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6. Laurent series and singularities
We have already seen that a holomorphic function f can be expressed as a Taylor series: i.e. if f is differentiable on a domain D and z 0 ∈ D then we can write
f (z) =
∑^ ∞
n=
an(z − z 0 )n^ (6.0.1)
for suitable constants an, and this expression is valid for |z − z 0 | < R, for some R > 0. The idea of Laurent series is to generalise (6.0.1) to allow negative powers of (z − z 0 ). This turns out to be a remarkably useful tool.
§6.1 Laurent series
Definition. A Laurent series is a series of the form
∑^ ∞
n=−∞
an(z − z 0 )n. (6.1.1)
As (6.1.1) is a doubly infinite sum, we need to take care as to what it means. We define (6.1.1) to mean ∑∞
n=
a−n(z − z 0 )−n^ +
∑^ ∞
n=
an(z − z 0 )n^ = Σ−^ + Σ+.
The first question to address is when does (6.1.1) converge? For this, we need both Σ− and Σ+^ to converge. Now Σ+^ converges for |z − z 0 | < R 2 for some R 2 ≥ 0, where R 2 is the radius of convergence of Σ+. We can recognise Σ−^ as a power series in (z − z 0 )−^1. This has a radius of convergence equal to, say, R− 1 1 ≥ 0. That is, Σ−^ converges when |(z − z 0 )−^1 | < R− 1 1. In other words, Σ−^ converges when |z − z 0 | > R 1. Combining these, we see that if 0 ≤ R 1 < R 2 ≤ ∞ then (6.1.1) converges in the annulus
{z ∈ C | R 1 < |z − z 0 | < R 2 }.
See Figure 6.1.1. The following theorem says that if we have a function f that is holomorphic on an annulus then it can be expressed as a Laurent series. (Compare this with Taylor’s theorem: if f is holomorphic on a disc then it can be expressed as a Taylor series.) Moreover, we can obtain an expression for the constants an in terms of the function f.
z 0 R 1 R 2
Figure 6.1.1: An annulus in C with centre z 0 and radii R 1 < R 2.
Theorem 6.1.1 (Laurent’s theorem) Suppose that f is holomorphic in the annulus {z ∈ C | R 1 < |z − z 0 | < R 2 }, where 0 ≤ R 1 < R 2 ≤ ∞. Then we can write f as a Laurent series: for R 1 < |z − z 0 | < R 2 we have
f (z) =
∑^ ∞
n=
an(z − z 0 )n^ +
∑^ ∞
n=
a−n(z − z 0 )−n^ (6.1.2)
Moreover, if Cr(t) = z 0 + reit^ with R 1 < r < R 2 , 0 ≤ t ≤ 2 π is a path around z 0 contained in this annulus then
an =
2 πi
Cr
f (z) (z − z 0 )n+^
dz
for n ∈ Z.
Remark. Note that in this case we cannot conclude that an = f (n)(z 0 )/n! as we do not know that f is differentiable at z 0 (indeed, it may not even be defined at z 0 ).
Remark. The proof is similar to the proof of Taylor’s theorem and can be found in Stewart and Tall’s book.
We call the series (6.1.2) the Laurent series of f (z) about z 0 or the Laurent expansion of f (z). We call ∑−^1
n=−∞
an(z − z 0 )n
the principal part of the Laurent series. Thus the principal part of a Laurent series is the part that contains all the negative powers of (z − z 0 ).
Example. Let f (z) = ez^ + e^1 /z^. Recall that ez^ =
n=0 z
n/n! for all z ∈ C. Hence
e^1 /z^ =
n=0 z
−n/n! for all z 6 = 0. Hence
f (z) =
∑^ ∞
n=−∞
cnzn
We can also write
1 z − 2
z
1 − (^2) z
z
∑^ ∞
n=
( (^) z
2
)−n
∑^ ∞
n=
2 n−^1 zn^
by recognising the middle term as the sum of a geometric progression with common ratio (z/2)−^1. This converges when |(2/z)−^1 | < 1, i.e. when |z| > 2. Using (6.1.3) and (6.1.5) we see that we can expand
f (z) = −
∑^ ∞
n=
zn^ +
∑^ ∞
n=
( (^) z 2
)n
∑^ ∞
n=
2 n+
zn
and this is valid in the annulus 0 ≤ |z| < 1. Using (6.1.4) and (6.1.5) we can expand
f (z) =
∑^ ∞
n=
zn^
∑^ ∞
n=
( (^) z
2
)n
zn^
z
z 22
zn 2 n+^
and this is valid in the annulus 1 < |z| < 2. Using (6.1.4) and (6.1.6) we can expand
f (z) =
∑^ ∞
n=
zn^
∑^ ∞
n=
2 n−^1 zn
∑^ ∞
n=
1 − 2 n−^1 zn
and this is valid in the annulus |z| > 2.
§6.2 Singularities
Definition. A singularity of a function f (z) is a point z at which f (z) is not differentiable. (Indeed, f may not even be defined at z.)
Example. If f (z) = 1/z then f is not defined at the origin (we are not allowed to divide by 0). Hence f has a singularity at z = 0.
Suppose that f has a singularity at z 0.
Definition. If there exists a punctured disc 0 < |z − z 0 | < R such that f is differentiable on this punctured disc then we say that z 0 is an isolated singularity of f.
Example. In the above example, 0 is an isolated singularity of f (z) = 1/z.
In this course we will only be interested in isolated singularities. Suppose that f has an isolated singularity at z 0. We expand f as a Laurent series around z 0 to obtain
f (z) =
∑^ ∞
n=
an(z − z 0 )n^ +
∑^ ∞
n=
bn(z − z 0 )−n,
and this is valid for 0 < |z − z 0 | < R. Consider the principal part of the Laurent series
∑^ ∞
n=
bn(z − z 0 )−n. (6.2.1)
There are three possibilities: the principal part of f may have
(i) no terms,
(ii) a finite number of terms,
(iii) an infinite number of terms.
§6.2.1 Removable singularities
Suppose that f has an isolated singularity at z 0 and that the principal part of the Laurent series (6.2.1) has no terms in it. In this case, for 0 < |z − z 0 | < R we have that
f (z) = a 0 + a 1 (z − z 0 ) + · · · + an(z − z 0 )n^ + · · ·.
The radius of convergence of this power series is at least R, and so f (z) extends to a function that is differentiable at z 0.
Example. Let
f (z) =
sin z z
z 6 = 0.
Then f has an isolated singularity at 0 as f (z) is not defined at z = 0. However, we know that sin z z
z^2 3!
z^4 5!
for z 6 = 0. Define f (0) = 1. Then f (z) is differentiable for all z ∈ C. Hence f has a removable singularity at z = 0.
§6.2.2 Poles
Suppose that f has an isolated singularity at z 0 and that the principal part of the Laurent series (6.2.1) has finitely many terms in it. In this case, for 0 < |z − z 0 | < R we can write
f (z) =
bm (z − z 0 )m^
b 1 z − z 0
∑^ ∞
n=
an(z − z 0 )n
where bm 6 = 0. In this case, we say that f has a pole of order m at z 0.
Example. Let
f (z) =
sin z z^4
, z 6 = 0.
Then f has an isolated singularity at z = 0. We can write
sin z z^4
z^3
z
z −
z^3 + · · ·.
Hence f has a pole of order 3 at z = 0.
Exercise 6. Describe the type of singularity at 0 of each of the following functions:
(i) sin(1/z) (z 6 = 0), (ii) z−^3 sin^2 z, (iii)
cos z − 1 z^2
