# Singularities, essential singularities, poles, simple poles

Could someone possible explain the differences between each of these;

Singularities, essential singularities, poles, simple poles.

I understand the concept and how to use them in order to work out the residue at each point, however, done fully understand what the difference is for each of these

As far as i understand a simple pole is a singularity of order $1$?

then we have poles of order $n$ which aren't simple?

not too sure about essential singularity

The point $$z_{0}$$ is an isolated singularity of $$f(z)$$ if $$f(z)$$ is analytic in $$0 \lt |z-z_{0}| \lt r$$ (a circle of radius r centered at $$z_{0}$$ with the point $$z_{0}$$ punched out). If one expands a function $$f(z)$$ in a Laurent series about the point $$z_{0}$$, $$f(z) = \sum\limits_{k=-\infty}^{\infty} a^{k} (z-z_{0})^{k}$$ we can classify isolated singularties into 3 cases:

1. If there are no negative powers of $$z-z_{0}$$, then $$z_{0}$$ is a removable singularity and the Laurent series is a power series.

• Example: $$\frac{\sin(z)}{z} = 1 - \frac{z^{2}}{3!} + \frac{z^{4}}{5!} - ...$$ has a removable singularity at 0.
2. $$f(z)$$ has a pole of order m at $$z_{0}$$ if m is the largest positive integer such that $$a_{-m} \ne 0$$. A pole of order one is a simple pole. A pole of order two is a double pole, etc.

• Example: $$f(z) = \frac{1}{(z-3i)^{7}}$$ has a pole of order 7 at $$z=3i$$
3. If there are an infinite number of negative powers of $$z-z_{0}$$, then $$z_{0}$$ is an essential singularity.

• Example: $$\mathrm{e}^{1/z} = 1 + \frac{1}{z} + \frac{1}{2!z^{2}} + ...$$ has an essential singularity at 0.

There are three kinds of singularities.

Removable singularity, which can be extended to a holomorphic function over that point.

Poles, which is removable after multiplying some $(z-a)^n$. The smallest $n$ is called the order of the pole, when $n=1$, it is called simple.

Essential singularity: neither of the above. For example $g(z)=e^{1/z}$ since $|g(z)z^l|$ is never bounded near $0$.

• could you explain what a holomorphic function is? May 16, 2015 at 3:02
• @smith See here: en.wikipedia.org/wiki/Holomorphic_function May 16, 2015 at 4:14
• I like to call removable singularities "poles of order $0$" Jul 8, 2016 at 3:48
• @reuns high five for removable singularities as poles of order 0. are essential singularities poles of order negative infinity? and zeroes are poles of order positive infinity? Oct 27, 2021 at 20:51
• @JohnSmithKyon Sure. Removable means that $f$ extends to an analytic function. Pole means that $1/f$ is analytic. Essential singularity means that... $f$ is given by a Laurent series around the point and (can you finish?) Oct 28, 2021 at 1:16

Singularity:

$\quad$ A point $a$ is said to be a singular point of a function $f$ if

i) f is not analytic at $a$ and

ii) if we can find a neighborhood of $f(a)$ such that there exists a point $b$ in which $f$ is analytic.

Essential Singularity:

$\quad$ A point $a$ is said to be a essential singular point of a function $f$ if

i) f is not analytic at $a$ and

ii) if every neighborhood of $f(a)$ contains infinte number of points in which $f$ is analytic.

Poles:

a point $a$ is said to be a pole if

i)it is a essential singularity and

ii)$\lim_{z \to a} f(z) = \infty$

A pole of order 1 is simple pole and double pole if it is order 2.

• well many people say a singularity for a point $a$ where $f(z)$ isn't analytic but it is analytic on an open $U$ with $a$ on the boundary (so $U$ doesn't have to contain a neighborhood of $a$, example a branch point) Oct 11, 2016 at 1:13