# Type of singularity in $\frac{1}{1-e^z}$

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?

## 3 Answers

$$\frac1{1-e^z}=\sum_{n\ge0}e^{zn}\tag1$$ 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 $$z=0$$.

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.

$$\\$$

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.

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 $$\lim_{z \rightarrow 0} \frac{z}{e^z - 1} < \infty$$.

• Why would that prove that 0 is a pole of order 1? – Eduardo Magalhães Jun 6 at 18:34
• 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 $-m + 1 < 0$. – Vickie Jun 6 at 18:37