PARTIAL FRACTION DECOMPOSITION: AHLFORS’
ALGORITHM
ALBERT MARDEN
1. Polynomials
A complex polynomial of degree n has the form
P (z) = an z n + an−1 z n−1 + · · · + a0 ,
an 6= 0.
A polynomial of degree n has exactly complex zeros (roots), counting multiplicity. Thus the zero of P (z) = z n has multiplexity n.
In principle, a polynomial can be completely factored into its roots:
P (z) = c(z − ζ1 )k1 (z − ζ2 )k2 · · · (z − ζr )kr ,
c 6= 0.
Here ζi is a zero and ki is its multiplicity. Since the polynomial has degree
n it must be that
r
X
n=
ki = k1 + k2 + · · · kr .
i=1
2. Rational functions
A rational function is the quotient of two polynomials:
an z n + an−1 z n−1 + · · · + a0
P (z)
=
, an , bm 6= 0.
Q(z)
bm z m + bm−1 z m−1 + · · · + b0
It is always assumed the the numerator P (z) and the denominator Q(z) have
no common zero—otherwise by cancellation the degrees of the numerator
and denominator can be reduced.
Here are the basic properties of rational functions:
• The finite zeros of R(z) are the zeros of P (z).
• The finite poles of R(z) are the zeros of Q(z).
• If R(∞) 6= 0, ∞ than m = n.
• If R(∞) = ∞ then m < n and R has a pole of multiplicity n − m at
∞.
• If R(∞) = 0 then m > n and R has a zero of multiplicity m − n at
∞.
• The number of zeros of R equals the number of its poles equals the
number of solutions of R(z) = a, any a ∈ C ∪ {∞}. This number,
p =max (m,n), is called the order of R.
R(z) =
Date: Written for my class Math 5583, October 1, 2012.
1
2
ALBERT MARDEN
3. The algorithm
3.1. Step 1. If deg(P ) >deg(Q), divide Q into P resulting in
(1)
R(z) = G(z) + H(z),
G(0) = 0.
We do the division so that G(z) is a polynomial without constant term. For
example,
z2 + 1
R(z) = 2
= G(z) + H(z), G(z) = 0,
z −1
as compared to
z3 + 1
z+1
=z+ 2
,
2
z −1
z −1
z4 + 7
z2 + 7
2
=
z
+
.
z2 − 1
z2 − 1
The degree of G is the multiplicity of the pole of R at ∞. And H(z) is a
rational function with H(∞) 6= ∞.
3.2. Step 2. Denote the distinct finite poles of R by b1 , b2 , . . . bq .
For each index j, 1 ≤ j ≤ q. set
1
z = bj + ,
ζ
(2)
1
Rj (ζ) = R(bj + ).
ζ
For these rational functions, Rj (∞) = R(bj ) = ∞.
Repeat Step 1) with Rj (ζ):
(3)
Rj (ζ) = Gj (ζ) + Hj (ζ).
Here Gj (ζ) is a polynomial in ζ without constant term so Gj (0) = 0. The
degree of Gj (ζ) is the multiplicity of the pole of Rj at ζ = ∞, And Hj (ζ) is
a rational function with Hj (∞) finite.
1
Return now to the original notation ζ = z−b
to get
j
(4)
R(z) = Gj (
1
1
) + Hj (
).
z − bj
z − bj
1
Note that when z = ∞ (ζ = 0) Gj is zero so that Gj is a polynomial in z−b
j
without a constant term. Moreover when ζ = ∞ (z = bj ), Hj (∞) 6= ∞.
3.3. Step 3. Consider the new rational function
(5)
∗
R (z) = R(z) − G(z) −
q
X
j=1
We will show that
(a) R∗ (bj ) 6= ∞, 1 ≤ j ≤ q.
(b) R∗ (∞) 6= ∞.
(c) R∗ (z) is a constant.
Gj (
1
).
z − bj
PARTIAL FRACTION DECOMPOSITION: AHLFORS’ ALGORITHM
3
Item (a) is true because from Equation (5)
q
X
1
1
1
∗
(6) lim R (z) = lim [Gj (
) + Hj (
)] − G(z) −
Gi (
)
z→bj
z→bj
z − bj
z − bj
z − bi
i=1
= lim Hj (
(7)
z→bj
X
1
1
) − G(z) −
Gi (
z − bj
z − bi
6= ∞.
i6=j
To give Equation (7), the two Gj terms in Equation (6) cancel.
Item (b) is true because from Equations (5,1)
q
X
(8)
lim R∗ (z) = lim [G(z) + H(z)] − G(z) −
Gj (
z→∞
z→∞
j=1
= lim H(z) −
z→∞
q
X
1
)
z − bj
Gj (0) = H(∞) 6= ∞.
j=1
4. The bottom line
Item (c): So Item (a) shows that R∗ has no poles at the bj ’s. Item (b)
shows that R∗ has no pole at ∞. What no poles? Yes, so R∗ (z) can only
be a constant c. Returning to the definition of R∗ (z) in Equation (5), this
formula becomes
q
X
1
(9)
R(z) = [G(z) + c] +
),
Gj (
z − bj
j=1
the last sum has no constant term. The first bracket encloses a polynomial
(possibly 0) with constant term c. The second bracket encloses what is
called the singular part of R(z).
To complete the picture,
(10)
Degree of polyn Gj (ζ), namely Gj (
1
), = Mulplicity of pole of R(z) at z = bj .
z − bj
For example
z2 + 1
R(z) = 3
=
z (z − 1)(z − 2)2
A B
C
+ 2+ 3
z
z
z
D
+
+
z−1
E
F
+
z − 2 (z − 2)2
To prove Equation (10), multiply both sides by (z − bj )k , where k is the
multiplicity of bj . On the left limz→bj R(z)(z − bj )k is finite and 6= 0, since
(z − bj )k cancels against this factor of the denometer. On the right side,
1
limz→bk (z − bj )k Gj ( z−b
) must likewise be 6= 0, ∞. That is, Gj (·) also has
j
degree exactly k.