Partial Fraction Expansion All the polynomials in this note are assumed to be complex polynomials

1. Partial Fraction Expansion

All the polynomials in this note are assumed to be complex polynomials.

A rational function R(z) = P (z)/Q(z) is a quotient of two polynomials P, Q with Q(z) 6=

0. By Fundamental Theorem of algebra,

Q(z) = q(z − a1)^m1· · · (z − a_r)^mr, where a₁, · · · , a_r are distinct complex numbers and m_i ≥ 1.

Proposition 1.1. Suppose Q(z) = Q1(z)Q2(z) is a product of polynomials such that Q₁(z), Q₂(z) are relatively prime, i.e. the greatest common divisor of Q₁(z), Q₂(z) is 1.

Then there exist P1(z), P2(z) so that P (z)

Q(z) = P₁(z)

Q1(z)+ P₂(z) Q2(z).

Proof. Since Q₁(z) and Q₂(z) are relatively prime, using division algorithm, we can find u₁(z), u₂(z) so that Q₁(z)u₁(z) + Q₂(z)u₂(z) = 1. Then

P (z)

Q(z) = 1 · P (z) Q(z)

= Q₁(z)u₁(z)P (z) + Q₂(z)u₂(z)P (z) Q₁(z)Q₂(z)

= u₂(z)P (z)

Q₁(z) +u₁(z)P (z) Q₂(z)

= P₁(z)

Q₁(z) + P₂(z) Q₂(z),

where P₁(z) = u₂(z)P (z) and P₂(z) = u₁(z)P (z). We complete the proof.  Corollary 1.1. Assume Q(z) = (z − a1)^m1· · · (z − a_r)^mr where a1, · · · , ar are distinct.

Then

(1.1) P (z)

Q(z) = P1(z)

(z − a₁)^m1 + · · · + P2(z) (z − a_r)^mr.

Proof. We will prove the statement by induction on deg Q. Let Q1(z) = (z − a1)^m1 and Q2(z) = (z − a2)^m2· · · (z − a_r)^mr. By Lemma 1.1, we can find P1(z) and P₁⁰(z) so that

P (z)

Q(z) = P₁(z)

(z − a1)^m1 + P₁⁰(z) Q2(z). Since deg Q₂ < deg Q, by induction hypothesis, we can write

P₁⁰(z)

Q2(z) = P₂(z)

(z − a2)^m2 + · · · + P_r(z) (z − ar)^mr

for some polynomials P₂(z), · · · , P_r(z). This implies (1.1).  Let us study the case P (z)/(z − a)^m first. We may assume that deg P ≤ m − 1. Using division algorithm, we can find a₀, · · · , a_m so that

(1.2) P (z) = a_m+ a_m−1(z − a) + · · · + a₁(z − a)^m−1. Then we obtain that

P (z)

Q(z) = a₁

z − a+ a₂

(z − a)² + · · · + a_m (z − a)^m.

1

In reality, how do we compute a0, · · · , am? In fact, a_i= 1

(i − 1)!

dⁱ⁻¹ dzⁱ⁻¹

P (z)

Q(z)(z − a)^m    z=a

.

Or, we can compute these numbers a1, · · · , am using residues. First, we observe that 1

2πi I

Cr(z)

(z − a)^m−1P (z)

Q(z)dz = am

where r is small. Denote

R1(z) = P (z)

Q(z) − am

(z − a)^m = P (z) − am

(z − a)^m . By (1.2),

P (z) − a_m = a_m−1(z − a) + · · · + a₁(z − a)^m−1. Hence we find

R1(z) = am−1+ am−2(z − a) + · · · + a1(z − a)^m−2

(z − a)^m−1 .

Denote P1(z) = am−1+ am−2(z − a) + · · · + a1(z − a)^m−2 and Q1(z) = (z − a)^m−1. Then we have

P1(z)

Q₁(z) = a1

z − a+ a2

(z − a)² + · · · + am−1

(z − a)^m−1.

Similarly, we can compute am−1 using residue. By induction, we can compute all ai. In general, we consider P (z)/Q(z) of the form (1.1). Let f₁(z) be the rational function

f₁(z) = P₂(z)

(z − a₂)^m2 + · · · + P_r(z) (z − a_r)^mr. Then we can write

P (z)

Q(z) = a¹₁ z − a1

+ a¹₂

(z − a1)² + · · · + a¹_m

(z − a1)^m1 + f₁(z).

Notice that a₁ is not a pole of f₁(z). For r small enough, f (z) is holomorphic on D_r(a₁).

For 0 ≤ i ≤ m − 1, one has

z→alim1

1 i!

dⁱ⁻¹

dzⁱ⁻¹f₁(z)(z − a₁)^m1 = 0.

Using the previous observation.

z→alim1

1 i!

dⁱ⁻¹ dzⁱ⁻¹

P (z)

Q(z)(z − a1)^m1 = a¹_i. Similarly, for 1 ≤ s ≤ r, one has

z→alims

1 i!

dⁱ⁻¹ dzⁱ⁻¹

P (z)

Q(z)(z − as)^ms = a^s_i. Here we write

R(z) =

ms

X

i=1

a^s_i

(z − as)ⁱ + f_s(z), f_s(z) =X

i6=s

P_i(z) (z − ai)^mi.

Or using the residue computation, for r small enough so that D_r(a_i) excludes all a_j for j 6= i. Then

1 2πi

I

Cr(ai)

P (z)

Q(z)(z − ai)^mi−1dz = aⁱ_mi.

If we set Hi(z) =Q

j6=i(z − aj)^mj, then P (z)

Q(z)(z − ai)^mi−1 = 1 z − a_i

P (z) H_i(z).

We know P (z)/H_i(z) is holomorphic on D_r(a). By Cauchy integral formula, we obtain 1

2πi I

Cr(a)

P (z)/H_i(z)

z − a_i dz = P (a_i) H_i(a_i).

This gives aⁱ_mi = P (a_i)/H_i(a_i). Now, subtract R(z) from aⁱ_mi/(z − a_i)^mi, we obtain R₁(z) = R(z) − aⁱ_mi

(z − a_i)^mi =

mi−1

X

j=1

aⁱ_j

(z − a_i)^j + f_i(z).

Inductively, we can find all aⁱ_s for 1 ≤ s ≤ m_i.

Example 1.1. Find the partial fraction expansion of R(z) = 3z²+ z + 1

(z − 1)²(z − 2)³. Let f1(z) and f2(z) be rational functions

f1(z) = A

z − 1+ B

(z − 1)², f2(z) = C

z − 2+ D

(z − 2)² + E (z − 2)³ so that R(z) = f₁(z) + f₂(z). Multiplying R(z) by z − 1, we find

R(z)(z − 1) = 1

z − 1·3z²+ z + 1

(z − 2)³ = A + B

z − 1+ f₂(z)(z − 1).

Let r be small enough so that 2 does not belong to Dr(1) and hence f2(z) is holomorphic on D_r(1). By Cauchy-integral formula,

B = 1 2πi

I

Cr(1)

R(z)(z − 1)dz = 3 · 1²+ 1 + 1 (1 − 2)³ = −5.

Subtracting R(z) from −5/(z − 2)², we obtain R(z) − −5

(z − 2)² = A

z − 1+ f2(z).

By simple calculation, R(z) − −5

(z − 2)² = 3z²+ z + 1 + 5(z − 2)³

(z − 1)²(z − 2)³ = 5z³− 27z²+ 61z − 39

(z − 1)²(z − 2)³ = 5z²− 22z + 39 (z − 1)(z − 3)². By Cauchy integral formula and by the fact that f₂(z) is holomorphic on D_r(1),

A = 1 2πi

I

Cr(1)

5z²− 22z + 39

(z − 1)(z − 2)³dz = 5 · 1 − 22 · 1 + 39

(1 − 2)³ = −22.

On the other hand, f₂(z) =



R(z) − −5 (z − 1)²



− −22

z − 1 = 5z²− 22z + 39

(z − 1)(z − 2)³ − −22 z − 1. Now, we know

f₂(z) = 5z²− 22z + 39

(z − 1)(z − 2)³ − −22

z − 1 = 22z²− 105z + 137 (z − 2)³ .

Denote P (z) = 22z²− 105z + 137. Note that P (2) = 15, P⁰(2) = −17 and P⁰⁰(2)/2! = 22.

We have (Taylor expansion of P (z) at z = 2)

P (z) = 15 − 17(z − 2) + 22(z − 2)². Hence we find

f₂(z) = P (z)

(z − 2)³ = 22

z − 2+ −17

(z − 2)² + 15 (z − 2)³. We conclude that

3z²+ z + 1

(z − 1)²(z − 2)³ = −22

z − 1+ −5 (z − 1)²

 +

 22

z − 2+ −17

(z − 2)² + 15 (z − 2)³



Proposition 1.2. (Special Case) Let a₁, · · · , a_r be distinct complex numbers and Q(z) = (z − a₁) · · · (z − a_r). Then

P (z) Q(z) =

r

X

i=1

P (a_i)/Q⁰(a_i) z − ai

. Proof. Let us assume that

P (z)

Q(z) = c₁

z − a₁ + · · · + c_r z − a_r. Then we know

(z − ai)P (z) Q(z) = c1

z − ai

z − a₁ + · · · + ci−1

z − ai

z − a_i−1 + ci+ ci+1

z − ai

z − a_i+1+ · · · + cr

z − ai

z − a_n. Taking limit z → ai, we obtain

z→alimi

(z − ai)P (z) Q(z) = ci. On the other hand, by Q(ai) = 0,

z→alimi

Q(z)

z − a_i = lim

z→ai

Q(z) − Q(a_i)

z − a_i = Q⁰(a_i).

As long as we can show that Q⁰(ai) 6= 0,

z→alimi

(z − ai)P (z)

Q(z) = P (a_i) Q⁰(ai). Notice that

Q(z) z − a_i =Y

j6=i

(z − a_j) ⇒ Q⁰(a_i) = lim

z→ai

Q(z) z − a_i =Y

j6=i

(a_i− a_j) 6= 0.

We prove that c_i= P (a_i)/Q⁰(a_i). 

In the above case, since Q(z) has simple pole at a_i, by residue calculus, P (ai)

Q⁰(a_i) = Resai

P (z)

Q(z) = P (ai) Q

j6=i(a_i− a_j). Example 1.2. Find the partial fraction expansion of

R(z) = z²+ z + 1 (z − 1)(z − 2)(z − 3).

Write

R(z) = A

z − 1+ B

z − 2+ C z − 3.

Then A, B, C are residues of R(z) at 1, 2, 3 respectively. Choose r small enough so that Dr(1) does not contain 2, 3 and Dr(2) does not contain 1, 3 and Dr(3) does not contain 1, 2.

By Cauchy integral formula, A = 1

2πi I

Cr(1)

R(z)dz = 1²+ 1 + 1 (1 − 2)(1 − 3) = 3

2 and

B = 1 2πi

I

Cr(2)

R(z)dz = 2²+ 2 + 1

(2 − 1)(2 − 3) = −7 and

C = 1 2πi

I

Cr(3)

R(z)dz = 3²+ 3 + 1

(3 − 1)(3 − 2) = 13 2 . Example 1.3. Find the partial fraction expansion of

R(z) = 1 z⁴+ 1.

The polynomial z⁴ + 1 has 4 distinct roots. We will denote them by {z₁, z₂, z₃, z₄} = {e^iπ/4, e^iπ/2, e^i3π/4, e^i3π/2}. Let us consider the partial fraction expansion of R(z) :

1

z⁴+ 1 = A₁ z − z1

+ A₂ z − z2

+ A₃ z − z3

+ A₄ z − z4

. The zero z_i of z⁴+ 1 is a simple pole of R(z), and thus

A_i = Res_zi(R(z)) = lim

z→zi

z − z_i z⁴+ 1 = 1

4z³_i.

Since z_i⁴ = −1, z_i³= −1/z_i. This gives A_i = −z_i/4. It is very interesting to see that if D is a closed disk containing all zi, then

1 2πi

I

C

R(z)dz =

4

X

i=1

Res_zi(R(z)) = −z₁+ z₂+ z₃+ z₄

4 = 0

using Vieta’s formula. Here C is the boundary of D.