Cauchy's Theorem
2.7
2001/08/08
2002/07/26
Doug
Daniels
rainking@alumni.rice.edu
Steven
Cox
cox@rice.edu
Doug
Daniels
rainking@alumni.rice.edu
Mark
Barrett
markb@alumni.rice.edu
Steven
Cox
cox@rice.edu
cauchy
cauchy's
complex
curves
functions
integration
linear algebra
theorem
In this module we begin by discussing integrations of complex functions over complex curves and end with Cauchy's Theorem.
Introduction
Our main goal is a better understanding of the partial
fraction expansion of a given transfer function. With respect
to the example that closed the discussion of complex
differentiation, see this equation - In this equation, we found
z
I
B
1
z
λ
1
P
1
1
z
λ
1
2
D
1
1
z
λ
2
P
2
where the
P
j
and
D
j
enjoy the amazing properties
-
B
P
1
P
1
B
λ
1
P
1
D
1
and
B
P
2
P
2
B
λ
2
P
2
-
P
1
P
2
I
P
1
2
P
1
P
2
2
P
2
and
D
1
2
0
-
P
1
D
1
D
1
P
1
D
1
and
P
2
D
1
D
1
P
2
0
In order to show that this always
happens, i.e., that it is not a quirk
produced by the particular B in
this
equation, we require a few additional tools from the
theory of complex variables. In particular, we need the fact
that partial fraction expansions may be carried out through
complex integration.
Integration of Complex Functions Over Complex Curves
We shall be integrating complex functions over complex
curves. Such a curve is parameterized by one complex valued
or, equivalently, two real valued, function(s) of a real
parameter (typically denoted by
t
). More precisely,
C
z
t
x
t
y
t
a
t
b
For example, if
x
t
y
t
t
while
a
0
and
b
1
, then
C
is the line segment joining
0
0
to
1
.
We now define
z
C
f
z
t
a
b
f
z
t
z
t
For example, if
C
t
t
0
t
1
as above and
f
z
z
then
z
C
z
t
0
1
t
t
1
t
0
1
t
t
2
t
while if
C
is the unit circle
t
0
t
2
then
z
C
z
t
0
2
t
t
t
0
2
2
t
t
0
2
2
t
2
t
0
Remaining with the unit circle but now integrating
f
z
1
z
we find
z
C
z
t
0
2
t
t
2
We generalize this calculation to arbitrary (integer) powers
over arbitrary circles. More precisely, for integer
m
and fixed complex
a
we integrate
z
a
m
over
C
a
r
a
r
t
0
t
2
the circle of radius
r
centered at
a
. We find
z
C
a
r
z
a
m
t
0
2
a
r
t
a
m
r
t
r
m
1
t
0
2
m
1
t
z
C
a
r
z
a
m
r
m
1
t
0
2
m
1
t
m
1
t
2
m
-1
0
When integrating more general functions it is often convenient
to express the integral in terms of its real and imaginary
parts. More precisely
z
C
f
z
x
C
u
x
y
v
x
y
y
C
u
x
y
v
x
y
z
C
f
z
x
C
u
x
y
y
C
v
x
y
x
C
v
x
y
y
C
u
x
y
z
C
f
z
t
a
b
u
x
t
y
t
x
t
v
x
t
y
t
y
t
t
a
b
u
x
t
y
t
y
t
v
x
t
y
t
x
t
The second line should invoke memories of:
Green's Theorem
If
C
is a closed curve and
M
and
N
are continuously differentiable real-valued functions on
C
in
, the region enclosed by
C
, then
x
C
M
y
C
N
y
x
C
in
x
N
y
M
Applying this to the situation above, we find, so long as
C
is closed, that
z
C
f
z
y
x
C
in
x
v
y
u
y
x
C
in
x
u
y
v
At first glance it appears that Green's Theorem only serves to
muddy the waters. Recalling the Cauchy-Riemann
equations however we find that each of these double
integrals is in fact identically zero! In brief, we have
proven:
Cauchy's Theorem
If
f
is differentiable on and in the closed curve
C
then
z
C
f
z
0
.
Strictly speaking, in order to invoke Green's Theorem we
require not only that
f
be differentiable but that its derivative in fact be
continuous. This however is simply a limitation of our simple
mode of proof; Cauchy's Theorem is true as stated.
This theorem, together with , permits us to integrate every proper rational
function. More precisely, if
q
f
g
where f
is a polynomial of degree at most
m
1
and
g is an
mth degree polynomial with
h distinct zeros at
λ
j
j
1
…
h
with respective multiplicities of
m
j
j
1
…
h
we found that
q
z
j
1
h
k
1
m
j
q
j
,
k
z
λ
j
k
Observe now that if we choose
r
j
so small that
λ
j
is the only zero of g encircled by
C
j
C
λ
j
r
j
then by Cauchy's Theorem
z
C
j
q
z
k
1
m
j
q
j
,
k
z
C
j
1
z
λ
j
k
In we found that each, save
the first, of the integrals under the sum is in fact zero.
Hence,
z
C
j
q
z
2
q
j
,
1
With
q
j
,
1
in hand, say from this equation or residue, one
may view as a means for
computing the indicated integral. The opposite reading,
i.e., that the integral is a convenient
means of expressing
q
j
,
1
, will prove just as useful. With that in mind, we
note that the remaining residues may be computed as integrals
of the product of q
and the appropriate factor. More precisely,
z
C
j
q
z
z
λ
j
k
1
2
q
j
,
k
One may be led to believe that the precision of this result is
due to the very special choice of curve and function. We
shall see ...