Laurent’s series in an annulus

At a Glance

• Frequency: 8 sub-parts across 8 of 14 years (2014, 2015, 2018, 2019, 2021, 2022, 2025, 2026)
• Priority tier: T2
• Marks (count): 10 (4), 15 (1), 20 (2)
• Average solve time: ~13 min
• Difficulty mix: medium 5, hard 2
• Section: A | Dominant type: computation

Why This Chapter Matters

Laurent series appears in every year from 2014 to 2025 (7 appearances) in Section A at 10–20 marks. The 20-mark questions (2015, 2021) require three separate expansions corresponding to three annuli around $z=0$. The 10-mark questions ask for a single annulus, typically the one nearest a specified singularity. The entire topic reduces to one skill: after partial-fraction decomposition, choose the correct geometric-series direction (in $z$ or in $1/z$) for each factor depending on which annulus you are in. Once that choice is systematic, the mechanics are routine.

Minimum Theory

Laurent series. A function analytic in an annulus $r<|z-z_0|<R$ has a unique Laurent expansion there: $f(z)=\sum_{n=-\infty}^{\infty}c_n(z-z_0)^n.$ The terms with $n<0$ form the principal part; the terms with $n\ge0$ form the analytic part.

Key tool: geometric series in two directions. For a simple factor $(z-a)^{-1}$:

• If $|z-z_0|<|a-z_0|$ (inside the singularity): factor out $(a-z_0)$ and expand in $z-z_0$. Example about $z_0=0$: $\frac{1}{z-a}=-\frac{1}{a-z}=-\frac{1}{a}\cdot\frac{1}{1-(z/a)}=-\frac{1}{a}\sum_{n=0}^\infty\!\left(\frac{z}{a}\right)^n,\quad|z|<|a|.$
• If $|z-z_0|>|a-z_0|$ (outside): factor out $(z-z_0)$ and expand in $(a-z_0)/(z-z_0)$. Example: $\frac{1}{z-a}=\frac{1}{z}\cdot\frac{1}{1-(a/z)}=\frac{1}{z}\sum_{n=0}^\infty\!\left(\frac{a}{z}\right)^n=\sum_{n=0}^\infty\frac{a^n}{z^{n+1}},\quad|z|>|a|.$

Coefficient formula — the tool for when partial fractions are unavailable. The Laurent expansion in a given annulus is unique, and its coefficients are contour integrals: for any positively oriented circle $C_r: |z-z_0|=r$ lying inside the annulus,

$c_n=\frac{1}{2\pi i}\oint_{C_r}\frac{f(w)}{(w-z_0)^{n+1}}\,dw,\qquad n\in\mathbb{Z}.$

The value does not depend on which $r$ you choose. Two consequences carry every question where the geometric-series route is not available (because $f$ is transcendental in $z+1/z$, say, and has no partial fractions at all):

• Uniqueness is a working tool, not a footnote. If two two-sided series represent the same $f$ on the same annulus, their coefficients agree term by term. So any symmetry of $f$ can be converted into a symmetry of the $c_n$ — see laurent-coefficient-integral.
• Choosing $r=1$ makes the coefficient real. On $|w|=1$ put $w=e^{i\theta}$: then $dw=ie^{i\theta}d\theta$, the $e^{i\theta}$ cancels against one power of $w^{-(n+1)}$, and $w+\tfrac1w=2\cos\theta$ is real. A complex contour integral becomes a real Fourier-type integral over $[0,2\pi]$.

Procedure (rational $f$). (1) Partial-fraction decompose $f$. (2) Identify the singular radii: the moduli $|a_j-z_0|$ for each singularity $a_j$, listed in increasing order — these divide the plane into concentric annuli. (3) In each annulus, choose the expansion direction for each factor: inside → power series in $(z-z_0)$; outside → power series in $1/(z-z_0)$.

Number of annuli. With $k$ distinct singular radii, there are $k+1$ regions: the disc $|z-z_0|<r_1$, then each annulus $r_j<|z-z_0|<r_{j+1}$, then the exterior $|z-z_0|>r_k$. The “about $z=0$” question with poles at $|z|=1$ and $|z|=2$ gives exactly three regions.

Question Archetypes

Two patterns, split by whether the series is computed or characterised.

laurent-expansion (7 question(s); 2014, 2015, 2018, 2019, 2021, 2022, 2025)

Recognition Cues

• “Find the Laurent series of $f(z)$ in [annulus description].”
• “Find all possible Taylor/Laurent expansions about $z=z_0$.”
• “Expand $f$ in the region $0<|z-a|<r$.”

Solution Template

1. Partial fractions (for rational functions). Write $f=\sum_j A_j/(z-a_j)^{k_j}$.
2. Identify the singular radii from $z_0$: compute $|a_j-z_0|$. Sort them: $r_1<r_2<\ldots<r_k$.
3. List the annuli: $|z-z_0|<r_1$, $r_1<|z-z_0|<r_2$, …, $|z-z_0|>r_k$.
4. In each annulus, expand each factor:
   • For $|z-z_0|<|a_j-z_0|$: expand $(z-a_j)^{-1}$ in powers of $(z-z_0)$.
   • For $|z-z_0|>|a_j-z_0|$: expand in powers of $1/(z-z_0)$.
5. Collect terms (sum the series). State the annulus of validity explicitly.
6. For transcendental functions (e.g. $1/(e^z-1)$): factor out the lowest-power zero of the denominator, expand the remaining factor as $1/(1+w)=\sum(-w)^n$ collecting terms to the required order.

Worked Example(s)

2015 Paper 2, 2015-P2-Q2c (20 marks)

Find all Taylor/Laurent expansions of $f(z)=(2z-3)/(z^2-3z+2)$ about $z=0$.

Partial fractions: $f=1/(z-1)+1/(z-2)$. Singular radii from $0$: $|1|=1$, $|2|=2$. Three regions:

Region I ($|z|<1$): both factors inside their singularities. $\frac{1}{z-1}=-\sum_{n=0}^\infty z^n,\quad \frac{1}{z-2}=-\frac12\sum_{n=0}^\infty\!\left(\frac z2\right)^n.$ $\boxed{f(z)=-\sum_{n=0}^\infty\!\left(1+\frac{1}{2^{n+1}}\right)z^n,\quad|z|<1.}$

Region II ($1<|z|<2$): first factor outside ($|1/z|<1$), second still inside ($|z/2|<1$). $\frac{1}{z-1}=\frac1z\sum_{n=0}^\infty z^{-n}=\sum_{n=1}^\infty z^{-n},\quad \frac{1}{z-2}=-\frac12\sum_{n=0}^\infty(z/2)^n.$ $\boxed{f(z)=\sum_{n=1}^\infty z^{-n}-\sum_{n=0}^\infty\frac{z^n}{2^{n+1}},\quad 1<|z|<2.}$

Region III ($|z|>2$): both factors outside. $\frac{1}{z-1}=\sum_{n=1}^\infty z^{-n},\quad \frac{1}{z-2}=\frac1z\sum_{n=0}^\infty(2/z)^n=\sum_{n=1}^\infty\frac{2^{n-1}}{z^n}.$ $\boxed{f(z)=\sum_{n=1}^\infty\frac{1+2^{n-1}}{z^n},\quad|z|>2.}$

2014 Paper 2, 2014-P2-Q1d (10 marks)

Laurent series of $1/(z^2(z-1))$ about $z=0$ and $z=1$.

About $z=0$, region $0<|z|<1$: $f=\frac{1}{z^2}\cdot\frac{1}{z-1}=-\frac{1}{z^2}\sum_{n=0}^\infty z^n.$ $\boxed{f(z)=-\frac{1}{z^2}-\frac{1}{z}-1-z-z^2-\cdots,\quad 0<|z|<1.}$

About $z=1$, region $0<|z-1|<1$: let $w=z-1$, $z=1+w$: $f=\frac{1}{w(1+w)^2}=\frac1w\sum_{n=0}^\infty(-1)^n(n+1)w^n=\frac{1}{w}-2+3w-4w^2+\cdots$ $\boxed{f(z)=\frac{1}{z-1}-2+3(z-1)-4(z-1)^2+\cdots,\quad 0<|z-1|<1.}$

2019 Paper 2, 2019-P2-Q4b (10 marks)

First three terms of the Laurent series of $1/(e^z-1)$ in $0<|z|<2\pi$.

$e^z-1=z(1+z/2+z^2/6+\cdots)=:z(1+w)$. So $f=\tfrac1z\cdot\tfrac{1}{1+w}=\tfrac1z(1-w+w^2-\cdots)$.

$1-w+w^2=1-(z/2+z^2/6)+z^2/4+\cdots=1-z/2+(1/4-1/6)z^2+\cdots=1-z/2+z^2/12+\cdots$

$\boxed{\frac{1}{e^z-1}=\frac{1}{z}-\frac{1}{2}+\frac{z}{12}-\cdots,\quad 0<|z|<2\pi.}$

2018 Paper 2, 2018-P2-Q4b (15 marks)

Laurent series of $1/((1+z^2)(z+2))$ in (i) $|z|<1$, (ii) $1<|z|<2$, (iii) $|z|>2$.

Partial fractions: $f=\tfrac15\cdot\tfrac{1}{z+2}+\tfrac15\cdot\tfrac{2-z}{z^2+1}$. Singular radii: $|{\pm i}|=1$, $|{-2}|=2$.

(i) $|z|<1$: both $1/(z+2)$ and $1/(z^2+1)$ expand in powers of $z$ (Taylor): $f=\tfrac15\sum_{n=0}^\infty\tfrac{(-1)^n}{2^{n+1}}z^n+\tfrac15(2-z)\sum_{n=0}^\infty(-1)^n z^{2n}.$

(ii) $1<|z|<2$: $1/(z+2)$ still in $z$; $1/(z^2+1)$ switches to $1/z^2$ (since $|z|>1$): $f=\tfrac15\sum_{n=0}^\infty\tfrac{(-1)^n}{2^{n+1}}z^n+\tfrac15(2-z)\sum_{n=0}^\infty(-1)^n z^{-2n-2}.$

(iii) $|z|>2$: both expand in $1/z$: $f=\tfrac15\sum_{n=0}^\infty(-1)^n 2^n z^{-n-1}+\tfrac15(2-z)\sum_{n=0}^\infty(-1)^n z^{-2n-2}.$

2025 Paper 2, 2025-P2-Q1d (10 marks)

Expand $f(z)=1/((z+1)(z+3))$ valid for $1<|z|<3$.

In this annulus $|z|>1$ (so $1/(z+1)$ expands in $1/z$) and $|z|<3$ (so $1/(z+3)$ expands in $z/3$): $\frac{1}{z+1}=\frac{1}{z}\cdot\frac{1}{1+1/z}=\frac{1}{z}\sum_{n=0}^\infty\frac{(-1)^n}{z^n}=\sum_{n=0}^\infty\frac{(-1)^n}{z^{n+1}}.$ $\frac{1}{z+3}=\frac{1}{3}\cdot\frac{1}{1+z/3}=\frac{1}{3}\sum_{n=0}^\infty\!\left(-\frac{z}{3}\right)^n.$

Partial fractions: $f=\tfrac12\big(\tfrac{1}{z+1}-\tfrac{1}{z+3}\big)$: $\boxed{f(z)=\tfrac12\sum_{n=0}^\infty\frac{(-1)^n}{z^{n+1}}-\frac{1}{6}\sum_{n=0}^\infty\!\left(-\frac{z}{3}\right)^n,\quad 1<|z|<3.}$

Common Traps

• Choosing the wrong direction: if you use $|z|<|a|$ expansion when in the $|z|>|a|$ region, you get a series that diverges. Always determine which inequality holds for each factor in the given annulus.
• Forgetting the minus sign in $1/(z-a)=-1/(a-z)$ for the $|z|<|a|$ expansion. Write the geometric-series step explicitly to avoid this.
• “About $z=0$” means expand in powers of $z$, not $z-a$ for some other centre. If the question says “about $z=1$”, substitute $w=z-1$ before expanding.
• $1/(1+w)^2$ is not the same as $1/(1-w^2)$. For repeated poles near the expansion centre, the expansion is $\sum_{n=0}^\infty(-1)^n(n+1)w^n$ (from differentiating the geometric series).
• Transcendental functions: factor out the zero of lowest order ($e^z-1=z(1+\text{higher})$), divide by $z$, then invert $1+w$ as a geometric series.

laurent-coefficient-integral (1 question(s); 2026)

This is the other kind of Laurent question, and the geometric-series machinery above is useless for it. Here the series is handed to you and the coefficients are presented as an integral; your job is to prove the representation and identify the coefficients. There is nothing to partial-fraction.

Recognition Cues

• “Prove that $f(z)=a_0+\sum a_n(z^n+z^{-n})$, where $a_n=\ldots$” — the expansion is stated, not requested.
• The coefficients are printed as a real integral over $[0,2\pi]$, typically with a $\cos n\theta$ weight.
• The two-sided series appears in the folded form $z^n+z^{-n}$ rather than as $\sum_{-\infty}^{\infty}c_nz^n$.
• The function is transcendental in $z+\tfrac1z$ (or in $z-\tfrac1z$) — a shape with no partial-fraction decomposition.

Solution Template

1. Name the annulus and justify it. Locate the singularities; for $f(z)=g(z+\tfrac1z)$ with $g$ entire the only one is $z=0$, so the annulus is the whole punctured plane $0<|z|<\infty$. Say why it is that annulus — the region of validity is examinable content, and it is also what licenses step 3.
2. Invoke Laurent’s theorem to write $f(z)=\sum_{n=-\infty}^{\infty}c_nz^n$ there, quoting the coefficient formula and the uniqueness clause explicitly. You will use uniqueness as a tool, so state it.
3. Fold the series using the functional symmetry. If $f(1/z)=f(z)$ and the annulus is invariant under $z\mapsto1/z$ (as $0<|z|<\infty$ is), substitute $1/z$ into the expansion, re-index $m=-n$ — legitimate because Laurent convergence is absolute — and compare with the original. Uniqueness forces $c_{-n}=c_n$, which is exactly what collapses $\sum_{-\infty}^{\infty}c_nz^n$ into $c_0+\sum_{n\ge1}c_n(z^n+z^{-n})$.
4. Evaluate the coefficient on $|z|=1$. Put $w=e^{i\theta}$ in the coefficient formula. The $\tfrac{1}{2\pi i}$ and the $dw=ie^{i\theta}d\theta$ combine to $\tfrac{1}{2\pi}d\theta$, and $w+\tfrac1w=2\cos\theta$.
5. Kill the imaginary part. The result is $c_n=\tfrac{1}{2\pi}\int_0^{2\pi}f\!\left(\text{real}\right)e^{-in\theta}d\theta$; split into $\cos$ and $\sin$ parts and show the $\sin$ integral vanishes — by the reflection $\theta\mapsto2\pi-\theta$, not by assertion. Only then is $c_n=a_n$.

Worked Example

2026 Paper 2, 2026-P2-Q3a (15 marks)

Prove that $\cosh\!\left(z+\dfrac1z\right)=a_0+\sum_{n=1}^{\infty}a_n\!\left(z^n+\dfrac{1}{z^n}\right)$, where $a_n=\dfrac{1}{2\pi}\displaystyle\int_0^{2\pi}\cos n\theta\,\cosh(2\cos\theta)\,d\theta$.

Source: analysis/solutions/2026-P2-Q3a.md

Step 1 — Annulus. $z\mapsto z+\tfrac1z$ is analytic on $\mathbb{C}\setminus\{0\}$ and $\cosh$ is entire, so $f(z)=\cosh(z+\tfrac1z)$ is analytic on the punctured plane $0<|z|<\infty$ — itself an annulus, with $r_1=0$ and $r_2=\infty$. ($z=0$ is an isolated essential singularity: $\cosh w\to\infty$ along real $w$ but oscillates boundedly along imaginary $w$, so $f$ has no limit there.)

Step 2 — Laurent. By Laurent’s theorem there is a unique representation $f(z)=\sum_{n=-\infty}^{\infty}c_nz^n$ on $0<|z|<\infty$, converging absolutely, with $c_n=\tfrac{1}{2\pi i}\oint_{|w|=r}f(w)w^{-n-1}dw$ for any $r>0$.

Step 3 — The folding symmetry. The argument $z+\tfrac1z$ is unchanged by $z\mapsto\tfrac1z$, so

$f\!\left(\tfrac1z\right)=f(z)\qquad(z\ne0).$

The annulus $0<|z|<\infty$ is invariant under inversion, so we may substitute and stay in the region of validity:

$f\!\left(\tfrac1z\right)=\sum_{n=-\infty}^{\infty}c_nz^{-n}=\sum_{m=-\infty}^{\infty}c_{-m}z^{m},$

the re-indexing being legitimate by absolute convergence. So $\sum_m c_{-m}z^m=\sum_m c_mz^m$ on the same annulus, and uniqueness gives

$c_{-n}=c_n\quad\text{for all }n.$

Splitting the series accordingly,

$f(z)=c_0+\sum_{n=1}^{\infty}c_nz^n+\sum_{n=1}^{\infty}c_{-n}z^{-n}=c_0+\sum_{n=1}^{\infty}c_n\!\left(z^n+\frac{1}{z^n}\right).$

The printed shape is now established; only the identification of $c_n$ remains.

Step 4 — Take $r=1$. With $w=e^{i\theta}$ we get $w+\tfrac1w=2\cos\theta$ and $dw=ie^{i\theta}d\theta$, so

$c_n=\frac{1}{2\pi i}\int_0^{2\pi}\frac{\cosh(2\cos\theta)}{e^{i(n+1)\theta}}\,ie^{i\theta}\,d\theta=\frac{1}{2\pi}\int_0^{2\pi}\cosh(2\cos\theta)\,e^{-in\theta}\,d\theta.$

Step 5 — The imaginary part vanishes. Let $I_n=\int_0^{2\pi}\cosh(2\cos\theta)\sin n\theta\,d\theta$. Substituting $\theta=2\pi-\phi$ and using $\cos(2\pi-\phi)=\cos\phi$, $\sin n(2\pi-\phi)=-\sin n\phi$ gives $I_n=-I_n$, so $I_n=0$. Hence

$c_n=\frac{1}{2\pi}\int_0^{2\pi}\cos n\theta\,\cosh(2\cos\theta)\,d\theta=a_n,$

and each $c_n$ is real. Combining with Step 3:

$\boxed{\cosh\!\left(z+\frac1z\right)=a_0+\sum_{n=1}^{\infty}a_n\!\left(z^n+\frac{1}{z^n}\right),\quad 0<|z|<\infty.}\;\blacksquare$

Worth one extra line: replacing $\theta$ by $\theta+\pi$ in the coefficient integral (a shift by a period) and using that $\cosh$ is even gives $a_n=(-1)^na_n$, so $a_n=0$ for every odd $n$. Equivalently $f(-z)=f(z)$, so only even powers occur. Noting this shows you are reading the object rather than reciting a method.

Common Traps

• Reaching for partial fractions. There are none. If the function is transcendental in $z+\tfrac1z$, the geometric-series route above does not apply at all and the coefficient formula is the only way in.
• Asserting $c_{-n}=c_n$ instead of proving it. Without it the folded form $a_n(z^n+z^{-n})$ is unearned, and that folding is a large share of the marks. Either route works — the uniqueness argument of Step 3, or the direct one-liner $c_{-n}=\tfrac{1}{2\pi}\int\cos(-n\theta)\cosh(2\cos\theta)d\theta=a_n$ once Step 5 is done — but one of them must appear.
• Discarding the imaginary part by fiat. $c_n$ is a priori complex; that $\int_0^{2\pi}\cosh(2\cos\theta)\sin n\theta\,d\theta=0$ needs the reflection argument. “By symmetry” without the substitution is not enough at 15 marks.
• Confusing the two normalising constants. The Laurent coefficient carries $\tfrac{1}{2\pi i}$; the final real formula carries $\tfrac{1}{2\pi}$. The $i$ is cancelled by the $i$ in $dw$ — show that cancellation.
• Omitting the annulus. "$0<|z|<\infty$" is not decoration here: it is the statement that $0$ is the only singularity, and its invariance under $z\mapsto1/z$ is precisely what makes Step 3 legal.

Practice Set

• 2021-P2-Q2c (20 m) — — three-region expansion for a rational function with a cubic denominator about $z=0$; the hardest mark-value variant.
• 2022-P2-Q1d (10 m) — — two annuli about $z=1$ and $z=3$; uses substitution $w=z-1$.