2Contour integration
IB Complex Analysis
2.7 Laurent series
If f is holomorphic at z
0
, then we have a local power series expansion
f(z) =
∞
X
n=0
c
n
(z − z
0
)
n
near
z
0
. If
f
is singular at
z
0
(and the singularity is not removable), then there
is no hope we can get a Taylor series, since the existence of a Taylor series would
imply f is holomorphic at z = z
0
.
However, it turns out we can get a series expansion if we allow ourselves to
have negative powers of z.
Theorem (Laurent series). Let 0 ≤ r < R < ∞, and let
A = {z ∈ C : r < |z −a| < R}
denote an annulus on C.
Suppose
f
:
A → C
is holomorphic. Then
f
has a (unique) convergent series
expansion
f(z) =
∞
X
n=−∞
c
n
(z − a)
n
,
where
c
n
=
1
2πi
Z
∂B(a,ρ)
f(z)
(z − a)
n+1
dz
for
r < ρ < R
. Moreover, the series converges uniformly on compact subsets of
the annulus.
The Laurent series provides another way of classifying singularities. In the
case where r = 0, we just have
f(z) =
∞
X
−∞
c
n
(z − a)
n
on B(a, R) \ {a}, then we have the following possible scenarios:
(i) c
n
= 0 for all
n <
0. Then
f
is bounded near
a
, and hence this is a
removable singularity.
(ii)
Only finitely many negative coefficients are non-zero, i.e. there is a
k ≥
1
such that
c
n
= 0 for all
n < −k
and
c
−k
6
= 0. Then
f
has a pole of order
k at a.
(iii)
There are infinitely many non-zero negative coefficients. Then we have an
isolated essential singularity.
So our classification of singularities fit nicely with the Laurent series expansion.
We can interpret the Laurent series as follows — we can write
f(z) = f
in
(z) + f
out
(z),
where
f
in
consists of the terms with positive power and
f
out
consists of those with
negative power. Then
f
in
is the part that is holomorphic on the disk
|z −a| < R
,
while
f
out
(
z
) is the part that is holomorphic on
|z − a| > r
. These two combine
to give an expression holomorphic on
r < |z − a| < R
. This is just a nice way
of thinking about it, and we will not use this anywhere. So we will not give a
detailed proof of this interpretation.
Proof.
The proof looks very much like the blend of the two proofs we’ve given for
the Cauchy integral formula. In one of them, we took a power series expansion
of the integrand, and in the second, we changed our contour by cutting it up.
This is like a mix of the two.
Let w ∈ A. We let r < ρ
0
< |w − a| < ρ
00
< R.
a
˜γ
˜
˜γ
w
ρ
0
ρ
00
We let ˜γ be the contour containing w, and
˜
˜γ be the other contour.
Now we apply the Cauchy integral formula to say
f(w) =
1
2πi
Z
˜γ
f(z)
z − w
dz
and
0 =
1
2πi
Z
˜
˜γ
f(z)
z − w
dz.
So we get
f(w) =
1
2πi
Z
∂B(a,ρ
00
)
f(z)
z − w
dz −
1
2πi
Z
∂B(a,ρ
0
)
f(z)
z − w
dz.
As in the first proof of the Cauchy integral formula, we make the following
expansions: for the first integral, we have w −a < z − a. So
1
z − w
=
1
z − a
1
1 −
w−a
z−a
!
=
∞
X
n=0
(w −a)
n
(z − a)
n+1
,
which is uniformly convergent on z ∈ ∂B(a, ρ
00
).
For the second integral, we have w − a > z − a. So
−1
z − w
=
1
w −a
1
1 −
z−a
w−a
!
=
∞
X
m=1
(z − a)
m−1
(w −a)
m
,
which is uniformly convergent for z ∈ ∂B(a, ρ
0
).
By uniform convergence, we can swap summation and integration. So we get
f(w) =
∞
X
n=0
1
2πi
Z
∂B(a,ρ
00
)
f(z)
(z − a)
n+1
dz
!
(w −a)
n
+
∞
X
m=1
1
2πi
Z
∂B(a,ρ
0
)
f(z)
(z − a)
−m+1
dz
!
(w −a)
−m
.
Now we substitute n = −m in the second sum, and get
f(w) =
∞
X
n=−∞
˜c
n
(w −a)
n
,
for the integrals
˜c
n
. However, some of the coefficients are integrals around the
ρ
00
circle, while the others are around the
ρ
0
circle. This is not a problem. For
any
r < ρ < R
, these circles are convex deformations of
|z − a|
=
ρ
inside the
annulus A. So
Z
∂B(a,ρ)
f(z)
(z − a)
n+1
dz
is independent of ρ as long as ρ ∈ (r, R). So we get the result stated.
Definition
(Principal part)
.
If
f
:
B
(
a, r
)
\ {a} → C
is holomorphic and if
f
has Laurent series
f(z) =
∞
X
n=−∞
c
n
(z − a)
n
,
then the principal part of f at a is
f
principal
=
−1
X
n=−∞
c
n
(z − a)
n
.
So
f − f
principal
is holomorphic near
a
, and
f
principal
carries the information
of what kind of singularity f has at a.
When we talked about Taylor series, if
f
:
B
(
a, r
)
→ C
is holomorphic
with Taylor series
f
(
z
) =
P
∞
n=0
c
n
(
z − a
)
n
, then we had two possible ways of
expressing the coefficients of c
n
. We had
c
n
=
1
2πi
Z
∂B(a,ρ)
f(z)
(z − a)
n+1
dz =
f
(n)
(a)
n!
.
In particular, the second expansion makes it obvious the Taylor series is uniquely
determined by f .
For the Laurent series, we cannot expect to have a simple expression of the
coefficients in terms of the derivatives of the function, for the very reason that
f
is not even defined, let alone differentiable, at
a
. So is the Laurent series unique?
Lemma. Let f : A → C be holomorphic, A = {r < |z −a| < R}, with
f(z) =
∞
X
n=−∞
c
n
(z − a)
n
Then the coefficients c
n
are uniquely determined by f.
Proof. Suppose also that
f(z) =
∞
X
n=−∞
b
n
(z − a)
n
.
Using our formula for c
k
, we know
2πic
k
=
Z
∂B(a,ρ)
f(z)
(z − a)
k+1
dz
=
Z
∂B(a,ρ)
X
n
b
n
(z − a)
n−k−1
!
dz
=
X
n
b
n
Z
∂B(a,ρ)
(z − a)
n−k−1
dz
= 2πib
k
.
So c
k
= b
k
.
While we do have uniqueness, we still don’t know how to find a Laurent
series. For a Taylor series, we can just keep differentiating and then get the
coefficients. For Laurent series, the above integral is often almost impossible to
evaluate. So the technique to compute a Laurent series is blind guesswork.
Example. We know
sin z = z −
z
3
3!
+
z
5
5!
− ···
defines a holomorphic function, with a radius of convergence of
∞
. Now consider
cosec z =
1
sin z
,
which is holomorphic except for
z
=
kπ
, with
k ∈ Z
. So
cosec z
has a Laurent
series near z = 0. Using
sin z = z
1 −
z
2
6
+ O(z
4
)
,
we get
cosec z =
1
z
1 +
z
2
6
+ O(z
4
)
.
From this, we can read off that the Laurent series has
c
n
= 0 for all
n ≤ −
2,
c
−1
= 1,
c
1
=
1
5
. If we want, we can go further, but we already see that
cosec
has a simple pole at z = 0.
By periodicity, cosec has a simple pole at all other singularities.
Example. Consider instead
sin
1
z
=
1
z
−
1
3!z
3
+
1
5!z
5
− ··· .
We see this is holomorphic on
C
∗
, with
c
n
6
= 0 for infinitely many
n <
0. So this
has an isolated essential singularity.
Example.
Consider
cosec
1
z
. This has singularities at
z
=
1
kπ
for
k ∈ N
=
{
1
,
2
,
3
, ···}
. So it is not holomorphic at any punctured neighbourhood
B
(0
, r
)
\
{
0
}
of zero. So this has a non-isolated singularity at zero, and there is no Laurent
series in a neighbourhood of zero.
We’ve already done most of the theory. In the remaining of the course, we
will use these techniques to do stuff. We will spend most of our time trying to
evaluate integrals, but before that, we will have a quick look on how we can use
Laurent series to evalue some series.
Example (Series summation). We claim that
f(z) =
∞
X
n=−∞
1
(z − n)
2
is holomorphic on C \ Z, and moreover if we let
f(z) =
π
2
sin
2
(πz)
,
We will reserve the name
f
for the original series, and refer to the function
z 7→
π
2
sin
2
(πz)
as g instead, until we have proven that they are the same.
Our strategy is as follows — we first show that
f
(
z
) converges and is holo-
morphic, which is not hard, given the Weierstrass
M
-test and Morera’s theorem.
To show that indeed we have
f
(
z
) =
g
(
z
), we first show that they have equal
principal part, so that
f
(
z
)
− g
(
z
) is entire. We then show it is zero by proving
f − g
is bounded, hence constant, and that
f
(
z
)
− g
(
z
)
→
0 as
z → ∞
(in some
appropriate direction).
For any fixed
w ∈ C \ Z
, we can compare it with
P
1
n
2
and apply the
Weierstrass
M
-test. We pick
r >
0 such that
|w −n| >
2
r
for all
n ∈ Z
. Then
for all z ∈ B(w; r), we have
|z − n| ≥ max{r, n −|w| − r}.
Hence
1
|z − n|
2
≤ min
1
r
2
,
1
(n − |w|−r)
2
= M
n
.
By comparison to
P
1
n
2
, we know
P
n
M
n
converges. So by the Weierstrass
M-test, we know our series converges uniformly on B(w, r).
By our results around Morera’s theorem, we see that
f
is a uniform limit of
holomorphic functions
P
N
n=−N
1
(z−n)
2
, and hence holomorphic.
Since
w
was arbitrary, we know
f
is holomorphic on
C \ Z
. Note that we
do not say the sum converges uniformly on
C \ Z
. It’s just that for any point
w ∈ C \ Z
, there is a small neighbourhood of
w
on which the sum is uniformly
convergent, and this is sufficient to apply the result of Morera’s.
For the second part, note that
f
is periodic, since
f
(
z
+ 1) =
f
(
z
). Also, at
0,
f
has a double pole, since
f
(
z
) =
1
z
2
+ holomorphic stuff near
z
= 0. So
f
has
a double pole at each
k ∈ Z
. Note that
1
sin
2
(πz)
also has a double pole at each
k ∈ Z.
Now, consider the principal parts of our functions — at
k ∈ Z
,
f
(
z
) has
principal part
1
(z−k)
2
. Looking at our previous Laurent series for cosec(z), if
g(z) =
π
sin πz
2
,
then
lim
z→0
z
2
g
(
z
) = 1. So
g
(
z
) must have the same principal part at 0 and
hence at k for all k ∈ Z.
Thus
h
(
z
) =
f
(
z
)
−g
(
z
) is holomorphic on
C \Z
. However, since its principal
part vanishes at the integers, it has at worst a removable singularity. Removing
the singularity, we know h(z) is entire.
Since we want to prove f(z) = g(z), we need to show h(z) = 0.
We first show it is boundedness. We know
f
and
g
are both periodic with
period 1. So it suffices to focus attention on the strip
−
1
2
≤ x = Re(z) ≤
1
2
.
To show this is bounded on the rectangle, it suffices to show that
h
(
x
+
iy
)
→
0
as
y → ±∞
, by continuity. To do so, we show that
f
and
g
both vanish as
y → ∞.
So we set z = x + iy, with |x| ≤
1
2
. Then we have
|g(z)| ≤
4π
2
|e
πy
− e
−πy
|
→ 0
as y → ∞. Exactly analogously,
|f(z)| ≤
X
n∈Z
1
|x + iy −n|
2
≤
1
y
2
+ 2
∞
X
n=1
1
(n −
1
2
)
2
+ y
2
→ 0
as
y → ∞
. So
h
is bounded on the strip, and tends to 0 as
y → ∞
, and is hence
constant by Liouville’s theorem. But if
h →
0 as
y → ∞
, then the constant
better be zero. So we get
h(z) = 0.