Laurent Series

In this lesson you’ll extend power series to allow negative exponents, learn that the region of validity is an annulus rather than a disc, see that one function can have several different Laurent series, and meet the coefficient $c_{-1}$ that the rest of the section revolves around.

Taylor series need the function to be analytic on a whole disc. That rules out any expansion around a singularity, which is exactly where the interesting behaviour is. Laurent’s fix is to allow negative powers and to shrink the region to an annulus that excludes the bad point.

Laurent’s theorem. If $f$ is analytic on the annulus $r < |z-z_0| < R$, then

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

converging on that annulus, with

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

for any circle $C$ in the annulus. The expansion is unique.

Notice the coefficient formula is the same one as in Cauchy’s integral formula for derivatives, now allowed to run over negative $n$ as well. For $n \ge 0$ and $f$ analytic on the full disc, $c_n = f^{(n)}(z_0)/n!$ and the negative coefficients all vanish, recovering the Taylor series. Laurent series generalize Taylor series rather than replacing them.

The two halves have names:

• $\sum_{n\ge0}c_n(z-z_0)^n$ - the analytic part, converging for $|z-z_0|<R$
• $\sum_{n<0}c_n(z-z_0)^n$ - the principal part, converging for $|z-z_0|>r$

Both conditions together give the annulus. The principal part is the new object and it is what encodes the singularity.

Because the region is an annulus and a function can have several, the expansion is not unique to the function - only to the function together with the annulus.

Take $f(z) = \frac{1}{(z-1)(z-2)}$ expanded about 0. The poles at 1 and 2 cut the plane into three regions, and each gets its own series:

• $|z|<1$: analytic here, so an ordinary Taylor series with only non-negative powers.
• $1<|z|<2$: negative powers from the pole at 1, positive powers from the pole at 2.
• $|z|>2$: both poles are “inside,” so only negative powers appear.

The figure computes the middle annulus’s coefficients directly from the integral formula on the circle $|z|=1.5$. Using partial fractions, that series should be

$$f(z) = -\sum_{k\ge1}z^{-k} - \sum_{k\ge0}\frac{z^k}{2^{k+1}}$$

so $c_{-1}=c_{-2}=\cdots=-1$ and $c_k = -1/2^{k+1}$ for $k \ge 0$, giving $c_0=-0.5$ and $c_1=-0.25$. Those are the numbers the figure measures.

Asking “what is the Laurent series of $f$?” is incomplete. You must say where.

Almost never from the integral. The three usable techniques:

Partial fractions plus geometric series. Split into terms $\frac{A}{z-a}$, then expand each in the form appropriate to the region. Inside, write $\frac{1}{z-a} = -\frac1a\cdot\frac{1}{1-z/a}$ and expand in $z/a$. Outside, write $\frac{1}{z-a} = \frac1z\cdot\frac{1}{1-a/z}$ and expand in $a/z$. Which factor you pull out is the whole decision, and it is decided by which quantity is less than 1 in your region.

Multiply a known series by a power. For $\frac{\sin z}{z^4}$, take the series for $\sin z$ and divide term by term.

Substitute into a known series. For $e^{1/z}$, put $1/z$ into the exponential series to get $\sum_{n\ge0}\frac{z^{-n}}{n!}$, an infinite principal part.

Integrate the series term by term around a circle in the annulus. From lesson 8, $\oint (z-z_0)^n dz = 0$ for every integer $n$ except $n=-1$, where it is $2\pi i$. So every term dies except one:

$$\oint_C f(z)\,dz = 2\pi i\,c_{-1}$$

That is the residue theorem in one line. The number $c_{-1}$ gets the name residue of $f$ at $z_0$, and computing contour integrals reduces entirely to finding it. Everything in the next three lessons is about extracting $c_{-1}$ efficiently.

The principal part also tells you what kind of singularity you have, which is the subject of the next lesson:

• No principal part - removable singularity.
• Finitely many terms, lowest power $(z-z_0)^{-m}$ - pole of order $m$.
• Infinitely many terms - essential singularity.

This is the cleanest of the several equivalent definitions, and it is why Laurent series come before the classification.

Example 1: Find the Laurent series of $\frac{e^z}{z^3}$ about 0.

Solution. Divide the exponential series by $z^3$:

$$\frac{1}{z^3}\left(1+z+\frac{z^2}{2}+\frac{z^3}{6}+\frac{z^4}{24}+\cdots\right)$$ $$= \frac{1}{z^3}+\frac{1}{z^2}+\frac{1}{2z}+\frac16+\frac{z}{24}+\cdots$$

Valid on $0<|z|<\infty$. The principal part has three terms, so this is a pole of order 3, and

$$\operatorname{Res} = c_{-1} = \frac12$$

∎

Example 2: Find the Laurent series of $e^{1/z}$ about 0.

Solution. Substitute $1/z$ into the exponential series:

$$e^{1/z} = \sum_{n=0}^{\infty}\frac{1}{n!z^n} = 1+\frac1z+\frac{1}{2z^2}+\frac{1}{6z^3}+\cdots$$

Infinitely many negative powers, so this is an essential singularity, and $c_{-1}=1$. ∎

Example 3: Expand $\frac{1}{(z-1)(z-2)}$ on $|z|<1$.

Solution. Partial fractions:

$$\frac{1}{(z-1)(z-2)} = \frac{-1}{z-1}+\frac{1}{z-2}$$

Here $|z|<1<2$, so expand both in powers of $z$:

$$\frac{-1}{z-1} = \frac{1}{1-z} = \sum_{n\ge0}z^n$$ $$\frac{1}{z-2} = \frac{-1}{2}\cdot\frac{1}{1-z/2} = -\sum_{n\ge0}\frac{z^n}{2^{n+1}}$$ $$f(z) = \sum_{n\ge0}\left(1-\frac{1}{2^{n+1}}\right)z^n = \frac12 + \frac34z+\frac78z^2+\cdots$$

No negative powers, as expected on a disc where $f$ is analytic. ∎

Example 4: Same function on $1<|z|<2$.

Solution. Now $|1/z|<1$ and $|z/2|<1$, so pull out different factors:

$$\frac{-1}{z-1} = \frac{-1}{z}\cdot\frac{1}{1-1/z} = -\sum_{n\ge0}\frac{1}{z^{n+1}}$$ $$\frac{1}{z-2} = -\sum_{n\ge0}\frac{z^n}{2^{n+1}} \quad\text{(unchanged)}$$ $$f(z) = -\sum_{n\ge1}z^{-n} - \sum_{n\ge0}\frac{z^n}{2^{n+1}}$$

So $c_{-1}=-1$, $c_0=-\frac12$, $c_1=-\frac14$, matching the figure’s measured values. ✓

Same function, same centre, completely different series.

Example 5: Same function on $|z|>2$.

Solution. Now both $|1/z|<1$ and $|2/z|<1$:

$$\frac{-1}{z-1} = -\sum_{n\ge0}\frac{1}{z^{n+1}}, \qquad \frac{1}{z-2} = \frac1z\cdot\frac{1}{1-2/z} = \sum_{n\ge0}\frac{2^n}{z^{n+1}}$$ $$f(z) = \sum_{n\ge0}\frac{2^n-1}{z^{n+1}} = \frac{0}{z}+\frac{1}{z^2}+\frac{3}{z^3}+\cdots$$

Note $c_{-1}=0$ here, which is consistent: a large circle encloses both poles, whose residues $-1$ and $+1$ cancel, exactly as computed in the Cauchy’s theorem lesson. ✓

Example 6: Find the residue of $\frac{\sin z}{z^4}$ at 0.

Solution.

$$\frac{1}{z^4}\left(z-\frac{z^3}{6}+\frac{z^5}{120}-\cdots\right) = \frac{1}{z^3}-\frac{1}{6z}+\frac{z}{120}-\cdots$$

The coefficient of $\frac1z$ is $-\frac16$, so

$$\operatorname{Res}_{z=0}\frac{\sin z}{z^4} = -\frac16$$

∎

Note the pole has order 3, not 4, because $\sin z$ contributes a zero of order 1 at the origin.

Example 7: Evaluate a contour integral from the series.

Compute $\displaystyle\oint_{|z|=1}e^{1/z}\,dz$.

Solution. From Example 2, $c_{-1}=1$. Integrating the series term by term, only that coefficient survives:

$$\oint_{|z|=1}e^{1/z}\,dz = 2\pi i\cdot1 = 2\pi i$$

∎

An essential singularity is no obstacle to integration. You do not need to understand the singularity, only to read off one coefficient - which is the practical power of Laurent series.

System response near a resonance. Expanding a transfer function in a Laurent series about a pole separates the resonant behaviour, which is the principal part, from the smooth background. Engineers use the leading coefficient to characterize a resonance’s strength.

Partial fraction expansion in circuit analysis. Decomposing a transfer function into pole terms is the first step in inverting a Laplace transform, and each term is a one-term Laurent series about its pole.

Asymptotic expansions. Behaviour of a function for large argument is a Laurent series in $1/z$ about infinity, which is how the growth of special functions and the tails of distributions are described.

Multipole expansions. In electrostatics and gravitation, the field far from a charge distribution is expanded in inverse powers of distance, with monopole, dipole and quadrupole terms. It is a Laurent series with physical names for the coefficients.

Digital filter design. The $z$-transform of a causal filter is a Laurent series, and the region of convergence - an annulus - determines whether the filter is stable and causal. Signal processing texts state stability directly in terms of that annulus.

Perturbation near a singular limit. Problems with a small parameter appearing in a denominator are handled by Laurent expansion in that parameter, standard in fluid dynamics and in quantum field theory regularization.