I need to find the singularities of $$f(z) = \frac{1-e^z}{2+e^z}$$ My effort: Poles of function are given by $$2+e^z=0\implies e^z = -2 \implies z = \log 2+i(2k+1)\pi$$ for k integer.

All these are singularities termed as simple poles. By definition, limit point of these which is $\infty$ is a non-isolated singularity.

Further, limit point of zeros is again infinity which is a isolated-essential singularity.

But if both of isolated and non isolated coincides we take it as a non-isolated singularity. Am i correct? These are the only singularities?


1 Answer 1


Every neighborhood of $\infty$ contains a pole, implying, as you correctly state, that $\infty$ is not an isolated singularity. What makes you believe that it is also an isolated singularity?

  • $\begingroup$ I have read somewhere else that limit point of zeros is such a singularity. $\endgroup$
    – Mittal G
    Nov 26, 2018 at 13:10
  • $\begingroup$ Probably you misremember what you read. $\endgroup$ Nov 26, 2018 at 15:31

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