# What's the difference between the different types of poles, zeroes and singularities in complex analysis?

I am trying to get an understanding on the difference between the different types of poles, zeroes and singularities in complex analysis and how to identify them. When is it a removable singularity, and why? When is it a simple pole? etc. So far I am having trouble with this, and would greatly appreciate some suggested ways of thinking/methods when trying to identify them.

I don't really have an example, as I just generally want to learn and understand it.

• It's a bit unclear what you mean with "different poles and zeros". Do you mean poles/zeros of different order? – Daniel Fischer Jul 31 '14 at 18:09
• I forgot to add singularities! Will edit. – trippelganger Jul 31 '14 at 18:14

In complex variables class, they teach you to use the Laurent series for this

A function with a simple pole at $z_0$ has a Laurent series $$\frac{b_{-1}}{z-z_0} + \sum_{n=0}^{\infty} a_n (z-z_0)^n$$ Roughly speaking, if you multiply this function by $(z-z_0)$, it becomes analytic at $z_0$.

You can have poles of higher order $$\frac{b_{-m}}{(z-z_0)^m} + ... + \frac{b_{-1}}{z-z_0} + \sum_{n=0}^{\infty} a_n (z-z_0)^n$$

In this case you'll have to multiply by $(z-z_0)^m$ to get an analytic function at $z_0$.

If the negative part of the series goes on forever, you have an essential singularity. Functions behave very badly in the neighborhood of such a singularity. $$\sum_{n=-\infty}^{\infty} a_n (z-z_0)^n$$

A zero is just a root of a function, if $F(z_0)=0$ then $z_0$ is a zero of the function.

A removable singularity is a "technical" singularity, e.g. $\sin(z)/z$ for $z_0 = (0,0)$. You can get rid of it by redefining the function suitably at $z_0$. Once you do this, $\sin(z)/z = 1 - \frac{z^2}{3!} + \ldots$.

In addition to poles and essential singularities, there are singularities which are in the shape of a line, like a branch cut.

• Can you give an example of essential singularities? – an offer can't refuse Oct 18 '15 at 4:56

This is how poles of different order looks like. If you are not sure, you can just plot it.

$\dfrac{z}{1-\cos z}$ is a good example. $1-\cos z$ has zero at $z=2πk$ but $z$ on the nominator removed the second order pole at $z=0$ by order 1. So, this function has a pole of order 1 at $z=0$ but of order 2 at other $z=2πk$.

Here is an illustration:

Just in case you are interested in my MATLAB code: