Math 185 Problem Set 5. Due Sunday 10/5 at 3pm
In the first 5 problems of this homework set, we will study rational functions. We will use
Liouville’s theorem and some polynomial alge
a to develop the partial fraction decomposition
theorem for rational functions and deduce some facts about complex integrals. I will begin by
giving you some definicions and facts about rational functions that you can use.
Definition. We will be working with complex coefficients. A rational function is a function
α(z) =
f(z)
g(z)
with f(z) = fdz
d + fd−1z
d−1 + · · · + f0 and g(z) = geze + ge−1ze−1 + · · · + g0(z) polynomials with complex
coefficients, and g 6= 0. We say that two rational functions are equal if their values are equal wherever both are
defined. So the function z1 is equal to the function
z2
z , even though the latter is (technically) only defined on C
∗.
Facts about polynomials and rational functions (you may assume these without proof; we will
prove some of them in class.)
• Every polynomial f(z) can be uniquely expressed as a product of linear factors, as follows:
f(z) = fd ·
k∏
j=1
(z − zj)nj ,
where fd is the leading term, the distinct complex numbers zj are called the roots and nj is the multiplicity
of the root zj . The degree d of the polynomial f (i.e., the largest d such that fd 6= 0) is equal to the sum
of all the multiplicities. (This sum is also called the “number of roots counted with multiplicity”.)
• For any pair of polynomials f, g with f, g 6= 0, there is a unique “division with remainder” expression
f(z)
g(z) = q(z) +
(z)
g(z) with q(z), the polynomial quotient, is a polynomial and r(z), the residue g mod f, is a
polynomial of degree less than the degree of f . The quotient r(z)g(z) is called the fractional part of the rational
function f(z)g(z) . Example:
z2 + 1
z + 1
= z − 1 + 2
z + 1
;
here z − 1 is the polynomial quotient and 2 is the remainder z2 + 1 mod z + 1.
• An expression f(z)g(z) for a rational function is said to be in lowest terms if g(z) is monic (i.e., g(z) =
1 · ze + ge−1ze−1 + · · · + g0z0) and f, g have no roots in common. For α(z) = f(z)g(z) a rational function in
lowest terms, roots of g are called the poles of the function α. The quotient formula for derivatives implies
that a complex derivative exists everywhere except where the denominator is zero, i.e., α(z) is holomorphic
everywhere except its poles.
Problem 1. (a) Show that every rational function f(z)g(z) can be expressed as a fraction of two polynomials
f ′(z)
g′(z)
for f ′, g′ in lowest terms (the prime here does not denote derivative). Hint: decompose f, g into linear factors.
1
(b) If fg is a rational function in lowest terms and γ is a simple closed curve which does not go through any
oots of g, use either Cauchy’s theorem (for example Cauchy II) or the fundamental theorem of line integrals to
show that ∫
γ
f(z)
g(z)
dz =
∫
γ
(z)
g(z)
dz
for r(z) = g(z) mod f(z) the remainder.
(c) Assume deg(f) = d,deg(g) = e and the degrees satisfy the inequality e ≥ d. Let α(z) = f(z)g(z) . Define
α(∞) = fe
ge
.
Note in particular that if the inequality is strict e > d then α(∞) = 0 since fe = 0 (the eth term is past the
highest term of f(z)). Show that
lim
|z|→∞
α(z) = α(∞)
Here the limit statement means that you must prove that for any real � > 0 there exists real N such that |z| > N
implies |α(z)− α(∞)| < �.
2. You may assume all parts of 1. You may also use the following theorem, which is the Chinese
emainder theorem for polynomials.
Theorem 1. If two nonzero polynomials p(z), q(z) have no common roots then any polynomial f(z) can be
expressed as a combination f(z) = a(z)p(z) + b(z)q(z), for a(z), b(z) two other polynomials.
You can find a proof of this fact online or in an abstract alge
a textbook.1
Let α(z) = f(z)g(z) , in reduced terms. Let zj be a root of g(z), with multiplicity nj . Then we can write
g(z) = g̃j(z) · (z − zj)nj ,
for g̃j a polynomial with no z−zj terms. Since the two factors have no roots in common, we can find polynomials
a(z), b(z) such that
f(z) = a(z)(z − zj)nj + b(z)g̃j(z)
y the Chinese Remainder Theorem above. Let
j(z) = b(z) mod (z − zj)nj .
Define
αj(z) :=
j(z)
(z − zj)nj
.
This is the singular part of g at the pole zj , equivalently the “partial fraction term at zj”.
2(a) Show that the rational function α(z)− αj(z) is holomorphic at zj (i.e., that it is equivalent to anothe
ational function with no pole at zj).
(b) Show that the function αj(z) is holomorphic everywhere except z0 and lim|z|→∞ α(z) = 0.
1Although you will not need this for the problems, I’ll say a couple of words about the proof here. Note that it is enough to find
suitable a(z), b(z) for f(z) = 1 the constant polynomial, since then f(z) = f(a(z)p(z) + b(z)q(z)) expands to a valid expression for f .
Now if one of p(z), q(z), say p(z) has degree 1, then dividing the remainder expression q(z) = a(z)p(z) + r by the constant r (nonzero
since p(z), q(z) are assumed to have no common roots) gives 1 = 1
q(z) − a(z)
(z). From here, you can induct on degree of p(z) to
give a full proof.
2
(c) Assume α(z) = f(z)g(z) with deg g(z) > deg f(z) (i.e., it is its own fractional part). Let z1, . . . , zk be the
oots of g with degrees n1, . . . , nk. Let α1, . . . , αk be the singular parts of α at z1, . . . , zk. Prove that the function
F (z) := α(z)−
k∑
j=1
αj
is an entire function with lim|z|→∞ F (z) = 0. Deduce using Liouville’s theorem that
α(z) =
k∑
j=1
αj(z).
(d) Now let α(z) = f(z)g(z) be any rational function in lowest terms (no degree inequalities on deg f, deg g). Let
z1, . . . , zk be roots of g with multiplicities n1, . . . , nk as before. Show that there is an expression
α(z) = q(z) +
∑ rj(z)
(z − zj)nj
,
for q(z) a polynomial function and rj(z) polynomial functions of degree deg rj < nj . This is the partial fraction
decomposition of α(z).
3. Find the partial fraction decompositions of the following functions (if you’re stuck, look at http:
www.
mesacc.edu/~scotz47781/mat150/notes/part_fract_decomp/Partial_Frac_Decomp_Notes.pdf or find an-
other online source for partial fraction decompositions).
(a) z
2−3
z2+1
(b) z
4+1
z2(z+i)
(c) 1
z2(z−1)
4. Use the partial fraction decomposition and Cauchy’s Integral Formula to compute the loop integrals of each
of the functions 3(a) - 3(c) above over the circles CR whenever they are defined (hint: you will only need to
consider the cases 0 < R < 1 and 1 < R <∞).
5. Assume that α(z) = r(z)g(z) be a rational function which is its own fractional part. Write g(z) = gez
e +
ge−1z
e−1 + · · · + g0, for ge 6= 0. Since the degree of r(z) is smaller than the degree of g(z), we can r(z) =
e−1z
e−1 + re−2z
e−2 + · · · + r0 (as usual, if the degree of r(z) is less than the degree of g(z) minus 1 then
e−1 = 0).
(a) Show (without using partial fraction decomposition) that
lim
R→∞
∫
CR
α(z)dz =
fe−1
ge
.
(Hint: write the integral as
∫ 2π
0 F (R exp(it))dt for F (z) a rational function and use 1(c)).
(b) Using (a), compute the R→∞ limit of
∫
CR
(z)
g(z) for the fractional part of 3(a) - 3(c) above and check that
it agrees with the values you obtained in 3. (Why do loop integrals only depend on the fractional part, again?)
6. Some computational practice with Harmonic functions.
Look at Gamelin, Exercise III.3.1 on page 86. For each u(x, y) defined in a-d,
1. Check that u(x, y) is harmonic,
2. Find the Harmonic conjugate (i.e., solve the Cauchy-Riemann equations for f(x, y) = u(x, y) + iv(x, y)
3
http:
www.mesacc.edu/~scotz47781/mat150/notes/part_fract_decomp/Partial_Frac_Decomp_Notes.pdf
http:
www.mesacc.edu/~scotz47781/mat150/notes/part_fract_decomp/Partial_Frac_Decomp_Notes.pdf