Residues and poles: Foundations — Maths deep dive

0. The plane where everything lives

Before any formula, you need to know what a complex number looks like.

Figure s01 — The point $z = x + i y$ drawn as the red dot at the tip of the blue arrow. The yellow dashed leg is $x = Re (z)$ (go right), the green dashed leg is $y = Im (z)$ (go up); the blue arrow's length is $∣ z ∣ = x^{2} + y^{2}$ . This is the "plane" every later picture lives on.

Why we need this. Residues live at points $z_{0}$ in this plane, and we integrate along paths drawn on this plane. If $z$ were just a number on a line, there would be no "going around" anything — you need two dimensions to loop.

1. What is a power series?

Several later definitions lean on the phrase "power series", so we pin it down before using it.

Why we need it. The complex exponential (next) is a power series, and "analytic" will mean "locally equal to a power series". So this is the vocabulary those definitions speak.

2. The complex exponential and polar form

To "go around" a point we need an angle. The tool that turns "angle" into "point on a circle" is the complex exponential, so we build it first.

Figure s02 — The blue arrow is $e^{i θ}$ , a point on the white dotted unit circle at the yellow angle $θ$ . The green arrow shows the direction of travel: hold $ρ$ fixed and let $θ$ increase from $0$ to $2 π$ and the red dot sweeps once, counter-clockwise, around the whole circle. This sweep is the "loop" we integrate over.

3. The complex derivative — what "analytic" really means

The word analytic rests on a limit, so we state that limit before using the word.

4. Functions of a complex number, and where they break

Figure s03 — The blue-shaded region is the punctured disk $0 < ∣ z - z_{0} ∣ < R$ : $f$ is analytic on the whole ring (green label) but the centre $z_{0}$ is poked out (red hollow dot). The yellow arrows point inward toward $z_{0}$ to suggest the values of $f$ shooting off to infinity as you approach the bad point.

5. The Laurent series — the fuel for everything

A Taylor series (a power series with only powers $z^{0}, z^{1}, z^{2}, \dots$ ) can only describe nice points. To describe a blow-up we also allow negative powers. Those negative powers only make sense in a ring-shaped region, so we first name that ring.

6. Two integrals-around-a-loop, and the theorems that tame them

7. The Residue Theorem — the derivation shown in full

We now compute $\oint_{C} f d z$ around a small circle of radius $ρ$ enclosing exactly the one singularity $z_{0}$ , step by step.

Step 1 — parametrize the loop. Let $C$ be the circle $∣ z - z_{0} ∣ = ρ$ traced counter-clockwise: $z = z_{0} + ρ e^{i θ}, θ \in [0, 2 π] .$ Differentiate this with respect to $θ$ to get the tiny step $d z$ : $\frac{d z}{d θ} = ρ \cdot i e^{i θ} ⟹ d z = i ρ e^{i θ} d θ .$ Why: it converts the abstract loop-integral into an ordinary integral in the single real variable $θ$ .

Step 2 — insert the Laurent series and swap sum with integral. Choose $ρ$ with $r < ρ < R$ so the circle lies inside the annulus. There the series converges uniformly, and — as noted in Section 5 — uniform convergence is precisely the theorem that permits swapping the infinite sum and the integral (finite errors stay finite when integrated term by term). Hence: $\oint_{C} f d z = \oint_{C} \sum_{n = - \infty}^{\infty} a_{n} (z - z_{0})^{n} d z = \sum_{n = - \infty}^{\infty} a_{n} \oint_{C} (z - z_{0})^{n} d z .$

Step 3 — evaluate one term. With $z - z_{0} = ρ e^{i θ}$ and $d z = i ρ e^{i θ} d θ$ : $\oint_{C} (z - z_{0})^{n} d z = \int_{0}^{2 π} (ρ e^{i θ})^{n} i ρ e^{i θ} d θ = i ρ^{n + 1} \int_{0}^{2 π} e^{i (n + 1) θ} d θ .$ Why: we collect all the $ρ$ 's and merge the two exponentials $e^{in θ} \cdot e^{i θ} = e^{i (n + 1) θ}$ .

Step 4 — apply the averaging fact with $k = n + 1$ . In Section 2's formula the exponent index was $k$ ; here $k = n + 1$ (and since $n$ is an integer, so is $k$ ). The formula splits into two cases — and we must keep both: $\int_{0}^{2 π} e^{i (n + 1) θ} d θ = {2 π 0 n + 1 = 0 (i.e. n = - 1) n + 1 \neq = 0 (i.e. n \neq = - 1) .$ Feeding each case back into Step 3: $\oint_{C} (z - z_{0})^{n} d z = {i ρ^{0} \cdot 2 π = 2 π i i ρ^{n + 1} \cdot 0 = 0 n = - 1 n \neq = - 1.$ Notice the radius $ρ$ vanished from the surviving case ( $ρ^{n + 1} = ρ^{0} = 1$ ) — the answer does not depend on how big the circle is.

Step 5 — collapse the sum. In the sum of Step 2, every single term is multiplied by $0$ except the one with $n = - 1$ , whose factor is $2 π i$ . That surviving term carries the coefficient $a_{- 1}$ : $\oint_{C} f (z) d z = \sum_{n \neq = - 1} a_{n} \cdot 0 + a_{- 1} \cdot 2 π i = 2 π i a_{- 1} = 2 π i Res (f, z_{0}) .$ So only the $a_{- 1}$ term survives, and the whole loop-integral equals $2 π i$ times that one number — the residue. That is the entire punchline.

How the foundations feed the topic

The map below reads top-to-bottom as a chain of dependence: each arrow means "you must own the upper box before the lower one is meaningful". The two theorem boxes (Cauchy, Residue) sit at the bottom because they consume everything above.

Equipment checklist

Cover the right side and test yourself before reading the parent note.

What does

z = x + i y

look like on paper?

A point / arrow in a 2D plane:

x

right,

y

up.

What does

∣ z - z_{0} ∣ < R

describe geometrically?

A filled disk of radius

R

centred at

z_{0}

What does the extra "

0 <

" in

0 < ∣ z - z_{0} ∣ < R

remove?

The centre point

z_{0}

itself — a punctured (poked-out) disk.

What is a power series, in one phrase?

An "infinite polynomial"

\sum a_{n} (z - c)^{n}

that converges inside a disk around its centre

c

How is

e^{w}

defined for complex

w

By the power series

\sum_{n \geq 0} w^{n} / n!

, which converges for every complex

w

Why is

e^{i θ} = cos θ + i sin θ

Put

w = i θ

in the series; the real and imaginary groups are exactly the

cos

and

sin

series.

z = z_{0} + ρ e^{i θ}

, what happens as

θ

runs

0 \to 2 π

You travel once counter-clockwise around a circle of radius

ρ

about

z_{0}

What is

\int_{0}^{2 π} e^{ik θ} d θ

for an integer

k

2 π

k = 0

, otherwise

0

(holds for negative

k

too).

State the limit definition of

f^{'} (z)

lim_{h \to 0} (f (z + h) - f (z)) / h

, the same value from every direction of approach.

What does "analytic on a region" mean?

Complex-differentiable at every point there, hence locally a convergent power series.

What is an annulus

r < ∣ z - z_{0} ∣ < R

The ring/washer of points whose distance from

z_{0}

is between the inner radius

r

and outer radius

R

Name the three isolated-singularity types.

Removable (no negative powers), pole of order

m

(finitely many), essential (infinitely many).

Why can we integrate the Laurent series term by term?

It converges uniformly on any circle inside its annulus, which licenses swapping sum and integral.

What is an antiderivative

F

f

A function with

F^{'} (z) = f (z)

; a closed loop returns

F

to its start, so the change is

0

State Cauchy's Integral Theorem with its hypotheses.

f

is analytic on and inside a positively oriented simple closed piecewise-smooth

C

, then

\oint_{C} f d z = 0

In the derivation, what is

d z

when

z = z_{0} + ρ e^{i θ}

d z = i ρ e^{i θ} d θ

Which index maps to the averaging fact, and which survives?

The exponent is

k = n + 1

; only

n = - 1

(so

k = 0

) survives, giving

2 π i

; all other terms are multiplied by

0

State the Residue Theorem with its hypotheses.

For

f

analytic on and inside a positively oriented simple closed piecewise-smooth

C

except at isolated singularities

z_{j}

inside,

\oint_{C} f d z = 2 π i \sum_{j} Res (f, z_{j})

Connections

Laurent series — the series whose coefficient $a_{- 1}$ is the residue.
Singularities — classifies the bad points (removable / pole / essential) where residues live.
Cauchy's Integral Theorem — the reason all non- $a_{- 1}$ terms die on a loop.
Cauchy's Integral Formula — the residue idea applied to $\frac{f ( z )}{z - z _{0}}$ .
Contour integration — using $\oint_{C}$ to evaluate real integrals.
Residues and poles — the parent topic these foundations build toward.