# holomorphic function in a punctured space

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

-
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

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 $\vert a_{n}\vert = 0$
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.