3Laurent series and singularities
IB Complex Methods
3.1 Taylor and Laurent series
If f is analytic at z
0
, then it has a Taylor series
f(z) =
∞
X
n=0
a
n
(z − z
0
)
n
in a neighbourhood of
z
0
. We will prove this as a special case of the coming
proposition. Exactly which neighbourhood it applies in depends on the function.
Of course, we know the coefficients are given by
a
n
=
f
(n)
(z
0
)
n!
,
but this doesn’t matter. All the standard Taylor series from real analysis apply
in C as well. For example,
e
z
=
∞
X
n=0
z
n
n!
,
and this converges for all z. Also, we have
(1 − z)
−1
=
∞
X
n=0
z
n
.
This converges for |z| < 1.
But if
f
has a singularity at
z
0
, we cannot expect such a Taylor series, since
it would imply
f
is non-singular at
z
0
. However, it turns out we can get a series
expansion if we allow ourselves to have negative powers of z.
Proposition
(Laurent series)
.
If
f
is analytic in an annulus
R
1
< |z −z
0
| < R
2
,
then it has a Laurent series
f(z) =
∞
X
n=−∞
a
n
(z − z
0
)
n
.
This is convergent within the annulus. Moreover, the convergence is uniform
within compact subsets of the annulus.
Proof.
(non-examinable) We wlog
z
0
= 0. Given a
z
in the annulus, we pick
r
1
, r
2
such that
R
1
< r
1
< |z| < r
2
< R
2
,
and we let
γ
1
and
γ
2
be the contours
|z|
=
r
1
,
|z|
=
r
2
traversed anticlockwise
respectively. We choose γ to be the contour shown in the diagram below.
z
0
r
1
r
2
z
γ
We now apply Cauchy’s integral formula (after a change of notation):
f(z) =
1
2πi
I
γ
f(ζ)
ζ −z
dζ.
We let the distance between the cross-cuts tend to zero. Then we get
f(z) =
1
2πi
I
γ
2
f(ζ)
ζ −z
dz −
1
2πi
I
γ
1
f(ζ)
ζ −z
dζ,
We have to subtract the second integral because it is traversed in the opposite
direction. We do the integrals one by one. We have
I
γ
1
f(ζ)
ζ −z
= −
1
z
I
γ
1
f(ζ)
1 −
ζ
z
dζ
Taking the Taylor series of
1
1−
ζ
z
, which is valid since
|ζ|
=
r
1
< |z|
on
γ
1
, we
obtain
= −
1
z
I
γ
1
f(ζ)
∞
X
m=0
ζ
z
m
dζ
= −
∞
X
m=0
z
−m−1
I
γ
1
f(ζ)ζ
m
dζ.
This is valid since
H
P
=
P
H
by uniform convergence. So we can deduce
−
1
2πi
I
γ
1
f(ζ)
ζ −z
dζ =
−1
X
n=−∞
a
n
z
n
,
where
a
n
=
1
2πi
I
γ
1
f(ζ)ζ
−n−1
dζ
for n < 0.
Similarly, we obtain
1
2πi
I
γ
2
f(ζ)
ζ −z
dζ =
∞
X
n=0
a
n
z
n
,
for the same definition of a
n
, except n ≥ 0, by expanding
1
ζ −z
=
1
ζ
1
1 −
z
ζ
=
∞
X
n=0
z
n
ζ
n+1
.
This is again valid since
|ζ|
=
r
2
> |z|
on
γ
2
. Putting these result together, we
obtain the Laurent series. The general result then follows by translating the
origin by z
0
.
We will not prove uniform convergence — go to IB Complex Analysis.
It can be shown that Laurent series are unique. Then not only is there a
unique Laurent series for each annulus, but if we pick two different annuli on
which
f
is analytic (assuming they overlap), they must have the same coefficient.
This is since we can restrict the two series to their common intersection, and
then uniqueness requires the coefficients to be must be the same.
Note that the Taylor series is just a special case of Laurent series, and if
f
is
holomorphic at
z
0
, then our explicit formula for the coefficients plus Cauchy’s
theorem tells us we have a
n
= 0 for n < 0.
Note, however, that we needed the Taylor series of
1
1−z
in order to prove
Taylor’s theorem.
Example.
Consider
e
z
z
3
. What is its Laurent series about
z
0
= 0? We already
have a Taylor series for e
z
, and all we need to do is to divide it by z
3
. So
e
z
z
3
=
∞
X
n=0
z
n−3
n!
=
∞
X
n=−3
z
n
(n + 3)!
.
Example.
What is the Laurent series for
e
1/z
about
z
0
? Again, we just use the
Taylor series for e
z
. We have
e
1
z
= 1 +
1
z
+
1
2!z
2
+
1
3!z
3
+ ··· .
So the coefficients are
a
n
=
1
(−n)!
for n ≤ 0.
Example. This is a little bit more tricky. Consider
f(z) =
1
z − a
,
where
a ∈ C
. Then
f
is analytic in
|z| < |a|
. So it has a Taylor series about
z
0
= 0 given by
1
z − a
= −
1
a
1 −
z
a
−1
= −
∞
X
n=0
1
a
n+1
z
n
.
What about in
|z| > |a|
? This is an annulus, that goes from
|a|
to infinity. So it
has a Laurent series. We can find it by
1
z − a
=
1
z
1 −
a
z
−1
=
∞
X
m=0
a
m
z
m+1
=
−1
X
n=−∞
a
−n−1
z
n
.
Example. Now consider
f(z) =
e
z
z
2
− 1
.
This has a singularity at
z
0
= 1 but is analytic in an annulus 0
< |z − z
0
| <
2
(the 2 comes from the other singularity at
z
=
−
1). How do we find its Laurent
series? This is a standard trick that turns out to be useful — we write everything
in terms of ζ = z − z
0
. So
f(z) =
e
ζ
e
z
0
ζ(ζ + 2)
= e
z
0
2ζe
ζ
1 +
1
2
ζ
−1
=
e
2ζ
1 + ζ +
1
2!
ζ
2
+ ···
1 −
1
2
ζ + ···
=
e
2ζ
1 +
1
2
ζ + ···
=
e
2
1
z − z
0
+
1
2
+ ···
.
This is now a Laurent series, with a
−1
=
1
2
e, a
0
=
1
4
e etc.
Example.
This doesn’t seem to work for
f
(
z
) =
z
−1/2
. The reason is that the
required branch cut of
z
−
1
2
would pass through any annulus about
z
= 0. So we
cannot find an annulus on which f is analytic.
The radius of convergence of a Laurent series is at least the size of the annulus,
as we that’s how we have constructed it. So how large is it?
It turns out the radius of convergence of a Laurent series is always the
distance from
z
0
to the closest singularity of
f
(
z
), for we may always choose
R
2
to be that distance, and obtain a Laurent series valid all the way to
R
2
(not inclusive). This Laurent series must be the same series as we would have
obtained with a smaller radius, because Laurent series are unique.
While this sounds just like some technicalities, this is actually quite useful.
When deriving a Laurent series, we can make any assumptions we like about
z − z
0
being small. Then even if we have derived a Laurent series for a really
small neighbourhood, we automatically know the series is valid up to the other
point where f is singular.