Laurent Series, Poles, and Residues

A residue is one Laurent coefficient, but it can determine an entire contour integral. This compression works because the integral of every integer power of $z-a$ around a small loop vanishes except the power $(z-a)^{-1}$. The result is local—compute one coefficient at each enclosed isolated singularity—and global—add those coefficients with the contour’s winding numbers.

Required background. Holomorphic Functions and Cauchy Theory supplies Cauchy’s formula, winding numbers, and the contour hypotheses used by the residue theorem.

This page develops the residue method with its hypotheses visible. It treats isolated singularities and finite contour integrals, then applies them to a regulated free-propagator frequency integral. Branch points and sheet changes are deferred to Branches, Sheets, Analytic Continuation, and Monodromy; thermal sums and KMS conditions belong to Thermal and Nonequilibrium QFT.

Laurent series remember the annulus

Let $a\in\mathbb C$, and suppose $f$ is holomorphic on the annulus

$$A(a;r,R) = \{z\in\mathbb C:r<|z-a|<R\}, \qquad 0\leq r<R\leq\infty.$$

Then $f$ has a unique Laurent expansion

$$f(z) = \sum_{n=-\infty}^{\infty}c_n(z-a)^n$$

that converges uniformly on every compact subannulus. If $r<\rho<R$ and the circle $|\zeta-a|=\rho$ is traversed counterclockwise, the coefficients are

$$c_n = \frac{1}{2\pi i} \oint_{|\zeta-a|=\rho} \frac{f(\zeta)} {(\zeta-a)^{n+1}} \,\mathrm d\zeta.$$

Cauchy’s theorem makes this value independent of the chosen radius $\rho$ as long as the circle remains in the same annulus. The nonnegative powers form the regular part,

$$\sum_{n=0}^{\infty}c_n(z-a)^n,$$

and the negative powers form the principal part,

$$\sum_{n=1}^{\infty}c_{-n}(z-a)^{-n}.$$

For the existence, uniqueness, and compact-subannulus convergence theorem, see Conway 1978, Chapter V.

The annulus is part of the answer. For example,

$$f(z)=\frac{1}{z(z-1)}$$

has, on $0<|z|<1$,

$$f(z) = -\frac{1}{z}\frac{1}{1-z} = -\frac{1}{z}-1-z-z^2-\cdots.$$

On $1<|z|<\infty$, however,

$$f(z) = \frac{1}{z^2}\frac{1}{1-z^{-1}} = \frac{1}{z^2}+\frac{1}{z^3}+\frac{1}{z^4}+\cdots.$$

Both series are correct, but only the first lives on a deleted neighborhood of $0$. It is therefore the one that supplies local data at $0$.

Isolated singularities and their local types

A point $a$ is an isolated singularity of $f$ if $f$ is holomorphic on some deleted disk

$$0<|z-a|<R.$$

The principal part of the Laurent expansion on this disk gives the classification.

• If every $c_{-n}$ vanishes, $a$ is a removable singularity. Defining $f(a)=c_0$ extends $f$ holomorphically.
• If $c_{-m}\neq0$ and $c_{-n}=0$ for every $n>m$, then $a$ is a pole of order $m$.
• If infinitely many negative-power coefficients are nonzero, $a$ is an essential singularity.

At a pole of order $m$ there is a holomorphic function $g$ with $g(a)\neq0$ such that

$$f(z)=\frac{g(z)}{(z-a)^m}.$$

This factorization is often more useful than computing a full Laurent series. It also separates zeros from poles: if $h$ has a zero of order $m$ at $a$, then

$$h(z)=(z-a)^m q(z), \qquad q(a)\neq0,$$

so $1/h$ has a pole of order $m$.

A branch point is not an isolated singularity of a single-valued holomorphic branch on a full deleted disk. For example, defining $\log(z-a)$ requires a cut in every neighborhood of $a$. A Laurent classification at $a$ is therefore the wrong tool.

The residue is the surviving coefficient

The residue of $f$ at an isolated singularity $a$ is

$$\operatorname{Res}_{z=a} f = c_{-1}.$$

Equivalently, for any sufficiently small counterclockwise circle around $a$,

$$\operatorname{Res}_{z=a} f = \frac{1}{2\pi i} \oint f(z)\,\mathrm dz.$$

Indeed,

$$\oint_{|z-a|=\rho}(z-a)^n\,\mathrm dz = \begin{cases} 2\pi i, & n=-1,\\ 0, & n\neq-1. \end{cases}$$

The residue exists for removable singularities and essential singularities as well as poles. It is zero at a removable singularity, but a zero residue does not imply removability: $1/(z-a)^2$ has residue zero and still has a double pole.

Simple poles

If $a$ is a simple pole, then

$$\operatorname{Res}_{z=a}f = \lim_{z\to a}(z-a)f(z).$$

More generally, if $g$ and $h$ are holomorphic near $a$,

$$h(a)=0, \qquad h'(a)\neq0,$$

then $g/h$ has at most a simple pole and

$$\operatorname{Res}_{z=a}\frac{g(z)}{h(z)} = \frac{g(a)}{h'(a)}.$$

If $g(a)=0$, write $g(z)=(z-a)g_1(z)$ and $h(z)=(z-a)h_1(z)$, where $h_1(a)=h'(a)\neq0$. Then $g/h=g_1/h_1$ extends holomorphically across $a$, so the apparent singularity is removable and the formula correctly gives residue zero.

Higher-order poles

If $f$ has a pole of order $m$, set

$$G(z)=(z-a)^m f(z).$$

Then $G$ is holomorphic at $a$, and the residue is the coefficient of $(z-a)^{m-1}$ in its Taylor series:

$$\operatorname{Res}_{z=a}f = \frac{1}{(m-1)!} \left. \frac{\mathrm d^{m-1}}{\mathrm dz^{m-1}} \bigl[(z-a)^m f(z)\bigr] \right|_{z=a}.$$

The actual pole order is the smallest exponent that makes $(z-a)^m f(z)$ holomorphic at $a$. Any integer $k\geq m$ also gives

$$\operatorname{Res}_{z=a}f = \frac{1}{(k-1)!} \left. \frac{\mathrm d^{k-1}}{\mathrm dz^{k-1}} \bigl[(z-a)^k f(z)\bigr] \right|_{z=a},$$

although it introduces unnecessary derivatives. An exponent below the pole order is invalid because the bracketed function remains singular at $a$.

The residue theorem

Let $\gamma$ be a closed piecewise smooth contour in an open set $\Omega$. Suppose $f$ is holomorphic in $\Omega$ except at finitely many isolated singularities $a_1,\ldots,a_N$, none on $\gamma$, and suppose $\gamma$ is null-homologous in $\Omega$. Then

$$\oint_\gamma f(z)\,\mathrm dz = 2\pi i \sum_{j=1}^{N} \operatorname{Ind}(\gamma,a_j) \operatorname{Res}_{z=a_j}f,$$

where

$$\operatorname{Ind}(\gamma,a) = \frac{1}{2\pi i} \oint_\gamma\frac{\mathrm dz}{z-a}$$

is the winding number. For a counterclockwise simple closed contour, $\operatorname{Ind}(\gamma,a)$ is $1$ inside and $0$ outside. Clockwise orientation changes the sign.

Choose disjoint small counterclockwise circles $C_j$ around the singularities. In $\Omega\setminus\{a_1,\ldots,a_N\}$, the cycle

$$\gamma - \sum_{j=1}^N \operatorname{Ind}(\gamma,a_j)C_j$$

is null-homologous. Cauchy’s theorem therefore sets its integral to zero, yielding the stated residue sum. For a positively oriented simple boundary, this reduces to the familiar picture in which the punctures occur as clockwise inner boundary circles. A singularity on $\gamma$ is excluded because neither its winding number nor the ordinary contour integral is then defined without an additional prescription.

A finite-contour calculation

Consider the counterclockwise circle $|z|=2$ and

$$f(z) = \frac{e^z}{z^2(z-1)}.$$

There is a double pole at $0$ and a simple pole at $1$. At the double pole,

$$\begin{aligned} \operatorname{Res}_{z=0}f &= \left. \frac{\mathrm d}{\mathrm dz} \left(\frac{e^z}{z-1}\right) \right|_{z=0}\\ &= \left. \left( \frac{e^z}{z-1} -\frac{e^z}{(z-1)^2} \right) \right|_{z=0}\\ &=-2. \end{aligned}$$

At the simple pole,

$$\operatorname{Res}_{z=1}f=e.$$

Therefore

$$\oint_{|z|=2} \frac{e^z}{z^2(z-1)} \,\mathrm dz = 2\pi i(e-2).$$

The answer uses only the two local coefficients; no antiderivative is needed.

A reliable residue-calculus workflow

For a parameter-dependent contour integral, use the following order.

1. State the contour and orientation. Include every finite segment, indentation, cut edge, and large arc.
2. Locate and classify singularities. Distinguish denominator zeros, canceled zeros, poles, essential singularities, and nonisolated branch points.
3. Determine winding numbers. “Inside” is insufficient for a self-intersecting or multiply wound contour.
4. Compute only the needed residues. Use factorization, the simple-pole quotient formula, or the higher-pole derivative formula.
5. Apply the residue theorem to the closed finite contour.
6. Justify every limit separately. A large arc vanishes only after an estimate; a small indentation has its own limit; a regulator is removed only in its declared mode of convergence.

Worked versions of this residue workflow appear in Orloff 2018, Topics 8–9, PDF.

For a semicircle of radius $R$, the elementary estimate

$$\left| \int_{\Gamma_R} f(z)\,\mathrm dz \right| \leq \pi R\sup_{z\in\Gamma_R}|f(z)|$$

shows that $f=O(R^{-1-\delta})$ with $\delta>0$ is enough for the arc to vanish. An oscillatory factor may improve the estimate in one half-plane and worsen it in the other; its sign must be checked rather than recalled from a diagram.

QFT application: a regulated free-propagator integral

This frequency-contour check follows the pole placement and Fourier conventions of Schwartz 2014, § 6.2, pp. 75–77.

Let

$$E_{\mathbf p} = \sqrt{|\mathbf p|^2+m^2}, \qquad E_{\mathbf p}>0,$$

and keep $\eta>0$ finite in

$$I_\eta(t,\mathbf p) = \int_{-\infty}^{\infty} \frac{\mathrm dp^0}{2\pi} \frac{i\,e^{-ip^0t}} {\bigl(p^0-E_{\mathbf p}+i\eta\bigr) \bigl(p^0+E_{\mathbf p}-i\eta\bigr)}.$$

This exact regulator places the poles at

$$z_+=E_{\mathbf p}-i\eta, \qquad z_-=-E_{\mathbf p}+i\eta.$$

Its denominator equals

$$(p^0)^2-E_{\mathbf p}^2 +2iE_{\mathbf p}\eta+\eta^2,$$

so its $\eta\to0^+$ boundary value is the usual Feynman prescription. At fixed $\eta$, the integrand is meromorphic and behaves as $O(|p^0|^{-2})$ apart from the exponential.

For $t>0$ and $p^0=x+iy$,

$$|e^{-ip^0t}|=e^{yt},$$

so the exponential decays in the lower half-plane. More generally, for either sign of $t$, the chosen closing half-plane has $yt\leq0$ and the exponential has modulus at most one there. Let

$$M=\max(|z_+|,|z_-|).$$

On a closing semicircle $|z|=R>2M$,

$$|(z-z_+)(z-z_-)| \geq (R-M)^2 \geq \frac{R^2}{4},$$

and hence

$$\left| \int_{\Gamma_R} \frac{i\,e^{-izt}} {(z-z_+)(z-z_-)} \,\mathrm dz \right| \leq \frac{4\pi}{R} \longrightarrow0.$$

Thus the lower arc vanishes for $t>0$. Closing there gives a clockwise contour and encloses $z_+$. Since

$$\operatorname{Res}_{p^0=z_+} \frac{i\,e^{-ip^0t}} {(p^0-z_+)(p^0-z_-)} = \frac{i\,e^{-iz_+t}}{z_+-z_-},$$

the orientation contributes a minus sign:

$$\begin{aligned} I_\eta(t,\mathbf p) &= \frac{-2\pi i}{2\pi} \frac{i\,e^{-iz_+t}}{z_+-z_-}\\ &= \frac{e^{-iz_+t}}{z_+-z_-}, \qquad t>0. \end{aligned}$$

For $t<0$, close counterclockwise in the upper half-plane and enclose $z_-$. Then

$$\begin{aligned} I_\eta(t,\mathbf p) &= \frac{2\pi i}{2\pi} \frac{i\,e^{-iz_-t}}{z_--z_+}\\ &= \frac{e^{-iz_-t}}{z_+-z_-}, \qquad t<0. \end{aligned}$$

Because $z_+-z_-=2(E_{\mathbf p}-i\eta)$,

$$\lim_{\eta\to0^+}I_\eta(t,\mathbf p) = \frac{e^{-iE_{\mathbf p}|t|}}{2E_{\mathbf p}}.$$

At $t=0$, the rational $O(|p^0|^{-2})$ decay makes either closure legal and gives the same continuous value. With the site’s Fourier convention, the inverse transform carries $e^{-ip^0t}$; this is why positive $t$ selects the lower half-plane. Sources using $e^{+i\omega t}$ close in the opposite half-plane without disagreeing physically.

The pole placement is not decorative. Moving both poles above or below the real axis changes the boundary condition and hence the time support of the result. The developed interpretation of the time-ordered two-point function belongs to Scalar Propagators, Ordered Correlators, and Sources.

Where the method stops

A pole on the contour needs a new definition. The ordinary contour integral is not defined. A Cauchy principal value, an upper or lower indentation, and a distributional boundary value are different prescriptions and must not be silently interchanged.

A branch point is not a pole. Mark a branch and its cut, then track the two boundary values. A small-loop residue cannot replace a cut discontinuity.

Closing a contour is an added argument. The residue theorem evaluates the closed contour. It does not prove that the auxiliary arc vanishes.

Parameters can move singularities. Before varying a mass, external energy, or regulator, check whether a pole crosses the contour, two poles coalesce, or the contour becomes pinched. A formula derived in one parameter region need not continue unchanged.

Residue zero does not mean regular. Higher-order poles and essential singularities can have no $(z-a)^{-1}$ term.

The $i0$ notation is a limit. Locate poles and establish bounds at positive regulator first. Removing the regulator can require distributional convergence rather than pointwise substitution.

Exercises

1. Classify the singularity of $(e^z-1-z)/z^3$ at $z=0$ and find its residue.

   Check

   Since

   $$e^z-1-z = \frac{z^2}{2} +\frac{z^3}{6} +O(z^4),$$

   division by $z^3$ gives

   $$\frac{e^z-1-z}{z^3} = \frac{1}{2z} +\frac16 +O(z).$$

   The point is a simple pole and the residue is $1/2$.

2. Explain why the outer Laurent expansion

   $$\frac{1}{z(z-1)} = \sum_{n=0}^{\infty}z^{-n-2}, \qquad |z|>1,$$

   cannot be used to conclude that the residue at $0$ is zero.

   Check

   A residue at $0$ is determined by a Laurent series on a deleted neighborhood $0<|z|<R$. The outer series is valid only for $|z|>1$ and does not approach $0$. On $0<|z|<1$, the correct expansion begins $-z^{-1}-1-z-\cdots$, so the residue at $0$ is $-1$.

3. For $t>0$, reproduce the sign in the regulated propagator calculation. Which pole is enclosed, and why does clockwise orientation not make the final answer negative?

   Check

   The factor $e^{-ip^0t}$ decays below, so the lower closure encloses $z_+=E_{\mathbf p}-i\eta$ and is clockwise. The contour contributes $-2\pi i$, while the integrand residue contains a numerator $i$. Their product is positive:

   $$\frac{-2\pi i}{2\pi} \frac{i\,e^{-iz_+t}}{z_+-z_-} = \frac{e^{-iz_+t}}{z_+-z_-}.$$

References

• John B. Conway, Functions of One Complex Variable I, 2nd ed., Chapter V, Springer, 1978. Book record. This is the structural source for isolated singularities, Laurent expansions, residues, and the homological form of the residue theorem.
• Jeremy Orloff, 18.04 Complex Variables with Applications, MIT OpenCourseWare, 2018: Topic 7, Taylor and Laurent Series, PDF, Topic 8, Residue Theorem, PDF, and Topic 9, Definite Integrals Using the Residue Theorem, PDF. These notes supply the elementary expansions, computational formulas, contour workflow, and large-arc estimates.
• Matthew D. Schwartz, Quantum Field Theory and the Standard Model, §6.2, pp. 75–77, Cambridge University Press, 2014. Book record. This is the QFT source for the free Feynman-propagator frequency integral, its pole placement, and the contour-orientation translation.