3 The residue theorem

Suppose $f\left(z\right)$ is a function which is analytic inside and on a closed contour $C$ , except for a pole of order $m$ at $z=z_0$ , which lies inside $C$ . To evaluate ${\oint }_{C}f\left(z\right)dz$ we can expand $f\left(z\right)$ in a Laurent series in powers of $\left(z-z_0\right)$ . If we let $\Gamma$ be a circle of centre $z_0$ lying inside $C$ then, as we saw in Section 262, ${\oint }_{C}f\left(z\right)dz={\int \; <!--nolimits\; -->}_{\Gamma }f\left(z\right)dz.$

From Key Point 1 in Section 26.4 we know that the integral of each of the positive and negative powers of $\left(z-z_0\right)$ is zero with the exception of $\frac{b_1}{z-z_0}$ and this has value $2\pi b_1$ . Since it is the only coefficient remaining after the integration, it is called the residue of $f\left(z\right)$ at $z=z_0$ . It is given by

$b_1=\frac{1}{2\pi i}{\oint }_{C}f\left(z\right)dz.$

Calculating the residue, for any given function $f\left(z\right)$ is an important task and we examine some results concerning its determination for functions with simple poles, double poles and poles of order $m$ .

Finding the residue

If $f\left(z\right)$ has a simple pole at $z=z_0$ then $f\left(z\right)=\frac{b_1}{z-z_0}+a_0+a_1\left(z-z_0\right)+a_2{\left(z-z_0\right)}^2+\dots$

so that $\left(z-z_0\right)f\left(z\right)=b_1+a_0\left(z-z_0\right)+a_1{\left(z-z_0\right)}^2+a_2{\left(z-z_0\right)}^3+\dots$

Taking limits as $z\to z_0$ , $\underset{z\to z_0}{lim}\left\{\left(z-z_0\right)f\left(z\right)\right\}=b_1.$

For example, let $f\left(z\right)=\frac{1}{z^2+1}\equiv \frac{1}{\left(z+i\right)\left(z-i\right)}\equiv \frac{-\frac{1}{2i}}{z+i}+\frac{\frac{1}{2i}}{z-i}$ .

There are simple poles at $z=-i$ and $z=i$ . The residue at $z=i$ is

$\underset{z\to i}{lim}\left\{\left(z-i\right)\frac{1}{\left(z+i\right)\left(z-i\right)}\right\}=\underset{z\to i}{lim}\left(\frac{1}{z+i}\right)=\frac{1}{2i}.$

Similarly, the residue at $z=-i$ is

$\underset{z\to i}{lim}\left\{\left(z+i\right)\frac{1}{\left(z+i\right)\left(z-i\right)}\right\}=\underset{z\to -i}{lim}\left(\frac{1}{z-i}\right)=\frac{-1}{2i}.$

Task!

This Task concerns $f\left(z\right)=\frac{1}{z^2+4}$ .

1. Identify the singularities of $f\left(z\right)$ :

   Answer

   $f\left(z\right)=\frac{1}{\left(z+2i\right)\left(z-2i\right)}=\frac{-\frac{1}{4i}}{z+2i}+\frac{\frac{1}{4i}}{z-2i}$ . There are simple poles at $z=-2i$ and $z=2i$ .

2. Now find the residues of $f\left(z\right)$ at $z=2i$ and at $z=-2i$ :

   Answer

   $\underset{z\to 2i}{lim}\left\{\left(z-2i\right)\frac{1}{\left(z+2i\right)\left(z-2i\right)}\right\}=\underset{z\to 2i}{lim}\left(\frac{1}{z+2i}\right)=\frac{1}{4i}$ .

   Similarly at $z=-2i$ .

   $\underset{z\to -2i}{lim}\left\{\left(z+2i\right)\frac{1}{\left(z+2i\right)\left(z-2i\right)}\right\}=\underset{z\to -2i}{lim}\left(\frac{1}{z-2i}\right)=-\frac{1}{4i}.$

In general the residue at a pole of order $m$ at $z=z_0$ is

$b_1=\frac{1}{\left(m-1\right)!}\underset{z\to z_0}{lim}\left\{\frac{d^{m-1}}{d{z}^{m-1}}\left[{\left(z-z_0\right)}^{m}f\left(z\right)\right]\right\}.$

As an example, if $f\left(z\right)=\frac{z^2+1}{{\left(z+1\right)}^3},f\left(z\right)$ has a pole of order 3 at $z=-1$ ( $m=3$ ).

We need first

$\frac{d^2}{d{z}^2}\left[{\left(z+1\right)}^3\frac{\left(z^2+1\right)}{{\left(z+1\right)}^3}\right]=\frac{d^2}{d{z}^2}\left[z^2+1\right]=\frac{d}{dz}\left[2z\right]=2.$

Then $b_1=\frac{1}{2!}\times 2=1$ .

We have a useful result (Key Point 5) which allows us to evaluate contour integrals quickly when $f\left(z\right)$ has only poles inside the contour.

Example 16

Let $f\left(z\right)=\frac{1}{z^2+1}$ . Find the integrals ${\oint }_{C_1}dz,{\oint }_{C_2}dz$ and ${\oint }_{C_3}dz$ in which $C_1$ is the circle $\left|z-i\right|=1$ , $C_2$ is the circle $\left|z+i\right|=1$ , and $C_3$ is any path enclosing both $z=i$ and $z=-i$ . See Figure 17.

Figure 17

Solution

Figure 17 shows that only the pole at $z=i$ lies inside $C_1$ . The residue at this pole is $\frac{1}{2i}$ , as we found earlier. Hence ${\oint }_{C_1}f\left(z\right)dz=2\pi i\times \frac{1}{2i}=\pi$ .

Also, the residue at $z=-i$ , the only pole inside $C_2$ , is $-\frac{1}{2i}$ . Hence

${\oint }_{C_2}f\left(z\right)dz=-2\pi i\times \frac{1}{2i}=-\pi .$

Note that the contour $C_3$ encloses both poles so that ${\oint }_{C_3}f\left(z\right)dz=2\pi i\left(\frac{1}{2i}-\frac{1}{2i}\right)=0$ .

Exercises

1. Identify the singularities of  $f\left(z\right)=\frac{1}{z^2\left(z^2+9\right)}$  and find the residue at each of them.
2. Find the integral   ${\oint }_{C}f\left(z\right)dz$  where $f\left(z\right)=\frac{1}{z^2+4}$  and $C$ is
   1. the circle  $\left|z-2i\right|=1;$
   2. the circle   $\left|z+2i\right|=1;$
   3. any closed path enclosing both $z=2i$  and   $z=-2i$ .

Answer