Laurent Series & Classification of Singularities

Derivation

Laurent Series & Classification of Singularities

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Statement

Let \(f\) be holomorphic on the open annulus \(A=\{z:R_1<|z-z_0|<R_2\}\) with \(0\le R_1<R_2\le\infty\). Then \(f\) admits a unique two-sided series expansion \(f(z)=\sum_{n=-\infty}^{\infty} a_n\,(z-z_0)^n\), convergent (absolutely, and uniformly on compact subsets) throughout \(A\), with coefficients \(a_n=\dfrac{1}{2\pi i}\displaystyle\oint_{\gamma}\dfrac{f(\zeta)}{(\zeta-z_0)^{\,n+1}}\,d\zeta\) for any positively oriented circle \(\gamma\subset A\) about \(z_0\). When \(z_0\) is an isolated singularity (\(R_1=0\)), the negative-index part — the principal part — classifies the singularity as removable, a pole of order \(m\), or essential.

Why it matters

The Taylor series describes a function near a point of analyticity; the Laurent series is its indispensable generalisation to a neighbourhood surrounding a bad point. It is the single tool that converts the local behaviour of \(f\) near an isolated singularity into a finite list of numbers — the principal-part coefficients — and in particular isolates the residue \(a_{-1}\), the object the residue theorem integrates.

Physically the same expansion organises multipole fields, the frequency content of causal response functions, scattering amplitudes with bound-state poles and branch structure, and perturbative expansions that fail to converge. Classifying singularities is how one reads the analytic structure of a physical amplitude off its formula.

Assumptions

\(f\) is single-valued and holomorphic on the whole open annulus \(A\).If \(f\) has further singularities inside \(A\), no single global Laurent series about \(z_0\) exists; one gets a different expansion in each singularity-free sub-annulus.

The two bounding radii satisfy \(R_1<R_2\), i.e. the annulus is non-degenerate.If \(R_1=R_2\) the region is empty (a circle) and there is nothing to expand; convergence of the two half-series would have no common domain.

The singularity under classification is isolated: \(f\) is holomorphic on some punctured disc \(0<|z-z_0|<R_2\).If singularities accumulate at \(z_0\) (as for \(1/\sin(1/z)\) at \(z_0=0\)), \(z_0\) is a non-isolated singularity, the principal part is not defined, and the removable/pole/essential trichotomy does not apply.

Boundedness/limit criteria (Riemann, Casorati–Weierstrass) are invoked for the classification, requiring the punctured-disc topology.Drop the isolated-point structure and the limit statements \(\lim_{z\to z_0}f\) and \(\lim_{z\to z_0}(z-z_0)^m f\) lose meaning, breaking the clean characterisation of pole order.

Derivation

\[ f(z)=\frac{1}{2\pi i}\oint_{\Gamma_2}\frac{f(\zeta)}{\zeta-z}\,d\zeta-\frac{1}{2\pi i}\oint_{\Gamma_1}\frac{f(\zeta)}{\zeta-z}\,d\zeta \]

Fix \(z\in A\) and pick radii \(r_1,r_2\) with \(R_1<r_1<|z-z_0|<r_2<R_2\); let \(\Gamma_j\) be the circle \(|\zeta-z_0|=r_j\), positively oriented. The boundary of the sub-annulus between them is \(\Gamma_2-\Gamma_1\). Cauchy's integral formula on this region (cut into simply-connected pieces so the crosscuts cancel) reproduces \(f(z)\) with the inner circle subtracted. B

\[ \frac{1}{\zeta-z}=\frac{1}{(\zeta-z_0)-(z-z_0)}=\frac{1}{\zeta-z_0}\cdot\frac{1}{1-\frac{z-z_0}{\zeta-z_0}}=\sum_{n=0}^{\infty}\frac{(z-z_0)^n}{(\zeta-z_0)^{\,n+1}} \]

On \(\Gamma_2\) we have \(|\zeta-z_0|=r_2>|z-z_0|\), so the ratio \(\left|\frac{z-z_0}{\zeta-z_0}\right|<1\) and the geometric series converges uniformly in \(\zeta\) on \(\Gamma_2\). Uniform convergence licenses termwise integration next. B

\[ -\frac{1}{\zeta-z}=\frac{1}{(z-z_0)-(\zeta-z_0)}=\frac{1}{z-z_0}\cdot\frac{1}{1-\frac{\zeta-z_0}{z-z_0}}=\sum_{k=0}^{\infty}\frac{(\zeta-z_0)^{k}}{(z-z_0)^{\,k+1}} \]

On \(\Gamma_1\) the inequality reverses: \(|\zeta-z_0|=r_1<|z-z_0|\), so now \(\left|\frac{\zeta-z_0}{z-z_0}\right|<1\) and this second geometric series converges uniformly on \(\Gamma_1\). The two kernels are expanded about opposite ends of the annulus. B

\[ f(z)=\sum_{n=0}^{\infty}(z-z_0)^n\underbrace{\frac{1}{2\pi i}\oint_{\Gamma_2}\frac{f(\zeta)\,d\zeta}{(\zeta-z_0)^{\,n+1}}}_{a_n}+\sum_{k=0}^{\infty}(z-z_0)^{-(k+1)}\underbrace{\frac{1}{2\pi i}\oint_{\Gamma_1}f(\zeta)(\zeta-z_0)^{k}\,d\zeta}_{b_{k}} \]

Insert the two expansions of Steps 2–3 into Step 1 and integrate term by term (permitted by uniform convergence). The \(\Gamma_2\) integral yields non-negative powers; the \(\Gamma_1\) integral yields negative powers. B

\[ \frac{1}{2\pi i}\oint_{\Gamma_1}\!\frac{f(\zeta)\,d\zeta}{(\zeta-z_0)^{-k}}=\frac{1}{2\pi i}\oint_{\gamma}\!\frac{f(\zeta)\,d\zeta}{(\zeta-z_0)^{\,n+1}}\quad\text{with }n=-(k+1) \]

Both \(\Gamma_1\) and \(\Gamma_2\) are homotopic in \(A\) to any circle \(\gamma=\{|\zeta-z_0|=r\}\), \(R_1<r<R_2\). Since \(f(\zeta)/(\zeta-z_0)^{n+1}\) is holomorphic in the annulus, Cauchy's theorem makes the contour integral independent of \(r\); relabel \(b_k\) as \(a_{-(k+1)}\). Both families now read off the same formula. A

\[ f(z)=\sum_{n=-\infty}^{\infty}a_n\,(z-z_0)^n,\qquad a_n=\frac{1}{2\pi i}\oint_{\gamma}\frac{f(\zeta)}{(\zeta-z_0)^{\,n+1}}\,d\zeta \]

Merge the two sums into one bilateral series indexed by \(n\in\mathbb{Z}\), with the single unified coefficient formula. This is the Laurent expansion. A

\[ \text{If } f=\sum c_n(z-z_0)^n \text{ on } A,\quad \oint_\gamma \frac{f(\zeta)\,d\zeta}{(\zeta-z_0)^{m+1}}=\sum_n c_n\oint_\gamma (\zeta-z_0)^{n-m-1}d\zeta=2\pi i\,c_m \]

Uniqueness: assume any convergent Laurent representation, multiply by \((z-z_0)^{-m-1}\) and integrate over \(\gamma\). Only the \(n=m\) term survives because \(\oint_\gamma(\zeta-z_0)^{p}\,d\zeta=2\pi i\,\delta_{p,-1}\). Hence \(c_m=a_m\): the coefficients are forced, so the expansion is unique. C

\[ \text{Principal part } P(z)=\sum_{n=-\infty}^{-1}a_n(z-z_0)^n \;\Longrightarrow\; \begin{cases} P\equiv 0 & \text{removable}\\[2pt] a_{-m}\ne0,\ a_n=0\ (n<-m) & \text{pole of order }m\\[2pt] \text{infinitely many }a_{n\to-\infty}\ne0 & \text{essential} \end{cases} \]

Specialise to \(R_1=0\) (isolated singularity). The classification is a statement purely about which negative coefficients vanish — a finite, terminating, or non-terminating principal part. Riemann's theorem identifies the removable case (bounded \(\Rightarrow a_{n<0}=0\)); \((z-z_0)^m f\) removable gives the pole case; Casorati–Weierstrass/Picard characterise the essential case. C

Result

\[ f(z)=\sum_{n=-\infty}^{\infty}a_n(z-z_0)^n,\qquad a_n=\frac{1}{2\pi i}\oint_{\gamma}\frac{f(\zeta)}{(\zeta-z_0)^{n+1}}\,d\zeta,\qquad \mathrm{Res}_{z_0}f=a_{-1} \]

Reading. Any function holomorphic on an annulus is exactly a sum of a Taylor-like part (non-negative powers, convergent inside \(R_2\)) and a principal part (negative powers, convergent outside \(R_1\)). The coefficients are unique and are contour integrals of \(f\). The lone coefficient \(a_{-1}\) is the residue. The tail of the principal part is a complete diagnostic: empty \(\Rightarrow\) the singularity was an illusion (removable); finite of length \(m\) \(\Rightarrow\) pole of order \(m\); infinite \(\Rightarrow\) essential, where \(f\) behaves as wildly as any holomorphic function can.

Units check. The relation is dimensionally homogeneous term by term. If \(f\) carries physical units \([f]\) and the coordinate \((z-z_0)\) carries \([L]\), then \(a_n\) must carry \([f][L]^{-n}\), so every term \(a_n(z-z_0)^n\) has units \([f]\), matching the left side. In the coefficient formula, \(d\zeta\) contributes \([L]\), \((\zeta-z_0)^{n+1}\) contributes \([L]^{n+1}\), and \(f\) contributes \([f]\), giving \([f][L]\,[L]^{-(n+1)}=[f][L]^{-n}\) — consistent. The prefactor \(1/2\pi i\) is dimensionless.

Limiting cases

\(R_1\to0\) with all \(a_{n<0}=0\): the annulus fills in to a disc and the Laurent series reduces to the ordinary Taylor series \(\sum_{n\ge0}a_n(z-z_0)^n\).
Single simple pole (\(a_{-1}\ne0\), \(a_{n<-1}=0\)): near \(z_0\), \(f(z)\approx a_{-1}/(z-z_0)\); the residue alone controls the leading divergence.
\(R_2\to\infty\): the annulus becomes the exterior of a disc; the non-negative-power part must terminate or the coefficients must decay, and one obtains the Laurent expansion "about infinity", the setting for the residue at \(\infty\).
Rational \(f=P/Q\) between two consecutive pole-moduli: the expansion is a partial-fractions sum, each factor expanded on the correct side of its pole, reproducing the two-sided structure explicitly.

Breaks when

Extra singularity inside the annulus. If \(f\) is singular at some \(z_1\) with \(R_1<|z_1-z_0|<R_2\), no single Laurent series about \(z_0\) represents \(f\) throughout \(A\). One must split \(A\) at \(|z-z_0|=|z_1-z_0|\); the coefficients differ across the divide (e.g. \(1/[(z)(z-1)]\) has different Laurent series for \(|z|<1\) and \(1<|z|\)).
Non-isolated singularity. If singular points accumulate at \(z_0\) — as with \(1/\sin(1/z)\) at \(0\), whose poles pile up at the origin — there is no punctured disc of analyticity, no principal part, and the removable/pole/essential trichotomy is undefined.
Branch point. At a branch point (e.g. \(\sqrt{z}\) or \(\log z\) at \(0\)) \(f\) is not single-valued on any punctured disc, so a single-valued Laurent series cannot exist; one needs a Puiseux/logarithmic expansion on a Riemann surface instead.
Boundary of convergence. The series can diverge exactly on \(|z-z_0|=R_1\) or \(=R_2\); it represents \(f\) only strictly inside the open annulus, so evaluating on the bounding circles is not guaranteed.

Failure modes

"One function, one Laurent series." Forgetting that the expansion depends on the annulus: \(\frac{1}{z-1}\) about \(z_0=0\) is \(-\sum_{n\ge0}z^n\) for \(|z|<1\) but \(\sum_{n\ge1}z^{-n}\) for \(|z|>1\). Students quote the wrong one.
Expanding the geometric series on the wrong side. Writing \(\frac{1}{1-w}=\sum w^n\) when \(|w|>1\); the series must be arranged so the ratio has modulus below one, which fixes whether you get positive or negative powers.
Reading off the residue as "the coefficient of \(1/(z-z_0)\) after any manipulation." The residue is \(a_{-1}\) of the Laurent series valid in the punctured disc around \(z_0\), not of an expansion valid in some other region.
Mislabelling a pole order. Concluding order \(m\) from the numerator power without cancelling common factors, e.g. calling \(\frac{\sin z}{z^2}\) a double pole when the \(z\) in \(\sin z\) reduces it to a simple pole.
Calling \(e^{1/z}\) a "pole of infinite order." There is no such thing; infinitely many negative coefficients means essential, a qualitatively different beast (Picard behaviour, not a divergent limit).
Assuming the series converges on the boundary circles. Using it to evaluate \(f\) at \(|z-z_0|=R_2\), where convergence can fail.

Discussion

The Laurent theorem is the exact statement that "holomorphic on an annulus" and "represented by a convergent two-sided power series there" are the same condition. The mechanism in the proof is worth internalising: Cauchy's formula surrounds \(z\) by two circles, and the single kernel \(1/(\zeta-z)\) is expanded in opposite geometric series on the two circles because the inequality \(|\zeta-z_0|\gtrless|z-z_0|\) flips between them. That flip is the entire origin of the two-sidedness — positive powers come from the outer circle, negative powers from the inner one.

Classification then becomes bookkeeping on the principal part. Riemann's removable-singularity theorem shows boundedness near \(z_0\) forces every negative coefficient to vanish, so a "singularity" where \(f\) merely fails to be defined but stays bounded is fictitious. A pole of order \(m\) is precisely the case where \((z-z_0)^m f\) is bounded but \((z-z_0)^{m-1}f\) is not; equivalently \(1/f\) has a zero of order \(m\). The essential case is the residual category — neither removable nor a pole — and Casorati–Weierstrass shows \(f\) comes arbitrarily close to every complex value in every neighbourhood of \(z_0\); Picard's great theorem sharpens this to: \(f\) attains every value, with at most one exception, infinitely often.

The deeper structural fact is that the two halves converge on complementary domains with independent radii: the analytic part \(\sum_{n\ge0}a_n(z-z_0)^n\) is a genuine power series with radius of convergence \(\ge R_2\), while \(\sum_{n<0}a_n(z-z_0)^n\), as a power series in \(1/(z-z_0)\), converges for \(|z-z_0|>R_1\). Their overlap is the annulus, and the annulus is maximal: \(R_1\) is the modulus of the nearest singularity inside, \(R_2\) of the nearest outside. This is why the Laurent expansion is the natural language for the resolvent \((A-z)^{-1}\) of an operator near an isolated point of its spectrum, where the principal part carries the spectral projection and the pole order the size of the Jordan block. In physics the residues of a Green's function or scattering amplitude at its poles give decay rates and coupling strengths, while essential singularities signal genuinely non-perturbative structure (e.g. \(e^{-1/g}\) instanton factors in a coupling \(g\)).

Common misconceptions. An essential singularity is not "a very bad pole"; it is a different class, and a function can be perfectly finite along some approach directions (\(e^{1/z}\to0\) as \(z\to0^-\) along the real axis) while blowing up along others. Also, "removable" does not mean \(f\) was already defined at \(z_0\) — it means it can be defined there to make \(f\) holomorphic; the value is forced to be \(\lim_{z\to z_0}f\).

Worked examples

Example 1 — Laurent series of \(f(z)=\dfrac{1}{z(z-2)}\) in the annulus \(0<|z|<2\), and its residue at \(0\).

\[ \frac{1}{z(z-2)}=\frac{1}{z}\cdot\frac{1}{z-2}=\frac{1}{z}\cdot\frac{-1}{2}\cdot\frac{1}{1-\frac{z}{2}} \]

Factor out the singular \(1/z\) and arrange the remaining factor with a unit constant term. A

\[ \frac{1}{1-\frac{z}{2}}=\sum_{n=0}^{\infty}\left(\frac{z}{2}\right)^n,\qquad \left|\frac{z}{2}\right|<1\iff|z|<2 \]

Geometric series, valid throughout the given annulus since \(|z|<2\). This is the "correct side" for the pole at \(z=2\). A

\[ f(z)=-\frac{1}{2z}\sum_{n=0}^{\infty}\frac{z^n}{2^n}=-\frac{1}{2z}-\sum_{n=0}^{\infty}\frac{z^{n}}{2^{n+2}}\cdot z^{0}=-\frac{1}{2z}-\frac14-\frac{z}{8}-\frac{z^2}{16}-\cdots \]

Multiply the series by \(-1/(2z)\); the \(n=0\) term supplies the single negative power. A

\[ f(z)=-\frac{1}{2z}-\sum_{n=0}^{\infty}\frac{z^{n}}{2^{\,n+2}},\qquad \mathrm{Res}_{0}f=a_{-1}=-\tfrac12 \]

Reading. Simple pole at \(0\) (principal part is one term), residue \(-\tfrac12\); consistent with the direct formula \(\mathrm{Res}_0=\lim_{z\to0}z\,f(z)=\frac{1}{0-2}=-\tfrac12\). Units: taking \(z\) dimensionless, \(f\) is dimensionless and each coefficient carries \(2^{-(n+2)}\), pure numbers, as required.

Example 2 — Classify the singularity of \(g(z)=z^2 e^{1/z}\) at \(z=0\) and give its residue.

\[ e^{1/z}=\sum_{k=0}^{\infty}\frac{1}{k!}\,z^{-k}=1+\frac{1}{z}+\frac{1}{2!\,z^2}+\frac{1}{3!\,z^3}+\cdots,\qquad 0<|z|<\infty \]

Substitute \(1/z\) into the entire exponential series, valid on the whole punctured plane. Infinitely many negative powers already appear. A

\[ g(z)=z^2 e^{1/z}=\sum_{k=0}^{\infty}\frac{z^{\,2-k}}{k!}=z^2+z+\frac12+\frac{1}{3!\,z}+\frac{1}{4!\,z^2}+\cdots \]

Multiply the series by \(z^2\), shifting every exponent up by two. The negative-power tail is still infinite. A

\[ a_{-1}=\text{coefficient of }z^{-1}\ \Rightarrow\ 2-k=-1\Rightarrow k=3\Rightarrow a_{-1}=\frac{1}{3!}=\frac16 \]

Read off the residue as the \(z^{-1}\) coefficient from the explicit series. A

\[ z=0\ \text{is an essential singularity of }g;\qquad \mathrm{Res}_{0}\,g=\frac{1}{3!}=\frac16 \]

Reading. The factor \(z^2\) removes only two negative powers; infinitely many remain, so multiplying by any finite power of \(z\) can never leave a bounded function — the hallmark of an essential singularity, not a pole. The residue is still a single well-defined number, \(1/6\). Units: \(z\) dimensionless throughout, so \(g\) and every coefficient are pure numbers.

Problems

Find the Laurent expansion of \(f(z)=\dfrac{1}{z-3}\) valid for \(|z|>3\), and state the region of convergence.
Solution
For \(|z|>3\), factor to make the ratio small: \(\frac{1}{z-3}=\frac{1}{z}\cdot\frac{1}{1-\frac{3}{z}}=\frac{1}{z}\sum_{n=0}^{\infty}\left(\frac{3}{z}\right)^n=\sum_{n=0}^{\infty}\frac{3^{n}}{z^{\,n+1}}=\frac{1}{z}+\frac{3}{z^2}+\frac{9}{z^3}+\cdots.\) Here \(|3/z|<1\iff|z|>3\), so the series converges for \(|z|>3\). It is a pure principal part (all negative powers), as expected for an expansion in the exterior region.
Expand \(f(z)=\dfrac{1}{z(z-2)}\) in the annulus \(|z|>2\) (contrast with Example 1).
Solution
Partial fractions: \(\frac{1}{z(z-2)}=\frac12\left(\frac{1}{z-2}-\frac1z\right).\) For \(|z|>2\), expand \(\frac{1}{z-2}=\frac1z\cdot\frac{1}{1-2/z}=\sum_{n=0}^\infty\frac{2^n}{z^{n+1}}=\frac1z+\frac2{z^2}+\frac4{z^3}+\cdots.\) Then \(f=\frac12\Big(\sum_{n\ge0}\frac{2^n}{z^{n+1}}-\frac1z\Big)=\frac12\Big(\frac1z+\frac2{z^2}+\frac4{z^3}+\cdots-\frac1z\Big)=\frac{1}{z^2}+\frac{2}{z^3}+\frac{4}{z^4}+\cdots=\sum_{n=2}^\infty\frac{2^{n-2}}{z^{n}}.\) The \(1/z\) terms cancel, so \(a_{-1}=0\): the residue "at infinity picture" differs from the \(0<|z|<2\) series, illustrating region-dependence.
Classify the singularity at \(z=0\) of \(h(z)=\dfrac{\sin z}{z^{3}}\) and find \(\mathrm{Res}_{0}h\).
Solution
Use \(\sin z=z-\frac{z^3}{3!}+\frac{z^5}{5!}-\cdots.\) Then \(h=\frac{1}{z^3}\left(z-\frac{z^3}{6}+\frac{z^5}{120}-\cdots\right)=\frac{1}{z^2}-\frac{1}{6}+\frac{z^2}{120}-\cdots.\) The most negative power is \(z^{-2}\) with nonzero coefficient, and nothing more negative, so \(z=0\) is a pole of order 2. The residue is the \(z^{-1}\) coefficient, which is absent: \(\mathrm{Res}_{0}h=0\). (The odd parity of \(\sin z/z^3\) guarantees no odd-power, hence no \(z^{-1}\), term.)
Determine all Laurent coefficients of \(f(z)=e^{z+1/z}\) about \(z=0\), expressing \(a_n\) as a series, and identify the singularity type.
Solution
Write \(e^{z+1/z}=e^{z}e^{1/z}=\Big(\sum_{j\ge0}\frac{z^j}{j!}\Big)\Big(\sum_{k\ge0}\frac{z^{-k}}{k!}\Big).\) The coefficient of \(z^{n}\) collects all \(j-k=n\): \(a_n=\sum_{k=\max(0,-n)}^{\infty}\frac{1}{(k+n)!\,k!}.\) (These are the modified Bessel coefficients: \(a_n=I_n(2)\), since the generating function of \(I_n\) is \(e^{(t+1/t)x/2}\) with \(x=2\).) Because \(e^{1/z}\) contributes infinitely many nonzero negative powers that are not cancelled, \(z=0\) is an essential singularity. In particular \(\mathrm{Res}_0 f=a_{-1}=I_{-1}(2)=I_1(2)\approx1.5906.\)
The function \(f(z)=\dfrac{1}{(z-1)(z-2)}\) has three natural annuli about \(z_0=0\): \(|z|<1\), \(1<|z|<2\), \(|z|>2\). Give the Laurent (or Taylor) series in the middle annulus \(1<|z|<2\).
Solution
Partial fractions: \(\frac{1}{(z-1)(z-2)}=\frac{1}{z-2}-\frac{1}{z-1}.\) In \(1<|z|<2\): for the pole at \(2\) we are inside its circle so expand in positive powers, \(\frac{1}{z-2}=\frac{-1}{2}\frac{1}{1-z/2}=-\sum_{n=0}^{\infty}\frac{z^n}{2^{n+1}};\) for the pole at \(1\) we are outside so expand in negative powers, \(-\frac{1}{z-1}=-\frac1z\frac{1}{1-1/z}=-\sum_{m=0}^{\infty}\frac{1}{z^{m+1}}.\) Adding, \(f(z)=-\sum_{n=0}^{\infty}\frac{z^{n}}{2^{\,n+1}}-\sum_{m=1}^{\infty}\frac{1}{z^{m}},\) i.e. \(a_n=-2^{-(n+1)}\) for \(n\ge0\) and \(a_n=-1\) for \(n\le-1\). Convergence: the positive part needs \(|z|<2\), the negative part needs \(|z|>1\); together, exactly the annulus \(1<|z|<2\).