# Why the Residue?

**Mathematics 113**

*October 23, 2003*

On Wednesday (October 22), we had a class discussion about the residue
theorem and why it is true (as opposed to a proof that it is true).
There were a lot of good reasons given, but they came somewhat out of
order; this is my attempt to organize them into a logical order
according to which question they try to answer.

Some of these questions will seem completely intuitive and not in
need of an explanation to some of you, but I wanted to be as complete
as possible. It is my hope that you will gain an intuition for
analytic functions so that all of the questions seem obvious!

Please let me know if I missed anything, or if you come up with some
new answers to these questions that aren't listed!

**Question 1**
Why is the residue theorem true? For a simple closed curve

and
analytic function

with isolated singularities, why is

There were two basic approaches to this question suggested.

**Reason 1.1**
To evaluate a curve along an arbitrary closed curve, possibly
enclosing some isolated singularities, we can shrink the curve by
deformations until it becomes a union of small circles around each
singularity and straight lines [connecting the circles to some base
point]. The integrals along the straight lines cancel out [since
each line is traversed twice in opposite directions], and the
integral around a small circle picks out the

coefficient at
that singularity, times

.

Of course, this reason naturally raises two other questions: ``Why can
we deform the curve?'' (Question 2) and
``Why does the integral around a small circle pick out the
coefficient?'' (Question 4).

**Reason 1.2**
The integral is a

*linear* function of the function

, so
we can split

into an analytic function

and a principal
part for each singularity. The integral of the analytic function

vanishes, and the integral of a function with a single
singularity picks out the residue at the singularity times the
winding number.

Again, we have two subsidiary questions: ``Why is the integral of an
analytic function on a closed curve zero?''
(Question 3) and, again, why we get the
residue in particular (Question 4).

**Question 2**
Why is the integral along a complex contour invariant under
deformations that fix the endpoints (and that don't pass over any
singularities)?

**Reason 2.1**
If we deform a smooth curve a little bit and superimpose the new
curve on the old curve, we generally see some crescent-shaped
regions cut out by the two curves. The difference between the
integral of

along the two curves will be the sum of the
integrals along the boundary of these crescent-shaped regions; but
the integral of an analytic function

, defined on a
simply-connected region

, along the boundary of

is equal to zero.

**Reason 2.2**
We have seen that complex differentiable functions are incredibly
conservative, and have all sorts of nice properties. In particular,
once differentiable functions are both infinitely differentiable,
and have an anti-derivative (at least in small regions): if

is analytic in a disk, then there is a function

so that

But the Fundamental Theorem of Calculus [which is also true in the
complex setting] says that

In particular, the integral does not depend on the path we take to
get from

to

, as long as we stay within a region for which

is defined.

**Question 3**
Why is the integral of an analytic function

along a simple
closed curve

equal to 0 (as long as

is analytic in the
whole interior of

)?

**Reason 3.1**
The interior of

is a disk, which is simply-connected, so we can
find an anti-derivative

to the function

defined on the
interior of

as well as

itself. But then, by the Fundamental
Theorem of Calculus (as in Reason

2.2), if

runs from the point

back to

,

For all the above reasons, we reduce the original question to an
integral around a single singularity. For simplicity, let's suppose
that we're integrating around a circle of radius that winds once
counterclockwise around a singularity at 0.

**Question 4**
If

has an isolated singularity at 0, why is

That is, why does the integral pick up

times the coefficient
of

?

By linearity of the integral, we can reduce this question to the
integral of a simple power:

**Question 5**
Why is

equal to 0 if

, and equal to

if

?

We came up with many different answers to this question, addressing
different possible values of .

**Reason 5.1**
For

,

is analytic on the disk, and so the integral
around a closed curve is 0 as in Question

3.

**Reason 5.2**
For

,

has an antiderivative defined on

,
namely

, and so the integral of

around a
closed curve is 0 by the Fundamental Theorem of Calculus, as in
Reason

3.1.

**Reason 5.3**
For

, as we integrate around the closed circle, the angle of

is

and the angle of

is

; adding these
up, we find that

has a constant angle of

, so we
don't get any cancellation and the integral is at angle

, i.e.,
some multiple of

.
For

, the angle of

is

so the angle of

is

, which rotates as we go around the circle (with

increasing from 0 to

). Since the magnitude of the vectors we
add remains constant, the integral is 0 for

.

**Reason 5.4**
For

, we expect that

The logarithm function is the inverse of the exponential function,
and so if we write

in polar coordinates, we have

However, the angle

is only defined up to adding a multiple of

, and if we move around the circle counterclockwise starting
at 0, the angle

increases continuously from 0 up to

, so

(The notation is a little loose here, but that's OK, since we're
giving a reason, not a proof.)

**Reason 5.5**
For

,

, consider taking the original circle
and shrinking it down until it has radius

.
The length of the curve we integrate along scales down by a factor
of

, while the function

which we
are integrating scales up by a factor of

. These two
cancel out, so the integral seems to stay constant, and it is
reasonable to suppose that the magnitude of the integral is equal to
the length of the original curve, which is

.

**Reason 5.6**
For

, if we shrink our original contour to a circle of small
radius

, the length of the curve scales down by a factor of

while the value of the function

scales down by a factor of

(or remains constant, if

). By the M-L theorem, the
total integral is bounded above by some constant times

;
since

can be as small as we like, the integral must be 0.

**Reason 5.7**
For

, if we expand the original contour to a large
circle of radius

, the length of the contour scales up by a
factor of

while the function

scales down by
a factor of

. The scaling down of

beats the scaling
up of the length of the curve, so by the M-L theorem the integral is
bounded above by a constant times

, which can be made as
small as we like for

sufficiently large.

**Reason 5.8**
For

, consider perturbing the function a little bit:

(The second step is just the partial fraction expansion of

.) The perturbed function has a simple pole with
residue

at

and a simple pole with residue

at

. For

sufficiently small, both poles
will be contained inside our circle

, but since the two residues
have equal magnitude but opposite signs, the integral around any
loop that encloses both poles will be 0. (This argument is closely
related to the electric field of a dipole, written in terms of the
fields of two close and opposite-charged particles. A similar
argument works for any

.)

Dylan Thurston
2003-10-23