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Classifying Singularities and Branch Points

Source lecture(s): PHY622 Ch. 4–5

Intuition

Everything useful about an analytic function is encoded in where it fails to be analytic. Poles give you the residue theorem; branch points give you multivaluedness and the cuts you must draw to control it. Classifying the singularity is always the first step in evaluating a contour integral, because it decides which tool applies.

Isolated singularities

Read off the Laurent expansion about \(z_0\):

Type Principal part Test Example
Removable none \(\lim_{z\to z_0}f\) exists and is finite \(\sin z/z\) at 0
Pole of order \(m\) terminates at \((z-z_0)^{-m}\) \((z-z_0)^mf\) is finite and non-zero \(1/(z-1)^3\)
Essential infinite neither limit exists \(e^{1/z}\) at 0

The practical test for a pole's order: multiply by \((z-z_0)^m\) and see what makes the limit finite and non-zero. For a rational function it is just the multiplicity of the root in the denominator.

Branch points are different

A branch point is not an isolated singularity, and the difference is what makes it awkward. Encircle a pole and the function returns to its starting value. Encircle a branch point and it does not.

Take \(f(z) = \sqrt{z} = r^{1/2}e^{i\theta/2}\). Start at \(\theta = 0\) and go once around the origin. Now \(\theta = 2\pi\), so

\[f \to r^{1/2}e^{i\pi} = -\sqrt{z}\]

You come back to the same point with the opposite value. The function is multivalued, and \(z=0\) is a branch point.

Similarly \(\log z = \ln r + i\theta\) gains \(2\pi i\) on each circuit — infinitely many branches.

Branch cuts

To get a single-valued function you forbid the offending circuits by cutting the plane: draw a curve from the branch point to another branch point (or to infinity) and declare it off limits.

  • \(\sqrt{z}\): one cut from 0 to ∞. Usually the negative real axis, but that is convention, not mathematics — you may put it anywhere, and the right choice is the one that keeps your contour out of trouble.
  • \(\log z\): same.
  • \(\sqrt{z^2-1}\): branch points at \(z = \pm1\). You may cut between them (a finite cut, leaving the function single-valued outside — often the useful choice) or from each out to infinity. These give genuinely different single-valued functions.

The cut is not a property of the function; it is a choice you make. What is intrinsic is the location of the branch points.

Why physicists care

  • Dispersion relations. A response function's branch cut on the real axis marks the continuum of dissipative excitations. Poles are discrete modes; cuts are continua. The distinction is physical: a pole is a long-lived resonance, a cut is a decay channel.
  • Scattering. The S-matrix's poles are bound states and resonances; its branch cuts are thresholds where new channels open. Reading the analytic structure is reading the physics.
  • Landau damping. The plasma dispersion function requires analytic continuation around a pole, and the correct prescription — which contour, on which side — is fixed by causality, not by convenience.
  • Renormalisation and regularisation. Contour rotations (Wick rotation) depend entirely on not crossing cuts.

Analytic continuation

If two analytic functions agree on any set with a limit point, they agree everywhere they are both defined. So an analytic function defined on a small patch has at most one extension — the continuation is unique.

This is remarkably strong. \(\sum z^n\) converges only for \(|z|<1\), but it equals \(1/(1-z)\) there, and \(1/(1-z)\) is defined everywhere except \(z=1\). The series is a partial glimpse of a function that exists far beyond its disc of convergence. The Riemann zeta function is defined this way outside \(\mathrm{Re}\,s > 1\), and the Casimir effect's \(\zeta(-3)\) is a continuation, not a sum.

Common mistakes

  • Treating a branch point like a pole and trying to take a residue at it. There is no Laurent expansion about a branch point.
  • Letting a contour cross a cut. Cauchy's theorem does not apply across it; the function is discontinuous there by construction.
  • Believing the cut is unique. Only the branch points are intrinsic.
  • Assuming a singularity is isolated. \(1/\sin(1/z)\) has poles accumulating at \(z=0\), so \(z=0\) is not isolated and none of the classification above applies to it.

Knowledge graph position

Prerequisites: analytic functions, Laurent series. Leads to: residue calculus, dispersion relations, S-matrix theory, analytic continuation.

Quiz

Q1 (conceptual). What distinguishes a branch point from a pole?

Answer

Encircling a pole returns the function to its original value; encircling a branch point does not — the function is multivalued there. Consequently a branch point has no Laurent expansion and no residue, and you must introduce a cut to make the function single-valued.

Q2 (computational). Classify the singularities of \(f(z) = \dfrac{\sin z}{z^2(z-\pi)}\).

Answer

At \(z=0\): \(\sin z \approx z - z^3/6\), so \(f \approx \frac{1}{z(z-\pi)}\) near 0 — the \(z^2\) is reduced to \(z^1\), giving a simple pole, not a double one. At \(z = \pi\): \(\sin z\) is non-zero there, so a simple pole. Nothing else in the finite plane.

Q3 (MCQ). The branch cut of \(\sqrt{z}\) is conventionally on the negative real axis because:

  • (a) the function is undefined there
  • (b) it is a convenient convention — the cut may be placed along any curve from 0 to ∞
  • (c) it is required by Cauchy's theorem
  • (d) the derivative diverges there
Answer

(b). Only the branch point at \(z=0\) is intrinsic. The cut's location is your choice, and in practice you place it wherever it keeps your contour clear.