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My complex analysis textbook uses the following definitions for removable singularity, essential singularity and for poles of a complex function:

If $a$ is an isolated singularity of $f$ and the numbers $c_n$ for $n \in \mathbb{Z}$ are the coefficients in the Laurent series of the function, $\sum_{n\in\mathbb{Z}} c_n(z-a)^n$, then:

  • $a$ is a removable singularity if $c_n = 0$ for $n < 0$

  • $a$ is a pole of order $m \in \mathbb{N}$ if $c_{-m} \neq 0$ and $c_n=0$ for $n < -m$

  • $a$ is an essential singularity if $c_n \neq 0$ for an infinite number of negative values of $n$

Then, based on this definition I am supposed to classifies the singularity of the function $f(z)= \frac{1}{1-e^z}$ at the point $z = 0$

So, first I began by finding the Laurent series of this funcion:

$$\begin{align} \frac{1}{1-e^z}&=\sum_{n \geq0}e^{zn} \end{align}$$

Because we have that $e^z=\sum_{k \geq 0}\frac{1}{k!}z^k$. So if we let $z=zn$ then we get:

$$\begin{align} \sum_{n \geq0}e^{zn} &= \sum_{n \geq0} \sum_{k \geq 0}\frac{n^k}{k!}z^k \\ \\ &= \sum_{k \geq0}\ \underbrace{ \sum_{n \geq 0} \frac{n^k}{k!}}_{:=a_k}\ z^k \\ \\ &= \sum_{k \geq0} a_k z^k \end{align}$$

So, according to the definition my book gave, this is a removable singularity. But when I checked the solutions it said that $z=0$ is a pole of order $1$. So did I do something wrong and if so what did I do wrong or is the did the book author make a mistake?

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$$ \frac1{1-e^z}=\sum_{n\ge0}e^{zn}\tag1 $$ is not the Laurent series for $\frac1{1-e^z}$. It is one series for $\frac1{1-e^z}$ that only converges for $\operatorname{Re}(z)\lt0$, so probably not good for use in finding the type of singularity of $\frac1{1-e^z}$ at $z=0$.

The Laurent series for $\frac1{1-e^z}$ at $z=0$ is a little messy to compute, but suppose we have the Laurent series at $z=0$ $$ \frac1{1-e^z}=\sum_{k=-n}^\infty a_kz^k\tag2 $$ The $n$ we want to find is the smallest $n$ so that $\lim\limits_{z\to0}\frac{z^n}{1-e^z}$ is finite.

If $\lim\limits_{z\to0}\frac1{1-e^z}$ were finite, then $n=0$. However, this limit is $\infty$.

Otherwise, if $\lim\limits_{z\to0}\frac z{1-e^z}$ were finite, then $n=1$. Using L'Hôpital, we get that $$ \begin{align} \lim_{z\to0}\frac z{1-e^z} &=\lim_{z\to0}\frac1{-e^z}\\ &=-1\tag3 \end{align} $$ Thus, $n=1$. So $\frac1{1-e^z}$ has a pole of order $1$ at $z=0$.

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1) An isolated singularity is removable in the sense that one can define limit of the function at that point where singularity is. In this case, $\lim_{z\to 0} \frac{1}{1-e^z}=\frac{1}{1-e^0}=\frac{1}{1-1} $ which you can't define.

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2) We say a point $z_0$ is a pole if by multiplying some power of $(z-z_0)$ with the function $f(z)$ , you can kill the singularity in the sense that multiplying $f(z)$ with some power of $(z-z_0)$ you can define the value at $z_0$ of the resulting function and the least power of $(z-z_0)$, you need to do multiply is called the order of the pole. In this case, $z=0$ is an isolated singularity and $\lim_{z\to 0} \frac{z}{1-e^z}=-1$ . Hence, $0$ is a pole of order 1.

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By your definition, $$a_0 = \sum_{n \geq 0} \frac{n^0}{0!} = \sum_{n \geq 0} 1 = \infty,$$ so $a_0$ can't be a coefficient of the Laurent series.

To show it's actually a pole of order 1, you could verify that $\lim_{z \rightarrow 0} \frac{z}{e^z - 1} < \infty$.

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  • $\begingroup$ Why would that prove that 0 is a pole of order 1? $\endgroup$ – Eduardo Magalhães Jun 6 at 18:34
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    $\begingroup$ If the Laurent series of $f$ has a $c_{-m}z^{-m}$ term with $m \geq 2$, then the Laurent series of $zf$ will have a $c_{-m} z^{-m + 1}$ term, which will go to infinity as $z \rightarrow 0$ since $-m + 1 < 0$. $\endgroup$ – Vickie Jun 6 at 18:37

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