Ullrich Chapter 4, Part II (through the proof of Thm. 4.11)
Math 325
Remarks
Now comes the first big theorem in one of its final forms: Cauchy’s Integral Formula (homology version),
which replaces the condition of convexity of the domain with a much weaker hypothesis:
Theorem 4.9 (CIFH1 ): If V ⊂ C is open and Γ is a cycle in V so that for all a ∈ C r V , Ind(Γ, a) = 0,
Z
1
f (w)
∗
dw = Ind(Γ, z)f (z) .
then for all f ∈ H(V ) and all z ∈ V r Γ ,
2πi Γ w − z
Note a number of significant improvements from our previous version of the CIF:
Our domain V need not be convex—the index hypothesis says that the shape of V doesn’t matter
as long as Γ doesn’t [algebraically] wrap all the way around any point in the complement of V .
The integral is around any cycle Γ in V , rather than just boundaries of circles or disks—an
important fact about those was that they wrapped around the point z exactly once in a counterclockwise direction, which is what the Ind(Γ, z) factor handles (it will be one for our boundaries
of circles or disks).
We now just require z ∈ V r Γ∗ —the points at which this works no longer need to be “inside”
the boundary of the disk as they did before. (This will be crucial when we later come to our
result about Laurent series on annuli, the proof of which will directly mirror the proof for disks.)
An immediate consequence will be Cauchy’s Theorem (homology version):
Thm. 4.10 (CauchyH1 ): If V ⊂ C is open and ΓZis a cycle in V so that for all a ∈ C r Γ∗ , Ind(Γ, a) = 0,
then for all f ∈ H(V ),
f (z) dz = 0.
Γ
Note that this answers our question of when we can guarantee for a given chain Γ in V that the integral
of every f ∈ H(V ) along Γ will be zero.
The proof of CIFH1 is involved; supposing the hypotheses:

 f (z) − f (w)
for z 6= w, and
Define g : V × V → C by g(z, w) =
z−w
f 0 (z) for z = w.
[Lemma 4.8]
Then g is continuous on V × V and for each w ∈ V , gw ∈ H(V ) (where gw (z) = g(z, w)).
Setting Ω = {z ∈ C r Γ∗ : Ind(Γ, z) = 0} (i.e., the points of C algebraically “outside” Γ), we
decompose C into two open sets: C = V ∪ Ω; we’ll define analytic functions on each of these domains
separately:
Z
1
Define FV : V → C by FV (z) =
g(z, w) dw. A bit of technical work involving continu2πi Γ
ity and Morera’s Theorem shows us that FV ∈ H(V ).
Z
1
f (w)
Define FΩ : Ω → C by FΩ (z) =
dw; a similar argument shows that FΩ ∈ H(Ω).
2πi Γ w − z
A quick but crucial computation shows that the functions FV and FΩ agree on the set V ∩ Ω, so that
together they give a function F ∈ H(C) (even though our original f was only defined on V !).
Finally, we note that limz→∞ F (z) = 0, meaning that F is bounded on C and thus constant! From this,
another quick computation yields the result.
Remarks, cont’d
For c ∈ C and 0 ≤ r < R ≤ ∞, we define the annulus around a with inner radius r and outer radius R by:
A(a, r, R) = {z ∈ C : r < |z − a| < R}.
Functions analytic on annuli can be represented by Laurent series there:
∞
X
Theorem 4.11: If f ∈ H(A(a, r, R)), then f (z) =
cn (z − a)n for all z ∈ A(a, r, R) ,
n=−∞
Z
1
f (w)
dw.
2πi ∂D(a,ρ) (w − a)n+1
The proof of this result is a clever twist on the proof of existence for power series:
where cn =
First, consider the cycle Γ = ∂D(a, R0 ) − ∂D(a, r0 ) for some r < r0 < R0 < R; because these circular
paths go in opposite directions, we can apply our new Cauchy Integral Formula.
In each integral, rewrite
1
w−z
as a power series (in
z−a
w−a
for the first, as usual, and in
w−a
z−a
A bit of algebra and switching limits, as in the power series proof, gets us the result!
Problem assignments 4.1–4.3
for the second).