
8.9: Poles


Poles refer to isolated singularities. So, we suppose $$f(z)$$ is analytic on $$0 < |z - z_0| < r$$ and has Laurent series

$f(z) = \sum_{n = 1}^{\infty} \dfrac{b_n}{(z - z_0)^n} + \sum_{n = 0}^{\infty} a_n (z - z_0)^n.$

Definition: poles

If only a finite number of the coefficients $$b_n$$ are nonzero we say $$z_0$$ is a finite pole of $$f$$. In this case, if $$b_k \ne 0$$ and $$b_n = 0$$ for all $$n > k$$ then we say $$z_0$$ is a pole of order $$k$$.

• If $$z_0$$ is a pole of order 1 we say it is a simple pole of $$f$$.
• If an infinite number of the $$b_n$$ are nonzero we say that $$z_0$$ is an essential singularity or a pole of infinite order of $$f$$.
• If all the $$b_n$$ are 0, then $$z_0$$ is called a removable singularity. That is, if we define $$f(z_0) = a_0$$ then $$f$$ is analytic on the disk $$|z - z_0| < r$$.

The terminology can be a bit confusing. So, imagine that I tell you that $$f$$ is defined and analytic on the punctured disk $$0 < |z - z_0| < r$$. Then, a priori, we assume $$f$$ has a singularity at $$z_0$$. But, if after computing the Laurent series we see there is no singular part we can extend the definition of $$f$$ to the full disk, thereby 'removing the singularity’.

We can explain the term essential singularity as follows. If $$f(z)$$ has a pole of order $$k$$ at $$z_0$$ then $$(z - z_0)^k f(z)$$ is analytic (has a removable singularity) at $$z_0$$. So, $$f(z)$$ itself is not much harder to work with than an analytic function. On the other hand, if $$z_0$$ is an essential singularity then no algebraic trick will change $$f(z)$$ into an analytic function at $$z_0$$.

Examples of Poles

We’ll go back through many of the examples from the previous sections.

Example $$\PageIndex{1}$$

The rational function

$f(z) = \dfrac{1 + 2z^2}{z^3 + z^5} \nonumber$

expanded to

$f(z) = \left(\dfrac{1}{z^3} + \dfrac{1}{z}\right) - \sum_{n = 0}^{\infty} (-1)^n z^{2n + 1}. \nonumber$

Thus, $$z = 0$$ is a pole of order 3.

Example $$\PageIndex{2}$$

Consider

$f(z) = \dfrac{z + 1}{z} = 1 + \dfrac{1}{z}. \nonumber$

Thus, $$z = 0$$ is a pole of order 1, i.e. a simple pole.

Example $$\PageIndex{3}$$

Consider

$f(z) = \dfrac{z}{z^2 + 1} = \dfrac{1}{2} \cdot \dfrac{1}{z - i} + g(z), \nonumber$

where $$g(z)$$ is analytic at $$z = i$$. So, $$z = i$$ is a simple pole.

Example $$\PageIndex{4}$$

The function

$f(z) = \dfrac{1}{z(z - 1)} \nonumber$

has isolated singularities at $$z = 0$$ and $$z = 1$$. Show that both are simple poles.

Solution

In a neighborhood of $$z = 0$$ we can write

$f(z) = \dfrac{g(z)}{z}, \text{ where } g(z) = \dfrac{1}{z - 1}. \nonumber$

Since $$g(z)$$ is analytic at 0, $$z = 0$$ is a finite pole. Since $$g(0) \ne 0$$, the pole has order 1, i.e. it is simple. Likewise, in a neighborhood of $$z = 1$$,

$f(z) = \dfrac{h(z)}{z - 1}, \text{ where } h(z) = \dfrac{1}{z}. \nonumber$

Since $$h$$ is analytic at $$z = 1$$, $$f$$ has a finite pole there. Since $$h(1) \ne 0$$ it is simple.

Example $$\PageIndex{5}$$

Consider

$e^{1/z} = 1 + \dfrac{1}{z} + \dfrac{1}{2! z^2} + \dfrac{1}{3! z^3} + \ ... \nonumber$

So, $$z = 0$$ is an essential singularity.

Example $$\PageIndex{6}$$

$$\log (z)$$ has a singularity at $$z = 0$$. Since the singularity is not isolated, it can’t be classified as either a pole or an essential singularity.

Residues

In preparation for discussing the residue theorem in the next topic we give the definition and an example here.

Note well, residues have to do with isolated singularites.

Definition: Residue

Consider the function $$f(z)$$ with an isolated singularity at $$z_0$$, i.e. defined on $$0 < |z - z_0| < r$$ and with Laurent series

$f(z) = \sum_{n = 1}^{\infty} \dfrac{b_n}{(z - z_0)^n} + \sum_{n = 0}^{\infty} a_n (z - z_0)^n.$

The residue of $$f$$ at $$z_0$$ is $$b_1$$. This is denoted

$\text{Res}(f, z_0) \ \ \ \ or \ \ \ \ \text{Res}_{z = z_0} f = b_1.$

What is the importance of the residue? If $$\gamma$$ is a small, simple closed curve that goes counterclockwise around $$z_0$$ then

$\int_{\gamma} f(z) = 2\pi i b_1.$

This is easy to see by integrating the Laurent series term by term. The only nonzero integral comes from the term $$b_1/z$$.

Example $$\PageIndex{7}$$

The function

$f(z) = e^{1/(2z)} = 1 + \dfrac{1}{2z} + \dfrac{1}{2(2z)^2} + \ ... \nonumber$

has an isolated singularity at 0. From the Laurent series we see that

$\text{Res}(f, 0) = \dfrac{1}{2}. \nonumber$

8.9: Poles is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Jeremy Orloff (MIT OpenCourseWare) via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.