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Let $f$ be a holomorphic function on the punctured plane $ 0 < z < \infty $. Assume that there exist a positive constant C and a real constant M such that: $ \lvert f(z)\rvert \le C \lvert z \rvert^M $ for $ 0 < \lvert z\rvert < \frac{1}{2} $ . Show that $z=0$ is either a pole or a removable singularity for $f$ , and find sharp bounds for the order of $f$ at 0

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What did you try? –  Davide Giraudo Oct 8 '12 at 21:24
    
I think you mean $0 < |z| < \infty$. –  Michael Albanese Oct 8 '12 at 22:01

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up vote 1 down vote accepted

Let's estimate the magnitude of the Laurent coefficients. Let $0 < \epsilon < 1/2$, then for any $n \in \mathbb{Z}$: $ \vert a_{n} \vert \leq \frac{1}{2\pi}\int_{\vert z \vert = \epsilon} \frac{\vert f(z)\vert}{\vert z\vert^{n+1}}\,\vert dz\vert \leq C\epsilon^{M-n}$

Suppose $n < M$, then since we can pick $\epsilon$ arbitrarily small we see that $\vert a_{n}\vert = 0$

Otherwise, we have no useful estimates.

Now if $M \geq 0$ then the Laurent series has no singular part, so $f(z)$ can be extended to zero. This would correspond to a removable singularity.

If $M < 0$ then $a_{n} = 0$ for every $n < M$ so $z = 0$ is a pole. The singularity could not be essential because $M$ was assumed real.

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