Laurent series — principal part, annulus of convergence: Question bank

Before we start, two words we lean on constantly, defined in plain language:

Spot the error

"For

\frac{1}{z - 1}

∣ z ∣ > 1

, write

\frac{1}{z - 1} = - \sum_{n \geq 0} z^{n}

Wrong direction —

\sum z^{n}

needs

∣ z ∣ < 1

. In

∣ z ∣ > 1

factor out the bigger quantity:

\frac{1}{z} \cdot \frac{1}{1 - 1/ z} = \sum_{n \geq 0} z^{- n - 1}

, a Geometric series in

1/ z

"Since

\frac{1}{( z - 1 ) ( z - 2 )}

is one function, its residue at

0

is a fixed number."

0

is not a singularity of this

f

, so there is no residue at

0

; the "Laurent series about

0

" question is about which annulus you expand in, not a residue.

"The principal part of

\frac{e ^{z}}{z ^{2}}

\frac{1}{2 !} + \frac{z}{3 !} + \dots

Backwards — those are the non-negative powers (the analytic part). The principal part is only

\frac{1}{z ^{2}} + \frac{1}{z}

, the negative powers.

e^{1/ z}

has infinitely many negative powers, so it must have a pole of infinite order."

There is no "infinite-order pole." Infinitely many negative terms is by definition an essential singularity — a different, worse category than any finite-order pole.

"To expand

f

1 < ∣ z ∣ < 2

, I'll expand both partial fractions the same way I did in

∣ z ∣ < 1

The singularity at

z = 1

is now inside the ring, so

\frac{- 1}{z - 1}

must be expanded in

1/ z

(negative powers), while

\frac{1}{z - 2}

stays in positive powers of

z

— each piece is expanded toward its own fence.

"The coefficient formula

c_{n} = \frac{1}{2 π i} \oint \frac{f}{( z - a ) ^{n + 1}} d z

needs

γ

to be a circle."

Any positively-oriented loop once around

a

inside the annulus works — deformation invariance (Cauchy integral formula) means the shape is irrelevant, only the winding is.

"The analytic part converges outside a disk."

The analytic part

\sum_{n \geq 0} c_{n} (z - a)^{n}

converges inside

∣ z - a ∣ < R

; it is the principal part that converges outside

∣ z - a ∣ > r

Why questions

Why must we state the annulus before writing a Laurent series?

Because the same

f

has different expansions on different rings; without naming the annulus the coefficients are ambiguous, so the answer is meaningless.

Why does only

c_{- 1}

survive when integrating

f

around a loop?

Because

\oint_{γ} (z - a)^{k} d z = 0

for every

k \neq = - 1

and equals

2 π i

only for

k = - 1

, so the loop integral filters out all powers except the residue term — this is the engine of the Residue theorem.

Why can a pole force the inner radius

r

to be

0

A pole is a singularity only at

a

with nothing else nearby inside; the ring's hole can shrink to the single point

a

, giving the punctured disk

0 < ∣ z - a ∣ < R

Why does the outer radius

R

equal the distance to the nearest outside singularity?

The analytic part is a genuine power series, so it converges up to the closest fence outside; the first singularity you meet moving outward stops convergence and sets

R

Why do we start from the known Taylor series of

e^{z}

when expanding

e^{z} / z^{2}

The numerator

e^{z}

is analytic, so its Taylor series is available; dividing by

z^{2}

just shifts every power down by two, instantly producing the negative-power (principal) part.

Why can't a plain Taylor series describe a function near a pole?

Taylor series contain only non-negative powers, which stay finite as

z \to a

; but a pole makes

f

blow up, so no combination of finite-behaving terms can match it — you need negative powers.

Edge cases

A function whose principal part is zero but which is not defined at

a

This is a removable singularity: the Laurent series has no negative powers, so redefining

f (a) = c_{0}

makes it analytic — the "hole" was cosmetic.

The Laurent series of a function with singularities at

z = 1

and

z = 2

expanded about

0

: how many annuli?

Three —

∣ z ∣ < 1

, the ring

1 < ∣ z ∣ < 2

, and

∣ z ∣ > 2

— one region per gap carved out by the fences.

What does the Laurent series look like in the outermost region

∣ z ∣ > 2

\frac{1}{( z - 1 ) ( z - 2 )}

Purely negative powers of

z

(all singularities are now "inside"), so the whole thing is principal part and behaves like

\sim 1/ z^{2}

for large

z

A singularity where the principal part is

\frac{1}{z} + \frac{1}{z ^{2}} + \frac{1}{z ^{3}} + \dots

(never stops).

Essential singularity — infinitely many negative terms; near

a

the function is wildly erratic and (Picard) hits almost every value.

Can

R = \infty

? Give the reading.

Yes — when there is no outer singularity (e.g.

e^{1/ z}

e^{z} / z^{2}

), the ring extends to infinity, giving

0 < ∣ z - a ∣ < \infty

Can the annulus be empty (no valid region)?

If two singularities coincide there is genuinely no gap, but for distinct fences

r < R

always, so a real annulus exists; an "empty" ring would signal you mislabelled which fence is inner vs outer.

Recall One-line self-test

Name the three singularity types by principal-part length. ::: Zero terms = removable; finitely many (deepest is $(z - a)^{- m}$ ) = pole of order $m$ ; infinitely many = essential. See Poles and singularities.

Question bank — Laurent series — principal part, annulus of convergence

True or false — justify

Spot the error

Why questions

Edge cases