Laurent series — principal part, annulus of convergence: Worked examples

This page is a drill. The parent note built the theory; here we push it through every kind of input the topic can throw at you. Before each example: forecast the answer yourself. The point of a drill is the surprise when your guess is wrong.

The scenario matrix

Every Laurent problem lands in exactly one cell of this grid. The columns are "how bad is the singularity", the rows are "what makes the problem awkward".

Awkwardness ↓ \ Singularity type →	Removable (no principal part)	Pole of order $m$ (finite principal part)	Essential (infinite principal part)
Simplest / textbook centre	Ex 1	Ex 2	Ex 6
Multiple annuli, one function	—	Ex 3 (three rings)	—
Centre ≠ singularity (shifted $a$ )	—	Ex 4	—
Degenerate / limiting input	Ex 5 (limit $z \to 0$ )	Ex 7 ( $m \to \infty$ boundary)	Ex 6
Real-world / word problem	—	Ex 8 (signal decay)	—
Exam twist (product of two blow-ups)	—	Ex 9	—

Cells marked "—" collapse into a neighbour (e.g. a removable singularity has only one interesting annulus, so it needs no multi-ring row). We hit all 9 examples, covering every reachable cell.

Before we start, one reusable tool that every example leans on.

Figure — Laurent series — principal part, annulus of convergence

The figure above shows why the same $\frac{1}{z - 1}$ splits two ways: if $z$ is inside the fence at $1$ we factor out the constant; if outside, we factor out $z$ . Keep this picture in mind — it is 80% of the work.

Removable singularities

Worked example

Ex 1 — Cell: Removable, textbook centre. $f (z) = \frac{sin z}{z}$ about $a = 0$ . Forecast: guess — does this have any $\frac{1}{z}$ terms?

Steps.

Write the Taylor series of the numerator: $sin z = z - \frac{z ^{3}}{3 !} + \frac{z ^{5}}{5 !} - \dots$ . Why this step? The numerator is analytic (a genuine Taylor citizen, see Taylor series); dividing later just shifts powers, which is trivial once we have the series.
Divide every term by $z$ : $\frac{s i n z}{z} = 1 - \frac{z ^{2}}{3 !} + \frac{z ^{4}}{5 !} - \dots .$ Why this step? Division by $z$ lowers each power by one. The lowest term was $z^{1}$ , so the lowest becomes $z^{0}$ — no negative power appears.
Read the principal part: there is none.

Result: principal part $= 0$ ⇒ removable singularity. The "hole" at $0$ is fake; defining $f (0) = 1$ makes $f$ analytic. Valid in $0 < ∣ z ∣ < \infty$ .

Verify: $lim_{z \to 0} \frac{s i n z}{z} = 1$ , which equals the constant term $c_{0} = 1$ we found. A removable singularity always has a finite limit — sanity check passes.

Worked example

Ex 5 — Cell: Removable, limiting/degenerate input. $f (z) = \frac{1 - cos z}{z ^{2}}$ about $0$ . Forecast: the denominator is $z^{2}$ — surely a double pole? Guess before reading.

Steps.

$cos z = 1 - \frac{z ^{2}}{2 !} + \frac{z ^{4}}{4 !} - \dots$ , so $1 - cos z = \frac{z ^{2}}{2 !} - \frac{z ^{4}}{4 !} + \dots$ . Why this step? The cancellation of the leading $1$ is the whole trick — the numerator secretly starts at $z^{2}$ , matching the denominator.
Divide by $z^{2}$ : $\frac{1 - c o s z}{z ^{2}} = \frac{1}{2 !} - \frac{z ^{2}}{4 !} + \frac{z ^{4}}{6 !} - \dots .$ Why this step? The $z^{2}$ in numerator and denominator kill each other. The "double pole" was an illusion.
Principal part $= 0$ again ⇒ removable.

Verify: $lim_{z \to 0} \frac{1 - c o s z}{z ^{2}} = \frac{1}{2}$ (a standard limit). Our constant term is $c_{0} = \frac{1}{2 !} = \frac{1}{2}$ . ✔ This is the degenerate case: a big denominator does not guarantee a pole — you must check whether the numerator vanishes to matching order.

Poles

Worked example

Ex 2 — Cell: Pole, textbook centre. $f (z) = \frac{cos z}{z ^{3}}$ about $0$ . Forecast: what order pole, and what is the residue $c_{- 1}$ ?

Steps.

$cos z = 1 - \frac{z ^{2}}{2 !} + \frac{z ^{4}}{4 !} - \dots$ . Why this step? Same reason as before — expand the analytic numerator, then shift.
Divide by $z^{3}$ : $\frac{c o s z}{z ^{3}} = principal part \frac{1}{z ^{3}} - \frac{1}{2 !} \frac{1}{z} + analytic part \frac{z}{4 !} - \dots .$ Why this step? Each term drops by 3 powers. The most negative is $z^{- 3}$ , and it stops there (nothing more negative) — that finiteness is what makes it a pole, not essential.
Most negative power is $z^{- 3}$ ⇒ pole of order 3. The residue is the coefficient of $z^{- 1}$ : $c_{- 1} = - \frac{1}{2 !} = - \frac{1}{2}$ .

Verify: For a pole of order $m = 3$ , $Res = \frac{1}{2 !} lim_{z \to 0} \frac{d ^{2}}{d z ^{2}} (z^{3} f (z)) = \frac{1}{2} cos^{''} (0) = \frac{1}{2} (- cos 0) = - \frac{1}{2}$ . ✔ Matches (this is the Residue theorem shortcut).

Worked example

Ex 3 — Cell: Pole, multiple annuli. $f (z) = \frac{1}{z ( z - 3 )}$ about $0$ — do all three rings. Forecast: how many distinct Laurent series does one function have here? Guess the number.

The singularities sit at $z = 0$ and $z = 3$ . They cut the plane (centred at $0$ ) into rings: $0 < ∣ z ∣ < 3$ and $∣ z ∣ > 3$ . See the figure.

Steps.

Partial fractions: $\frac{1}{z ( z - 3 )} = \frac{- 1/3}{z} + \frac{1/3}{z - 3}$ . Why this step? Each simple piece is a single fence, so it becomes a clean geometric series once we choose the right expansion direction (the Ex-1 picture).
Ring A: $0 < ∣ z ∣ < 3$ . The $\frac{- 1/3}{z}$ piece is already a single negative power (keep it). For the other, $∣ z ∣ < 3$ so $∣ z /3∣ < 1$ : $\frac{1/3}{z - 3} = \frac{1/3}{- 3} \cdot \frac{1}{1 - z /3} = - \frac{1}{9} \sum_{n \geq 0} (\frac{z}{3})^{n} .$ Why this step? We factored out the larger quantity $3$ so the ratio $z /3$ has modulus $< 1$ — that is the "wrong way round" pitfall from the parent note, avoided. Full series: $f (z) = - \frac{1}{3 z} - \frac{1}{9} n \geq 0 \sum \frac{z ^{n}}{3 ^{n}} = - \frac{1}{3 z} - \frac{1}{9} - \frac{z}{27} - \dots$ . Principal part $= - \frac{1}{3 z}$ ⇒ simple pole at $0$ , residue $c_{- 1} = - \frac{1}{3}$ .
Ring B: $∣ z ∣ > 3$ . Now $∣3/ z ∣ < 1$ , so factor out $z$ : $\frac{1/3}{z - 3} = \frac{1/3}{z} \cdot \frac{1}{1 - 3/ z} = \frac{1}{3 z} \sum_{n \geq 0} (\frac{3}{z})^{n} = \sum_{n \geq 0} \frac{3 ^{n}}{3} z^{- n - 1} .$ Why this step? In the outer ring $z$ is the big quantity, so $3/ z$ is the small ratio. Add the untouched $- \frac{1}{3 z}$ : $f (z) = - \frac{1}{3 z} + n \geq 0 \sum \frac{3 ^{n - 1}}{1} z^{- n - 1} = n \geq 1 \sum (\frac{3 ^{n - 1}}{1}) z^{- n - 1} + \dots$ . The $n = 0$ term is $\frac{1}{3 z}$ , cancelling the $- \frac{1}{3 z}$ exactly, leaving only deeper negative powers: $f (z) = \frac{1}{z ^{2}} + \frac{3}{z ^{3}} + \frac{9}{z ^{4}} + \dots$ . All negative powers.

Result: two distinct Laurent series (rings A and B) for one function.

Verify: Residue at $0$ (from ring A, the ring that surrounds $0$ ) is $c_{- 1} = - \frac{1}{3}$ . Cross-check: $Res_{z = 0} = lim_{z \to 0} z \cdot \frac{1}{z ( z - 3 )} = \frac{1}{0 - 3} = - \frac{1}{3}$ . ✔ In ring B the coefficient of $z^{- 1}$ is $0$ — correct, because ring B does not wind around the residue in the residue-theorem sense used for that ring.

Worked example

Ex 4 — Cell: Pole, centre ≠ singularity. $f (z) = \frac{1}{z - 2}$ about $a = 1$ , in the ring $0 < ∣ z - 1∣ < 1$ . Forecast: the singularity is at $2$ , but we expand about $1$ . Will there be negative powers of $(z - 1)$ ?

Steps.

Rewrite everything in the variable $w = z - 1$ (so the centre becomes the origin): $f = \frac{1}{( w + 1 ) - 2} = \frac{1}{w - 1} .$ Why this step? Laurent is always "about the centre"; renaming makes the centre $w = 0$ so we can use our one tool directly.
The nearest singularity is at $w = 1$ , i.e. $∣ z - 1∣ = 1$ . In $0 < ∣ z - 1∣ < 1$ we have $∣ w ∣ < 1$ , so factor out the constant: $\frac{1}{w - 1} = \frac{- 1}{1 - w} = - \sum_{n \geq 0} w^{n} = - 1 - w - w^{2} - \dots .$ Why this step? $∣ w ∣ < 1$ makes $u = w$ a valid ratio; factoring out $- 1$ produces the $\frac{1}{1 - u}$ shape.
Back-substitute $w = z - 1$ : $f (z) = - \sum_{n \geq 0} (z - 1)^{n} .$

Result: no principal part in this ring — the centre $1$ is an ordinary analytic point of $f$ , so about it we get a pure Taylor series. The singularity lives on the boundary $∣ z - 1∣ = 1$ , not inside.

Verify: At $z = 1$ the sum is $- 1$ , and $f (1) = \frac{1}{1 - 2} = - 1$ . ✔ The lesson: a pole of the function is only a "principal-part" pole in a ring that actually encircles it.

Essential singularities and limiting behaviour

Worked example

Ex 6 — Cell: Essential. $f (z) = z^{2} e^{1/ z}$ about $0$ . Forecast: finite or infinite principal part? And what is the residue?

Steps.

Expand $e^{1/ z} = \sum_{n \geq 0} \frac{1}{n !} z^{- n} = 1 + \frac{1}{z} + \frac{1}{2 ! z ^{2}} + \frac{1}{3 ! z ^{3}} + \dots$ . Why this step? $e^{w} = \sum w^{n} / n!$ everywhere; substituting $w = 1/ z$ is legal for $z \neq = 0$ and instantly gives infinitely many negative powers.
Multiply by $z^{2}$ : $z^{2} e^{1/ z} = z^{2} + z + \frac{1}{2 !} + \frac{1}{3 ! z} + \frac{1}{4 ! z ^{2}} + \dots .$ Why this step? Multiplying by $z^{2}$ raises each power by 2. It shifts the whole tail up but cannot cut off the infinitely long negative tail.
Still infinitely many negative powers ⇒ essential singularity. Residue $c_{- 1} = \frac{1}{3 !} = \frac{1}{6}$ .

Verify: $c_{- 1}$ is the coefficient of $z^{- 1}$ : it came from the $\frac{1}{3 ! z ^{3}}$ term of $e^{1/ z}$ times $z^{2}$ , giving $\frac{1}{3 !} z^{- 1}$ . So $c_{- 1} = \frac{1}{6}$ . ✔ Note multiplying by $z^{2}$ did not demote an essential singularity to a pole — only finite truncation could, and $e^{1/ z}$ never truncates.

Worked example

Ex 7 — Cell: Boundary between pole and essential. Contrast $g (z) = \frac{1}{z ^{5}}$ vs. $h (z) = \frac{1}{sin ( 1/ z )}$ -style thinking about $0$ . Forecast: which one is "just a big pole" and which is "genuinely essential"? Guess.

Steps.

$g (z) = z^{- 5}$ . Principal part is the single term $z^{- 5}$ ; it stops. Why this step? A pole of order $m$ is defined by the most negative power being finite. Here $m = 5$ , done.
For a function like $sin (1/ z) = \sum_{n \geq 0} \frac{( - 1 ) ^{n}}{( 2 n + 1 )!} z^{- (2 n + 1)} = \frac{1}{z} - \frac{1}{3 ! z ^{3}} + \dots$ the negative tail never stops. Why this step? This is the limiting case $m \to \infty$ : a pole is "an essential singularity that happened to be finite." Once the order would be infinite, we rename it essential.
So the classification is a clean dichotomy: largest negative power finite ⇒ pole of that order; not finite ⇒ essential. Removable is just "order $0$ ".

Verify: For $g$ , residue $c_{- 1} = 0$ (no $z^{- 1}$ term in $z^{- 5}$ ), consistent with a pure power. For $sin (1/ z)$ , residue $c_{- 1} = 1$ (the $\frac{1}{z}$ term). Both read straight off the series. ✔

Real-world and exam-twist

Worked example

Ex 8 — Cell: Word problem. A stable signal amplifier has transfer function $H (z) = \frac{1}{z - \frac{1}{2}}$ . For the region $∣ z ∣ > \frac{1}{2}$ (the "causal" region outside the pole), expand $H$ in powers of $1/ z$ and read off the impulse-response coefficients $h_{n}$ (the coefficient of $z^{- n - 1}$ ). Forecast: will the coefficients grow or decay as $n$ increases? (A stable amplifier should decay.)

Steps.

We are in $∣ z ∣ > \frac{1}{2}$ , so $\frac{1/2}{z} < 1$ . Factor out the big quantity $z$ : $H (z) = \frac{1}{z} \cdot \frac{1}{1 - \frac{1}{2 z}} = \frac{1}{z} \sum_{n \geq 0} (\frac{1}{2 z})^{n} = \sum_{n \geq 0} \frac{1}{2 ^{n}} z^{- n - 1} .$ Why this step? Outside the pole, $z$ dominates, so $\frac{1}{2 z}$ is the small ratio — this is the ring that encircles the singularity, exactly where the physical (causal) expansion lives.
Read off $h_{n} = \frac{1}{2 ^{n}}$ : $h_{0} = 1, h_{1} = \frac{1}{2}, h_{2} = \frac{1}{4}, \dots$ . Why this step? The coefficient of $z^{- n - 1}$ is the $n$ -th impulse-response sample in this convention.

Result: $h_{n} = (1/2)^{n}$ — a decaying response ⇒ stable. The residue $c_{- 1} = h_{0} = 1$ is $Res_{z = 1/2} H = 1$ .

Verify: $\sum_{n \geq 0} h_{n} = \sum (1/2)^{n} = \frac{1}{1 - 1/2} = 2 = H (1)$ ? Check: $H (1) = \frac{1}{1 - 1/2} = 2$ . ✔ The series summed at $z = 1$ reproduces the function's value — and a finite total energy confirms stability.

Worked example

Ex 9 — Cell: Exam twist, product of two blow-ups. $f (z) = \frac{e ^{z}}{z ^{2} ( z - 1 )}$ about $0$ , in the ring $0 < ∣ z ∣ < 1$ . Find the residue at $0$ . Forecast: the pole at $0$ has order $2$ ; guess the residue before grinding.

Steps.

In $0 < ∣ z ∣ < 1$ expand the two analytic ingredients: $e^{z} = 1 + z + \frac{z ^{2}}{2} + \dots$ and $\frac{1}{z - 1} = \frac{- 1}{1 - z} = - \sum_{n \geq 0} z^{n} = - 1 - z - z^{2} - \dots$ (valid since $∣ z ∣ < 1$ ). Why this step? Both are honest Taylor series here; the only singular factor is $\frac{1}{z ^{2}}$ , which we hold aside and apply last.
Multiply the two power series, keeping terms up to $z^{1}$ (we only need enough to reach $z^{- 1}$ after dividing by $z^{2}$ ): $e^{z} \cdot \frac{1}{z - 1} = (1 + z + \dots) (- 1 - z - \dots) = - 1 - 2 z + O (z^{2}) .$ Why this step? The residue is the coefficient of $z^{- 1}$ in $f$ ; dividing by $z^{2}$ shifts powers down by 2, so we need the coefficient of $z^{1}$ in this product.
Divide by $z^{2}$ : $f (z) = \frac{- 1 - 2 z + O ( z ^{2} )}{z ^{2}} = - \frac{1}{z ^{2}} - \frac{2}{z} + O (1) .$ Why this step? Now the principal part is visible: order-2 pole, and $c_{- 1} = - 2$ .

Result: residue at $0$ is $c_{- 1} = - 2$ .

Verify: Order-2 formula: $Res = lim_{z \to 0} \frac{d}{d z} (z^{2} f) = lim_{z \to 0} \frac{d}{d z} \frac{e ^{z}}{z - 1} = lim \frac{e ^{z} ( z - 1 ) - e ^{z}}{( z - 1 ) ^{2}} = \frac{1 \cdot ( - 1 ) - 1}{1} = - 2$ . ✔ (This is the Cauchy integral formula machinery in disguise.)

Recall One-line summary of the whole drill

The two questions ::: (1) removable / pole-of-order- $m$ / essential — read from where the negative tail stops; (2) which ring — pick $u$ so $∣ u ∣ < 1$ there, then expand. Why one function can have several series ::: Different rings, separated by singularities, force different valid geometric expansions. Fast residue when it's a pole ::: $Res = \frac{1}{( m - 1 )!} lim_{z \to a} \frac{d ^{m - 1}}{d z ^{m - 1}} [(z - a)^{m} f (z)]$ .

Active recall

\frac{s i n z}{z}

about

0

, what is the principal part?

Zero — it's a removable singularity, lowest power is

z^{0}

Residue of

\frac{c o s z}{z ^{3}}

0

c_{- 1} = - \frac{1}{2}

How many distinct Laurent series does

\frac{1}{z ( z - 3 )}

have about

0

Two — one in

0 < ∣ z ∣ < 3

, one in

∣ z ∣ > 3

Expanding

\frac{1}{z - 2}

about

1

0 < ∣ z - 1∣ < 1

, is there a principal part?

No — it's a pure Taylor series; the singularity is on the boundary, not encircled.

Why does multiplying

e^{1/ z}

z^{2}

not remove the essential singularity?

The negative tail is infinite; a finite power shift cannot truncate it.

Impulse coefficients of

\frac{1}{z - 1/2}

∣ z ∣ > \frac{1}{2}

h_{n} = (1/2)^{n}

— decaying, hence stable.

Residue of

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

0

- 2