\[ \newcommand{\Ann}{\mathsf{Ann}} \newcommand{\Arg}{\mathsf{Arg}} \newcommand{\C}{\mathbb{C}} \newcommand{\CC}{\mathbb{C}} \newcommand{\R}{\mathbb{R}} \newcommand{\N}{\mathbb{N}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\Im}{\mathsf{Im}} \newcommand{\intd}{\,\mathsf{d}} \newcommand{\Re}{\mathsf{Re}} \newcommand{\ball}{\mathsf{B}} \newcommand{\Log}{\mathsf{Log}} \newcommand{\exp}{\mathsf{exp}} \newcommand{\sin}{\mathsf{sin}} \newcommand{\cos}{\mathsf{cos}} \newcommand{\Res}{\mathsf{Res}} \newcommand{\wind}{\mathsf{wind}} \]

10.3 Calculating Residues

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In order to apply Cauchy's Residue Theorem we need to be able to easily calculate residues. In some cases, ad hoc manipulations have to be used to calculate the Laurent series, but there are many cases where one can calculate them more systematically. Recall that if $f(z)$ has Laurent series \[ f(z) = \frac{a_{-m}}{(z-b)^m} + \cdots + \frac{a_{-1}}{(z-b)} + \sum_{n=0}^{\infty} a_{n}(z-b)^n \] with $a_m \ne 0$ then we say that $f$ has a pole of order $m$ at $b$. It is easy to calculate the residue at a simple pole.

Lemma

  1. If $f$ has a simple pole at $b$ then $\Res(f,b) = \lim\limits_{z\to b} (z-b) f(z)$.
  2. If $f(z) = p(z)/q(z)$ where $p,q$ are holomorphic, $p(b) \ne 0$, $q(b)=0$ and $q'(b) \ne 0$, then $\Res(f,b) = \dfrac{p(b)}{q'(b)}$.
Proof:

  1. If $f$ has a simple pole at $b$ then it has a Laurent series \[ f(z) = \frac{a_{-1}}{z-b} + \sum_{n=0}^{\infty} a_n (z-b)^n \] valid on some punctured disc $\Ann(b,0,R)$. Hence \[ (z-b)f(z) = a_{-1} + \sum_{n=0}^{\infty} a_n (z-b)^{n+1} \] so that $\lim\limits_{z \to b} (z-b)f(z) = a_{-1}$.
  2. The hypotheses imply that $f$ has a simple pole at $b$. By the first part and the fact that $q(b)=0$, the residue is \[ \lim_{z \to b} \frac{(z-b)p(z)}{q(z)} = \lim_{z \to b} \frac{p(z)}{\left( \frac{q(z)-q(b)}{z-b} \right)} = \dfrac{p(b)}{q'(b)}. \] from the definition of th derivative.
Example

For example, let \[ f(z) = \frac{\cos(\pi z)}{(1-z^3)} \] on $\mathbb{C} \setminus \{1,e^{2i \pi / 3}, e^{4 i \pi / 3} \}$. This has a simple pole at $z=1$ and satisfies the hypothesis of the lemma. Hence \[ \Res(f,1) = \frac{\cos(\pi)}{(-3) \cdot 1^{2}} = \frac{1}{3}. \]

We can generalise the first part of the lemma to poles of order $m$.

Lemma

If $f$ has a pole of order $m$ at $b$ then \[\Res(f,b) = \lim\limits_{z\to b} \left( \dfrac{1}{(m-1)!} \dfrac{\mathsf{d}^{m-1}}{\mathsf{d} z^{m-1}} \left( (z-b)^{m}f(z) \right) \right) \]

Proof:

If $f$ has a pole of order $m$ at $b$ then it has a Laurent series \[ f(z) = \frac{a_{-m}}{(z-b)^m} + \cdots + \frac{a_{-1}}{z-b} + \sum_{n=0}^{\infty} a_n (z-b)^n \] on $\Ann(b,0,R)$. Hence \[ \begin{align*} & (z-b)^m f(z) \\ ={} & a_{-m} + (z-b) a_{-(m-1)} + \cdots + (z-b)^{m-1} a_{-1} + \sum_{n=0}^{\infty} a_n (z-b)^{m+n} \end{align*} \] on $\Ann(b,0,R)$. Differentiating this $m-1$ times gives \[ \begin{align*} & \frac{\mathsf{d}^{m-1}}{\mathsf{d} z^{m-1}} (z-b)^m f(z) \\ ={} & (m-1)! a_{-1} + \sum_{n=0}^{\infty} \frac{(m+n)!}{(n+1)!} a_{n} (z-b)^{n+1} \end{align*} \] and, after dividing by $(m-1)!$ and letting $z \to b$, we get the desired result. $\square$

Example

Let \[ f(z) = \left( \frac{z+1}{z-1} \right)^3 \] on $\C \setminus \{1\}$. This has a pole of order three at $1$. To calculate the residue we note that $(z-1)^{3} f(z) = (z+1)^{3}$. Hence \[ \frac{1}{2!} \frac{\mathsf{d}^2}{\mathsf{d}z^2} \left( (z-1)^{3} f(z) \right) = \frac{6}{2!}(z+1) \to \frac{6}{2!} \times 2 = 6 \] as $z \to 1$. Hence $\Res(f,1)=6$.

Let us check this by calculating the Laurent series. First let us change variables by writing $w=z-1$. Then $z=w+1$ and we can write \[ \begin{aligned} \left( \frac{z+1}{z-1} \right)^3 & = \frac{(w+2)^3}{w^3} \\ & = \frac{w^3+6w^2+12w+8}{w^3} \\ & = \frac{8}{w^3} + \frac{12}{w^2} + \frac{6}{w} + 1 \\ & = \frac{8}{(z-1)^3} + \frac{12}{(z-1)^2} + \frac{6}{(z-1)} + 1 \end{aligned} \] Hence $f$ has a pole of order three at $z=1$ and we can read off $\Res(f,1)=6$ as the coefficient of $1/(z-1)$.

In other cases, one has to manipulate the formula for $f$ to calculate the residue.

Example

Let \[ f(z) = \frac{1}{z \sin(z)}. \] on $\C \setminus \{ k \pi : k \in \Z \}$. This has singularities whenever the denominator is zero. Hence the singularities are at $k \pi$ for $k \in \Z$.

For $k \ne 0$ the denominator has a simple zero at $k \pi$ so $f$ has a simple pole $k \pi$. Thus \[ \Res(f,k\pi) = \dfrac{1}{\sin(k\pi) + k \pi \cos(k \pi)} = \dfrac{(-1)^k}{k \pi} \] using the rule for the residue of a ratio at a simple pole.

The derivative of the denominator is zero at $0$ but its second derivative is not, so the denominator has a zero of order two at $0$. We can calculate the residue at $0$ with the $m=2$ case of the rule for calculating residues at higher-order poles. Thus \[ \Res(f,0) = \lim_{z \to 0} ( z^2 f(z) )' = \lim_{z \to 0} \left( \dfrac{z}{\sin(z)} \right)' = \lim_{z \to 0} \dfrac{\sin(z) - z \cos(z)}{(\sin z)^2} \] but how can we calculate this limit? We can calculate the orders of the zero at $0$ of the numerator and the denominator. From their power series \[ \begin{aligned} \sin(z) - z \cos(z) & = \left( z - \dfrac{z^3}{3!} + \dfrac{z^5}{5!} - \cdots \right) - z \left( 1 - \dfrac{z^2}{2!} + \dfrac{z^4}{4!} - \cdots \right) \\ & = z^3 \left( \dfrac{1}{2!} - \dfrac{1}{3!} \right) + \cdots \\ & = \dfrac{1}{3} z^3 + \cdots = \dfrac{1}{3} z^3 g(z) \end{aligned} \] where $g$ is holomorphic on some $\ball(0,r)$ and $g(0) \ne 0$. Similarly \[ (\sin z)^2 = \dfrac{1}{2} (1 - \cos(2z)) = \dfrac{1}{2} \left( \dfrac{4z^2}{2!} + \cdots \right) = z^2 h(z) \] where $h$ is holomorphic on some $\ball(0,s)$ and $h(0) \ne 0$. Thus \[ \Res(f,0) = \lim_{z \to 0} \dfrac{\frac{1}{3} z^3 g(z)}{z^2 h(z)} = \lim_{z \to 0} \dfrac{z h(z)}{3 g(z)} = 0 \] so the residue at $0$ is $0$.