Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am stuck on a problem which is given as an exercise in my book. I really would like to solve it primarily by myself, but unfortunately I don't "see" the solution intuitively and therefore I have no good starting point. The problem is the following: Show that if $f$ has an isolated pole singularity at $c$, every neighborhood of $c$ that lies in the domain of f contains all complex numbers $z$ with $|z|>r$ , for some $r$ dependent on the choice of neighborhood. A small hint would be appreciated. It will allow me to get some sleep.

share|cite|improve this question

Use the definition of pole and the fact that holomorphic functions map open sets to open sets. Full solution is written below.

By definition, if $(1/f)(c)$ is defined to be zero, $1/f$ is holomorphic at $c$ and has a zero there. Since holomorphic functions map open set to open sets, image of a neighborhood of $c$, say $V$, under $1/f$ contains an open disc centered around $0$, say with radius $R$. This means that image of $V$ under $f$ contains all points whose distance from origin is farther than $1/R$.

share|cite|improve this answer

You should simpy consider $1/f$. This function is holomorphic around $c$ and also at $c$ with $f(c)=0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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