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
zI−B-1=1z−
λ
1
P
1
+1z−
λ
1
2
D
1
+1z−
λ
2
P
2
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
P
j
and
D
j
D
j
enjoy the amazing properties
-
B
P
1
=
P
1
B=
λ
1
P
1
+
D
1
B
P
1
P
1
B
λ
1
P
1
D
1
(1)
and
B
P
2
=
P
2
B=
λ
2
P
2
B
P
2
P
2
B
λ
2
P
2
-
P
1
+
P
2
=I
P
1
P
2
I
(2)
P
1
2=
P
1
P
1
2
P
1
P
2
2=
P
2
P
2
2
P
2
and
D
1
2=0
D
1
2
0
-
P
1
D
1
=
D
1
P
1
=
D
1
P
1
D
1
D
1
P
1
D
1
(3)
and
P
2
D
1
=
D
1
P
2
=0
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 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.
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
t). More precisely,
C≡{zt=xt+ⅈyt|a≤t≤b}
C
z
t
x
t
y
t
a
t
b
For example, if
xt=yt=t
x
t
y
t
t
while
a=0
a
0
and
b=1
b
1
, then
C
C
is the line segment joining
0+ⅈ0
0
0
to
1+ⅈ
1
.
We now define
∫Cfzdz≡∫abfzt
z
′tdt
z
C
f
z
t
a
b
f
z
t
z
t
For example, if
C={t+ⅈt|0≤t≤1}
C
t
t
0
t
1
as above and
fz=z
f
z
z
then
∫Czdz=∫01t+ⅈt1+ⅈdt=∫01t−t+ⅈ2tdt=ⅈ
z
C
z
t
0
1
t
t
1
t
0
1
t
t
2
t
while if
C
C
is the unit circle
{ⅇⅈt|0≤t≤2π}
t
0
t
2
then
∫Czdz=∫02πⅇⅈtⅈⅇⅈtdt=ⅈ∫02πⅇⅈ2tdt=ⅈ∫02πcos2t+ⅈsin2tdt=0
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
fz=1z
f
z
1
z
we find
∫Cz-1dz=∫02πⅇ-ⅈtⅈⅇⅈtdt=2πⅈ
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
m
and fixed complex
a
a
we integrate
z−am
z
a
m
over
Car≡{a+rⅇⅈt|0≤t≤2π}
C
a
r
a
r
t
0
t
2
the circle of radius
r
r
centered at
a
a. We find
∫Carz−amdz=∫02πa+rⅇⅈt−amrⅈⅇⅈtdt=ⅈrm+1∫02πⅇⅈm+1tdt
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
(4)
∫Carz−amdz=ⅈrm+1∫02πcosm+1t+ⅈsinm+1tdt=2πⅈifm=-10otherwise
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
∫Cfzdz=∫Cuxy+ⅈvxydx+ⅈ∫Cuxy+ⅈvxydy
z
C
f
z
x
C
u
x
y
v
x
y
y
C
u
x
y
v
x
y
∫Cfzdz=∫Cuxydx−∫Cvxydy+ⅈ∫Cvxydx+ⅈ∫Cuxydy
z
C
f
z
x
C
u
x
y
y
C
v
x
y
x
C
v
x
y
y
C
u
x
y
∫Cfzdz=∫abuxtyt
x
′t−vxtyt
y
′tdt+ⅈ∫abuxtyt
y
′t+vxtyt
x
′tdt
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:
If
C
C
is a closed curve and
M
M
and
N
N
are continuously differentiable real-valued functions on
C
in
C
in
, the region enclosed by
C
C, then
∫CMdx+∫CNdy=∫∫
C
in
∂∂xN−∂∂yMdxdy
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
C is closed, that
∫Cfzdz=-∫∫
C
in
∂∂xv+∂∂yudxdy+ⅈ∫∫
C
in
∂∂xu+∂∂yvdxdy
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:
If
f
f
is differentiable on and in the closed curve
C
C then
∫Cfzdz=0
z
C
f
z
0
.
Strictly speaking, in order to invoke Green's Theorem we
require not only that
f
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 Equation 4, permits us to integrate every proper rational
function. More precisely, if
q=fg
q
f
g
where f
f is a polynomial of degree at most
m−1
m
1
and
gg is an
mmth degree polynomial with
hh distinct zeros at
{
λ
j
|j=1…h}
λ
j
j
1
…
h
with respective multiplicities of
{
m
j
|j=1…h}
m
j
j
1
…
h
we found that
qz=∑j=1h∑k=1
m
j
q
j
,
k
z−
λ
j
k
q
z
j
1
h
k
1
m
j
q
j
,
k
z
λ
j
k
(5)
Observe now that if we choose
r
j
r
j
so small that
λ
j
λ
j
is the only zero of
gg encircled by
C
j
≡C
λ
j
r
j
C
j
C
λ
j
r
j
then by Cauchy's Theorem
∫
C
j
qzdz=∑k=1
m
j
q
j
,
k
∫
C
j
1z−
λ
j
kdz
z
C
j
q
z
k
1
m
j
q
j
,
k
z
C
j
1
z
λ
j
k
In
Equation 4 we found that each, save
the first, of the integrals under the sum is in fact zero.
Hence,
∫
C
j
qzdz=2πⅈ
q
j
,
1
z
C
j
q
z
2
q
j
,
1
(6)
With
q
j
,
1
q
j
,
1
in hand, say from
this equation or
residue, one
may view
Equation 6 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
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
q and the appropriate factor. More precisely,
∫
C
j
qzz−
λ
j
k−1dz=2πⅈ
q
j
,
k
z
C
j
q
z
z
λ
j
k
1
2
q
j
,
k
(7)
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 ...
