# The Residue of an Analytic Function at a Pole Singularity

Recall from The Residue of an Analytic Function at a Point page that if $f$ is analytic on the annulus $A(z_0, 0, r)$ and if $\displaystyle{\sum_{n=-\infty}^{\infty} a_n(z – z_0)^n}$ is the Laurent series expansion of $f$ on $A(z_0, 0, r)$ then the residue of $f$ at $z_0$ is defined as:

(1)

We proved in a simple proposition that if $z_0$ is a removable singularity of $f$ then $\mathrm{Res} (f, z_0) = 0$. We will now look at another proposition which characterizes the residues $\mathrm{Res} (f, z_0)$ for which $z_0$ is a pole singularity of order $k$.

Proposition 1: If $z_0$ is a pole singularity of order $k$ of $f$ then $\displaystyle{\mathrm{Res} (f, z_0) = \lim_{z \to z_0} \frac{[(z – z_0)^k f(z)]^{(k-1)}}{(k-1)!}}$. |

**Proof:**Let $z_0$ be a pole singularity of order $k$ of $f$. Then there exists a function, $g$, which is analytic, $g(z_0) \neq 0$, such that:

(2)

- Hence for any positively oriented closed piecewise smooth curve $\gamma$ in $A(z_0, 0, r)$ ($r > 0$) we have that:

(3)

- By reversing Cauchy’s Integral Formula for Derivatives we have that:

(4)

- From $(*)$ we see that:

(5)

- Thereforefore:

(6)

For example, consider the following function:

(7)

Suppose that we want to find $\mathrm{Res} (f, 0)$. Note that $z_0 = 0$ is an isolated singularity of $f$. If we can show that $z_0 = 0$ is a pole singularity of order $k$ of $f$ then we can apply the theorem above to compute $\mathrm{Res} (f, 0)$.

We have that:

(8)

Therefore:

(9)

Hence $z_0 = 0$ is a pole singularity of order $k = 1$ of $f$. We also see from the expansion above that $\mathrm{Res} (f, 0) = 1$ (the coefficient attached to the term, $\displaystyle{\frac{1}{z}}$, but let’s use the formula in the previous proposition to compute this. Since $z_0 = 0$ is a pole of order $k = 1$ we have that:

(10)

By L’Hospital’s rule:

(11)

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