3Residue calculus
IB Complex Analysis
3.6 Rouch´es theorem
We now want to move away from computing integrals, and look at a different
application — Rouch´es theorem. Recall one of the first applications of complex
analysis is to use Liouville’s theorem to prove the fundamental theorem of algebra,
and show that every polynomial has a root. One might wonder — if we know
a bit more about the polynomial, can we say a bit more about how the roots
behave?
To do this, recall we said that if
f
:
B
(
a
;
r
)
→ C
is holomorphic, and
f
(
a
) = 0,
then f has a zero of order k if, locally,
f(z) = (z −a)
k
g(z),
with g holomorphic and g(a) 6= 0.
Analogously, if
f
:
B
(
a, r
)
\ {a} → C
is holomorphic, and
f
has at worst a
pole at a, we can again write
f(z) = (z −a)
k
g(z),
where now
k ∈ Z
may be negative. Since we like numbers to be positive, we say
the order of the zero/pole is |k|.
It turns out we can use integrals to help count poles and zeroes.
Theorem
(Argument principle)
.
Let
U
be a simply connected domain, and let
f
be meromorphic on
U
. Suppose in fact
f
has finitely many zeroes
z
1
, ··· , z
k
and finitely many poles
w
1
, ··· , w
`
. Let
γ
be a piecewise-
C
1
closed curve such
that z
i
, w
j
6∈ image(γ) for all i, j. Then
I(f ◦ γ, 0) =
1
2πi
Z
γ
f
0
(z)
f(z)
dz =
k
X
i=1
ord(f; z
i
)I
γ
(z
i
) −
`
X
j=1
ord(f, w
j
)I(γ, w
j
).
Note that the first equality comes from the fact that
I(f ◦ γ, 0) =
1
2πi
Z
f◦γ
dw
w
=
1
2πi
Z
γ
df
f(z)
=
1
2πi
Z
γ
f
0
(z)
f(z)
dz.
In particular, if
γ
is a simple closed curve, then all the winding numbers of
γ
about points
z
i
, w
j
lying in the region bound by
γ
are all +1 (with the right
choice of orientation). Then
number of zeroes − number of poles =
1
2π
(change in argument of f along γ).
Proof. By the residue theorem, we have
1
2πi
Z
γ
f
0
(z)
f(z)
dz =
X
z∈U
Res
f
0
f
, z
I(γ, z),
where we sum over all zeroes and poles of
z
. Note that outside these zeroes and
poles, the function
f
0
(z)
f(z)
is holomorphic.
Now at each
z
i
, if
f
(
z
) = (
z − z
j
)
k
g
(
z
), with
g
(
z
j
)
6
= 0, then by direct
computation, we get
f
0
(z)
f(z)
=
k
z −z
j
+
g
0
(z)
g(z)
.
Since at
z
j
,
g
is holomorphic and non-zero, we know
g
0
(z)
g(z)
is holomorphic near
z
j
. So
Res
f
0
f
, z
j
= k = ord(f, z
j
).
Analogously, by the same proof, at the w
i
, we get
Res
f
0
f
, w
j
= −ord(f; w
j
).
So done.
This might be the right place to put the following remark — all the time, we
have assumed that a simple closed curve “bounds a region”, and then we talk
about which poles or zeroes are bounded by the curve. While this seems obvious,
it is not. This is given by the Jordan curve theorem, which is actually hard.
Instead of resorting to this theorem, we can instead define what it means to
bound a region in a more convenient way. One can say that for a domain
U
, a
closed curve γ ⊆ U bounds a domain D ⊆ U if
I(γ, z) =
(
+1 z ∈ D
0 z 6∈ D
,
for a particular choice of orientation on
γ
. However, we shall not worry ourselves
with this.
The main application of the argument principle is Rouch´es theorem.
Corollary
(Rouch´es theorem)
.
Let
U
be a domain and
γ
a closed curve which
bounds a domain in
U
(the key case is when
U
is simply connected and
γ
is a
simple closed curve). Let
f, g
be holomorphic on
U
, and suppose
|f| > |g|
for all
z ∈ image
(
γ
). Then
f
and
f
+
g
have the same number of zeroes in the domain
bound by γ, when counted with multiplicity.
Proof.
If
|f| > |g|
on
γ
, then
f
and
f
+
g
cannot have zeroes on the curve
γ
.
We let
h(z) =
f(z) + g(z)
f(z)
= 1 +
g(z)
f(z)
.
This is a natural thing to consider, since zeroes of
f
+
g
is zeroes of
h
, while
poles of h are zeroes of f . Note that by assumption, for all z ∈ γ, we have
h(z) ∈ B(1, 1) ⊆ {z : Re z > 0}.
Therefore
h◦γ
is a closed curve in the half-plane
{z
:
Re z >
0
}
. So
I
(
h◦γ
; 0) = 0.
Then by the argument principle,
h
must have the same number of zeros as poles
in
D
, when counted with multiplicity (note that the winding numbers are all
+1).
Thus, as the zeroes of
h
are the zeroes of
f
+
g
, and the poles of
h
are the
poles of f, the result follows.
Example.
Consider the function
z
6
+ 6
z
+ 3. This has three roots (with
multiplicity) in {1 < |z| < 2}. To show this, note that on |z| = 2, we have
|z|
4
= 16 > 6|z| + 3 ≥ |6z + 3|.
So if we let
f
(
z
) =
z
4
and
g
(
z
) = 6
z
+ 3, then
f
and
f
+
g
have the same number
of roots in {|z| < 2}. Hence all four roots lie inside {|z| < 2}.
On the other hand, on |z| = 1, we have
|6z| = 6 > |z
4
+ 3|.
So 6
z
and
z
6
+ 6
z
+ 3 have the same number of roots in
{|z| <
1
}
. So there is
exactly one root in there, and the remaining three must lie in
{
1
< |z| <
2
}
(the
bounds above show that |z| cannot be exactly 1 or 2). So done.
Example. Let
P (x) = x
n
+ a
n−1
x
n−1
+ ··· + a
1
x + a
0
∈ Z[x],
and suppose a
0
6= 0. If
|a
n−1
| > 1 + |a
n−2
| + ··· + |a
1
| + |a
0
|,
then
P
is irreducible over
Z
(and hence irreducible over
Q
, by Gauss’ lemma
from IB Groups, Rings and Modules).
To show this, we let
f(z) = a
n−1
z
n−1
,
g(z) = z
n
+ a
n−2
z
n−2
+ ··· + a
1
z + a
0
.
Then our hypothesis tells us |f| > |g| on |z| = 1.
So f and P = f + g both have n −1 roots in the open unit disc {|z| < 1}.
Now if we could factor
P
(
z
) =
Q
(
z
)
R
(
z
), where
Q, R ∈ Z
[
x
], then at least
one of
Q, R
must have all its roots inside the unit disk. Say all roots of
Q
are
inside the unit disk. But we assumed
a
0
6
= 0. So 0 is not a root of
P
. Hence it
is not a root of
Q
. But the product of the roots
Q
is a coefficient of
Q
, hence an
integer strictly between 0 and 1. This is a contradiction.
The argument principle and Rouch´es theorem tell us how many roots we
have got. However, we do not know if they are distinct or not. This information
is given to us via the local degree theorem. Before we can state it, we have to
define the local degree.
Definition
(Local degree)
.
Let
f
:
B
(
a, r
)
→ C
be holomorphic and non-
constant. Then the local degree of
f
at
a
, written
deg
(
f, a
) is the order of the
zero of f(z) − f(a) at a.
If we take the Taylor expansion of
f
about
a
, then the local degree is the
degree of the first non-zero term after the constant term.
Lemma. The local degree is given by
deg(f, a) = I(f ◦ γ, f (a)),
where
γ(t) = a + re
it
,
with 0 ≤ t ≤ 2π, for r > 0 sufficiently small.
Proof.
Note that by the identity theorem, we know that,
f
(
z
)
− f
(
a
) has an
isolated zero at
a
(since
f
is non-constant). So for sufficiently small
r
, the
function
f
(
z
)
−f
(
a
) does not vanish on
B(a, r) \{a}
. If we use this
r
, then
f ◦γ
never hits f (a), and the winding number is well-defined.
The result then follows directly from the argument principle.
Proposition
(Local degree theorem)
.
Let
f
:
B
(
a, r
)
→ C
be holomorphic and
non-constant. Then for
r >
0 sufficiently small, there is
ε >
0 such that for any
w ∈ B
(
f
(
a
)
, ε
)
\ {f
(
a
)
}
, the equation
f
(
z
) =
w
has exactly
deg
(
f, a
) distinct
solutions in B(a, r).
Proof.
We pick
r >
0 such that
f
(
z
)
−f
(
a
) and
f
0
(
z
) don’t vanish on
B
(
a, r
)
\{a}
.
We let
γ
(
t
) =
a
+
re
it
. Then
f
(
a
)
6∈ image
(
f ◦ γ
). So there is some
ε >
0 such
that
B(f(a), ε) ∩ image(f ◦γ) = ∅.
We now let
w ∈ B
(
f
(
a
)
, ε
). Then the number of zeros of
f
(
z
)
− w
in
B
(
a, r
) is
just
I
(
f ◦ γ, w
), by the argument principle. This is just equal to
I
(
f ◦ γ, f
(
a
)) =
deg(f, a), by the invariance of I(Γ, ∗) as we move ∗ in a component C \Γ.
Now if
w 6
=
f
(
a
), since
f
0
(
z
)
6
= 0 on
B
(
a, r
)
\{a}
, all roots of
f
(
z
)
−w
must
be simple. So there are exactly deg(f; a) distinct zeros.
The local degree theorem says the equation
f
(
z
) =
w
has
deg
(
f, a
) roots for
w
sufficiently close to
f
(
a
). In particular, we know there are some roots. So
B(f(a), ε) is contained in the image of f. So we get the following result:
Corollary
(Open mapping theorem)
.
Let
U
be a domain and
f
:
U → C
is
holomorphic and non-constant, then
f
is an open map, i.e. for all open
V ⊆ U
,
we get that f(V ) is open.
Proof.
This is an immediate consequence of the local degree theorem. It suffices
to prove that for every
z ∈ U
and
r >
0 sufficiently small, we can find
ε >
0
such that
B
(
f
(
a
)
, ε
)
⊆ f
(
B
(
a, r
)). This is true by the local degree theorem.
Recall that Liouville’s theorem says every holomorphic
f
:
C → B
(0
,
1) is
constant. However, for any other simply connected domain, we know there are
some interesting functions we can write down.
Corollary.
Let
U ⊆ C
be a simply connected domain, and
U 6
=
C
. Then there
is a non-constant holomorphic function U → B(0, 1).
This is a weak form of the Riemann mapping theorem, which says that there
is a conformal equivalence to
B
(0
,
1). This just says there is a map that is not
boring.
Proof.
We let
q ∈ C \ U
, and let
φ
(
z
) =
z − q
. So
φ
:
U → C
is non-vanishing.
It is also clearly holomorphic and non-constant. By an exercise (possibly on the
example sheet), there is a holomorphic function
g
:
U → C
such that
φ
(
z
) =
e
g(z)
for all
z
. In particular, our function
φ
(
z
) =
z − q
:
U → C
∗
can be written as
φ(z) = h(z)
2
, for some function h : U → C
∗
(by letting h(z) = e
1
2
g(z)
).
We let
y ∈ h
(
U
), and then the open mapping theorem says there is some
r >
0 with
B
(
y, r
)
⊆ h
(
U
). But notice
φ
is injective by observation, and that
h
(
z
1
) =
±h
(
z
2
) implies
φ
(
z
1
) =
φ
(
z
2
). So we deduce that
B
(
−y, r
)
∩ h
(
U
) =
∅
(note that since y 6= 0, we have B(y, r) ∩ B(−y, r) = ∅ for sufficiently small r).
Now define
f : z 7→
r
2(h(z) + y)
.
This is a holomorphic function f : U → B(0, 1), and is non-constant.
This shows the amazing difference between C and C \{0}.