Chapter 1: Laurent Expansion
Expansion on an annulus, the principal and regular parts
Motivation
A Taylor expansion is valid around a point where a function is holomorphic, but it cannot be used around a singularity. The Laurent expansion extends the Taylor expansion by allowing negative-power terms, and so describes the behavior of a function on a region that contains a singularity.
A motivating example
$f(z) = \dfrac{e^z}{z}$ has a singularity at $z = 0$. We cannot Taylor-expand it around $z = 0$, but if we Taylor-expand the numerator $e^z$ and divide by $z$,
$$f(z) = \dfrac{e^z}{z} = \dfrac{1}{z}\left(1 + z + \dfrac{z^2}{2!} + \dfrac{z^3}{3!} + \cdots\right) = \dfrac{1}{z} + 1 + \dfrac{z}{2!} + \dfrac{z^2}{3!} + \cdots$$we obtain a series made up of several terms. Unlike a Taylor expansion, a negative-power term $\dfrac{1}{z}$ (the principal part) appears — this is the Laurent expansion. The remaining $1 + \dfrac{z}{2!} + \cdots$ is an ordinary power series (the regular part).
The Annulus
Definition: Annulus
For $0 \leq r_1 < r_2 \leq \infty$, the set
$$A(a; r_1, r_2) = \{z \in \mathbb{C} : r_1 < |z - a| < r_2\}$$is called the annulus centered at the point $a$.
A Taylor expansion requires the function to be holomorphic on a disk containing the center $a$, but an annulus excludes that inner disk. Hence a function holomorphic only on the annulus need not admit a Taylor expansion about $a$ (the center may even be a singularity). It can, however, always be expanded into a Laurent series.
Laurent Series
Definition: Laurent series
A Laurent series is a series of the following form:
$$\displaystyle\sum_{n=-\infty}^{\infty} a_n (z - a)^n = \cdots + \dfrac{a_{-2}}{(z-a)^2} + \dfrac{a_{-1}}{z-a} + a_0 + a_1(z-a) + a_2(z-a)^2 + \cdots$$Definition: Principal part and regular part
A Laurent series is split into two parts:
- Principal part: $\displaystyle\sum_{n=-\infty}^{-1} a_n (z-a)^n = \displaystyle\sum_{n=1}^{\infty} \dfrac{a_{-n}}{(z-a)^n}$
- Regular part: $\displaystyle\sum_{n=0}^{\infty} a_n (z-a)^n$
On convergence
The convergence of a Laurent series is understood as the convergence of its principal and regular parts separately:
- the regular part $\displaystyle\sum_{n=0}^{\infty} a_n (z-a)^n$ converges for $|z-a| < R_2$;
- the principal part $\displaystyle\sum_{n=1}^{\infty} \dfrac{a_{-n}}{(z-a)^n}$ converges for $|z-a| > R_1$.
Hence the whole series converges on the annulus $R_1 < |z-a| < R_2$.
The Laurent Theorem
Theorem: Laurent expansion
If $f(z)$ is holomorphic on the annulus $A(a; r_1, r_2) = \{z : r_1 < |z-a| < r_2\}$, then on this annulus $f(z)$ has a unique expansion
$$f(z) = \displaystyle\sum_{n=-\infty}^{\infty} a_n (z-a)^n$$where the coefficients are given by the integral formula
$$a_n = \dfrac{1}{2\pi i} \oint_C \dfrac{f(\zeta)}{(\zeta - a)^{n+1}} d\zeta$$and $C$ is any circle centered at $a$ lying in the annulus (positively oriented).
Proof
Fix a point $z$ in the annulus, and choose radii $\rho_1,\ \rho_2$ with $r_1 < \rho_1 < |z-a| < \rho_2 < r_2$. Let $C_+$ be the outer circle (radius $\rho_2$) and $C_-$ the inner circle (radius $\rho_1$), both positively oriented.
Step 1: Cauchy's integral formula on the annulus. Introduce a single crosscut joining $C_-$ and $C_+$, turning the closed annulus $\rho_1 \le |\zeta-a| \le \rho_2$ into a simply connected region; its boundary then consists of $C_+$ (positive orientation), $C_-$ (negative orientation), and the crosscut traversed in both directions. Since the integrand $\dfrac{f(\zeta)}{\zeta-z}$ is holomorphic there except at $\zeta=z$, the crosscut contributions cancel, and the ordinary Cauchy integral formula gives
$$f(z) = \frac{1}{2\pi i} \oint_{C_+} \frac{f(\zeta)}{\zeta - z}\, d\zeta \;-\; \frac{1}{2\pi i} \oint_{C_-} \frac{f(\zeta)}{\zeta - z}\, d\zeta \tag{1}$$Write the first term on the right as $I_1$ and the second (including its minus sign) as $I_2$.
Step 2: The outer integral $I_1$ (regular part). On $C_+$ we have $|\zeta-a| = \rho_2 > |z-a|$, so the ratio $\dfrac{z-a}{\zeta-a}$ has absolute value bounded by the constant $\theta_2 := \dfrac{|z-a|}{\rho_2} < 1$ (independent of $\zeta$), and
$$\frac{1}{\zeta-z} = \frac{1}{(\zeta-a)\left(1 - \frac{z-a}{\zeta-a}\right)} = \sum_{n=0}^{\infty} \frac{(z-a)^n}{(\zeta-a)^{n+1}}$$On $C_+$ each term has absolute value $\dfrac{\theta_2^{\,n}}{\rho_2}$, and $\displaystyle\sum_{n} \theta_2^{\,n} < \infty$, so by the Weierstrass $M$-test the series converges uniformly on $C_+$. Since $f$ is bounded and continuous on $C_+$, term-by-term integration is justified, giving
$$I_1 = \sum_{n=0}^{\infty} a_n (z-a)^n, \qquad a_n = \frac{1}{2\pi i} \oint_{C_+} \frac{f(\zeta)}{(\zeta-a)^{n+1}}\, d\zeta \quad (n \ge 0)$$Step 3: The inner integral $I_2$ (principal part). On $C_-$ we have $|\zeta-a| = \rho_1 < |z-a|$, so now $\left|\dfrac{\zeta-a}{z-a}\right| = \dfrac{\rho_1}{|z-a|} < 1$ is bounded by a constant, and
$$-\frac{1}{\zeta-z} = \frac{1}{(z-a)\left(1 - \frac{\zeta-a}{z-a}\right)} = \sum_{n=0}^{\infty} \frac{(\zeta-a)^n}{(z-a)^{n+1}}$$This too converges uniformly on $C_-$, so term-by-term integration gives
$$I_2 = \sum_{n=0}^{\infty} \left[\frac{1}{2\pi i} \oint_{C_-} f(\zeta)\,(\zeta-a)^n\, d\zeta\right] \frac{1}{(z-a)^{n+1}}$$Substituting $m = -(n+1)$ (as $n: 0 \to \infty$, $m: -1 \to -\infty$), with $\dfrac{1}{(z-a)^{n+1}} = (z-a)^m$ and $(\zeta-a)^n = \dfrac{1}{(\zeta-a)^{m+1}}$, we get
$$I_2 = \sum_{m=-\infty}^{-1} a_m (z-a)^m, \qquad a_m = \frac{1}{2\pi i} \oint_{C_-} \frac{f(\zeta)}{(\zeta-a)^{m+1}}\, d\zeta \quad (m \le -1)$$Step 4: Combining the coefficients into a single contour integral. For every integer $n$, the integrand $\dfrac{f(\zeta)}{(\zeta-a)^{n+1}}$ is holomorphic throughout the annulus $r_1 < |\zeta-a| < r_2$ (since $f$ is holomorphic there, and the singularity $\zeta=a$ lies inside the hole, outside the region). Hence, since the integrand is also holomorphic on the region between any two concentric circles, the difference of the integrals along them equals the integral along the boundary of that (smaller) annulus, which is $0$ by Cauchy's integral theorem; therefore the integral along any circle centered at $a$ lying in the annulus is the same. In particular the integrals over the outer $C_+$ and the inner $C_-$ agree, so the coefficient is given by this common value (the contour may be any concentric circle in the annulus):
$$a_n = \frac{1}{2\pi i} \oint_{C} \frac{f(\zeta)}{(\zeta-a)^{n+1}}\, d\zeta \qquad (n \in \mathbb{Z})$$Substituting these into $(1)$ gives
$$f(z) = I_1 + I_2 = \sum_{n=0}^{\infty} a_n (z-a)^n + \sum_{n=-\infty}^{-1} a_n (z-a)^n = \sum_{n=-\infty}^{\infty} a_n (z-a)^n$$Since $z$ was an arbitrary point of the annulus, this proves the existence of the expansion and the coefficient formula. Uniqueness is treated in §Uniqueness below. $\blacksquare$
Worked Examples
Example 1: Laurent expansion of $f(z) = \dfrac{1}{z(z-1)}$ around $z = 0$
By partial fractions:
$$f(z) = \dfrac{1}{z(z-1)} = -\dfrac{1}{z} + \dfrac{1}{z-1} = -\dfrac{1}{z} - \dfrac{1}{1-z}$$Region $0 < |z| < 1$:
Since $|z| < 1$, we may use $\dfrac{1}{1-z} = \displaystyle\sum_{n=0}^{\infty} z^n$.
$$f(z) = -\dfrac{1}{z} - \displaystyle\sum_{n=0}^{\infty} z^n = -\dfrac{1}{z} - 1 - z - z^2 - \cdots$$Region $|z| > 1$:
Since $|z| > 1$, use $\dfrac{1}{z-1} = \dfrac{1}{z} \cdot \dfrac{1}{1-\dfrac{1}{z}} = \dfrac{1}{z}\displaystyle\sum_{n=0}^{\infty} \dfrac{1}{z^n}$.
$$f(z) = -\dfrac{1}{z} + \displaystyle\sum_{n=0}^{\infty} \dfrac{1}{z^{n+1}} = -\dfrac{1}{z} + \dfrac{1}{z} + \dfrac{1}{z^2} + \dfrac{1}{z^3} + \cdots = \displaystyle\sum_{n=2}^{\infty} \dfrac{1}{z^n}$$Example 2: Laurent expansion of $f(z) = e^{1/z}$ around $z = 0$
Setting $w = \dfrac{1}{z}$ in $e^w = \displaystyle\sum_{n=0}^{\infty} \dfrac{w^n}{n!}$:
$$e^{1/z} = \displaystyle\sum_{n=0}^{\infty} \dfrac{1}{n! z^n} = 1 + \dfrac{1}{z} + \dfrac{1}{2! z^2} + \dfrac{1}{3! z^3} + \cdots$$This converges for $z \neq 0$. Because the principal part continues indefinitely, $z = 0$ is an essential singularity.
Example 3: Laurent expansion of $f(z) = \dfrac{\sin z}{z^3}$ around $z = 0$
From $\sin z = z - \dfrac{z^3}{3!} + \dfrac{z^5}{5!} - \cdots$:
$$\dfrac{\sin z}{z^3} = \dfrac{1}{z^2} - \dfrac{1}{3!} + \dfrac{z^2}{5!} - \dfrac{z^4}{7!} + \cdots$$Here $a_{-2} = 1$, $a_{-1} = 0$, $a_0 = -\dfrac{1}{6}$.
Caution: some singularities have no Laurent expansion (branch points)
Each example above could be expanded in integer powers $\displaystyle\sum_{n} a_n z^n$. However, $f(z) = \sqrt{z}$ at $z = 0$ admits neither a Taylor expansion nor a Laurent expansion.
The Laurent theorem requires $f$ to be single-valued and holomorphic on an annulus $0 < |z| < R$. But writing $z = \rho e^{i\theta}$ and going once around the origin ($\theta \to \theta + 2\pi$),
$$\sqrt{z} = \rho^{1/2} e^{i\theta/2} \;\longrightarrow\; \rho^{1/2} e^{i(\theta + 2\pi)/2} = -\sqrt{z}$$the sign flips, so the function is not single-valued near the origin (it is multivalued). Such a point $z = 0$ is neither a pole nor an essential singularity; it is called a branch point, and it falls outside the very framework of the Laurent expansion (a series in integer powers). Handling half-integer powers such as $\sqrt{z} = z^{1/2}$ requires a generalization that allows fractional powers, such as a Puiseux series.
Note that if we avoid the branch point and expand around a point $z \neq 0$ (e.g., $z = 1$) with a fixed branch, $\sqrt{z}$ does have an ordinary Taylor expansion: $\sqrt{z} = 1 + \dfrac{1}{2}(z-1) - \dfrac{1}{8}(z-1)^2 + \cdots$ ($|z-1| < 1$).
Uniqueness
Theorem: Uniqueness of the Laurent expansion
The Laurent expansion of a function $f(z)$ holomorphic on an annulus $A(a; r_1, r_2)$ is unique. That is, if
$$f(z) = \displaystyle\sum_{n=-\infty}^{\infty} a_n (z-a)^n = \displaystyle\sum_{n=-\infty}^{\infty} b_n (z-a)^n$$holds, then $a_n = b_n$ for every $n$.
Proof
Suppose $f(z) = \displaystyle\sum_{n=-\infty}^{\infty} a_n (z-a)^n$ converges on the annulus. Take a circle $C : |z-a| = \rho$ with $r_1 < \rho < r_2$ (positively oriented). A convergent Laurent series converges uniformly on every closed sub-annulus $\rho_1 \le |z-a| \le \rho_2$ (because the regular and principal parts each converge uniformly on compact subsets of their regions of convergence); in particular it converges uniformly on $C$.
Fix an arbitrary integer $k$, multiply both sides by $(z-a)^{-k-1}$, and integrate over $C$. Uniform convergence justifies term-by-term integration:
$$\oint_C \frac{f(z)}{(z-a)^{k+1}}\, dz = \sum_{n=-\infty}^{\infty} a_n \oint_C (z-a)^{\,n-k-1}\, dz$$Here we use the following basic integral. With $z = a + \rho e^{i\theta}$ ($0 \le \theta \le 2\pi$), we have $dz = i\rho\, e^{i\theta}\, d\theta$, so for any integer $m$,
$$\oint_C (z-a)^{m}\, dz = i\rho^{\,m+1} \int_0^{2\pi} e^{i(m+1)\theta}\, d\theta = \begin{cases} 2\pi i & (m = -1) \\ 0 & (m \ne -1) \end{cases}$$(for $m \ne -1$, the integral of $e^{i(m+1)\theta}$ over one period is $0$). Among the terms of the sum, $m = n-k-1$ satisfies $m = -1$ only when $n = k$, so only the $n = k$ term survives, and
$$\oint_C \frac{f(z)}{(z-a)^{k+1}}\, dz = 2\pi i\, a_k \qquad\Longrightarrow\qquad a_k = \frac{1}{2\pi i} \oint_C \frac{f(z)}{(z-a)^{k+1}}\, dz$$Thus each coefficient $a_k$ is determined uniquely by $f$ and $C$ and does not depend on how the series is written (this is exactly the coefficient formula obtained in the existence theorem). Therefore, even if $f(z) = \displaystyle\sum_n a_n (z-a)^n = \displaystyle\sum_n b_n (z-a)^n$ in two ways, both $a_k$ and $b_k$ equal this same integral, so $a_k = b_k$ for every $k$. $\blacksquare$
A practical consequence
By this uniqueness, one need not use the integral formula to find a Laurent expansion: any series expansion obtained by whatever method is the correct Laurent expansion. It is efficient to combine known Taylor expansions and algebraic manipulations to read off the coefficients.
Example: for $f(z) = \dfrac{e^z}{z^2}$, simply dividing the known series $e^z = \displaystyle\sum_{n=0}^{\infty} \dfrac{z^n}{n!}$ by $z^2$ gives
$$\dfrac{e^z}{z^2} = \dfrac{1}{z^2} + \dfrac{1}{z} + \dfrac{1}{2} + \dfrac{z}{6} + \dfrac{z^2}{24} + \cdots \qquad (0 < |z|)$$and by uniqueness this is already the Laurent expansion. There is no need to evaluate any coefficient via $\dfrac{1}{2\pi i}\displaystyle\oint_C \dfrac{f(\zeta)}{(\zeta-a)^{n+1}} d\zeta$, and the residue $a_{-1} = 1$ can be read off directly.
Summary
- Laurent series: $\displaystyle\sum_{n=-\infty}^{\infty} a_n (z-a)^n$ (includes negative powers)
- Principal part: the negative-power terms; regular part: the non-negative-power terms
- A function holomorphic on an annulus has a unique Laurent expansion
- Coefficients: $a_n = \dfrac{1}{2\pi i} \displaystyle\oint_C \dfrac{f(\zeta)}{(\zeta - a)^{n+1}} d\zeta$
- In particular, $a_{-1}$ is called the residue and plays a key role in contour integration
Frequently Asked Questions
Q1: What is a Laurent expansion?
A: A function $f(z)$ holomorphic on a punctured neighborhood of an isolated singularity $z_0$ can be expanded around $z_0$ as $f(z) = \displaystyle\sum_{n=-\infty}^{\infty} c_n (z - z_0)^n$ (the Laurent expansion). The presence of negative powers (the principal part) is what distinguishes it from a Taylor expansion.
Q2: How are the coefficients of a Laurent expansion found?
A: The coefficients are given by $c_n = \dfrac{1}{2\pi i} \displaystyle\oint_C \dfrac{f(z)}{(z-z_0)^{n+1}} dz$. In particular, $c_{-1}$ is the residue. In practice the coefficients are usually read off from known Taylor expansions or partial-fraction decompositions.
Q3: How is the Laurent expansion related to the classification of singularities?
A: If the principal part (the negative-power part) has finitely many terms, the point is a removable singularity (removable when all $c_{-n} = 0$) or a pole of order $n$; if the principal part has infinitely many terms, it is an essential singularity.
Q4: Why expand on an annulus?
A: If the center $a$ is a singularity, the function is not holomorphic on a disk containing $a$, so a Taylor expansion in non-negative powers alone cannot be used. The negative-power terms (the principal part) get smaller away from the center and converge outside, while the non-negative-power terms (the regular part) converge inside. The two converge simultaneously only on an annulus with the inner hole removed, so the Laurent expansion is considered on an annulus.