Find and classify the singularities of the following functions in $\mathbb{C}$:

  1. $\frac{1}{z(e^{\frac{1}{z}}+1)}$

  2. $\frac{1}{(z^2+1)(z-1)^2}-\frac{1}{4(z-i)}$

OK, so I think the first is the easier (perhaps). There's clearly an essential singularity at the origin caused by the exponential. However, I think there are also singularities where $e^{\frac{1}{z}}=-1$, which occurs when $z=\frac{1}{(2n+1) \pi}$ for $n \in \mathbb{Z}$, though I am not sure how to classify there. Help with that would be very appreciated.

For the second, we can split it into $\frac{1}{4(z+i)}-\frac{1}{2(z-1)}+\frac{1}{2(z-1)^2}$, which makes the position of the poles clear; at $-i, 1$. Is it the case that the pole at $-i$ is simple, and the pole at $1$ is a double pole. That seems to be the case.

Any help/verification would be very helpful. Thanks in advance.


2 Answers 2


Note that $0$ is not an isolated singularity in 1). The classification into removable singularities, poles and essential singularities are usually only defined for isolated singularities. (If you're not convinced, what kind of singularity would $\sqrt{z}$ have at the origin?)


Your calculation of the second exercise is correct. The partial fractions indicate, as you said, that $-i$ is a pole of first order, while $1$ is a pole of order $2$. The term $-\frac{1}{2(z-1)}$ does not matter because there is the one with $-\frac{1}{2(z-1)^\color{red}{2}}$.

Concerning the first one, you are also right about the essential singularity. The factor $z$ is too weak to counteract $e^{\frac{1}{z}}$.

You can then do a power series expansion around, for example, $z=\frac{1}{i\pi}$ of $e^\frac{1}{z}$. It starts

$$e^\frac{1}{z} = -1 - \pi^2(z-\frac{1}{i\pi}) \ldots$$

The $-1$ is countered by the $+1$, then you can factor out $(z-\frac{1}{i\pi})$ and get a pole of first order. Because of the periodicity, all $\frac{1}{(2n+1)\pi i}$ are then poles of first order.

  • $\begingroup$ I am not so sure about that removable singularity of $\frac{1}{e^\frac{1}{z}}$. I thought about it with Picard's great theorem, and when $e^\frac{1}{z}$ takes almost every value in every neighborhood of $0$, then $\frac{1}{e^\frac{1}{z}}$ should too. $\endgroup$ Commented May 28, 2012 at 1:04
  • $\begingroup$ Gregor, you're right. I was mistaken. $\endgroup$ Commented May 28, 2012 at 1:10

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