# 9.4: Residues

- Page ID
- 6525

In this section we’ll explore calculating residues. We’ve seen enough already to know that this will be useful. We will see that even more clearly when we look at the residue theorem in the next section.

We introduced residues in the previous topic. We repeat the definition here for completeness.

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

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

The **residue **of \(f\) at \(z_0\) is \(b_1\). This is denoted

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

What is the importance of the residue? If \(\gamma\) is a small, simple closed curve that goes counterclockwise around \(b_1\) 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 \(b_1/z\).

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

has an isolated singularity at 0. From the Laurent series we see that \(\text{Res} (f, 0) = 1/2\).

(i) Let

\[f(z) = \dfrac{1}{z^3} + \dfrac{2}{z^2} + \dfrac{4}{z} + 5 + 6z. \nonumber\]

\(f\) has a pole of order 3 at \(z = 0\) and \(\text{Res} (f, 0) = 4.\)

(ii) Suppose

\[f(z) = \dfrac{2}{z} + g(z), \nonumber\]

where \(g\) is analytic at \(z = 0\). Then, \(f\) has a simple pole at 0 and \(\text{Res} (f, 0) = 2\).

(iii) Let

\[f(z) = \cos (z) = 1 - z^2/2! + \ ... \nonumber\]

Then \(f\) is analytic at \(z = 0\) and \(\text{Res} (f, 0) = 0.\)

(iv) Let

\[\begin{align*} f(z) &= \dfrac{\sin (z)}{z} \\[4pt] &= \dfrac{1}{z} (z - \dfrac{z^3}{3!} + \ ...) = 1 - \dfrac{z^2}{3!} + \ ... \end{align*}\]

So, \(f\) has a removable singularity at \(z = 0\) and \(\text{Res} (f, 0) = 0.\)

Let

\[f(z) = \dfrac{z}{z^2 + 1}. \nonumber\]

Find the poles and residues of \(f\).

**Solution**

Using partial fractions we write

\[\begin{align*} f(z) &= \dfrac{z}{(z - i)(z + i)} \\[4pt] &= \dfrac{1}{2} \cdot \dfrac{1}{z - i} + \dfrac{1}{2} \cdot \dfrac{1}{z + i}. \end{align*}\]

The poles are at \(z = \pm i\). We compute the residues at each pole:

At \(z = i\):

\[f(z) = \dfrac{1}{2} \cdot \dfrac{1}{z - i} + \text{ something analytic at } i. \nonumber\]

Therefore the pole is simple and \(\text{Res} (f, i) = 1/2\).

At \(z = -i\):

\[f(z) = \dfrac{1}{2} \cdot \dfrac{1}{z + i} + \text{ something analytic at } -i. \nonumber\]

Therefore the pole is simple and \(\text{Res} (f, -i) = 1/2\).

Let

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

then we have the following Laurent expansions for \(f\) around \(z = 0\).

On \(0 < |z| < 1\):

\[\begin{align*} f(z) &= -\dfrac{1}{z} \cdot \dfrac{1}{1 - z} \\[4pt] &= -\dfrac{1}{z} (1 + z + z^2 +\ ...). \end{align*}\]

Therefore the pole at \(z = 0\) is simple and \(\text{Res}(f, 0) = -1\).

On \(1 < |z| < \infty\):

\[\begin{align*} f(z) &= \dfrac{1}{z^2} \cdot \dfrac{1}{1 - 1/z} \\[4pt] &= \dfrac{1}{z} (1 + \dfrac{1}{z} + \dfrac{1}{z^2} + \ ...). \end{align*}\]

Even though this is a valid Laurent expansion you must not use it to compute the residue at 0. This is because the definition of residue requires that we use the Laurent series on the region \(0 < |z - z_0| < r\).

Let

\[f(z) = \log (1 + z). \nonumber\]

This has a singularity at \(z = -1\), but it is not isolated, so not a pole and therefore there is no residue at \(z = -1\).

## Residues at Simple Poles

Simple poles occur frequently enough that we’ll study computing their residues in some detail. Here are a number of ways to spot a simple pole and compute its residue. The justification for all of them goes back to Laurent series.

Suppose \(f(z)\) has an isolated singularity at \(z = z_0\). Then we have the following properties.

If the Laurent series for \(f(z)\) has the form

\[\dfrac{b_1}{z - z_0} + a_0 + a_1 (z - z_0) + \ ...\]

then \(f\) has a simple pole at \(z_0\) and \(\text{Res} (f, z_0) = b_1\).

If

\[g(z) = (z - z_0) f(z)\]

is analytic at \(z_0\) then \(z_0\) is either a simple pole or are movable singularity. In either case \(\text{Res} (f, z_0) = g(z_0)\). (In the removable singularity case the residue is 0.)

**Proof**-
Directly from the Laurent series for \(f\) around \(z_0\).

If \(f\) has a simple pole at \(z_0\) then

\[\lim_{z \to z_0} (z - z_0) f(z) = \text{Res} (f, z_0)\]

This says that the limit exists and equals the residue. Conversely, if the limit exists then either the pole is simple, or \(f\) is analytic at \(z_0\). In both cases the limit equals the residue.

**Proof**-
Directly from the Laurent series for \(f\) around \(z_0\).

If \(f\) has a simple pole at \(z_0\) and \(g(z)\) is analytic at \(z_0\) then

\[\text{Res} (fg, z_0) = g(z_0) \text{Res} (f, z_0).\]

If \(g(z_0) \ne 0\) then

\[\text{Res}(f/g, z_0) = \dfrac{1}{g(z_0)} \text{Res} (f, z_0).\]

**Proof**-
Since \(z_0\) is a simple pole,

\[f(z) = \dfrac{b_1}{z - z_0} + a_0 + a_1 (z - z_0)\]

Since \(g\) is analytic,

\[g(z) = c_0 + c_1 (z - z_0) + \ ...,\]

where \(c_0 = g(z_0)\). Multiplying these series together it is clear that

\[\text{Res} (fg, z_0) = c_0 b_1 = g(z_0) \text{Res} (f, z_0). \ \ \ \ \ \ \ \ \ \text{QED}\]

The statement about quotients \(f/g\) follows from the proof for products because \(1/g\) is analytic at \(z_0\).

If \(g(z)\) has a simple zero at \(z_0\) then \(1/g(z)\) has a simple pole at \(z_0\) and

\[\text{Res} (1/g, z_0) = \dfrac{1}{g'(z_0)}.\]

**Proof**-
The algebra for this is similar to what we’ve done several times above. The Taylor expansion for \(g\) is

\[g(z) = a_1 (z - z_0) + a_2 (z - z_0)^2 + \ ...,\]

where \(a_1 = g'(z_0)\). So

\[\dfrac{1}{g(z)} = \dfrac{1}{a_1 (z - z_0)} (\dfrac{1}{1 + \dfrac{a_2}{a_1} (z - z_0) + \ ...})\]

The second factor on the right is analytic at \(z_0\) and equals 1 at \(z_0\). Therefore we know the Laurent expansion of \(1/g\) is

\[\dfrac{1}{g(z)} = \dfrac{1}{a_1 (z - z_0)} (1 + c_1 (z - z_0) + \ ...)\]

Clearly the residue is \(1/a_1 = 1/g'(z_0)\). \(\text{QED}\).

Let

\[f(z) = \dfrac{2 + z + z^2}{(z - 2)(z - 3)(z - 4)(z - 5)}. \nonumber\]

Show all the poles are simple and compute their residues.

**Solution**

The poles are at \(z = 2, 3, 4, 5\). They are all isolated. We'll look at \(z = 2\) the others are similar. Multiplying by \(z - 2\) we get

\[g(z) = (z - 2)f(z) = \dfrac{2 + z + z^2}{(z - 3) (z - 4) (z - 5)}. \nonumber\]

This is analytic at \(z = 2\) and

\[g(2) = \dfrac{8}{-6} = -\dfrac{4}{3}. \nonumber\]

So the pole is simple and \(\text{Res} (f, 2) = -4/3\).

Let

\[f(z) = \dfrac{1}{\sin (z)}. \nonumber\]

Find all the poles and their residues.

**Solution**

The poles of \(f(z)\) are the zeros of \(\sin (z)\), i.e. \(n \pi\) for \(n\) an integer. Since the derivative

\[\sin '(n\pi) = \cos (n \pi) \ne 0, \nonumber\]

the zeros are simple and by Property 5 above

\[\text{Res} (f, n\pi) = \dfrac{1}{\cos (n \pi)} = (-1)^n. \nonumber\]

Let

\[f(z) = \dfrac{1}{z(z^2 + 1)(z - 2)^2}. \nonumber\]

Identify all the poles and say which ones are simple.

**Solution**

Clearly the poles are at \(z = 0\), \(\pm i\), 2.

At \(z = 0\):

\[g(z) = zf(z) \nonumber\]

is analytic at 0 and \(g(0) = 1/4\). So the pole is simple and the residue is \(g(0) = 1/4\).

At \(z = i\):

\[g(z) = (z - i) f(z) = \dfrac{1}{z(z + i)(z - 2)^2} \nonumber\]

is analytic at \(i\), the pole is simple and the residue is \(g(i)\).

At \(z = -i\): This is similar to the case \(z = i\). The pole is simple.

At \(z = 2\):

\[g(z) = (z - 2) f(z) = \dfrac{1}{z(z^2 + 1)(z - 2)} \nonumber\]

is not analytic at 2, so the pole is not simple. (It should be obvious that it’s a pole of order 2.)

Let \(p(z)\), \(q(z)\) be analytic at \(z = z_0\). Assume \(p(z_0) \ne 0\), \(q(z_0) = 0\), \(q'(z_0) \ne 0\). Find

\[\text{Res}_{z = z_0} \dfrac{p(z)}{q(z)}. \nonumber\]

**Solution**

Since \(q'(z_0) \ne 0\), \(q\) has a simple zero at \(z_0\). So \(1/q\) has a simple pole at \(z_0\) and

\[\text{Res} (1/q, z_0) = \dfrac{1}{q'(z_0)}\nonumber\]

Since \(p(z_0) \ne 0\) we know

\[\text{Res} (p/q, z_0) = p(z_0) \text{Res} (1/q, z_0) = \dfrac{p(z_0)}{q'(z_0)}. \nonumber\]

## Residues at finite poles

For higher-order poles we can make statements similar to those for simple poles, but the formulas and computations are more involved. The general principle is the following

If \(f(z)\) has a pole of order \(k\) at \(z_0\) then

\[g(z) = (z - z_0)^k f(z)\]

is analytic at \(z_0\) and if

\[g(z) = a_0 + a_1 (z - z_0) + \ ...\]

then

\[\text{Res} (f,z_0) = a_{k - 1} = \dfrac{g^{(k - 1)} (z_0)}{(k - 1)!}.\]

**Proof**-
This is clear using Taylor and Laurent series for \(g\) and \(f\).

Let

\[f(z) = \dfrac{\sinh (z)}{z^5}\]

and find the residue at \(z = 0\).

**Solution**

We know the Taylor series for

\[\sinh (z) = z + z^3/3! + z^5/5! + \ ...\]

(You can find this using \(\sinh (z) = (e^z - e^{-z})/2\) and the Taylor series for \(e^z\).) Therefore,

\[f(z) = \dfrac{1}{z^4} + \dfrac{1}{3! z^2} + \dfrac{1}{5!} + \ ...\]

We see \(\text{Res} (f, 0) = 0.\)

Note, we could have seen this by realizing that \(f(z)\) is an even function.

Let

\[f(z) = \dfrac{\sinh (z) e^z}{z^5}.\]

Find the residue at \(z = 0\).

**Solution**

It is clear that \(\text{Res} (f, 0)\) equals the coefficient of \(z^4\) in the Taylor expansion of \(\sinh (z) e^z\). We compute this directly as

\[\sinh (z) e^z = (z + \dfrac{z^3}{3!} + \ ...) (1 + z + \dfrac{z^2}{2} + \dfrac{z^3}{3!} + \ ...) = \ ... + (\dfrac{1}{4!} + \dfrac{1}{3!}) z^4 + \ ...\]

So

\[\text{Res} (f, 0) = \dfrac{1}{3!} + \dfrac{1}{4!} = \dfrac{5}{24}.\]

Find the residue of

\[f(z) = \dfrac{1}{z(z^2 + 1) (z - 2)^2}\]

at \(z = 2\).

**Solution**

\(g(z) = (z - 2)^2 f(z) = \dfrac{1}{z(z^2 + 1)}\) is analytic at \(z = 2\). So, the residue we want is the \(a_1\) term in its Taylor series, i.e. \(g' (2)\). This is easy, if dull, to compute

\[\text{Res} (f, 2) = g'(2) = -\dfrac{13}{100}\]

## \(\cot (z)\)

The function \(\cot (z)\) turns out to be very useful in applications. This stems largely from the fact that it has simple poles at all multiples of \(\pi\) and the residue is 1 at each pole. We show that first.

\(f(z) = \cot (z)\) has simple poles at \(n \pi\) for \(n\) an integer and \(\text{Res} (f, n\pi) = 1\).

**Proof**-
\[f(z) = \dfrac{\cos (z)}{\sin (z)}.\]

This has poles at the zeros of sin, i.e. at \(z = n \pi\). At poles \(f\) is of the form \(p/q\) where \(q\) has a simple zero at \(z_0\) and \(p(z_0) \ne 0\). Thus we can use the formula

\[\text{Res} (f, z_0) = \dfrac{p(z_0)}{q'(z_0)}.\]

In our case, we have

\[\text{Res} (f, n\pi) = \dfrac{\cos (n \pi)}{\cos (n \pi)} = 1,\]

as claimed.

Sometimes we need more terms in the Laurent expansion of \(\cot (z)\). There is no known easy formula for the terms, but we can easily compute as many as we need using the following technique.

Compute the first several terms of the Laurent expansion of \(\cot (z)\) around \(z = 0\).

**Solution**

Since \(\cot (z)\) has a simple pole at 0 we know

\[\cot (z) = \dfrac{b_1}{z} + z_0 + a_1 z + a_2 z^2 + \ ...\]

We also know

\[\cot (z) = \dfrac{\cos (z)}{\sin (z)} = \dfrac{1 - z^2/2 + z^4/4! - \ ...}{z - z^3/3! + z^5/5! - \ ...}\]

Cross multiplying the two expressions we get

\[(\dfrac{b_1}{z} + a_0 + a_1 z + a_2 z^2 + \ ...) (z - \dfrac{z^3}{3!} + \dfrac{z^5}{5!} - \ ...) = 1 - \dfrac{z^2}{2} + \dfrac{z^4}{4!} - \ ...\]

We can do the multiplication and equate the coefficients of like powers of \(z\).

\[b_1 + a_0 z + (-\dfrac{b_1}{3!} + a_1) z^2 + (-\dfrac{a_0}{3!} + a_2) z^3 + (\dfrac{b_1}{5!} - \dfrac{a_1}{3!} + a_3) z^4 = 1 - \dfrac{z^2}{2!} + \dfrac{z^4}{4!}\]

So, starting from \(b_1 = 1\) and \(a_0 = 0\), we get

\[\begin{array} {rclcl} {-b_1/3! + a_1} & = & {-1/2!} & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ & {\Rightarrow \ \ \ \ \ a_1 = -1/3} \\ {-a_0/3! + a_2} & = & {0} & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ & {\Rightarrow \ \ \ \ \ a_2 = 0} \\ {b_1/5! - a_1/3! + a_3} & = & {1/4!} & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ & {\Rightarrow \ \ \ \ \ a_3 = -1/45} \end{array}\]

As noted above, all the even terms are 0 as they should be. We have

\[\cot (z) = \dfrac{1}{z} - \dfrac{z}{3} - \dfrac{z^3}{45} + ...\]