1.2.10
The Principle of the Argument.
Theorem 6. Suppose that γ is a simple closed contour and suppose that f is analytic on
and inside γ except for poles interior to γ. Suppose further f has no zeros on γ. Then
Z ′
f (z)
1
dz = Z − P
2πi γ f (z)
where Z is the number of zeros of f inside γ and P is the number of poles of f inside γ,
counted according to multiplicity.
Remark: At this stage one might worry about whether Z and P are necessarily finite. For
the moment you might add this as a hypothesis to the theorem, but as we shall see shortly,
this is always the case. Note however that you need to assume that f is analytic at the points
of γ, as well as in the interior, for this to be true. A function like f (z) = sin(1/(1 − z)) has
infinitely many zeros inside the unit circle for example. It is analytic in the open unit disk
(indeed, everywhere except z = 1).
Example: Let f (z) = z k with k ∈ Z, and let γ be the unit circle (oriented positively). Then
Z ′
Z
Z
1
f (z)
kz k−1
1
1
k
dz =
dz =
dz = k
k
2πi γ f (z)
2πi γ z
2πi γ z
which is of course correct! (What if k 6∈ Z?)
The proof is a straightforward application of the residue theorem. The result is called
the principle of the argument since the right hand side is the winding number I(f ◦ γ, 0),
that is, the winding number of 0 with respect to the transformed contour f ◦ γ. This is easily
verified by a change of variable.
1.2.11
Rouché’s Theorem
A consequence of the Principle of the Argument is a useful result called Rouché’s Theorem.
Theorem 7. Suppose that f and g are analytic on and inside a simple closed contour γ and
that |f (z)| > |g(z)| for all z on γ. Then f and f + g have the same number of zeros, counted
according to multiplicity, inside γ.
Example: Let h(z) = 3z 3 + cos z. Then h = f + g where f (z) = 3z 3 and g(z) = cos z.
Obviously f has a zero of order 3 inside the unit circle. If |z| = 1 then
|eiz | = ei Re z e− Im z = e− Im z ≤ e.
9
Thus
iz
e + e−iz e + e
≤
= e < |f (z)| = 3
|g(z)| = 2
2
Therefore, by Rouché’s Theorem, h also has three zeros inside the unit disk, counted according to multiplicity.
1.2.12
The maximum modulus theorem
Analytic functions have another elementary property that might have been skipped during
your first course.
Suppose that Ω is a bounded region in C. Let Ω denote the closure of Ω, that is, Ω
along with its boundary. Then Ω is closed and bounded, and hence it is a compact subset
of the plane. Recall that a continuous real-valued function on a compact set always attains
a maximum (and a minimum) value on that set. In particular, if f is analytic on some open
neighbourhood of Ω, then |f (z)| is continuous on Ω and hence attains a maximum value.
The remarkable thing is that this maximum always occurs on the boundary of Ω, not the
interior. Said more carefully:
Theorem 8. Suppose that the nonconstant function f is analytic on a bounded region Ω
and continuous on Ω. Then |f | attains a maximum value on Ω, and this maximum occurs
on the boundary of Ω.
Thus, for example, if f is entire, there is no way that you can draw a simple closed curve
γ in the plane so that |f | has its maximum inside γ rather than on γ. That is, the graph of
|f | has no hills!
Question: Can |f | have a minimum inside a curve γ? Obviously if f has a zero at z0 , then
|f | will have a minimum there. What if f is non-zero inside γ?
The proof of the Maximum Modulus Theorem has two parts. The first is to show that f
would need to be constant in a neighbourhood of any point where |f | has a local maximum.
Suppose that |f | had a maximum value at z0 ∈ Ω. As Ω is open, there is a little disk
D = {z : |z − z0 | ≤ ǫ} sitting entirely inside Ω. Further, |f | has a maximum value on D
at z0 . Let γ denote the circular boundary of D. Then, parametrizing the circle in the usual
way, γ(θ) = z0 + ǫeiθ ,
1
f (z0 ) =
2πi
Z
γ
f (z)
1
dz =
z − z0
2π
10
Z
2π
f (z0 + ǫeiθ ) dθ.
0
Thus
1
|f (z0 )| ≤
2π
Z
0
2π
|f (z0 + ǫeiθ )| dθ ≤ max |f (z0 + ǫeiθ )|.
0≤θ≤2π
But |f (z0 + ǫeiθ )| ≤ |f (z0 )| for all θ and so this can only happen if we have equality. That
is |f (z0 )| = |f (z0 + ǫeiθ )| for all θ — and all small ǫ.
Exercise: Prove that if f is analytic and |f | is constant, then f is contant.
The second part then shows that if |f | had a local maximum at any point interior to a
domain Ω, then it must be constant on Ω. This uses the fact that Ω is connected and so any
point z1 in Ω can be connected to z0 by a polygonal path in Ω. One then constructs a finite
sequence of overlapping little disks along this path on which f is constant in order to show
that f (z1 ) = f (z0 ). (See the full proof in Brown and Churchill).
2
Defining functions via integrals: part 1
Cauchy’s integral formula says that for a function f analytic on and inside γ
Z
1
f (w)
f (z) =
dw.
2πi γ w − z
(4)
That is, if you know f on γ, this completely determines f inside γ.
The right-hand side of this formula is an expression of the form
Z
F (z, w) dw.
γ
There are very many important functions which are defined by such a formula, where the
integral is generally taken over a circle, a line, or a half-line. For example, the gamma
function will be defined as
Γ(z) =
Z
∞
e−t tz−1 dt,
0
(where we use t rather than w to remind you that it is real). There are two issues here:
1. What is the domain of such a function? For which z ∈ C does the integral converge?
2. Where is such a function analytic? How do you differentiate the integral?
Z The first question is usually,
Z but not always, the easiest. Typically one uses the fact that
if |g(z)| dz converges then g(z) dz does too. Check that you know why this is true. The
γ
γ
11
converse
of this is of course false! Some of the trickier integrals that one might try such as
Z ∞
sin x
dx are not absolutely convergent.
x
0
For the example above, writing z = x + iy,
|e−t tz−1 | = e−t |tx+iy−1 | = e−t tx−1
Whatever x is, this goes to zero fast enough at ∞Z for the integral to converge. At t = 0
∞
however, this blows up if x < 1, and in particular,
|e−t tz−1 | dt diverges if x = Re z < 0.
0
Thus, the above expression for the gamma function certainly converges if Re z < 0.
We would certainly like to have more general theorems that say something like
Z
Z
∂F (z, w)
d
F (z, w) dw =
dw.
dz γ
∂z
γ
A slightly different question concerns whether you can use something like (4) to construct
analytic functions. Suppose, for example, that f is just defined on γ. Under what conditions
can one extend f in a nice way to a function fˆ which is analytic inside γ?
12
3
Analytic continuation
3.1
Extending functions - uniqueness
Even for real functions, one can have formulas which have different domains, but which give
the same output where they are both defined. For example:
f : R → R,
f (x) = x,
g : [0, ∞) → R,
g(x) =
√ 2
x .
In this example, f is an infinitely differentiable extension of g to a larger set. There are
actually lots of different infinitely differentiable extensions of g to R (find another one!).
With analytic functions this can’t happen! We start with a simple property of analytic
functions.
Lemma 9. Let f be analytic and suppose a is a zero of f . Then either a is an isolated zero
of f or f ≡ 0 in a neighbourhood of a.
Proof. There is an r > 0 with
f (z) =
∞
X
f (k) (a)(z − a)k
k=1
k!
,
|z − a| < r.
If f 6≡ 0 on |z − a| < r there is a least k such that f (k) (a) 6= 0 for otherwise the Taylor
expansion is identically zero. Therefore we can write
f (z) = (z − a)k (A + (z − a)g(z))
where A 6= 0 and g is analytic on |z − a| < r. Let M = max {|g(z)| : z ∈ D} where
D = {z : |z − a| ≤ r/2}, say. [M exists since g is continuous and D is closed and bounded,
that is, D is compact.] Then, by the triangle inequality
|f (z)| ≥ |z − a|k |A| − |z − a||g(z)|
so that for |z − a| small enough (in particular, for |z − a| <
|A|
)
M
we have
|f (z)| ≥ |z − a|k (|A| − |z − a|M) > 0.
Therefore f (z) 6= 0 for |z − a| small enough, so that a is an isolated zero, as required.
This lemma has dramatic consequences.
13
Theorem 10 (Analytic Continuation). Suppose that f, g are analytic functions on a region
Ω which coincide on a subset of Ω which has a limit point in Ω. Then f, g coincide on all of
Ω.
Remark: As you can see from the proof, it is important that Ω is connected!
Proof.
Let h = f − g and let a be the limit point. Then there is a sequence of points
an → a with each an 6= a and h(an ) = 0 for all n. Since h is continuous h(a) = 0. Let b be
any other point in Ω and fix a path in Ω joining a to b. [Regions are pathwise connected.]
Let z = z(t), 0 ≤ t ≤ 1 be a parametrization of this path. [This means z(t) is continuous,
z(0) = a and z(1) = b.]
Ω
b
b
z = z(t)
b
a
Let
T = {t : 0 ≤ t ≤ 1 and such that h(z(s)) = 0 for 0 ≤ s ≤ t}
Clearly T is nonempty as 0 ∈ T . And of course T is a bounded set, so the least upper bound
axiom for R tells us that t0 = sup T exists.
Since a is not an isolated zero of h (being the limit point of zeros) the lemma implies
h ≡ 0 on a neighbourhood N of a. Therefore h(z(s)) = 0 for s sufficiently close to 0 to
ensure that z(s) ∈ N. Consequently t0 is strictly greater than 0, and so we can choose an
sequence in [0, t0 ) which converges to t0 . This shows that c = z(t0 ) is the limit points of a
sequence of zeros of h and hence, by the lemma, h ≡ 0 on a neighbourhood of c.
But if t0 < 1, this implies that there is some number a little bit bigger that t0 which is
in T ! This is impossible by the definition of T and hence t0 = 1. Thus h(z(1)) = h(b) = 0.
14
It clearly follows then that f (z) = g(z) for all z ∈ Ω.
Corollaries/consequences
(i) The zeros of a non-constant analytic function on a region Ω are isolated.
(ii) A nonzero entire function can only have a finite number of zeros in any closed (and
bounded) disk [for otherwise, by compactness, one of its zeros must be a limit point.] Consequently an entire function can only have a countable (or denumerable) number of zeros. [If infinite in number then we can list them as a sequence a1 , a2 , a3 , . . . (where, say,
|an+1 | ≥ |an |).]
(iii) If f and g are analytic on a region Ω and agree on a line segment, or curve or a
non-empty open set then f ≡ g on Ω. Thus, if f is defined on some curve S (such as R) in
a region Ω, there is at most one analytic extension of f to all of Ω.
Note however that if one chooses two different regions Ω1 and Ω2 containing S, then
the extensions are only guaranteed to agree on the connected component of Ω1 ∩ Ω2 which
contains S.
Exercise: Find an example (where Ω1 ∩Ω2 is not connected) of where you can get extensions
that disagree.
(iv) The usual classical real valued functions (ex , sin x, . . .) have at most one analytic
extension to C. Thus it is safe to retain the notation ez , sin z, . . .. Moreover the usual
algebraic and analytic identities must extend. To see this, consider the identity
sin2 x + cos2 x = 1,
x ∈ R.
We know that sin x and cos x do have analytic extensions to all of C and hence sin2 x + cos2 x
extends analytically to all of C too. Of course the right-hand side has the obvious analytic
extension — the constant function 1. The analytic continuation theorem says that these two
analytic functions must now agree on all of C. Note that in this case the identity follows
easily from the formulas
sin z =
eiz − e−iz
,
2i
cos z =
eiz + e−iz
2
for the analytic extensions. The point, which was probably skipped when you first saw these
definitions, is that we actually didn’t have any choice. There is only one way to do the
extensions and they have to satisfy all the identities that hold in R.
15
Keep in mind that inequalities do not extend (for example, | sin x| ≤ 1 for all x, yet | sin z|
is unbounded on C).
3.2
Existence of extensions
Quite often in complex analysis, functions are defined via a series or an integral. Consider
for example
f1 (z) = 1 + z + z 2 + · · · ,
Z ∞
f2 (z) =
e−t(1−z) dt,
|z| < 1,
Re z < 1.
0
In both these cases, the given formula only makes sense for the given domain. Less obviously,
f2 is an analytic extension of f1 . That is f2 (z) = f1 (z) for all |z| < 1. And of course we
could extend further to
f3 (z) =
1
,
1−z
z 6= 1.
An obvious question to ask is how one might in general extend an analytic function f
defined on some region Ω to a larger region?
The standard method of continuation involves the use of power series.
z1
b
b
z0
ρ
r
For example, suppose that the function f is analytic on the disk |z − z0 | < r. Choose
a point z1 in this disk. Then f certainly has a power series expansion about z1 which
converges at least on |z − z1 | < r1 = r − |z1 − z0 |. If the power series actually converges
on |z − z1 | < ρ where ρ > r1 then we have extended f0 analytically from |z − z0 | < r to
|z − z0 | < r ∪ |z − z1 | < ρ. This can then be repeated. The problem is in doing this in
actually having a usable formula for f (z) and a good description of where ut works.
Not every analytic function can be extended to a larger domain. An interesting example,
P n!
due to Weierstrass, is the function f (z) = ∞
which diverges everywhere on |z| = 1, the
0 z
16
boundary of its disk of convergence. A cut is a curve of singularities and f (z) cannot be
extended beyond its cut |z| = 1.
Specific functions can sometimes be extended by very specific techniques. An example,
which we will study in detail later in the session, is the Riemann zeta function initially
defined by
∞
X
1
.
ζ(z) =
z
n
n=1
(5)
Recall that |nz | = nRe z , and so we are only sure of convergence if Re z > 1. The Riemann
zeta function can be extended to the region Re z > 0 by
Z 1
Z ∞
z−1
1
t
−1
−1
−1
t
−1 z−1
ζ(z) =
(e − 1) − t
t dt + (z − 1) +
(e − 1) t dt
Γ(z) 0
1
then by
1
ζ(z) =
Γ(z)
Z
0
1
Z ∞
t
1
1 z−1
t
−1
−1
−1
−1 z−1
t dt −
(e − 1) − t +
+
(e − 1) − t
t dt
2
2z
1
for −1 < Re z < 1. Finally, one can use Riemann’s functional equation
ζ(z) = 2(2π)z−1Γ(1 − z)ζ(1 − z) sin
πz 2
z 6= 0, −1, −2, . . .
on Re z ≤ −1.
At each stage one has to check that the new formula agrees with the previous one on some
suitable set, and that the new function is analytic on its domain. That clearly involves quite
a bit of work here! The bottom line however is that there is an analytic function ζ which is
defined except at z = 1 which extends (5).
Remark. It can be checked that the only zeros of ζ outside the strip 0 ≤ Re z ≤ 1 occur at
the integers −2, −4, −6, . . .. Riemann conjectured that the zeros inside the strip all occur on
Re z = 21 . This is the Riemann hypothesis; by universal agreement the outstanding unsolved
problem in mathematics.
R∞
The standard approach to the gamma function is to define Γ(x) = 0 tx−1 e−t dt, x > 1,
R∞
a real function with Γ(n + 1) = n. Next define Γ(z) = 0 tz−1 e−t dt, Re z > 0. Next one
shows that Γ(z)Γ(1 − z) =
π
,
sin πz
0 < Re z < 1. Then this formula can be used to define
Γ(z) for all complex z except 0 and the negative integers.
17