have the following theorem on my notes of a course on complex analysis I'm taking:

If f(z) is not bounded in a neighborhood of its isolated singular point z0, then z0 is a pole.

I think this cannot be right, for example sin(1/z) at z0=0 is not bounded, but has an essential singularity, not a pole. Just need some confirmation. Thanks.

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By neighborhood I mean a restricted neighborhood (that is, it doesn't include z0). And sin(1/z) is not bounded in any restricted neighborhood of z0. –  user6043 Jan 20 '11 at 4:05
    
@PEV: Consider $z=1/(in)$, $n=1,2,\ldots$. –  Jonas Meyer Jan 20 '11 at 4:08
    
Perhaps it was a slip of the tongue, "$|f(z)|\to\infty$" being (incorrectly) replaced with "$f$ is unbounded". –  Jonas Meyer Jan 20 '11 at 4:10

3 Answers 3

up vote 6 down vote accepted

Yes you are right. Great Picard's theorem states that if $w$ is an essential singularity of the analytic function $f$, then in any open set containing $w$, $f$ takes all possible values (with at most one exception), infinitely often.

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Riemann's removable singularity theorem says that $f$ has a removable singularity at $a$ if and only if $\lim_{z\to a}(z-a)f(z)=0$. This condition is easily satisfied if $f$ is bounded in a neighborhood of $a$, and therefore boundedness in a neighborhood of $a$ is equivalent to having a removable singularity.

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Let $G$ be an open connected set in $\mathbb{C}$. If $f$ is analytic on $G-\{w\}$ for some $w \in G$ then $w$ is a pole of $f$ if and only $\lim_{z \rightarrow w} f(z) = \infty$. So $\sin(1/z)$ does not have a pole at zero, since $\sin(1/z)$ does not approach $\infty$ at $z=0$

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