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Laurent Expansions and Singularities
Adrian Down
November 01, 2005
1 Laurent expansions
1.1 Statement
Theorem. Consider the annulus
A = { z : R 1 < | z − z 0 | < R 2 }
If f : A → C is differentiable, then:
f (z) =
∑^ ∞
n=−∞
an(z − z 0 )n^ ∀z ∈ A
where the series converges absolutely uniformly on any closed bounded subset of A.
- The coefficients of the series are given by the formula
an =
2 πı
Cr
f (w) (w − z 0 )n+^
dw
Cr = z 0 + reıt, 0 ≤ t ≤ 2 π r ∈ (R 1 , R 2 )
- If
f (z) =
∑^ ∞
n=−∞
bn(z − z 0 )n
with absolute convergence uniformly on any closed bounded subset of A, then bn = an
Note. • We allow for the degenerate annulus,
0 ≤ R 1 ≤ R 2 ≤ ∞
- The last conclusion says that we cannot construct a new series by changing the size of the annulus, as long as the radii of the annulus are within (R 1 , R 2 ).
- Laurent series are similar to Taylor series, but are somewhat more general. Laurent series allow for a “hole” in the center of the region in which the series is defined, and they also allow for negative exponents. For the proof, we assume z 0 = 0. For the more general case, see the text. We need some preliminary results to prove the general theorem.
1.2 Series with negative powers outside a given radius
1.2.1 Statement
Lemma. Given
- ρ > 0
- Cρ = { z : | z| = ρ }
- h : Cρ → C continuous
Then
∫
Cρ
h(w) w − z
dw = −
∑^ −^1
n=−∞
anzn^ ∀|z| > ρ
where an =
Cρ
h(w) wn+^
dw
with absolute convergence uniformly on any closed bounded subset.
Note. We are not assuming h is differentiable, only continuous. We did not say that the series converges to h, only that it converges to the integral expression above.
To get back to positive indices,
let n = −k − 1 ⇒ k = −n − 1
The series becomes
∑
k>N +
|h(w)||w|k |z|k+
Because we assumed that h is continuous, |h| is continuous and thus bounded.
|z|
max w∈Cρ |h|
k>N +
ρ |z|
)k
The summation is independent of w and tends to 0 as N tends to ∞. Thus uniform convergence is obtained by taking sufficiently large N. We can then write the desired result, ∫
Cρ
h(w) w − z
dw = −
∑^ −^1
n=−∞
zn
Cρ
h(w) wn+^
dw
1.3 Existence of Laurent series
Lemma. Laurent series exist
Proof. Given z, choose ρ 1 , ρ 2 such that
R 1 < ρ 1 < |z| < ρ 2 < R 2
Write
f (z) =
2 πı
Cρ 2
f (w) w − z
dz −
2 πı
Cρ 1
f (w) w − z
dw
The first integral is the same as that in the previous theorem about the existence of Taylor expansions. It has coefficients only positive values of n. The second is what we just did, and we saw that it can be expanded in terms of only negative n coefficients. Thus it is possible to construct a series in the stated form of the Laurent series.
1.4 Coefficients of Laurent series
Lemma. If
f (z) =
∑^ ∞
n=∞
bnzn
with absolute convergence uniformly in any closed bounded subset of an an- nulus,
{ z : r 1 < | z| < r 2 }
then
bn =
2 πı
Cr
f (w) wn+^
dw
for any r ∈ (r 1 , r 2 ).
Note. The integral does not depend on r for r ∈ (R 1 , R 2 ).
Proof. Let k ∈ Z. Using the expression for the Laurent expansion of f (z) proved above, ∫
Cr
f (w) wk+^
dw =
Cr
wk+
∑^ ∞
n=−∞
bnwndw
We can interchange the integral and the sum, provided that the sum is abso- lutely convergent. Absolute convergence is ensured by the assumption that the function converges uniformly on the circle. ∫
Cr
f (w) wk+^
dw =
∑^ ∞
n=−∞
bn
Cr
wn−k−^1 dw
We need the integral of a power function.
Fact.
p ∈ Z
∫ (^) W p
Cr
dw =
2 πı p = − 1 0 p 6 = − 1
Using this fact, there is only one nonzero term in the sum. ∑^ ∞
n=−∞
bn
Cr
wn−k−^1 dw = bk · 2 πı
2.2.2 Comparison to real analysis
Consider the function
F : R r { 0 } → R
defined such that
F (x) =
1 x > 0 − 1 x < 0
F satisfies the hypothesis of the theorem, but the singularity at z = 0 is not removable
2.2.3 Proof
Proof. We have a function f that is differentiable in a punctured disk. From previous results today, ∃ a Laurent expansion for this function on this disk.
∃an ∈ C st ∀z ∈ { z : 0 < | z − z 0 | < δ }
f (z) =
∑^ ∞
n=−∞
an(z − z 0 )n
We will show that
an = 0 ∀n < 0
Then f will be expressible as a Taylor series with only positive powers of z,
g(z) =
∑^ ∞
n=
an(z − z 0 )n
and hence is differentiable on { z : | z − z 0 | < δ }. We have an integral formula for the coefficients of the desired Laurent series. For any r ∈ (0, δ),
an =
2 πı
Cr
f (w) wn+^
dw
As usual, when we have something that doesn’t depend on a parameter, take a limit of that parameter to gain information. Let r → 0. We want to show
that in this limit, bn = 0 ∀n < 0. Use the usual tool for bounding integrals, the estimation lemma.
|an| <
2 π
L(Cr) max |w|=r
|f (w)| rn+
Canceling with the length of the circle, we have an upper bound on an
|an| ≤
max|w|<r |f (w)| rn
The maximum of |f (w)| is uniformly bounded as r → 0. If n < 0, then r−n goes to 0 as r → 0. Thus
|an| → 0 as r → 0
So an = 0 ∀n < 0
Thus ∑^ ∞
n=
an(z − z 0 )n
is an extension of f to a function that is differentiable on the whole disk.
Note. • This proof could not give us any information about coefficients for which n > 0, since the bound would diverge as r → 0.
- This theorem says that if a function is bounded in the neighborhood of a singularity, that singularity is removable.
2.3 Other types of singularities
2.3.1 Motivation
We have already said that if f has an isolated singularity at z 0 , then ∃ a Laurent series representation such that
f (z) =
∑^ ∞
n=−∞
an(z − z 0 )n
in some punctured neighborhood of z 0. This is the unique series representa- tion of f (z), and thus gives genuine information about the function. With this series, we can classify the other two types of singularities.
2.3.3 Essential singularities
Definition (Essential singularities). An isolated singularity is said to be essential if it is not removable and is not a pole.
Proof in a moment. First the last kind of singularity.
Theorem. Assume f has an essential singularity at z 0. Then for any w ∈ C, for any δ > 0 , for any > 0 ,
∃z st |z − z 0 | < δ and |f (z) − w| <
Note. • This theorem says that for any w ∈ C, ∃ a sequence zj → z 0 such that f (zj ) → w
- At an essential singularity, the function approaches all possible values.
- This theorem demonstrates the extreme difficulties posed by essential singularities.
Example. The canonical example of an essential singularity is
f (z) = e
1 z
Given w ∈ C, assume w 6 = 0. Choose ζ ∈ C such that
eζ^ = w
let zj =
ζ + 2πıj zj → 0 as j → ∞
Then
f (zj ) = e
1 zj (^) = eζ+2πıj^ = eζ^ = w
Thus we can define a function that takes any value near the essential singu- larity at z = 0.
The proof of this theorem will also be given in the next lecture.
