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# 8.9: Poles

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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

$\int_{\gamma} f(z) = 2\pi i b_1.$ Figure $$\PageIndex{1}$$: $$\gamma$$ is small enough to be inside $$|z - z_0| < r$$, and surround $$z_0$$. (CC BY-NC; Ümit Kaya)

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.