# A function as a fraction of two functions, where the denominator have the same zero as the numerator - Pole or removable singularity?

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 $$g$$?

• Yes that would make a lot of sense ... – Ethan Bolker May 29 at 20:00
• This isn’t necessarily a rational function question, as your first sentence implies. – Thomas Andrews May 29 at 20:30
• Specifically, a “rational function” is a ratio of two polynomials, while this function is a ratio of arbitrary holomorphic functions. – Thomas Andrews May 29 at 20:56

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 $$k=0.$$
• OP writes "rational function", but also that $f,g$ are (merely) holomorphic. So $f_1,g_1$ are perhaps also (merely) holomorphic -- without change to the argumentation. – Hagen von Eitzen May 29 at 20:21
• I corrected my mistake - thanks for pointing it out! The answer really helped me, with both the pole argument and the limit. However, I still don't fully understand the argument for why it is a removable singularity, when $n\geq m$. Is it because it is bounded? – Nukgi May 30 at 10:07