Definition 8.3.1.
A Laurent series centered at \(z_0\) is a double series of the form
1
After Pierre Alphonse Laurent (1813–1854).
\begin{equation*}
\ds \sz c_k ( z - z_0 )^k\text{.}
\end{equation*}
Section 8.3 Laurent Series
Theorem 8.1.8 gives a powerful way of describing holomorphic functions. It is, however, not as general as it could be. It is natural, for example, to think about representing \(\exp ( \frac 1 z )\) as
\begin{equation*}
\exp \left( \frac 1 z \right) \ = \ \s \frac 1 {k!} \left( \frac 1
z \right)^k \ = \ \s \frac 1 {k!} \, z^{-k} \text{,}
\end{equation*}
a “power series” with negative exponents. To make sense of expressions like the above, we introduce the concept of a double series
\begin{equation*}
\sz a_k \ := \ \s a_k + \sum_{k \geq 1} a_{-k} \,\text{.}
\end{equation*}
Here \(a_k \in \C\) are terms indexed by the integers. The double series above converges if and only if the two series on the right-hand side do. Absolute and uniform convergence are defined analogously. Equipped with this, we can now introduce the following central concept.
Example 8.3.2.
The series that started this section is the Laurent series of \(\exp ( \frac 1 z )\) centered at \(0\text{.}\)
Example 8.3.3.
We should pause for a minute and ask for which \(z\) a general Laurent series can possibly converge. By definition
\begin{equation*}
\sz c_k \left( z - z_0 \right)^k \ = \ \s c_k \left( z - z_0
\right)^k + \sum_{k \geq 1} c_{-k} \left( z - z_0 \right)^{-k} \text{.}
\end{equation*}
The first series on the right-hand side is a power series with some radius of convergence \(R_2\text{,}\) that is, with Theorem 7.4.3, it converges in \(\left\{ z \in \C : \, | z-z_0 | \lt R_2
\right\}\text{,}\) and the convergence is uniform in \(\left\{ z \in \C : \, | z-z_0
| \le r_2 \right\}\text{,}\) for any fixed \(r_2 \lt R_2\text{.}\) For the second series, we invite you (in Exercise 8.4.30) to revise our proof of Theorem 7.4.3 to show that this series converges for
\begin{equation*}
\frac{ 1 }{ |z-z_0|} \ \lt \ \frac 1 {R_1}
\end{equation*}
for some \(R_1\text{,}\) and that the convergence is uniform in \(\left\{ z \in \C : \, |
z-z_0 | \ge r_1 \right\}\text{,}\) for any fixed \(r_1 > R_1\text{.}\) Thus the Laurent series converges in the annulus
\begin{equation*}
A \ := \ \left\{ z \in \C : \, R_1 \lt | z-z_0 | \lt R_2
\right\}
\end{equation*}
(assuming this set is not empty, i.e., \(R_1 \lt R_2\)), and the convergence is uniform on any set of the form
\begin{equation*}
\left\{ z \in \C : \, r_1 \leq | z-z_0 | \leq r_2 \right\}
\qquad \text{ for } \ R_1 \lt r_1 \lt r_2 \lt R_2 \, \text{.}
\end{equation*}
Example 8.3.4.
We’d like to compute the start of a Laurent series for \(\frac 1 {\sin(z)}\) centered at \(z_0 = 0\text{.}\) We start by considering the function \(g: D[0,\pi] \to \C\) defined by
\begin{equation*}
g(z) := \begin{cases}\frac 1 { \sin(z) } - \frac 1 z \amp
\text{ if } z \ne 0 \, , \\ 0
\amp \text{ if } z=0 \, . \end{cases}
\end{equation*}
A quick application of L’Hôpital’s Rule (lhospital) shows that \(g\) is continuous (see Exercise 8.4.31). Even better, another round of L’Hôpital’s Rule proves that
\begin{equation*}
\lim_{ z \to 0 } \frac{ \frac 1 { \sin(z) } - \frac 1 z }{ z
} \ = \ \frac 1 6 \, \text{.}
\end{equation*}
But this means that
\begin{equation*}
g'(z) = \begin{cases}- \frac{ \cos(z) }{ \sin^2(z) } + \frac
1 {z^2} \amp \text{ if } z \ne 0 \, , \\ \frac 1 6
\amp \text{ if } z=0 \, , \end{cases}
\end{equation*}
in particular, \(g\) is holomorphic in \(D[0,\pi]\text{.}\) By Theorem 8.1.8, \(g\) has a power series expansion at 0, which we may compute using Corollary 8.1.5. It starts with
2
This is a (simple) example of analytic continuation: the function \(g\) is holomorphic in \(D[0,\pi]\) and agrees with \(\frac 1 { \sin(z) } - \frac 1 z\) in \(D[0,\pi]
\setminus \{ 0 \}\text{,}\) the domain in which the latter function is holomorphic. When we said, in Footnote 7.2.1, that the Riemann zeta function \(\zeta(z) = \sum_{ k \ge 1 } \frac 1 {k^z}\) can be extended to a function that is holomorphic on \(\C
\setminus \{ 1 \}\text{,}\) we were also talking about analytic continuation.
\begin{equation*}
g(z) \ = \ \frac 1 6 \, z + \frac{ 7 }{ 360 } \, z^3 + \frac{
31 }{ 15120 } \, z^5 + \cdots
\end{equation*}
and it converges, by Corollary 8.1.10, for \(|z| \lt \pi\text{.}\) But this gives our sought-after Laurent series
\begin{equation*}
\frac 1 { \sin(z) } \ = \ z^{ -1 } + \frac 1 6 \, z + \frac{ 7
}{ 360 } \, z^3 + \frac{ 31 }{ 15120 } \, z^5 + \cdots
\end{equation*}
which converges for \(0 \lt |z| \lt \pi\text{.}\)
SageMath has a hard time with Laurent series with infinitely many negative exponents:
But it can handle finitely many negative exponents:
Theorem 8.1.1 implies that a Laurent series represents a function that is holomorphic in its annulus of convergence. The fact that we can conversely represent any function holomorphic in such an annulus by a Laurent series is the substance of the next result.
Theorem 8.3.5.
Suppose \(f\) is a function that is holomorphic in \(A :=
\left\{ z \in \C : \, R_1 \lt |z-z_0| \lt R_2 \right\}\text{.}\) Then \(f\) can be represented in \(A\) as a Laurent series centered at \(z_0\text{,}\)
\begin{equation*}
f(z) \ = \ \sz c_k \left( z - z_0 \right)^k
\end{equation*}
with
\begin{equation*}
c_k \ = \ \frac 1 {2 \pi i} \int_{ C[z_0,r] } \frac{
f(w) }{ (w-z_0)^{k+1} } \, \diff{w} \, \text{,}
\end{equation*}
where \(R_1 \lt r \lt R_2\text{.}\)
By Cauchy’s Theorem 4.3.4 we can replace the circle \(C[z_0,r]\) in the formula for the Laurent coefficients by any path \(\gg \sim_A C[z_0,r]\text{.}\)
Proof.
Let \(g(z) = f(z+z_0)\text{;}\) so \(g\) is a function holomorphic in \(\left\{ z \in \C :
\, R_1 \lt |z| \lt R_2 \right\}\text{.}\) Fix \(R_1 \lt r_1 \lt |z| \lt r_2 \lt R_2\text{,}\) and let \(\gg\) be the path in Figure 8.3.6, where \(\gg_1 := C[0,r_1]\) and \(\gg_2 := C[0,r_2]\text{.}\) By Cauchy’s Integral Formula (Theorem 4.4.5),
\begin{align}
g(z) \amp \ = \ \frac 1 {2 \pi i} \int_\gg \frac{ g(w) }{ w-z } \,
\diff{w}\notag\\
\amp \ = \ \frac 1 {2 \pi i} \int_{ \gg_2 } \frac{ g(w) }{
w-z } \, \diff{w} - \frac 1 {2 \pi i} \int_{ \gg_1 } \frac{
g(w) }{ w-z } \, \diff{w} \,\text{.}\tag{8.1}
\end{align}
For the integral over \(\gg_2\) we play exactly the same game as in our proof of Theorem 8.1.8. The factor \(\frac 1 {w-z}\) in this integral can be expanded into a geometric series (note that \(w \in \gg_2\) and so \(| \frac z w | \lt 1\))
\begin{equation*}
\frac 1 {w-z} \ = \ \frac 1 w \, \frac 1 { 1 - \frac z w } \ = \
\frac 1 w \s \left( \frac z w \right)^k \text{,}
\end{equation*}
which converges uniformly in the variable \(w \in \gg_2\) by Exercise 7.5.30. Hence Proposition 7.3.6 applies:
\begin{equation*}
\int_{\gg_2} \frac{ g(w) }{ w-z } \, \diff{w} \ = \ \int_{\gg_2}
g(w) \, \frac 1 w \s \left( \frac z w \right)^k \diff{w} \ = \ \s
\left( \int_{\gg_2} \frac{ g(w) }{ w^{k+1} } \, \diff{w}
\right) z^k \text{.}
\end{equation*}
The integral over \(\gg_1\) is computed in a similar fashion; now we expand the factor \(\frac 1 {w-z}\) into the following geometric series (note that \(w \in \gg_1\) and so \(| \frac w z | \lt 1\))
\begin{equation*}
\frac 1 {w-z} \ = \ - \frac 1 z \, \frac 1 { 1 - \frac w z } \ = \ -
\frac 1 z \s \left( \frac w z \right)^k \text{,}
\end{equation*}
which converges uniformly in the variable \(w \in \gg_1\text{.}\) Again Proposition 7.3.6 applies:
\begin{align*}
\int_{\gg_1} \frac{ g(w) }{ w-z } \, \diff{w} \amp \ = \ -
\int_{\gg_1} g(w) \, \frac 1 z \s \left( \frac w z \right)^k
\diff{w}\\
\amp \ = \ - \s \left( \int_{\gg_1} g(w) w^k \, \diff{w}
\right) z^{-k-1}\\
\amp \ = \ - \sum_{k \leq -1} \left( \int_{\gg_1} \frac{
g(w) }{ w^{k+1} } \, \diff{w} \right) z^k \text{.}
\end{align*}
Putting everything back into (8.1) gives
\begin{equation*}
g(z) \ = \ \frac 1 {2 \pi i} \left( \s \left( \int_{\gg_2} \frac{
g(w) }{ w^{k+1} } \, \diff{w} \right) z^k + \sum_{k \leq -1}
\left( \int_{\gg_1} \frac{ g(w) }{ w^{k+1} } \, \diff{w}
\right) z^k \right) \text{.}
\end{equation*}
By Cauchy’s Theorem 4.3.4, we can now change both \(\gg_1\) and \(\gg_2\) to \(C[0,r]\text{,}\) as long as \(R_1 \lt r \lt R_2\text{,}\) which finally gives
\begin{equation*}
g(z) \ = \ \frac 1 {2 \pi i} \sz \left( \int_{C[0,r]} \frac{ g(w)
}{ w^{k+1} } \, \diff{w} \right) z^k \text{.}
\end{equation*}
The statement follows now with \(f(z) = g(z-z_0)\) and a change of variables in the integral.
This theorem, naturally, has several corollaries that have analogues in the world of Taylor series. Here are two samples:
Corollary 8.3.7.
If \(\sz c_k (z-z_0)^k\) and \(\sz d_k (z-z_0)^k\) are two Laurent series that both converge, for \(R_1 \lt |z-z_0| \lt R_2\text{,}\) to the same function, then \(c_k=d_k\) for all \(k
\in \Z\text{.}\)
Corollary 8.3.8.
If \(G\) is a region, \(z_0 \in G\text{,}\) and \(f\) is holomorphic in \(G \setminus \{ z_0 \}\text{,}\) then \(f\) can be expanded into a Laurent series centered at \(z_0\) that converges for \(0 \lt |z - z_0 | \lt R\) where \(R\) is at least the distance of \(z_0\) to \(\partial G\text{.}\)
Finally, we come to the analogue of Corollary 7.4.10 for Laurent series. We could revisit its proof, but the statement that would follow is actually the special case \(k = -1\) of Theorem 8.3.5, read from right to left:
Corollary 8.3.9.
Suppose \(f\) is a function that is holomorphic in \(A :=
\left\{ z \in \C : \, R_1 \lt |z-z_0| \lt R_2 \right\}\text{,}\) with Laurent series
\begin{equation*}
f(z) \ = \ \sz c_k \left( z - z_0 \right)^k\text{.}
\end{equation*}
If \(\gg\) is any simple, closed, piecewise smooth path in \(A\text{,}\) such that \(z_0\) is inside \(\gg\text{,}\) then
\begin{equation*}
\int_\gg f(z) \, \diff{z} \ = \ 2 \pi i \, c_{ -1 } \,\text{.}
\end{equation*}
This result is profound: it says that we can integrate (at least over closed curves) by computing Laurent series—in fact, we “only” need to compute one coefficient of a Laurent series. We will have more to say about this in the next chapter; for now, we give just one application, which might have been bugging you since the beginning of Chapter 7.
Example 8.3.10.
We will (finally!) compute (7.1), the integral \(\int_{ C[2,3] } \frac{ \exp(z) }{ \sin(z) } \,
\diff{z}\text{.}\) Our plan is to split up the integration path \(C[2,3]\) as in Figure 7.0.1, which gives, say,
\begin{equation*}
\int_{ C[2,3] } \frac{ \exp(z) }{ \sin(z) } \, \diff{z} \ = \
\int_{ C[0,1] } \frac{ \exp(z) }{ \sin(z) } \, \diff{z} +
\int_{ C[\pi,1] } \frac{ \exp(z) }{ \sin(z) } \, \diff{z} \, \text{.}
\end{equation*}
To compute the two integrals on the right-hand side, we can use Corollary 8.3.9, for which we need the Laurent expansions of \(\frac{ \exp(z) }{ \sin(z) }\) centered at 0 and \(\pi\text{.}\)
By Example 8.1.3 and Example 8.3.4,
\begin{align*}
\frac{ \exp(z) }{ \sin(z) } \amp \ = \ \left( 1 + z +
\frac 1 2 \, z^2 + \cdots \right) \left(
z^{ -1 } + \frac 1 6 \, z + \frac{ 7 }{ 360 } \, z^3 +
\cdots \right)\\
\amp \ = \ z^{ -1 } + 1 + \frac 2 3 \, z + \cdots
\end{align*}
and Corollary 8.3.9 gives \(\int_{ C[0,1] } \frac{ \exp(z) }{ \sin(z) } \,
\diff{z} = 2 \pi i\text{.}\)
For the integral around \(\pi\text{,}\) we use the fact that \(\sin(z) = \sin(\pi - z)\text{,}\) and so we can compute the Laurent expansion of \(\frac 1 { \sin(z) }\) at \(\pi\) also via Example 8.3.4:
\begin{align*}
\frac 1 { \sin(z) } \ \amp= \ - \frac 1 { \sin(z-\pi) } \\
\amp = \ -(z-\pi)^{ -1 } - \frac 1 6 \, (z-\pi) - \frac{ 7 }{ 360 }
\, (z-\pi)^3 - \cdots
\end{align*}
Adding Example 8.1.7 to the mix yields
\begin{align*}
\frac{ \exp(z) }{ \sin(z) } \amp \ = \ \left( e^\pi +
e^\pi (z-\pi) + \cdots
\right) \left( -(z-\pi)^{ -1 } - \frac 1 6 \, (z-\pi) -
\cdots \right)\\
\amp \ = \ {} -e^\pi (z-\pi)^{ -1 } - e^\pi - \frac 2 3
\, e^\pi (z-\pi) + \cdots
\end{align*}
and now Corollary 8.3.9 gives \(\int_{ C[\pi,1] } \frac{ \exp(z) }{ \sin(z) } \,
\diff{z} = -2 \pi i \, e^\pi\text{.}\) Putting it all together yields the integral we’ve been after for two chapters:
\begin{equation*}
\int_{ C[2,3] } \frac{ \exp(z) }{ \sin(z) } \, \diff{z} \ = \ 2
\pi i \left( 1 - e^\pi \right) \,\text{.}
\end{equation*}
