0
$\begingroup$

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 $g$?

$\endgroup$
3
  • $\begingroup$ 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
1
$\begingroup$

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 $k=0.$

$\endgroup$
3
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.