and when r > 0, the annulus looks
like the following.
PICT
The annulus consists of the points between the two circles.
In the following picture, let there be two parametrizations γ_{R} for the large circle and
ˆγ
_{r} for the small
one with orientation as shown. There are also two line segments oriented as shown which miss
z ∈ann
(z0,r,R )
and constitute the intersection of the two simple closed curves Γ_{1},Γ_{2}. These two simple
closed curves are oriented as shown. Thus each of Γ_{i} is positively oriented. Let f be continuous on
ann
(z0,r,R )
and be analytic on ann
(z0,r,R )
.
PICT
It follows from Theorem 15.6.1, that for z in the annulus,
∫ ∫
-1- f-(w) -1- f (w-)
2πi γRw − zdw + 2πi ˆγr w − zdw = f (z)
This is because the contributions to the line integrals along those straight lines is 0 since they cancel off
because of opposite orientations. Let γ_{r} be the opposite orientation from
ˆγ
_{r}. Then this reduces
to
∫ ∫
f-(w-)dw − f-(w-)dw = 2πif (z)
γR w − z γr w − z
Thus
[∫ ∫ ]
f (z) =-1 -----f-(w-)-----dw + -----f-(w)------dw
2πi γR w − z0 − (z − z0) γr (z − z0)− (w − z0)
Now note that for z in the annulus between the two circles and w ∈ γ_{R}^{∗},
| |
||z−-z0-||
w−z0
< 1, and for
w ∈ γ_{r}^{∗},
|| ||
|wz−−zz00|
< 1. In fact, in each case, there is b < 1 such that
||z − z || ||w − z ||
w ∈ γ∗R,||----0|| < b < 1,w ∈ γ∗r,||---0|| < b < 1 (16.3)
w− z0 z − z0
(16.3)
Thus you can use the formula for the sum of an infinite geometric series and conclude
⌊ ( ) ⌋
1 ∫ f (w )--1-∑ ∞ z−z0 n dw
f (z) =-- ⌈ γ∫R w−z0 ∑n=0∞ w(−wz−0z )n ⌉
2πi + γr f (w)(z1−z0) n=0 z−z00 dw
Then from the uniform estimates of 16.3, one can conclude uniform convergence of the partial sums for
w ∈ γ_{R}^{∗} or γ_{r}^{∗}, and so by the Weierstrass M test, Theorem 2.2.40, one can interchange the summation
with the integral and write
( )
∑∞ -1- ∫ ----1----- n
f (z) = 2πi γ f (w )(w− z )n+1dw (z − z0)
n=0∞ ( R∫ 0 )
+ ∑ -1- f (w)(w − z)ndw ----1----,
n=0 2πi γr 0 (z − z0)n+1
both series converging absolutely. Thus there are a_{n},b_{n}∈ X such that
∑∞ n ∞∑ −n
f (z) = an(z − z0) + bn(z − z0)
n=0 n=1
This proves most of the following theorem.
Theorem 16.2.2Let z ∈ann
(z0,r,R)
and let f : ann
(z0,r,R)
→ X be analytic onann
(z0,r,R )
andcontinuous on ann
(z0,r,R)
. Then for any z ∈ann
(z0,r,R )
,
∑∞ n ∑∞ −n
f (z) = an(z − z0) + bn(z − z0) (16.4)
n=0 n=1
(16.4)
where
∫
a = -1- f (w )-----1----dw
n 2πi γR (w − z0)n+1
∫
bn = -1- f (w )(w − z0)n−1dw
2πi γr
and both of these series in 16.4converge absolutely. If r <
ˆr
<
ˆR
< R, then convergence of both series isabsolute and uniform for z ∈ann
( )
z0,ˆr,R ˆ
.
Proof:Consider the sum with the negative exponents. The other is similar. Let
|f (w)|
≤ M on the
closure of the annulus.
( ∫ )
∑∞ −n -1- n
bn (z − z0) , bn = 2πi γr f (w)(w − z0) dw
n=1
Therefore,
∥bn∥
≤ 2πrMr^{n} and
|z − z0|
≥
ˆr
> r. Thus
∑q ∑q rn
∥bn∥|z − z0|− n ≤ 2πˆrM ˆrn < ε
n=p n=p
if p is large enough. Therefore, the partial sums are a uniformly Cauchy sequence so the sum converges
absolutely and uniformly on the set
{ }
z : ˆr ≤ |z − z0| ≤ ˆR
. ■
Note that for arbitrary α with r ≤ α ≤ R,
∫ ∫
-1- f (w)-----1----dw = -1- f (w )-----1----dw (16.5)
2πi γR (w − z0)n+1 2πi γα (w − z0)n+1
(16.5)
This is a simple application of the Cauchy integral theorem, Theorem 15.4.7 applied to the union of two
simple closed curves of the sort used to prove Theorem 16.2.2. You consider the annulus ann
(z0,α,R )
and
the following diagram.
PICT
The integrand is analytic on the inside of the two simple closed curves Γ_{1} and Γ_{2}. Letting γ_{1} and γ_{2} be
oriented parametrizations for these and using the argument that the integrals over the straight lines cancel,
this yields
∫ ∫
-1- ----1----- -1- -----1----
2πi γR f (w) (w − z0)n+1dw + 2πi ˆγα f (w)(w − z0)n+1dw = 0
and let γ_{α} be a parametrization of the circle centered at z_{0} of radius α which is
counterclockwise. We have
∑∞ ∑∞
f (w ) = an (w − z0)n + bn(w − z0)− n
n=0 n=1
for w in the annulus. Then for k ≥ 1,
∞ ∞
f (w )(w− z )k−1 = ∑ a (w− z )n+k−1 + ∑ b (w − z )− n+k−1
0 n=0 n 0 n=1 n 0
By uniform convergence,
∫ k−1 ∑∞ ∫ n+k−1
γα f (w )(w − z0) dw = an γα (w − z0) dw
n=0∞ ∫
+ ∑ b (w − z)−n+k− 1dw
n=1 n γα 0
Now in the sums, all integrals are 0 except the one when n = k in the second sum. Therefore,
∫ k−1 ∫ −1
f (w)(w − z0) dw = bk (w− z0) dw = 2πibk
γα γα
This shows that for any α,r < α < R,
-1- ∫ k− 1
bk = 2πi γ f (w )(w − z0) dw
α
Similar reasoning gives
∫
-1- ----1-----
an = 2πi γα f (w )(w − z0)n+1dw
and as explained above, nothing changes when α is changed. ■
Definition 16.2.4For f continuous on the closure of an annulus as just described and analyticon the annulus, it follows that on the annulus, f can be written as the sum of a power series anda series involving