Taylor and Laurent Series
Source lecture(s): PHY622 Ch. 4
Intuition
A Taylor series describes an analytic function near a point where it is well behaved. A Laurent series does the same near a point where it is not, by allowing negative powers:
The negative-power part is called the principal part, and it is where all the interesting information lives — the classification of the singularity, and the residue, are both read straight off it.
Taylor: the radius of convergence is set by the nearest singularity
For \(f\) analytic in a disc, \(a_n = f^{(n)}(z_0)/n!\) and the series converges inside a circle reaching to the nearest singularity in the complex plane.
This resolves a mystery from real calculus. Why does the Taylor series of
diverge for \(|x| > 1\), when the function is perfectly smooth there and nothing goes wrong on the real line? Because the nearest singularities are at \(z = \pm i\), off the real axis, at distance 1. A real function's convergence radius is controlled by complex singularities you cannot see from the real line. This single observation is the best advertisement for complex analysis in the whole course.
Laurent: an annulus, not a disc
If \(f\) has a singularity at \(z_0\), expand in an annulus \(r < |z-z_0| < R\) instead:
Two consequences worth stating:
- The expansion is not unique — it depends on which annulus. \(1/(z(z-1))\) has different Laurent series in \(0<|z|<1\) and in \(|z|>1\). Asking "the" Laurent series of a function is ill-posed; you must say where.
- \(a_{-1}\) is the residue. Set \(n = -1\) in the formula above and it reads \(a_{-1} = \frac{1}{2\pi i}\oint f\,dz\) — which is the residue theorem for a single pole. The residue is not a separate concept; it is one coefficient of the Laurent series.
Classifying the singularity by the principal part
| Principal part | Name | Example | Residue |
|---|---|---|---|
| none | removable | \(\sin z/z\) at 0 | 0 |
| finitely many terms, lowest \((z-z_0)^{-m}\) | pole of order \(m\) | \(1/z^2\) | \(a_{-1}\) |
| infinitely many terms | essential | \(e^{1/z}\) at 0 | \(a_{-1}\), still |
Essential singularities are genuinely wild: Picard's theorem says that in any neighbourhood of one, the function takes every complex value (with at most one exception) infinitely often. Watch \(e^{1/z}\) as \(z\to0\): along the positive reals it blows up, along the negative reals it goes to zero, along the imaginary axis it oscillates forever on the unit circle. There is no limit in any sense.
Worked example
Expand \(f(z) = \dfrac{1}{z(z-1)}\) in (a) \(0<|z|<1\) and (b) \(|z|>1\).
(a) Write \(\frac{1}{z-1} = -\frac{1}{1-z} = -(1 + z + z^2 + \cdots)\) for \(|z|<1\). Then
A simple pole at \(z=0\) with residue \(a_{-1} = -1\).
(b) For \(|z|>1\) factor differently: \(\frac{1}{z-1} = \frac{1}{z}\frac{1}{1-1/z} = \frac1z\left(1 + \frac1z + \frac1{z^2}+\cdots\right)\), so
No \(1/z\) term at all — the "residue" read from this expansion is 0. Which is correct? Both: they are expansions in different regions, and only the one valid in an annulus touching \(z_0\) gives the residue at \(z_0\). Getting this wrong is the classic Laurent-series error.
Common mistakes
- Asking for "the" Laurent series without specifying the annulus.
- Reading a residue from the wrong region's expansion. See above.
- Assuming a divergent Taylor series means the function misbehaves on the real line. It means there is a complex singularity within that radius.
- Treating an essential singularity as a very high-order pole. It is qualitatively different; no finite order describes it.
Related concepts
- Analytic functions · Cauchy's integral formula
- Residue theorem — \(a_{-1}\), promoted
- Classifying singularities & branch points
- Contour integration
- Contour integrator
Knowledge graph position
Prerequisites: analytic functions, Cauchy's integral formula. Leads to: residue theorem, asymptotic analysis, dispersion relations.
Quiz
Q1 (conceptual). Why does the Taylor series of \(1/(1+x^2)\) diverge for \(|x|>1\) despite the function being smooth everywhere on \(\mathbb{R}\)?
Answer
The radius of convergence is the distance to the nearest singularity in the complex plane, and \(1/(1+z^2)\) has poles at \(z = \pm i\), distance 1 from the origin. The real function's behaviour is controlled by singularities invisible from the real axis.
Q2 (computational). Find the residue of \(e^{1/z}\) at \(z = 0\).
Answer
\(e^{1/z} = \sum_{n\ge0}\frac{1}{n!z^n} = 1 + \frac1z + \frac{1}{2z^2} + \cdots\), so \(a_{-1} = 1\). Note this is an essential singularity — infinitely many negative powers — yet the residue is perfectly well defined and easy to read off.
Q3 (MCQ). A function with infinitely many negative powers in its Laurent expansion has:
- (a) a pole of infinite order
- (b) an essential singularity, near which it takes almost every complex value infinitely often
- (c) a removable singularity
- (d) a branch point
Answer
(b). By Picard's great theorem the behaviour is far wilder than any pole: every value except at most one is attained infinitely often in any neighbourhood.