Laurent series — principal part, annulus of convergence

WHAT is a Laurent series?

Why this formula? (Derivation from scratch.) We want $c_n$ for the expansion $f(z)=\sum_m c_m (z-a)^m$ . Multiply by $(z-a)^{-n-1}$ and integrate around a loop $\gamma$ in the annulus: $\oint_\gamma \frac{f(z)}{(z-a)^{n+1}}dz=\sum_m c_m\oint_\gamma (z-a)^{m-n-1}\,dz.$ Why this step? We're using the orthogonality of powers under contour integration. Now recall the key fact (derive it!): $\oint_\gamma (z-a)^{k}\,dz=\begin{cases}2\pi i & k=-1\\[2pt]0 & k\neq -1\end{cases}$ Proof of the key fact: parametrise $z=a+\rho e^{i\theta}$ , $dz=i\rho e^{i\theta}d\theta$ , so the integral becomes $\int_0^{2\pi}\rho^{k}e^{ik\theta}\,i\rho e^{i\theta}d\theta = i\rho^{k+1}\int_0^{2\pi}e^{i(k+1)\theta}d\theta$ . For $k\ne -1$ the $\theta$ -integral is $0$ ; for $k=-1$ it is $2\pi$ . ✔

So only the term with $m-n-1=-1$ , i.e. $m=n$ , survives: $\oint_\gamma \frac{f(z)}{(z-a)^{n+1}}dz=c_n\cdot 2\pi i\ \Rightarrow\ c_n=\frac{1}{2\pi i}\oint_\gamma\frac{f(z)}{(z-a)^{n+1}}dz.$ This also proves uniqueness: the coefficients are forced.

WHY an annulus, not a disk?

The analytic part $\sum_{n\ge0} c_n(z-a)^n$ converges inside a disk $|z-a|<R$ (like a normal power series). $R=$ distance to the nearest singularity outside.
The principal part $\sum_{n\ge1} c_{-n}(z-a)^{-n}$ is a power series in $w=\frac{1}{z-a}$ , so it converges for $|w|<\frac{1}{r}$ , i.e. $|z-a|>r$ . $r=$ distance to the nearest singularity at or inside $a$ .

The Laurent series converges where both converge: the annulus $r<|z-a|<R$ .

Figure — Laurent series — principal part, annulus of convergence

Classifying the singularity by the principal part

The coefficient $c_{-1}$ is the residue of $f$ at $a$ — the single number that survives integration (since only $k=-1$ gives $2\pi i$ ). That's why residues run the whole theory of contour integration.

Worked examples

Worked example 2.

$f(z)=\dfrac{1}{(z-1)(z-2)}$ — depends on the annulus Partial fractions first: $\frac{1}{(z-1)(z-2)}=\frac{-1}{z-1}+\frac{1}{z-2}$ . Why? Each simple piece is a geometric series we can expand in the right direction.

Region A: $|z|<1$ (both expanded as $|z|$ small): $\frac{-1}{z-1}=\frac{1}{1-z}=\sum_{n\ge0}z^n$ , and $\frac{1}{z-2}=\frac{-1}{2}\cdot\frac{1}{1-z/2}=-\sum_{n\ge0}\frac{z^n}{2^{n+1}}$ . Sum: $\sum_{n\ge0}\big(1-\tfrac{1}{2^{n+1}}\big)z^n$ . No principal part — pure Taylor (no singularity inside).

Region B: $1<|z|<2$ — Why split here? The singularity at $1$ is now inside, so we need negative powers for that piece. $\frac{-1}{z-1}=\frac{-1}{z}\cdot\frac{1}{1-1/z}=-\sum_{n\ge0}z^{-n-1}$ (valid $|z|>1$ ), keep $\frac{1}{z-2}=-\sum_{n\ge0}\frac{z^n}{2^{n+1}}$ (valid $|z|<2$ ). Now there is a principal part: $-\sum_{n\ge1}z^{-n}$ .

Region C: $|z|>2$ — both singularities inside; expand both in $1/z$ : $\frac{-1}{z-1}=-\sum_{n\ge0}z^{-n-1}$ , $\frac{1}{z-2}=\frac1z\cdot\frac{1}{1-2/z}=\sum_{n\ge0}2^n z^{-n-1}$ . Sum $=\sum_{n\ge0}(2^n-1)z^{-n-1}$ . All negative powers.

Recall Feynman: explain to a 12-year-old

Imagine a song that gets louder and louder as you walk toward a speaker — eventually unbearably loud at the speaker itself. A normal (Taylor) recipe can only describe quiet, polite sounds, so it fails right at the speaker. The Laurent recipe adds special "scream terms" ( $\frac1z$ , $\frac{1}{z^2}$ , …) that get huge near the speaker — so it can describe the song even very close to it, but only in the ring-shaped region between "too close" and "another speaker further away." The number of scream terms tells you how nasty the speaker is: none = fake speaker, a few = a normal loud one (a pole), infinitely many = a wildly crazy one (essential).

Active recall

What extra terms does a Laurent series have that a Taylor series lacks?

Negative-power terms

(z-a)^{-n}

, forming the principal part.

Define the principal part of a Laurent series.

The sum of all negative-power terms

\sum_{n\ge1}c_{-n}(z-a)^{-n}

Formula for the Laurent coefficient

c_n

c_n=\frac{1}{2\pi i}\oint_\gamma \frac{f(z)}{(z-a)^{n+1}}dz

Which contour integral fact makes the coefficient formula work?

\oint_\gamma (z-a)^k dz=2\pi i

k=-1

, else

0

Why a region of convergence shaped like an annulus and not a disk?

The analytic part converges inside

|z-a|<R

; the principal part converges outside

|z-a|>r

; both hold only on the ring

r<|z-a|<R

How do you read off the type of singularity from the principal part?

Zero terms ⇒ removable; finitely many (

\le m

) ⇒ pole of order

m

; infinitely many ⇒ essential.

What is the residue in terms of Laurent coefficients?

The coefficient

c_{-1}

Is a Laurent series of a function unique?

Unique per annulus; the same function can have different Laurent series in different annuli.

For

1/(z-1)

in the region

|z|>1

, how do you expand?

Factor out

z

\frac1z\cdot\frac{1}{1-1/z}=\sum_{n\ge0}z^{-n-1}

Principal part of

e^{1/z}

about 0, and what it implies?

\sum_{n\ge1}\frac{1}{n!z^n}

(infinitely many terms) ⇒ essential singularity.

Connections

Taylor series — special case with empty principal part.
Residue theorem — uses $c_{-1}$ ; Laurent is its foundation.
Poles and singularities — classified by the principal part.
Geometric series — the workhorse for building Laurent expansions.
Cauchy integral formula — generalised to give the $c_n$ formula.
Annulus of convergence — the two-radius analogue of the radius of convergence.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, Taylor series tab kaam karti hai jab function ek point ke around "bhala" yaani analytic hota hai — sirf positive powers $(z-a)^n$ hoti hain. Lekin agar function kisi point par phat jaata hai (jaise $1/z$ at $z=0$ ), toh Taylor fail ho jaati hai. Yahaan aati hai Laurent series: ye positive powers ke saath negative powers $(z-a)^{-1}, (z-a)^{-2}, \dots$ bhi allow karti hai. Yehi negative-power wala hissa principal part kehlata hai, aur yahi singularity ko handle karta hai.

Convergence ka region ek annulus (donut jaisi ring) $r<|z-a|<R$ hota hai. Logic simple hai: analytic part (positive powers) ek normal power series hai jo disk $|z-a|<R$ ke andar converge karti hai, jahan $R$ = bahar wali nearest singularity tak ki doori. Principal part actually $1/(z-a)$ mein power series hai, isliye wo $|z-a|>r$ par converge karti hai. Dono ek saath sirf ring mein satisfy hote hain — isliye annulus.

Bahut zaroori baat: ek hi function ke alag-alag annulus mein alag-alag Laurent series ho sakti hain. Jaise $\frac{1}{(z-1)(z-2)}$ ki $0$ ke around teen expansions hain: $|z|<1$ , $1<|z|<2$ , aur $|z|>2$ . Isliye pehle yeh decide karo ki kaunse ring mein expand kar rahe ho, fir geometric series ko sahi taraf se kholo — bada term factor out karo taaki ratio ka modulus $<1$ ho.

Principal part se singularity ka type pata chalta hai: koi negative term nahi = removable, finite negative terms (say $m$ tak) = order- $m$ pole, infinite negative terms = essential singularity. Aur sabse magical number hai $c_{-1}$ = residue, jo aage chalke contour integration aur Residue Theorem ka heart ban jaata hai. Yehi reason hai ki Laurent series elite-level complex analysis ki backbone hai.

Laurent series — principal part, annulus of convergence

WHAT is a Laurent series?

WHY an annulus, not a disk?

Classifying the singularity by the principal part

Worked examples

Active recall

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Advanced Topics (Elite Level)

Connections