After Cauchy's
Theorem eqn6 and Cauchy's Theorem eqn7 perhaps the most
useful consequence of Cauchy's
Theorem is the
Suppose that
C
2
C
2
is a closed curve that lies inside the region encircled by
the closed curve
C
1
C
1
. If f
is differentiable in the annular region outside
C
2
C
2
and inside
C
1
C
1
then
∫fzdz=∫fzdz
z
C
1
f
z
z
C
2
f
z
.
With reference to Figure 1 we
introduce two vertical segments and define the closed
curves
C
3
=abcda
C
3
a
b
c
d
a
(where the
bc
bc arc is clockwise and the
da da arc is counterclockwise)
and
C
4
=adcba
C
4
a
d
c
b
a
(where the
ad
ad
arc is counterclockwise and the
cb
cb
arc is clockwise). By merely following the arrows we learn that
∫fzdz=∫fzdz+∫fzdz+∫fzdz
z
C
1
f
z
z
C
2
f
z
z
C
3
f
z
z
C
4
f
z
As Cauchy's Theorem implies that
the integrals over
C
3
C
3
and
C
4
C
4
each vanish, we have our result.
This Lemma says that in order to integrate a function it
suffices to integrate it over regions where it is singular,
i.e. nondifferentiable.
Let us apply this reasoning to the integral
∫z(z−
λ
1
)(z−
λ
2
)d
z
z
C
z
z
λ
1
z
λ
2
where CC encircles both
λ
1
λ
1
and
λ
2
λ
2
as depicted in Figure 2. We find that
∫z(z−
λ
1
)(z−
λ
2
)d
z
=∫z(z−
λ
1
)(z−
λ
2
)d
z
+∫z(z−
λ
1
)(z−
λ
2
)d
z
z
C
z
z
λ
1
z
λ
2
z
C
1
z
z
λ
1
z
λ
2
z
C
2
z
z
λ
1
z
λ
2
Developing the integrand in partial fractions we find
∫z(z−
λ
1
)(z−
λ
2
)d
z
=
λ
1
λ
1
−
λ
2
∫1z−
λ
1
d
z
+
λ
2
λ
2
−
λ
1
∫1z−
λ
2
d
z
=2πi
λ
1
λ
1
−
λ
2
z
C
1
z
z
λ
1
z
λ
2
λ
1
λ
1
λ
2
z
C
1
1
z
λ
1
λ
2
λ
2
λ
1
z
C
1
1
z
λ
2
2
λ
1
λ
1
λ
2
Similarly,
∫z(z−
λ
1
)(z−
λ
2
)d
z
=2πi
λ
2
λ
2
−
λ
1
z
C
2
z
z
λ
1
z
λ
2
2
λ
2
λ
2
λ
1
Putting things back together we find
∫z(z−
λ
1
)(z−
λ
2
)d
z
=2πi(
λ
1
λ
1
−
λ
2
+
λ
2
λ
2
−
λ
1
)=2πi
z
C
z
z
λ
1
z
λ
2
2
λ
1
λ
1
λ
2
λ
2
λ
2
λ
1
2
(1)
We may view Equation 1 as a special
instance of integrating a rational function around a curve
that encircles all of the zeros of its denominator. In
particular, recalling that cauchy's theorem eqn5 and Cauchy's theorem
eqn6, we find
∫qzd
z
=∑
j
=1h∑
k
=1
m
j
∫
q
j
,
k
z−
λ
j
kd
z
=2πi∑
j
=1h
q
j
,
1
z
C
q
z
j
1
h
k
1
m
j
z
C
j
q
j
,
k
z
λ
j
k
2
j
1
h
q
j
,
1
(2)
To take a slightly more complicated example let us integrate
fzz−a
f
z
z
a
over some closed curve CC inside
of which ff is differentiable and
aa resides. Our Curve Replacement
Lemma now permits us to claim that
∫fzz−adz=∫fzz−adz
z
C
f
z
z
a
z
C
a
r
f
z
z
a
It appears that one can go no further without specifying
ff. The alert reader however
recognizes that in the integral over
Car
C
a
r
is independent of rr and so
proceeds to let
r→0
r
0
, in which case
z→a
z
a
and
fz→fa
f
z
f
a
. Computing the integral of
1z−a
1
z
a
along the way we are led to the hope that
∫fzz−adz=fa2πi
z
C
f
z
z
a
f
a
2
In support of this conclusion we note that
∫fzz−adz=∫fzz−a+faz−a−faz−adz=fa∫1z−adz+∫fz−faz−adz
z
C
a
r
f
z
z
a
z
C
a
r
f
z
z
a
f
a
z
a
f
a
z
a
f
a
z
C
a
r
1
z
a
z
C
a
r
f
z
f
a
z
a
Now the first term is
fa2πi
f
a
2
regardless of rr while, as
r→0
r
0
, the integrand of the second term approaches
dd
a
fa
a
f
a
and the region of integration approaches the point
aa. Regarding this second term,
as the integrand remains bounded as the perimeter of
Car
C
a
r
approaches zero the value of the integral must itself be zero.
This result if typically known as
If ff is differentiable on
and in the closed curve CC
then
fa=12πi∫fzz−adz
f
a
1
2
z
C
f
z
z
a
(3)
for each a lying inside
CC.
The consequences of such a formula run far and deep. We shall
delve into only one or two. First, we note that, as a does
not lie on CC, the right hand
side is a perfectly smooth function of
aa. Hence, differentiating each
side, we find
dd
a
fa=12πi∫fzz−a2dz
a
f
a
1
2
z
C
f
z
z
a
2
(4)
for each a lying inside
CC.
Applying this reasoning
nn times
we arrive at a formula for the nth derivative of
ff at
aa,
d
n
da
n
fa=n!2πi∫fzz−a1+nd
z
a
n
f
a
n
2
z
C
f
z
z
a
1
n
(5)
for each a lying inside
CC. The
upshot is that once
ff is shown
to be differentiable it must in fact be infinitely
differentiable. As a simple extension let us consider
12πi∫fz(z−
λ
1
)z−
λ
2
2d
z
1
2
z
C
f
z
z
λ
1
z
λ
2
2
where
ff is still assumed
differentiable on and in
CC and
that
CC encircles both
λ
1
λ
1
and
λ2
λ2
. By the curve replacement
lemma this integral is the sum
12πi∫fz(z−
λ
1
)z−
λ
2
2d
z
+12πi∫fz(z−
λ
1
)z−
λ
2
2d
z
1
2
z
C
1
f
z
z
λ
1
z
λ
2
2
1
2
z
C
2
f
z
z
λ
1
z
λ
2
2
where
λ
j
λ
j
now lies in only
C
j
C
j
. As
fzz−
λ
2
f
z
z
λ
2
is well behaved in
C
1
C
1
we may use
Equation 3 to
conclude that
12πi∫fz(z−
λ
1
)z−
λ
2
2d
z
=f
λ
1
λ
1
−
λ
2
2
1
2
z
C
1
f
z
z
λ
1
z
λ
2
2
f
λ
1
λ
1
λ
2
2
Similarly, as
fzz−
λ
1
f
z
z
λ
1
is well behaved in
C
2
C
2
we may use
Equation 4 to conclude that
12πi∫fz(z−
λ
1
)z−
λ
2
2d
z
=dd
a
faa−
λ
1

a
=
λ2
1
2
z
C
2
f
z
z
λ
1
z
λ
2
2
a
λ2
a
f
a
a
λ
1
These calculations can be read as a concrete instance of
If gg is a polynomial with roots
λ
j
j=1…h
λ
j
j
1
…
h
of degree
d
j
j=1…h
d
j
j
1
…
h
and CC is a closed curve
encircling each of the
λ
j
λ
j
and ff is differentiable on
and in CC then
∫fzgzd
z
=2πi∑
j
=1hres
λ
j
z
C
f
z
g
z
2
j
1
h
res
λ
j
where
res
λ
j
=limit
z
→
λ
j
1(
d
j
−1)!d
d
j
−1
z−
λ
j
d
j
fzgzd
z
d
j
−1
res
λ
j
z
λ
j
1
d
j
1
z
d
j
1
z
λ
j
d
j
f
z
g
z
is called the residue of
fg
f
g
at
λ
j
λ
j
.
One of the most important instances of
this theorem is the formula for the Inverse Laplace Transform.