With the tools of
complex functions
we may produce complex polynomials:
fz=zm+
c
m
-
1
zm−1+…+
c
1
z+
c
0
f
z
z
m
c
m
-
1
z
m
1
…
c
1
z
c
0
(1)
We say that such an
ff is of
order mm.
We shall often find it convenient to represent polynomials as
the product of their factors, namely:
fz=z−
λ
1
d
1
z−
λ
2
d
2
…z−
λ
h
d
h
f
z
z
λ
1
d
1
z
λ
2
d
2
…
z
λ
h
d
h
(2)
Each
λ
j
λ
j
is a
root of
degree
d
j
d
j
. Here
hh is the number of
distinct roots of
ff. We call
λ
j
λ
j
a
simple root when
d
j
=1
d
j
1
. We can observe the appearance of ratios of
polynomials or so called
rational functions.
Suppose
qz=fzgz
q
z
f
z
g
z
is rational, that
f
f
is of order at most
m−1
m
1
while
g
g
is of order
m
m
with the simple roots
λ
1
…
λ
m
λ
1
…
λ
m
.
It should come as no surprise that such a
q
q
should admit a Partial Fraction Expansion
qz=∑j=1m
q
k
z−
λ
j
q
z
j
1
m
q
k
z
λ
j
(3)
One uncovers the
q
j
q
j
by first multiplying each side by
z−
λ
j
z
λ
j
and then letting
z
z
tend to
λ
j
λ
j
. For example, if
1z2+1=
q
1
z+i+
q
2
z−i
1
z
2
1
q
1
z
i
q
2
z
i
(4)
then multiplying each side by
z+i
z
i
produces
1z−i=
q
1
+
q
2
z+iz−i
1
z
i
q
1
q
2
z
i
z
i
(5)
Now, in order to isolate
q
1
q
1
it is clear that we should set
z=-i
z
i
.
So doing we find
q
1
=i2
q
1
i
2
.
In order to find
q
1
q
1
we multiply
the above
equality by
z−i
z
i
and then set
z=i
z
i
. So doing we find
q
2
=-i2
q
2
i
2
, and so
1z2+1=i2z+i+-i2z−i
1
z
2
1
i
2
z
i
i
2
z
i
(6)
Returning to the general case, we encode the above in the
simple formula
q
j
=limz→
λ
j
z−
λ
j
qz
q
j
z
λ
j
z
λ
j
q
z
(7)
You should be able to use this to confirm that
zz+12=1/2z+1+1/2z−1
z
z
1
2
12
z
1
12
z
1
(8)
q
1
,
1
q
1
,
1
and
q
1
,
2
q
1
,
2
in the expansion
z+2z+12=
q
1
,
1
z+1+
q
1
,
2
z+12
z
2
z
1
2
q
1
,
1
z
1
q
1
,
2
z
1
2
(9)
Arguing as above it seems wise to multiply through by
z+12
z
1
2
and so arrive at
z+2=
q
1
,
1
z+1+
q
1
,
2
z
2
q
1
,
1
z
1
q
1
,
2
(10)
On setting
z=-1
z
-1
this gives
q
1
,
2
=1
q
1
,
2
1
.
With
q
1
,
2
q
1
,
2
computed the previous equation takes the simple form
z+1=
q
1
,
1
z+1
z
1
q
1
,
1
z
1
(11)
and so
q
1
,
1
=1
q
1
,
1
1
as well. Hence:
z+2z+12=1z+1+1z+12
z
2
z
1
2
1
z
1
1
z
1
2
(12)
This latter step grows more cumbersome for roots of higher
degree. Let us consider
z+22z+13=
q
1
,
1
z+1+
q
1
,
2
z+12+
q
1
,
3
z+13
z
2
2
z
1
3
q
1
,
1
z
1
q
1
,
2
z
1
2
q
1
,
3
z
1
3
(13)
The first step is still correct: multiply through by the factor
at its highest degree, here
3
3. This leaves us with
z+22=
q
1
,
1
z+12+
q
1
,
2
z+1+
q
1
,
3
z
2
2
q
1
,
1
z
1
2
q
1
,
2
z
1
q
1
,
3
(14)
Setting
z=-1
z
-1
again produces the last coefficient, here
q
1
,
3
=1
q
1
,
3
1
. We are left however with one equation in two
unknowns. Well, not really one equation, for the previous
equation is to hold for
all
z
z.
We exploit this by taking two derivatives, with respect to
z
z, of this equation. This produces
2z+2=2
q
1
,
1
z+1+
q
1
,
2
2
z
2
2
q
1
,
1
z
1
q
1
,
2
(15)
and
2=
q
1
,
1
2
q
1
,
1
(16)
The latter of course needs no comment. We derive
q
1
,
2
q
1
,
2
from the former by setting
z=-1
z
-1
. This example will permit us to derive a simple
expression for a partial fraction expansion.
- Definition 1: Partial Fraction Expansion
The partial
fraction expansion of a general proper rational function
q=fg
q
f
g
where
g
g
has
h
h
distinct roots
λ
1
…
λ
h
λ
1
…
λ
h
of respective degrees
d
1
…
d
h
d
1
…
d
h
can be written as
qz=∑j=1h∑k=1
d
j
q
j
,
k
z−
λ
j
k
q
z
j
1
h
k
1
d
j
q
j
,
k
z
λ
j
k
and it can be noted, as above, that
q
j
,
k
q
j
,
k
is the coefficient of
z−
λ
j
d
j
−k
z
λ
j
d
j
k
in the rational function
r
j
z≡qzz−
λ
j
d
j
r
j
z
q
z
z
λ
j
d
j
. Hence,
q
j
,
k
q
j
,
k
may be computed by setting
z=
λ
j
z
λ
j
in the ratio of the
d
j
−k
d
j
k
th derivative of
r
j
r
j
to
d
j
−k!
d
j
k
. That is,
q
j
,
k
=limz→
λ
j
1
d
j
−k!d
d
j
−kdz
d
j
−kz−
λ
j
d
j
qz
q
j
,
k
z
λ
j
1
d
j
k
z
d
j
k
z
λ
j
d
j
q
z
.
With respect to the previous definition observe that if we choose
r
j
r
j
so small that
λ
j
λ
j
is the only zero of
g
g
encircled by
R
j
≡C
λ
j
r
j
R
j
C
λ
j
r
j
then by Cauchy's Theorem
