Isolated Singularities | Sahithyan's Notes

A point $z_0 \in \mathbb{C}$ is a singular point, or a singularity, of $f$ if $f$ fails to be analytic at $z_0$ and $f$ is analytic at some point in every neighborhood of $z_0$ .

A singular point $z_0$ of $f$ is an isolated singularity iff $f$ is analytic in some punctured disk $D_r(z_0) \setminus \{z_0\}$ . Otherwise it is a non-isolated singularity.

Examples:

$f(z) = \frac{1}{1-z}$ has an isolated singularity at $z = 1$

Classification

Let $\alpha$ be an isolated singularity of $f$ with the Laurent series

f(z) = \sum_{k=-\infty}^{\infty} a_k (z-\alpha)^k, \quad 0 < |z-\alpha| < R

Then $\alpha$ is

Removable singularity
iff $a_k = 0$ for all $k < 0$ .
Pole of order $n$
iff $a_{-n} \neq 0$ and $a_k = 0$ for all $k < -n$ .
Essential singularity
iff $a_k \neq 0$ for infinitely many $k < 0$ .

A pole of order 1 is called a simple pole.

Examples:

$\frac{\sin z}{z}$ and $\frac{\cos z - 1}{z^2}$ have removable singularities at $z = 0$
$\frac{\sin z}{z^3}$ has a pole of order 2 at $z = 0$
$\frac{e^z}{z}$ has a simple pole at $z = 0$
$z^2 \sin\left(\frac{1}{z}\right)$ has an essential singularity at $z = 0$

Residue

The coefficient $a_{-1}$ is called the residue of $f$ at $\alpha$ .

\text{Res}(f, \alpha) = a_{-1}

Worked Examples

Isolated Singularity

Show that $z = 1$ is an isolated singularity of $f(z) = \frac{1}{1-z}$ .

Solution Reveal Hide

$f$ is analytic everywhere except at $z = 1$ , so it is analytic in the punctured disk $D_1(1) \setminus \{1\}$ . Every neighborhood of $1$ contains points where $f$ is analytic, so $z = 1$ is a singular point.

The Laurent series centered at $1$ is $f(z) = -\frac{1}{z-1}$ . Only $a_{-1} = -1$ is non-zero, so $z = 1$ is a simple pole with $\text{Res}(f, 1) = -1$ .

Non-Isolated Singularity

Show that $z = 0$ is a non-isolated singularity of $f(z) = \frac{1}{\sin(1/z)}$ .

Solution Reveal Hide

$f$ is not analytic at $z_k = \frac{1}{k\pi}$ for each non-zero integer $k$ , since $\sin(1/z_k) = 0$ .

$z_k \to 0$ as $k \to \infty$ , so every punctured disk centered at $0$ contains a singular point. $f$ is not analytic in any punctured disk centered at $0$ , so $z = 0$ is a non-isolated singularity.

Removable Singularity

Show that $\frac{\sin z}{z}$ and $\frac{\cos z - 1}{z^2}$ have removable singularities at $z = 0$ .

\frac{\sin z}{z} = 1 - \frac{z^2}{6} + \frac{z^4}{120} - \cdots

\frac{\cos z - 1}{z^2} = -\frac{1}{2} + \frac{z^2}{24} - \cdots

Both series have $a_k = 0$ for all $k < 0$ .

Pole

Show that $\frac{\sin z}{z^3}$ has a pole of order 2 and $\frac{e^z}{z}$ has a pole of order 1 at $z = 0$ .

Solution Reveal Hide

\frac{\sin z}{z^3} = \frac{1}{z^2} - \frac{1}{6} + \frac{z^2}{120} - \cdots

$a_{-2} = 1 \neq 0$ and $a_k = 0$ for all $k < -2$ . The residue is $a_{-1} = 0$ .

\frac{e^z}{z} = \frac{1}{z} + 1 + \frac{z}{2} + \cdots

$a_{-1} = 1 \neq 0$ and $a_k = 0$ for all $k < -1$ . The residue is $1$ .

Essential Singularity

Show that $z^2 \sin\left(\frac{1}{z}\right)$ has an essential singularity at $z = 0$ .

Solution Reveal Hide

z^2 \sin\left(\frac{1}{z}\right) = z^2 \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!\, z^{2n+1}} = z - \frac{1}{6z} + \frac{1}{120z^3} - \cdots

$a_k \neq 0$ for $k = -1, -3, -5, \ldots$ , which is infinitely many $k < 0$ . The residue is $-\frac{1}{6}$ .

Classify from the Series

Classify the singularity of $\frac{e^z - 1 - z}{z^3}$ at $z = 0$ and find the residue.

Solution Reveal Hide

\frac{e^z - 1 - z}{z^3} = \frac{\frac{z^2}{2} + \frac{z^3}{6} + \frac{z^4}{24} + \cdots}{z^3} = \frac{1}{2z} + \frac{1}{6} + \frac{z}{24} + \cdots

$z = 0$ is a simple pole with $\text{Res} = \frac{1}{2}$ .

Classify the singularity of $z^3 \cos\left(\frac{1}{z}\right)$ at $z = 0$ and find the residue.

Solution Reveal Hide

z^3 \cos\left(\frac{1}{z}\right) = z^3 - \frac{z}{2} + \frac{1}{24z} - \frac{1}{720z^3} + \cdots

$z = 0$ is an essential singularity with $\text{Res} = \frac{1}{24}$ .