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 $g$?

  • $\begingroup$ Yes that would make a lot of sense ... $\endgroup$ – Ethan Bolker May 29 at 20:00
  • $\begingroup$ This isn’t necessarily a rational function question, as your first sentence implies. $\endgroup$ – Thomas Andrews May 29 at 20:30
  • $\begingroup$ Specifically, a “rational function” is a ratio of two polynomials, while this function is a ratio of arbitrary holomorphic functions. $\endgroup$ – 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 $k=0.$

  • $\begingroup$ 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. $\endgroup$ – Hagen von Eitzen May 29 at 20:21
  • $\begingroup$ Thanks. In fact, only the first sentence mentions rational function, and I think the OP just used it incorrectly. But it led me partially astray. @HagenvonEitzen $\endgroup$ – Thomas Andrews May 29 at 20:32
  • $\begingroup$ 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? $\endgroup$ – Nukgi May 30 at 10:07

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