Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I don't really understand how one can find the type of singularities for a given function. Say if $$f(z) = \frac{z}{e^z-1}$$ then I know that the singularities are at $z = 2n\pi i$

However, how do I find the type? If I try to write out Laurent series for this by using $$e^z = 1 + z + z^2/2! + z^3/3! + \cdots $$, then I got a summation in the denominator, which I don't know how to rearrange as a Laurent series.

Can anyone please give me a hint?

share|cite|improve this question
Then $z=0$ is removable. – Ragnar May 31 '13 at 8:12
Oh great! I see it now. Thanks. – Shar May 31 '13 at 8:14
See here. – Mhenni Benghorbal May 31 '13 at 9:48
Related problem. – Mhenni Benghorbal May 31 '13 at 9:56
up vote 4 down vote accepted

On $z=0$ you can remove the singularity.

For the other use that a singularity at $z_0$ is a pole of order $k$ when $f(z)\cdot (z-z_0)^k$ does have a removable singularity at $z_0$.

share|cite|improve this answer
That is, to study the singularity at $2n\pi i$, you merely have to compute the order of the zero in the denominator. – GEdgar May 31 '13 at 13:56

Some of key points to decide the type of the singularity

I am not claiming that these points are complete. But certainly these point may help you to decide about the types of singularity of a function at a point.

1: A singular point $z_0 $is called an isolated singular point of an analytic function $f(z)$ if there exists a deleted $\epsilon$-spherical neighborhood of $z_0$ that contains no singularity. If no such neighborhood can be found, $z_0 $is called a non-isolated singular point. Thus an isolated singular point is a singular point that stands completely by itself, embedded in regular points.

2. An isolated singular point $z_0$ such that $f$ can be defined, or redefined, at $z_0$ in such a way as to be make it analytic at $z_0$ is called removable singularity. A singular point $z_0$ is removable if $\lim _{z\to z_{0}} f(z)$ exists. For example The singular point $z = 0$ is a removable singularity of $f(z) = \frac{\sin z}{z}$ since $\lim _{z\to z_{0}} \frac{\sin z}{z} = 1$. Dominic's answer says about when a singular point $z_0$ is a pole of order $k$.

3. A singular point that is not a pole or removable singularity is called an essential singular point.

4 Isolated essential singularity

Limit points of the sequence of a zeros of the nonzero analytic function $f(z)$ is an isolated singularity. Since it is not a pole it is an isolated essential singularity. For example consider the function $\sin \frac{1}{z}$ it has zeros at $z = \frac{1}{k\pi}$ . Limit points of the zeros is the zeros is the point $z =0$ hence, $z = 0$ is an isolated essential singularity of $f(z)$.

5 . Non -isolated essential singularity

Let $z_1, z_2,\ldots, z_n, \ldots$ be the set of points having limit point $z_0$ in a region $D$. Let $f$ be analytic in $D$ except at the poles $z_1, z_2,\ldots, z_n, \ldots$, then $z_0$ is also singularity of $f(z)$. The reason is that $f$ is unbounded in the neighborhood of $z_0$. But $z_0$ is not isolated because it is limit points of the poles. Hence $f(z)$ has a non-isolated singularity at $z_0$. For example consider the function $\frac{1}{\cos(1/z)}$ this function has poles at $k = \frac{2}{(2k+1)\pi}$ the point $z =0$ is the limit point of these poles hence given function has non -isolated essential singularity at $z = 0$.

share|cite|improve this answer
@srijan.... nice explanation sir... – monalisa May 31 '13 at 9:22
Thank you for the explanation. – Shar Jun 11 '13 at 3:45
you are welcome. – srijan Jun 12 '13 at 17:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.