Laurent’s series in an annulus

At a Glance

• Frequency: 7 sub-parts across 7 of 13 years (2014, 2015, 2018, 2019, 2021, 2022, 2025)
• 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: $\frac{1}{z-a}=\frac{1}{z}\cdot\frac{1}{1-(a/z)}=\frac{1}{z}\sum_{n=0}^\infty\!\left(\frac{a}{z}\right)^n=\sum_{n=0}^\infty\frac{a^n}{z^{n+1}},\quad|z|>|a|.$

Procedure. (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

One pattern covers all Laurent-series questions in the corpus.

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.

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 $w=z-1$.