Supplement: Laurent Series Development 1
Supplement: Laurent Series Development
(Theorem V.1.11)
Note. In this section, we give a detailed proof of Theorem V.1.11, the “Laurent
Series Development” theorem. First, we recall some previous results which are
needed in the proof of this theorem.
∞
X
Theorem III.1.3. If an (z − a)n , define the number R as 1
R
= lim |an |1/n (so
n=0
0 ≤ R ≤ ∞). Then
(a) if |z − a| < R, the series converges absolutely,
(b) if |z − a| > R, the series diverges, and
(c) if 0 < r < R then the series converges uniformly on |z − a| ≤ r. Moreover,
R is the only number having properties (a) and (b). R is called the radius of
convergence of the power series.
∞
X
Theorem IV.2.8. Let f be analytic in B(a; R). Then f (z) = an (z − a)n for
n=0
(n)
|z − a| < R where an = f (a)/n! and this series has radius of convergence ≥ R.
Lemma IV.5.1. Let γ be a rectifiable curve and suppose ϕ is a function defined
R
and continuous on {γ}. For each m ≥ 1 let Fm(z) = γ ϕ(w)(w − z)−m dw for
/ {γ}. Then each Fm is analytic on C \ {γ} and Fm0 (z) = mFm+1 (z).
z∈
Supplement: Laurent Series Development 2
Corollary IV.5.9. (Theorem 5.8 with one curve.) Let G be an open set and
f : G → C analytic. If γ is a closed rectifiable curve in G such that n(γ; w) = 0 for
all w ∈ C \ G then for a ∈ G \ {γ}
Z
(k) k! f (z)
f (a)n(γ; a) = dz.
2πi γ (z − a)k+1
Theorem V.1.2. If f has an isolated singularity at a then z = a is a removable
singularity if and only if lim (z − a)f (z) = 0.
z→a
Note. Now for our main result.
Theorem V.1.11. Laurent Series Development.
Let f be analytic in ann(a; R1, R2). Then
∞
X
f (z) = an (z − a)n
n=−∞
where the convergence is absolute and uniform over the closure of ann(a; r1, r2 ) if
R1 < r1 < r2 < R2. The coefficients an are given by
Z
1 f (w)
an = dw (1.12)
2πi γ (w − a)n+1
where γ is the circle γ(t) = a + reit where t ∈ [0, 2π] for any r with R1 < r < R2 .
Moreover, this series is unique.
Proof. If R1 < r1 < r2 < R2 and γ1 , γ2 are the circles γ1 (t) = a + r1 eit for
t ∈ [0, 2π] and γ2 (t) = a + r2 eit for t ∈ [0, 2π], then γ1 ∼ γ2 in ann(a; R1, R2 ).
Then by Cauchy’s Theorem (Third Version, Theorem IV.6.7) for any analytic g on
Supplement: Laurent Series Development 3
R R
ann(a; R1, R2 ), we have γ1
g= γ2
g. So the integral in (1.12) is independent of r,
so for each n ∈ Z, an is a constant (i.e., independent of γ).
Claim. Definition of f1 and f2 ; f (z) = f1 (z) + f2 (z).
Moreover, function f2 : B(a; R2) → C given as
Z
1 f (w)
f2 (z) = dw,
2πi γ2 w − z
where |z − a| < r2 and R1 < r2 < R2, is a well defined function (independent of
r2 ). By Lemma IV.5.1 (with m = 1 and ϕ(w) = f (w)), f2 is analytic in B(a; r2 ).
Similarly, if G = {z | |z − a| > R1} then f1 : G → C defined as
Z
−1 f (w)
f1 (z) = dw,
2πi γ1 w − z
where |z − a| > r1 (so r1 is chosen such that z ∈
/ {γ1 }) and R1 < r1 < R2 is analytic
in G.
If R1 < |z − a| < R2, fix r1 and r2 so that R1 < r1 < |z − a| < r2 < R2 . Let
γ1 (t) = a + r1 eit for t ∈ [0, 2π] and γ2 (t) = a + r2 eit for t ∈ [0, 2π], as above. Let
λ be a line segment from a point on γ1 radially to a point on γ2 which does not
Supplement: Laurent Series Development 4
contain z:
Since γ1 ∼ γ2 on ann(a; R1, R2 ), we have that the closed curve γ = γ2 − λ − γ1 + λ
is homotopic to 0. Next, n(γ2 ; z) = 1 and n(γ1 ; z) = 0 gives:
Z
1 f (w)
f (z) = dw by the Cauchy’s Integral Formula
2πi γ w − z
Z Z
1 f (w) 1 f (w)
= dw − dw since γ is piecewise smooth
2πi γ2 w − z 2πi γ1 w − z
= f2 (z) + f1 (z) by the definitions of f1 and f2.
We now create series for f1 (with negative powers of (z − a)) and f2 (with the usual
positive powers of (z − a)).
Claim. The an are as claimed for n ≥ 0.
Since f2 is analytic in B(a; R2) then by Theorem IV.2.8
∞
X
n f (n) (a)
f2(z) = an(z − a) and an = .
n=0
n!
1
R f(w)
Since f2 (z) = 2πi γ2 w−z dw, in Lemma IV.5.1 with m = 1 we have F1 (z) =
1
R ϕ(w)
2πi γ2 w−z dw where ϕ(w) = f (w), and therefore by induction
Z
(n) n! ϕ(w)
f2 (z) = n!Fn+1 (z) = dw.
2πi γ2 (w − z)n+1
Supplement: Laurent Series Development 5
We then have
Z
f (n) (a) 1 f (w)
an = = Fn+1 (a) = dw,
n! 2πi γ2 (w − a)n+1
as claimed.
Claim. Definition of g(z); g(z) is analytic for z ∈ B(0; 1/R1 ).
Now define g(z) for 0 < |z| < 1/r1 as g(z) = f1 (a + 1/z). Notice that a + 1/z =
a − (−1/z) and | − 1/z| > r1 , so a + 1/z ∈ ann(a; r1 , ∞). Since f1 is defined as
Z
−1 f (w)
f1 (z) = dw
2πi γ1 w − z
where γ1 (t) = a + r1 eit with t ∈ [0, 2π], then f1(a + 1/z) is defined as long as
/ {γ1 }. We have:
a + 1/z ∈
Since f1 (z) is analytic for all z with |z − a| > r1 (from the definition of f1), then
g(z) is analytic for all a + 1/z with |(a + 1/z) − a| > r1 , or |1/z| > r1 or 0 < |z| <
1/r1 . So g has an isolated singularity at z = 0. We now show g has a removable
singularity at z = 0 by showing limz→0 g(z) exists (and hence limz→0 zg(z) = 0,
as required by Theorem V.1.2 for g to have a removable singularity at z = 0).
Suppose R2 > r > r1 , z ∈ ann(a; r, ∞), and ρ(z) = d(z, C) where C is the circle
Supplement: Laurent Series Development 6
{w | |w − a| = r}:
Define M = max{|f (w)| | w ∈ C} and give C a counterclockwise orientation. Then
for such z “outside” of C,
Z
−1 f (w)
|f1(z)| = dw
2πi γ1 w − z
Z
−1 f (w)
= dw since γ1 ∼ C on ann(a; R1, R2 )
2πi C w − z
Z
1 |f (w)| 1 M 2πr Mr
≤ |dw| ≤ = .
2π C |w − z| 2π ρ(z) ρ(z)
Now limz→∞ ρ(z) = ∞ and M, r are constants, so limz→∞ f1 (z) = 0. Therefore,
lim g(z) = lim f1 (a + 1/z) = lim f1 (z) = 0.
z→0 z→0 z→∞
Since limz→0 g(z) = 0, g has a removable singularity at z = 0 by Theorem V.1.2. We
define g(0) = 0 and then g is analytic on B(0; 1/r1 ) (by the definition of removable
singularity). Notice that r1 > R1 was arbitrary above, and we can conclude that
each claim is valid for r1 replace with R1 (as in the text). So g can be written
P P∞
g(z) = ∞ n=1 Bn z n
(B0 = 0 since g(0) = 0). So, f 1 (a + 1/z) = g(z) = n=1 Bn a
n
Supplement: Laurent Series Development 7
for z ∈ B(0; 1/R1 ).
Claim. The an are as claimed for n ≤ −1.
−1
R f(x)
We have by definition that f1(z) = 2πi γ1 w−z dw for |z − a| > R1 where γ1 (t) =
a + r1 eit for t ∈ [0, 2π]. So for |a| < R1 we have
Z
1 −1 f (w)
f1 a + = dw.
z 2πi γ1 w − a + z1
Replacing w with a + 1/w, dw with −w −2 dw, and using the inverse of this mapping
to take γ1 to γ10 (so that the integrand receives the same values of w) we have
1 1 1 −it
γ10 (t) = = = e for t ∈ [0, 2π].
γ1 (t) − a (a + r1 eit ) − a r1
Then f1 (a + 1/z) becomes
Z Z
1 −1 f (a + 1/w) −2 1 f (a + 1/w)
f1 a + = [−w ] dw = dw
z 2πi γ10 a + w1 − a + 1z 2πi γ10 w 1 − wz
Z Z
1 zf (a + 1/w) 1 zf (a + 1/w)
= dw = dw
2πi γ10 w(z − w) 2πi −γ10 w(w − z)
Z
z f (a + 1/w)/w
= dw = g(z) for 0 < |z| < 1/R1 .
2πi −γ10 w−z
P
As established above, g(z) is analytic on B(0; 1/R1 ), so g(z) = ∞ n
n=1 Bn z (notice
that B0 = 0 since g(0) = 0). Also, by Theorem IV.2.8, Bn = g (n) (0)/n!. We now
calculate g (n) (0) using Lemma IV.5.1 and the representation of g(z) as
Z
z f (a + 1/w)/w
g(z) = dw
2πi −γ10 w−z
1 it
(notice that −γ10 (t) = r1
e for t ∈ [0, 2π]). From the Product Rule, Lemma IV.5.1
(with ϕ(w) = f (a + 1/w)/w), and Math Induction we can show that
Z Z
(n) n! f (a + 1/w)/w n!z f (a + 1/w)/w
g (z) = n
dw + dw for |z| < R1.
2πi −γ10 (w − z) 2πi −γ10 (w − z)n+1
Supplement: Laurent Series Development 8
n!
R f(a+1/w) g(n)(0) 1
R f(a+1/w)
So g (n) (0) = 2πi −γ10 w n+1
dw, so Bn = n!
= 2πi −γ10 w n+1
dw. Replacing w
with 1/(w − a) and dw with −(w − a)−2 dw and using the inverse of this mapping
to take γ10 to γ1 (similar to above)
Z Z
1 f (a + 1/w) 1 f (w) −1
Bn = dw = dw
2πi −γ10 w n+1 2πi −γ1 1 n+1 (w − a)2
Z Z w−a
−1 1 f (w)
= f (w)(w − a)n−1 dw = dw.
2πi −γ1 2πi γ1 (w − a)−n+1
Defining a−n as Bn (and replacing n with −n in Bn ) we have
Z
1 f (w)
a−n = dw for − n ≥ 1
2πi γ1 (w − a)n+1
P
and f1(a + 1/z) = ∞ n
n=1 Bn z for 0 < |z| < 1/R1 or
∞
X
f1 (z) = Bn (z − a)−n for z ∈ ann(a; R1, ∞)
n=1
X∞ −1
X
−n
= a−n (z − a) = an (z − a)n.
n=1 n=−∞
So
f (z) = f1 (z) + f2(z)
−1
X ∞
X
n
= an (z − a) + an (z − a)n
n=−∞
| {z } |n=0 {z }
for |z−a|>R1 for |z−a|<R2
∞
X
= an (z − a)n for z ∈ ann(a; R1, R2 )
n=−∞
where the an are as claimed.
Claim. The Laurent series converges absolutely and uniformly on r1 ≤ |a| ≤ r2 .
P
Since f2(z) = ∞ n
n=0 an(z −a) for z ∈ B(a; R2 ) then this series converges absolutely
for |z − a| < R2 by Theorem III.1.3(a) and converges uniformly for |z − a| ≤
Supplement: Laurent Series Development 9
P∞ n
r2 for any 0 < r2 < R2 by Theorem III.1.3(c). Similarly, g(z) = n=1 Bn z
converges absolutely for |z − a| < 1/R1 and converges uniformly for |z − a| ≤ 1/r1
P
for any r1 > R1 by Theorem III.1.3(a) and (c). So f1(z) = ∞ n=1 a−n (z − a)
−n
converges absolutely for |z − a| > R1 and uniformly for |z − a| ≥ r1 . Therefore
P
f (z) = ∞ n −
n=−∞ an (z − a) converges absolutely and uniformly on ann(a; r1 , r2 ) if
R1 < r1 < r2 < R2.
Claim. The Laurent series representation is unique for given f .
Now for the uniqueness. If we were dealing with an analytic function on B(a; R),
we could deduce that the coefficients are unique, in fact an = f (n) (a)/n! as given
in Theorem IV.2.8. For a Laurent series, we integrate. Let R1 < r1 < r2 < R2
and let γ = ((r1 + r2 )/2)eit for t ∈ [0, 2π]. Notice that the Laurent series converges
P
uniformly on ann(a; r1 , r2 )− as given above. So for f (z) = ∞ n
n=−∞ an (z − a) we
have for k ≥ 0:
Z Z X ∞
f (w)
k+1
dw = an (w − a)n−(k+1) dw
γ (w − a) γ n=−∞
Z −1 ∞
!
X X
= an (w − a)n−(k+1) + an (w − a)n−(k+1) dw
γ n=−∞ n=0
since the convergence is absolute—see Definition V.1.10
Z −1
! Z X ∞
!
X
= an (w − a)n−(k+1) dw + an (w − a)n−(k+1) dw
γ n=−∞ γ n=0
−1 Z
X ∞ Z
X
= an (w − a)n−(k+1) dw + an (w − a)n−(k+1) dw
n=−∞ γ n=0 γ
since the convergence is uniform; Lemma IV.2.7
∞ Z X ∞ Z
X a−n
= n+k+1
dw + an(w − a)n−(k+1) dw
n=1 γ (w − a) n=0 γ
Z
ak
= 0+0+ dw since each integrand has a primitive on
γ w−a
Supplement: Laurent Series Development 10
G = ann(a; r1 , r2 ) for k 6= n
= ak 2πi n(γ; a) = ak 2πi.
1
R f(w)
So for k ≥ 0 it must be that ak = 2πi γ (w−a)k+1 dw. The arbitrary nature of γ is
explained above. For k ≤ −1, consider
Z Z X ∞
f (w)
k+1
dw = an (w − a)n−k−1 dw
γ (w − a) γ n=−∞
Z −1 ∞
!
X X
= an (w − a)n−k−1 + an(w − a)n−k−1 dw
γ n=−∞ n=0
Z −1
! Z ∞
!
X X
= an (w − a)n−k−1 dw + an (w − a)n−k−1 dw
γ n=−∞ γ
n=0
−1
X Z ∞
X Z
n−k−1) n−k−1
= an (w − a) dw + an (w − a) dw
n=−∞ γ n=0 γ
∞ Z ∞ Z
X a−n X
= dw + an(w − a)n−k−1 dw
n=1 γ (w − a)n+k+1 n=0 γ
Z
ak
= 0+ dw + 0 = ak 2πi n(γ; a) = ak 2πi.
γ w−a
1
R f(w)
So for k ≤ −1 it must be that ak = 2πi γ (w−a)k+1 dw. The arbitrary nature of γ is
explained above.
Revised: 4/4/2018
