In terms of the real and imaginary parts of
ff this result brings the
Cauchy-Riemann equations.
∂∂xu=∂∂yv
x
u
y
v
(7)
and
∂∂xv=-∂∂yu
x
v
y
u
(8)
Regarding the converse proposition we note that when
ff has continuous partial
derivatives in region obeying the Cauchy-Reimann equations
then
ff is in fact differentiable
in the region.
We remark that with no more energy than that expended on their
real cousins one may uncover the rules for differentiating
complex sums, products, quotients, and compositions.
As one important application of the derivative let us attempt
to expand in partial fractions a rational function whose
denominator has a root with degree larger than one. As a
warm-up let us try to find
q
1
,
1
q
1
,
1
and
q
1
,
2
q
1
,
2
in the expression
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
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
(9)
On setting
z=-1
z
-1
this gives
q
1
,
2
=1
q
1
,
2
1
.
With
q
1
,
2
q
1
,
2
computed,
Equation 9 takes the simple form
z+1=
q
1
,
1
z+1
z
1
q
1
,
1
z
1
and so
q
1
,
2
=1
q
1
,
2
1
as well. Hence,
z+2z+12=1z+11z+12
z
2
z
1
2
1
z
1
1
z
1
2
This latter step grows more cumbersome for roots of higher
degrees. 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
The first step is still correct: multiply through by the
factor at its highest degree, here
33. 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
(10)
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
Equation 10 is to hold for
all
zz. We exploit this by taking
two derivatives, with respect to
zz, of
Equation 10. 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
and
2=
q
1
,
1
2
q
1
,
1
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
the
partial fraction expansion of the
general proper rational function
q=fg
q
f
g
where
gg has
hh distinct roots
λ1…λh
λ1
…
λh
of respective degrees
d1…dh
d1
…
dh
.
We write
qz=∑j=1h∑k=1dj
q
j
,
k
z-λjk
q
z
j
1
h
k
1
dj
q
j
,
k
z
λj
k
(11)
and note, as above, that
q
j
,
k
q
j
,
k
is the coefficient of
z-djdj-k
z
dj
dj
k
in the rational function
rjz≡qzz-λjdj
rj
z
q
z
z
λj
dj
Hence,
q
j
,
k
q
j
,
k
may be computed by setting
z=λj
z
λj
in the ratio of the
dj-k
dj
k
th
derivative of
rjrj to
dj-k!
dj
k
.
That is,
q
j
,
k
=limz→λj1dj-k!ddj-kdzdj-kz-λjdjqz
q
j
,
k
z
λj
1
dj
k
d
dj
k
d
z
dj
k
z
λj
dj
q
z
(12)
As a second example, let us take
B=100130011
B
1
0
0
1
3
0
0
1
1
(13)
and compute the
Φ
j
,
k
Φ
j
,
k
matrices in the expansion
zI-B-1=1z-1001z-1z-31z-301z-12z-31z-1z-31z-1=1z-1
Φ
1
,
1
+1z-12
Φ
1
,
2
+1z-3
Φ
2
,
1
z
I
B
-1
1
z
1
0
0
1
z
1
z
3
1
z
3
0
1
z
1
2
z
3
1
z
1
z
3
1
z
1
1
z
1
Φ
1
,
1
1
z
1
2
Φ
1
,
2
1
z
3
Φ
2
,
1
The only challenging term is the
3
1
3
1
element. We write
1z-12z-3=
q
1
,
1
z-1+
q
1
,
2
z-12+
q
2
,
1
z-3
1
z
1
2
z
3
q
1
,
1
z
1
q
1
,
2
z
1
2
q
2
,
1
z
3
It follows from
Equation 12 that
q
1
,
1
=ddz1z-31=-1/4
q
1
,
1
z
1
z
3
1
14
(14)
and
q
1
,
2
=1z-31=-1/4
q
1
,
2
1
z
3
1
14
(15)
and
q
2
,
1
=1z-323=1/4
q
2
,
1
1
z
3
2
3
14
(16)
It now follows that
zI-B-1=1z-1100-1/200-1/4-1/21+1z-12000000-1/200+1z-30001/2101/41/20
z
I
B
-1
1
z
1
1
0
0
-12
0
0
-14
-12
1
1
z
1
2
0
0
0
0
0
0
-12
0
0
1
z
3
0
0
0
12
1
0
14
12
0
(17)
In closing, let us remark that the method of partial fraction
expansions has been implemented in Matlab. In fact,
Equation 14,
Equation 15, and
Equation 16 all
follow from the single command:
[r,p,k]=residue([0 0 0 1],[1 -5 7 -3]).
The first input argument is Matlab-speak for the polynomial
fz=1
f
z
1
while the second argument corresponds to the denominator
gz=z-12z-3=z3-5z2+7z-3
g
z
z
1
2
z
3
z
3
5
z
2
7
z
3
.
