Pole and removable sigularity I try to solve the following problem. I do not sure how to begin :
Let $f$ be a holomorphic function on $\mathbb{C}\setminus \{0\}$. Assume that there exists a constant $C > 0$ and a real constant $M$ such that $$|f(z)| \leq C|z|^M$$ for $0 < |z| < \frac{1}{2}.$ Show that $z=0$ is either a pole or a removable singularities for $f$, and find sharp bound for $O_0(f)$, the order of $f$ at $0$ (I think that the order means the of $z$ in $f(z)$. I think that $f$ might should take the form $f(z) = \frac{g(z)}{z^{O_0(f)}}$ where $g$ is entire.)
I am not sure I should if I should write $f := \frac{g}{z^n}$ for some $n \in \mathbb{N} \cup \{0\}$ where $g$ is entire.(I have no particular reasons for writing $f$ in this form except that the most of the questions ask about poles or singularity of $f$, $f$ is often written in this form) I guess that the condition, $|f(z)| \leq C|z|^M$ for $0 < |z| < \frac{1}{2}$, might connect to that $f$ is a polynmial of degree at most $M$ on $0 < |z| < \frac{1}{2}$, but it seems that this contradict the form $f = g/ z^n$. So I am confusing how to start.

Suppose first that $M\ge0$. Then $f$ is bounded on a neighborhood of $z=0$, $z=0$ is a removable singularity and $f$ has a zero at $z=0$ of order $\lceil M\rceil$.
If $M<0$ consider $g(z)=z^{\lceil -M\rceil}\,f(z)$.
