3Residue calculus
IB Complex Analysis
3.3 Cauchy’s residue theorem
We finally get to Cauchy’s residue theorem. This in some sense a mix of all the
results we’ve previously had. Simply-connected Cauchy tells us the integral of a
holomorphic
f
around a closed curve depends only on its homotopy class, i.e. we
can deform curves by homotopy and this preserves the integral. This means the
value of the integral really only depends on the “holes” enclosed by the curve.
We also had the Cauchy integral formula. This says if
f
:
B
(
a, r
)
→ C
is
holomorphic, w ∈ B(a, ρ) and ρ < r, then
f(w) =
1
2πi
Z
∂B(a,ρ)
f(z)
z −w
dz.
Note that
f
(
w
) also happens to be the residue of the function
f(z)
z−w
. So this
really says if
g
has a simple pole at
a
inside the region bounded by a simple
closed curve γ, then
1
2π
Z
γ
g(z) dz = Res(g, a).
The Cauchy’s residue theorem says the result holds for any type of singularities,
and any number of singularities.
Theorem
(Cauchy’s residue theorem)
.
Let
U
be a simply connected domain,
and
{z
1
, ··· , z
k
} ⊆ U
. Let
f
:
U \ {z
1
, ··· , z
k
} → C
be holomorphic. Let
γ
: [
a, b
]
→ U
be a piecewise
C
1
-smooth closed curve such that
z
i
6
=
image
(
γ
)
for all i. Then
1
2πi
Z
γ
f(z) dz =
k
X
j=1
I(γ, z
i
) Res(f; z
i
).
The Cauchy integral formula and simply-connected Cauchy are special cases
of this.
Proof. At each z
i
, f has a Laurent expansion
f(z) =
X
n∈Z
c
(i)
n
(z −z
i
)
n
,
valid in some neighbourhood of z
i
. Let g
i
(z) be the principal part, namely
g
i
(z) =
−1
X
n=−∞
c
(i)
n
(z −z
i
)
n
.
From the proof of the Laurent series, we know
g
i
(
z
) gives a holomorphic function
on U \ {z
i
}.
We now consider
f −g
1
−g
2
−···−g
k
, which is holomorphic on
U \{z
1
, ··· , z
k
}
,
and has a removable singularity at each z
i
. So
Z
γ
(f − g
1
− ··· − g
k
)(z) dz = 0,
by simply-connected Cauchy. Hence we know
Z
γ
f(z) dz =
k
X
j=1
Z
γ
g
j
(z) dz.
For each
j
, we use uniform convergence of the series
P
n≤−1
c
(j)
n
(
z − z
j
)
n
on
compact subsets of U \ {z
j
}, and hence on γ, to write
Z
γ
g
j
(z) dz =
X
n≤−1
c
(j)
n
Z
γ
(z −z
j
)
n
dz.
However, for
n 6
=
−
1, the function (
z − z
j
)
n
has an antiderivative, and hence
the integral around γ vanishes. So this is equal to
c
(j)
−1
Z
γ
1
z −z
j
dz.
But
c
(j)
−1
is by definition the residue of
f
at
z
j
, and the integral is just the integral
definition of the winding number (up to a factor of 2πi). So we get
Z
γ
f(z) dz = 2πi
k
X
j=1
Res(f; z
j
)I(γ, z
j
).
So done.