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9.4: Residues

( \newcommand{\kernel}{\mathrm{null}\,}\)

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.

Definition: Residue

Consider the function f(z) with an isolated singularity at z0, i.e. defined on the region 0<|zz0|<r and with Laurent series (on that region)

f(z)=n=1bn(zz0)n+n=0an(zz0)n.

The residue of f at z0 is b1. This is denoted

Res(f,z0)=b1     or     Resz=z0f=b1.

What is the importance of the residue? If γ is a small, simple closed curve that goes counterclockwise around b1 then

γf(z)=2πib1.

8.9.1.svg
Figure 9.4.1: γ small enough to be inside |zz0|<r, surround z0 and contain no other singularity of f. (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 b1/z.

Example 9.4.1

f(z)=e1/2z=1+12z+12(2z)2+ ...

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

Example 9.4.2

(i) Let

f(z)=1z3+2z2+4z+5+6z.

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

(ii) Suppose

f(z)=2z+g(z),

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

(iii) Let

f(z)=cos(z)=1z2/2!+ ...

Then f is analytic at z=0 and Res(f,0)=0.

(iv) Let

f(z)=sin(z)z=1z(zz33!+ ...)=1z23!+ ...

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

Example 9.4.3 Using partial fractions

Let

f(z)=zz2+1.

Find the poles and residues of f.

Solution

Using partial fractions we write

f(z)=z(zi)(z+i)=121zi+121z+i.

The poles are at z=±i. We compute the residues at each pole:

At z=i:

f(z)=121zi+ something analytic at i.

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

At z=i:

f(z)=121z+i+ something analytic at i.

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

Example 9.4.4 Mild warning!

Let

f(z)=1z(1z)

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

On 0<|z|<1:

f(z)=1z11z=1z(1+z+z2+ ...).

Therefore the pole at z=0 is simple and Res(f,0)=1.

On 1<|z|<:

f(z)=1z2111/z=1z(1+1z+1z2+ ...).

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<|zz0|<r.

Example 9.4.5

Let

f(z)=log(1+z).

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=z0. Then we have the following properties.

Property 1

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

b1zz0+a0+a1(zz0)+ ...

then f has a simple pole at z0 and Res(f,z0)=b1.

Property 2

If

g(z)=(zz0)f(z)

is analytic at z0 then z0 is either a simple pole or are movable singularity. In either case Res(f,z0)=g(z0). (In the removable singularity case the residue is 0.)

Proof

Directly from the Laurent series for f around z0.

Property 3

If f has a simple pole at z0 then

lim

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.

Property 4

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). \nonumber

If g(z_0) \ne 0 then

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

Proof

Since z_0 is a simple pole,

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

Since g is analytic,

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

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} \nonumber

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

Property 5

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)}. \nonumber

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 + \ ..., \nonumber

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) + \ ...}) \nonumber

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) + \ ...) \nonumber

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

Example \PageIndex{6}

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.

Example \PageIndex{7}

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

Example \PageIndex{8}

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.)

Example \PageIndex{9}

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

Higher order poles

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

g(z) = (z - z_0)^k f(z) \nonumber

is analytic at z_0 and if

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

then

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

Proof

This is clear using Taylor and Laurent series for g and f.

Example \PageIndex{10}

Let

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

and find the residue at z = 0.

Solution

We know the Taylor series for

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

(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!} + \ ... \nonumber

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

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

Example \PageIndex{11}

Let

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

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 + \ ... \nonumber

So

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

Example \PageIndex{12}

Find the residue of

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

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} \nonumber

\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.

Fact

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)}. \nonumber

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)}. \nonumber

In our case, we have

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

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.

Example \PageIndex{13}

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 + \ ... \nonumber

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! - \ ...} \nonumber

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!} - \ ... \nonumber

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!} \nonumber

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} \nonumber

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} + ... \nonumber


This page titled 9.4: Residues 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 the style and standards of the LibreTexts platform.

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