I have this question about singularities for a function defined as a fraction of two holomorphic functions.
Let $f,g: U \to \mathbb C$ be holomorphic functions on a subset $U \subseteq \mathbb C$. Let $z_0 \in U$ be a zero for both functions: of order $n$ for $f$ and order $m$ for $g$. Show, for the function $h(z) = \frac{f(z)}{g(z)}$ that
(a) If $n \geq m$, then $h$ has a removable singularity in $z_0$. In addition: $\lim_{z \to z_0} h(z) = \frac{f^{(m)}(z_0)}{g^{(m)}(z_0)}$.
(b) If $n < m$, then $h$ has pole of order $m-n$ in $z_0$.
I feel like I understand the concepts of poles and removable singularities pretty well. However, I am not sure how to formulate an answer for this question. Would it makes sense to create a Laurent series around $z_0$ for both $f$ and $g$?
Hint: Write it as $f(z)=(z-z_0)^nf_1(z)$ and $g(z)=(z-z_0)^mg_1(z),$ where $f_1,g_1$ are holomorphic, and $f_1(z_0)\neq 0$ and $g_1(z_0)\neq0.$
The only hard part is the limit. For the limit, show that $$\begin{align}f^{(m)}(z_0)&=\frac{(z_0-z_0)^{n-m}f_1(z_0)}{m!}\\g^{(m)}(z_0)&=\frac{g_1(z_0)}{m!}\end{align}$$
Where $(z_0-z_0)^k$ is just zero when $k>0$ and $1$ when $k=0.$