Proposition

Find the poles or removable singularities of the following functions and determine their orders:

$(z^2 + 1)^{-3}(z - 1)^{-4}$.
$z\cot(z)$.
$z^{-5}\sin z$.
$\frac{z}{1 - \exp(z)}$.
$\frac{1}{1 - \exp(z)}$.

Solution

1

The order of the pole at $i$ is $3$.
The order of the pole at $1$ is $4$.

2

$\sin(z) = 0$ if and only if $z = k\pi$.

For any $k \in \mathbb{Z}$,

\[\begin{align*} \frac{z - k\pi}{\sin(z)} &= \frac{z - k\pi}{\sin(z - k\pi)} \\ &= \frac{z - k\pi}{(z - k\pi) - \frac{(z - k\pi)^3}{3!} + \frac{(z - k\pi)^5}{5!} - \cdots} \\ &= \frac{1}{1 - \frac{(z - k\pi)^2}{3!} + \frac{(z - k\pi)^4}{5!} - \cdots}. \end{align*}\]

The singularity at 0 is removable because the given function is $z\cot(z) = \cos(z)\frac{z}{\sin(z)}$.
The order of the pole at $k\pi$ for a nonzero $k \in \mathbb{Z}$ is 1.

3

\[\begin{align*} \frac{\sin(z)}{z^5} &= \frac{z - \frac{z^3}{3!} + \frac{z^5}{5!} - \cdots}{z^5} \\ &= \frac{1 - \frac{z^2}{3!} + \frac{z^4}{5!} - \cdots}{z^4}. \end{align*}\]

The order of the pole at $0$ is 4.

4

$\exp(z) = 1$ if and only if $z = 2k\pi i$.

For any $k \in \mathbb{Z}$,

\[\begin{align*} \frac{z - 2k\pi i}{1 - \exp(z)} &= \frac{z - 2k\pi i}{1 - \exp(z - 2k\pi i)} \\ &= \frac{z - 2k\pi i}{1 - (1 + \frac{(z - 2k\pi i)}{1!} + \frac{(z - 2k\pi i)^2}{2!} + \cdots)} \\ &= \frac{-1}{\frac{1}{1!} + \frac{(z - 2k\pi i)^1}{2!} + \cdots}. \end{align*}\]

The order of the pole at $2k \pi i$ is 1 as shown above.

5

Using the same argument as above, we can conclude that

The singularity at 0 is removable because the given function is $\frac{z}{1 - \exp(z)}$.
The order of the pole at $2k \pi i$ with $k \ne 0$ is 1.