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I have to classify the singularities (removable, pole and essential) of $\frac{1}{e^z-1}$.

I know that $e^z-1=0 \iff e^z=1 \iff z = 2\pi k i$ for each $k \in \mathbb{Z}$. Is there an easy to find out all type of singularity without passing through the Laurent series directly? (Please explain in details.)

Thanks!

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2 Answers 2

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For each $k\in\mathbb{Z}$, the expression $$ \frac{z-2\pi ik}{e^z-1} $$ is the reciprocal of the quotient which appears in the definition of the derivative of $e^z$ at $z=2\pi ik$. Since this derivative is equal to $1$, it follows that $$\lim_{z\to 2\pi ik}\frac{z-2\pi ik}{e^z-1}=1 $$ hence $\frac{1}{e^z-1}$ has a simple pole at $2\pi ik$.

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  • $\begingroup$ I am not so good with the notion of classification of singularities. I don't understand at all what you did. Could you explain to me in details please? Why do you take the derivative? Why do you use $\lim_{z\to 2\pi ik}\frac{z-2\pi ik}{e^z-1}=1$ to conclude? How do you get $\frac{z-2\pi ik}{e^z-1}$? $\endgroup$ Apr 2, 2016 at 19:31
  • $\begingroup$ Since $\frac{1}{|e^z-1|}\to\infty$ as $z\to 2\pi ik$, there must be a pole. The question is what is the order of the pole. This can be found by finding the minimal $m$ such that $\frac{(z-2\pi ik)^m}{e^z-1}$ is holomorphic near $2\pi ik$, and the limit computation shows that $m=1$ works. $\endgroup$ Apr 2, 2016 at 19:42
  • $\begingroup$ Why do you think in term of the limit? I don't understand why $\frac{1}{|e^z-1|}\to\infty$ as $z\to 2\pi ik$, there must be a pole? Which result allow you this fact? $\endgroup$ Apr 2, 2016 at 19:50
  • $\begingroup$ That's one way to define a pole: $f(z)$ has a pole at $z_0$ if $z_0$ is an isolated singularity and $\lim_{z\to z_0}|f(z)|=\infty$. $\endgroup$ Apr 2, 2016 at 19:51
  • $\begingroup$ I lot of books have as defintion : The singularity $z_0$ is a pole, of ordre $k$ iif $a_{-k} \not=0$ and $a_n=0$ for all $n < -k$. Could you tell me why both definitions are equivalent? $\endgroup$ Apr 2, 2016 at 19:56
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The series expansion for $e^z - 1$ is $$ e^z - 1 = \sum_{j=1}^\infty \frac{z^j}{j!} $$ so it has a simple zero (divide by $z$ to see this) at every $z = 2\pi i k$ by the periodicity of $\exp$. Hence the reciprocal $\frac{1}{e^z - 1}$ has a simple pole at every $z = 2\pi i k$.

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  • $\begingroup$ Which step does you not understand? $\endgroup$ Apr 2, 2016 at 19:35
  • $\begingroup$ Do you know how to obtain the series expansion for $e^z -1$? $\endgroup$ Apr 2, 2016 at 19:37
  • $\begingroup$ So which part does you not understand? $\endgroup$ Apr 2, 2016 at 19:43
  • $\begingroup$ Let do it, I'll do it another way. Thanks $\endgroup$ Apr 2, 2016 at 19:44
  • $\begingroup$ Sorry, i am replying on a mobile device and it is difficult to read mathjax there. $\endgroup$ Apr 2, 2016 at 19:44

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