Contour Integration, Laurent Series & Residues

A contour integral accumulates a complex-valued field along an oriented path. Holomorphic functions make these integrals unusually rigid: on suitable domains, closed integrals vanish; boundary values determine every interior derivative; and when isolated singularities are present, the integral depends only on one Laurent coefficient at each singularity. Geometry, orientation, analyticity, and topology are all part of the calculation.

Parameterize the path, including its orientation #

For a piecewise $C^1$ contour $C$ parameterized by $z(t)$, $a\le t\le b$,

$$\int_C f(z)\,dz =\int_a^b f(z(t))z'(t)\,dt.$$

Reversing orientation changes the sign. Concatenating paths adds their integrals. A circle $|z-z_0|=R$ traversed counterclockwise is

$$z(t)=z_0+Re^{it},\qquad 0\le t\le2\pi.$$

The estimation lemma is the first sanity check:

$$\left|\int_C f(z)\,dz\right| \le \max_{z\in C}|f(z)|\,\mathop{\mathrm{length}}(C).$$

It explains why auxiliary arcs vanish: the decay of the integrand must beat the growth of the arc length.

Primitives make integrals path-independent #

If $F'=f$ on a domain containing a path from $z_a$ to $z_b$, then

$$\int_C f(z)\,dz=F(z_b)-F(z_a).$$

Therefore every closed contour integral is zero. Conversely, path independence supplies a primitive by defining $F(z)$ as an integral from a fixed base point.

Cauchy's theorem supplies primitives under common topological conditions: if $f$ is holomorphic on a simply connected domain, then its integral around every closed contour in that domain is zero. On a domain with a hole, holomorphicity alone is not enough. The function $1/z$ is holomorphic on $\mathbb C\setminus\{0\}$, yet

$$\int_{|z|=1}\frac{dz}{z}=2\pi i.$$

The loop encloses the missing point, so it cannot be contracted within the domain.

Cauchy's integral formula turns boundaries into values #

If $f$ is holomorphic on and inside a positively oriented simple closed contour $C$, and $z_0$ lies inside, then

$$f(z_0)=\frac{1}{2\pi i}\int_C\frac{f(z)}{z-z_0}\,dz.$$

More generally,

$$f^{(n)}(z_0)=\frac{n!}{2\pi i} \int_C\frac{f(z)}{(z-z_0)^{n+1}}\,dz.$$

These formulas explain the rigidity of holomorphic functions. Interior values and every derivative are encoded on an arbitrarily small surrounding circle. Cauchy's estimates, power-series expansions, the identity theorem, and Liouville's theorem all follow from this mechanism.

Laurent series are indexed by an annulus #

Around $z_0$, a holomorphic function on an annulus $r<|z-z_0|<R$ has a unique Laurent series

$$f(z)=\sum_{n=-\infty}^{\infty}a_n(z-z_0)^n,$$

with coefficients

$$a_n=\frac{1}{2\pi i}\int_C\frac{f(\zeta)}{(\zeta-z_0)^{n+1}}\,d\zeta$$

where $C$ is a positively oriented circle $|\zeta-z_0|=\rho$ centered at $z_0$, with $r<\rho<R$.

The same algebraic function may have different Laurent expansions in different annuli. For example,

$$\frac1{z-1} =-\sum_{n=0}^{\infty}z^n,\qquad |z|<1,$$

but

$$\frac1{z-1} =\frac1z\frac1{1-1/z} =\sum_{n=0}^{\infty}z^{-n-1},\qquad |z|>1.$$

An expansion without its annulus is incomplete.

Isolated singularities are read from the principal part #

Suppose $f$ is holomorphic on $0<|z-z_0|<R$.

The residue is the coefficient of $(z-z_0)^{-1}$:

$$\mathop{\mathrm{Res}}(f,z_0)=a_{-1}.$$

At a simple pole,

$$\mathop{\mathrm{Res}}(f,z_0)=\lim_{z\to z_0}(z-z_0)f(z).$$

If $f=g/h$, $g$ and $h$ are holomorphic, $h(z_0)=0$, and $h'(z_0)\ne0$, then

$$\mathop{\mathrm{Res}}(f,z_0)=\frac{g(z_0)}{h'(z_0)}.$$

For a pole of order $m$,

$$\mathop{\mathrm{Res}}(f,z_0) =\frac1{(m-1)!} \lim_{z\to z_0}\frac{d^{m-1}}{dz^{m-1}} \left[(z-z_0)^mf(z)\right].$$

The residue theorem localizes a global integral #

If $f$ is holomorphic inside and on a positively oriented simple closed contour $C$ except for finitely many isolated singularities $z_k$ inside it, then

$$\int_C f(z)\,dz =2\pi i\sum_k\mathop{\mathrm{Res}}(f,z_k).$$

In inverse Laplace and Fourier contour calculations, these same residues become coefficients of exponential or oscillatory modes. A time-domain response that looks global can therefore be assembled from local pole data.

For multiply connected boundaries, use positive boundary orientation: the outer boundary is counterclockwise and each hole boundary is clockwise, so the domain stays on the left while traversing every component.

Worked example: a real integral #

For $a>0$, evaluate

$$I=\int_{-\infty}^{\infty}\frac{dx}{x^2+a^2}.$$

Integrate $f(z)=1/(z^2+a^2)$ over the real segment $[-R,R]$ and the upper semicircle. The poles are $z=ia$ and $z=-ia$; only $ia$ lies inside. Its residue is

$$\mathop{\mathrm{Res}}(f,ia) =\frac1{2ia}.$$

On the arc, $|f(z)|\le1/(R^2-a^2)$ for $R>a$, while the arc length is $\pi R$, so the arc integral tends to zero. Therefore

$$I=2\pi i\frac1{2ia}=\frac\pi a.$$

The sign is fixed by the counterclockwise orientation. Closing in the lower half-plane instead reverses the contour orientation and encloses the other pole; both changes must be handled together.

Choosing contours for real integrals #

Diagnostic checklist #

• Draw the contour and arrows; state which singularities are enclosed.
• Verify holomorphicity on the deformation region, not only on the final path.
• Keep the integration variable distinct from the evaluation point in Cauchy's formula.
• Name the annulus for every Laurent expansion.
• Classify each singularity before choosing a residue formula.
• Include the factor $2\pi i$ exactly once.
• Estimate every discarded arc or segment; a picture is not a convergence proof.
• For $z=e^{it}$, transform the bounds, orientation, integrand, and $dz=iz\,dt$ together.