2C1Y Complex Analysis
Shaun Stevens
Spring Semester 2007
Contents
1 Topology
2
1.1
Revision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
1.2
Paths and contours . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
1.3
Connectedness and simple connectedness . . . . . . . . . . . . . . . . . . . . . . . . .
8
2 Integration
11
2.1
Complex integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.2
Integration along a path . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
3 Cauchy’s Theorem
17
3.1
The proof of Cauchy’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
3.2
Consequences of Cauchy’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
3.3
Zeros of holomorphic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
4 Singularities and Residues
29
4.1
Laurent Expansions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
4.2
Singularities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
4.3
Residues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
4.4
Applications to real integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
4.5
Application to summation of series . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
1
1
Topology
1.1
Revision
Let C be the field of complex numbers. For any z ∈ C, we write z = x + iy, with x, y ∈ R; so
x = Re(Z) is the real part of z and y = Im(z) is the imaginary part of z. We for x, y ∈ R, we often
abbreviate
x = x + i0,
a real number,
iy = 0 + iy,
a purely imaginary numbers,
0 = 0 + i0.
Geometrically, we interpret C as a 2-dimensional space
over R, where x + iy corresponds to the point (x, y) in
the coordinate plane (called the complex plane).
The modulus of z = x + iy is defined to be
p
|z| = x2 + y 2 .
We also have the complex conjugate z = x − iy of z;
notice that z = z and that zz = |z|2 . Hence, for z 6= 0,
z
−1
z
y
z
= 2.
|z|
|z|
0
x
Similarly, we have
z + z = 2 Re(z),
and
z − z = 2i Im(z).
Remark. Notice that, geometrically, |z| represents the distance from z to 0. More generally, for
w, z ∈ C, the modulus |z − w| represents the distance between z and w. This will be very useful.
Lemma 1.1. Let z, w ∈ C.
(i) |z + w| ≤ |z| + |w| (the triangle inequality).
(ii) |z + w| ≥ ||z| − |w|| (the reverse triangle inequality).
Topology
For a ∈ C and r ∈ R, r > 0, we define
D(a; r) = {z ∈ C : |z − a| < r},
the open disc with centre a and radius r. Similarly, for r ≥ 0, we define
D(a; r) = {z ∈ C : |z − a| ≤ r},
the closed disc with centre a and radius r.
Definition 1.2. A subset S of C is called open if, for any z ∈ S, there exists δ > 0 such that
D(z; δ) ⊆ S.
2
Note that, in the definition of open set, the value of δ depends on the particular point z. Intuitively
(though you should never use this as a proof) open sets are sets with “no boundary points”. (This
can be made rigorous if we define what we mean by a boundary point.)
Examples.
(i) For a ∈ C and r > 0, the open disc D(a; r) is open.
(ii) The open upper half plane H = {z ∈ C : Im(z) > 0} is open.
(iii) For a ∈ C and r ≥ 0, the closed disc D(a; r) is not open.
Definition 1.3. A subset S of C is closed if its complement C \ S is open.
It may be that the definition of closed given last semester was: S is closed if it contains all its limit
points. These two definitions are equivalent (Exercise). Intuitively, closed sets are those which
“contain all their boundary points”.
Example. For a ∈ C and r ≥ 0, the closed disc D(a; r) is closed.
Remark. Note that subsets of C may be neither open nor closed: for example, the annulus
S = {z ∈ C : 1 < |z| ≤ 2}.
S is not open as 2 ∈ S but, for any δ > 0, the disc D(2, δ) is not contained in S since 2 +
D(2; r) \ S.
δ
2
∈
The complement C \ S is not open as 1 ∈ C \ S but, for any δ > 0, the disc D(1, δ) is not contained
in C \ S since 1 − min2 δ,1 ∈ D(1; r) but is not in C \ S. Hence S is not closed.
On the other hand, it is not clear at this stage whether or not a proper subset of C can be both
open and closed. (Certainly C and ∅ are both open and closed.)
Definition 1.4. (i) A subset S of C is bounded if there exists a constant M ∈ R such that
|z| ≤ M , for all z ∈ S. (Equivalently, S ⊆ D(0, M ).)
(ii) A subset S of C is compact if it is both closed and bounded.
Examples. (i) For a ∈ C and r ≥ 0, the closed disc D(a; r) is compact: it is certainly closed;
now put M = |a| + r and notice that, for all z ∈ D(a; r),
|z| = |(z − a) + a| ≤ |z − a| + |a| ≤ r + |a| = M,
where we have used the triangle inequality and the fact that |z − a| ≤ r since z ∈ D(a; r).
(ii) The closed upper half-plane H = {z ∈ C : Im(z) ≥ 0} is not compact. It is closed (Exercise)
but, for any M ∈ R, the point z = (|M | + 1)i is in H but has |z| = |M | + 1 > M . Hence H
is not bounded.
(iii) For any a, b ∈ C, the line segment
[a, b] := {z = (1 − t)a + tb : t ∈ [0, 1]}
is compact (Exercise).
You should think of compact sets as being the equivalent in C of closed intervals in R (more
precisely, of finite unions of closed intervals).
3
Continuity
Definition 1.5. Let S be a subset of C, let f : S → C be a function, and let a ∈ S.
(i) f is continuous at a if, for all ε > 0 there exists δ > 0 such that
z ∈ D(a; δ) ∩ S ⇒ f (z) ∈ D(f (a); ε),
that is
if z ∈ S with |z − a| < δ then |f (z) − f (a)| < ε.
Equivalently, we can write this as limz→a f (z) = f (a). Recall the crucially important thing
about the limit here: since we are working in C, we can approach a from any direction.
(ii) f is continuous in S if it is continuous at every point of S.
For example, the functions f (z) = z and g(z) = |z| are continuous in C. Moreover, the algebra of
continuous functions is valid so we immediately get things like
h(z) =
3z 2 − |z| + 1
is continuous in C \ − 21 .
2z + 1
Remark. It is also possible to characterize continuous functions in terms of open subsets of C. If
S is open then a function f : S → C is continuous if and only if, for all open U ⊆ C, the set
f −1 (U ) := {z ∈ S : f (z) ∈ U }
is open (Exercise).
The following Proposition will be quite useful. You should compare it to the Theorem from real
analysis which says that any continuous function on a closed interval f : [a, b] → R is bounded and
attains its bounds.
Proposition 1.6. Let S be a compact subset of C and let f : S → C be a continuous function.
Then the image f (S) = {f (z) : z ∈ S} of S is compact.
The proof, while not difficult, does not really fit in with the rest of the material in this course.
Holomorphy
Definition 1.7. Let S be an open subset of C, let f : S → C be a function and let s ∈ S.
(i) f is differentiable at z if the limit
lim
h→0
f (z + h) − f (z)
h
exists, in which case we write f ′ (z) for this limit.
(ii) f is holomorphic at z if there exists δ > 0 such that f is differentiable at every point of
D(z; δ).
(iii) f is holomorphic in Sif f is holomorphic at every point of S. Since we have assumed S is
open, this is actually the same as asking that f be differentiable at every point of S.
4
It is easy to check that f (z) = z is holomorphic in C. Then, since the product, quotient and chain
rules are all valid, we get that all rational functions are holomorphic where they are defined. For
example,
z 3 − 3Z + 2
g(z) =
is holomorphic in C \ − 12 , i .
(2z + 1)(z − i)
However, other functions such as h(z) = z and k(z) = |z| are nowhere holomorphic. It is actually
easy to check directly that the limit in Definition 1.7(i) does not exist in any open disc for these
functions, or we can just check the Cauchy-Riemann equations:
Theorem 1.8 (Cauchy-Riemann Equations). Suppose f : S → C is differentiable at z0 = x0 + iy0 .
Write
f (z) = u(x, y) + iv(x, y),
where z = x + iy.
Then u, v have first order partial derivatives at (x0 , y0 ) and, denote these by ux , uy , vx , vy , they
satisfy
ux = vy
and
uy = −vx .
Proving this is actually straightforward: we just compare the limit in Definition 1.7(i) taken firstly
as h → 0 along the real axis, then as h → 0 along the imaginary axis. But note that this does not
necessarily mean that the limit taken as h → 0 in some other way is also the same; that is, the
Cauchy-Riemann equations do not give an if and only if condition for differentiability – there are
functions which satisfy the Cauchy-Riemann equations at some point z0 but are not differentiable
there. However, if we also have that the partial derivatives are continuous in some disc around z0 ,
then the Cauchy-Riemann equations are enough to guarantee that f is differentiable at z0 .
The following Theorem will probably seem a little strange. It tells us what happens when we have
a function with zero derivative but, perhaps surprisingly, the conclusion is not that f is constant.
Theorem 1.9. Let S be an open set and let f : S → C be a holomorphic function. Then f ′ (z) = 0
for all z ∈ S if and only if f is locally constant, i.e. for all z ∈ S there is a δ > 0 such that f is
constant on D(z : δ).
Proof If f is locally constant then the definition immediately gives f ′ ≡ 0: Let z ∈ S and, for
any ε > 0, choose δ > 0 such that f is constant on D(z; δ). Then, for all h with |h| < δ, we have
z + h ∈ D(z; δ) so f (z + h) = f (z) and
0
f (z + h) − f (z)
− 0 = = 0 < ε.
h
h
Conversely, the proof of the Cauchy-Riemann equations shows that f ′ (z) = ux + ivx = vy − iuy , so
we immediately have ux = uy = vx = vy = 0 everywhere in S. Now fix z0 = x0 + iy0 ∈ S and let
δ > 0 be such that D(z0 ; δ)⊆S (remember that S is open). Let z = x + iy ∈ D(z0 ; δ) – we need to
show that f (z) = f (z0 ).
We put w = x + iy0 and notice that w ∈ D(z0 ; δ) also so that the line segments [z0 , w] and [w, z]
are both contained in D(z0 ; δ)⊆S.
5
Now the map r 7→ u(r, y0 ) is a real-valued function of a real variable [x0 , x] → R, with zero
derivative, so it is constant. Hence u(x, y0 ) = u(x0 , y0 ). Similarly, by considering s 7→ u(x0 , s),
we obtain u(x, y) = u(x, y0 ). Thus we have Re(f (z)) = Re(f (z0 )), and the same works for the
imaginary parts.
That there actually are locally constant functions which are not constant may seem strange, so
here is an example. Let S = D(2; 1) ∪ D(−2; 1) and define f : S → C by
(
0
if z ∈ D(−2; 1),
f (z) =
1
if z ∈ D(2; 1).
Notice that this has only been possible because S has a particular form, namely it is “in two
separate pieces”. We will return to this later.
P
The crucial example of holomorphic functions is that given by power series
an z n , with an ∈ C.
They have the following nice property:
P
There is an R (called the radius of convergence of the power series) such that
an z n converges
absolutely if |z| < R but diverges if |z| > R. The open disc D(0; R) is called the disc of convergence
of the power series.
Note that we allow R = 0 (in which case the disc of convergence is empty) and R = ∞ (in which
case the disc on convergence is all of C).
generally we can have power series centred at
P Also, more
n
any point a ∈ C; such a power series
an (z − a) has disc of convergence D(a; R) centred at a.
P
Theorem 1.10. Let P an z n be a power series with radius of convergence R > 0 and define a
n
function f by f (z) = ∞
n=0 an z , for z ∈ D(0; R). Then f is holomorphic in D(0; R) and
f ′ (z) =
∞
X
nan z n−1 ,
n=1
for z ∈ D(0; R).
Corollary 1.11. In the same situation as Theorem 1.10, f is infinitely differentiable on D(0; R)
and f (n) (0) = n!an .
Note that this shows that the coefficients in thePpower series
P are nuniquely determined by the
function: an = f (n) (0)/n!. So, if two power series
an z n and
bn z define the same function on
D(0; R) then, in fact, an = bn for all n ≥ 0.
P
zn
d
z
z
Examples. (i) The exponential function ez = ∞
n=0 n! is holomorphic in C and dz (e ) = e . It
satisfies all the usual properties of the (real) exponential function.
(ii) The trigonometric functions sin(z) and cos(z) are holomorphic in C, as are sinh(z) and
cosh(z).
(iii) From the sum of a geometric progression,
∞
X
1
zn,
=
1−z
n=0
Similarly,
for z ∈ D(0; 1).
∞
X
1
1
(−1)n (z − 1)n ,
=
=
z
1 + (z − 1)
n=0
6
for z ∈ D(1; 1).
(iv) The function
log(z) =
∞
X
(−1)n−1
n
n=1
(z − 1)n
is holomorphic in D(1; 1). Taking its derivative term-by-term, we see that
1.2
d
dz
log(z) = 1z .
Paths and contours
Here we introduce the notion of a path in C. We will soon see that the theory of such paths is
actually remarkably complicated but, for now, we will essentially only be interested in the definitions
and examples. Paths and contours are important to the notion of integration which we will meet
later.
Definition 1.12. Let [a, b] be an interval in R. A curve is a continuous function γ : [a, b] → C.
• The image set of γ is γ ∗ := {γ(t) : a ≤ t ≤ b}. Since [a, b] is compact and γ is continuous, γ ∗
is compact by Proposition 1.6. It also comes equipped with an orientation, corresponding to
t increasing from a to b.
• The initial point of γ is γ(a) and the final point of γ is γ(b).
• The curve is closed if γ(a) = γ(b), i.e. γ ∗ forms a “loop” in C. (Do not confuse this with the
notion of a closed set).
• The curve is simple if, for a ≤ s < t ≤ b, γ(s) 6= γ(t) unless s = a, t = b, i.e. γ ∗ does not
cross itself, except that we allow it to be closed.
• The curve is smooth if γ has a continuous derivative on [a, b]. Here γ is a complex-valued
function of a real variable and differentiability is defined exactly as for real-valued functions;
we have γ ′ (t) = (Reγ)′ (t) + i(Imγ)′ (t).
There are several simple operations that we can carry out on curves:
• Inversion: given a curve γ : [a, b] → C, there is a curve −γ : [a, b] → C with the same image
set but the opposite orientation:
(−γ)(t) = γ(a + b − t),
for t ∈ [a, b].
• Restriction: given a curve γ : [a, b] → C and a ≤ a1 < b1 ≤ b, we can restrict the function γ
to [a1 , b1 ] get γ|[a1 ,b1 ] : [a1 , b1 ] → C
γ|[a1 ,b1 ] (t) = γ(t),
for t ∈ [a1 ,1 b].
• Concatenation: given curves γ1 : [a, c] → C and γ2 : [c, b] → C with γ1 (c) = γ2 (c), we can
form the join γ of γ1 and γ2 :
(
γ1 (t), for t ∈ [a, c],
γ(t) =
γ2 (t), for t ∈ [c, b].
Note that, in this situation, we have γ1 = γ|[a,c] and γ2 = γ|[c,b] .
7
Definition 1.13. A path is a curve γ : [a, b] → C which is the join of finitely many smooth curves,
i.e. there exist a = a0 < a1 < · · · < an = b such that, for 0 ≤ k ≤ n − 1, γ|[ak ,ak+1 ] is smooth.
Remark. Note that two paths can have the same image set but be different paths. For example,
the path γ : [0, 1] → C given by γ(t) = t + it and the path η : [0, 2] → C given by η(t) = 41 (t2 + it2 )
both have image set the straight line segment [0, 1 + i].
In general, the image set γ ∗ of a path can be extremely complicated but we will mostly consider
two fairly simple types of path:
• Straight lines: for any w, z ∈ C, the image of the path γ given by
γ(t) = (1 − t)w + tz,
for t ∈ [0, 1]
is the line segment [w, z], traced from w to z. We will usually use the notation [w, z] for this
path, as well as its image.
• Circular arcs: any circular arc traced clockwise is the image of a path γ given by
γ(t) = a + reit ,
for t ∈ [θ1 , θ2 ],
for some a ∈ C, r > 0 and 0 < θ1 − θ2 ≤ 2π. In particular, γ(a; r) will denote the circular
path with centre a ∈ C and radius r > 0 given by
γ(a; r)(t) = a + reit ,
for t ∈ [0, 2π].
Definition 1.14. A contour is a simple closed path which is the finite join of straight lines and
circular arcs.
Note: it is really the image set which is a union of straight lines and circular arcs but the meaning
is clear.
1.3
Connectedness and simple connectedness
In this section, we return to topological notions, which we need to know what sorts of subset of C
are “good” for differentiating and integrating on.
Recall that we saw that, on certain sorts of set S, it is possible to have a function which is locally
constant but is not constant – namely, sets which are in “two pieces”. To avoid this problem, we
introduce the following definition:
Definition 1.15. An subset S of C is disconnected if there exist open subsets U, V of C such that
(i) U ∩ V = ∅;
(ii) S⊆U ∪ V ;
(iii) both S ∩ U and S ∩ V are non-empty.
Otherwise, S is connected. A connected open subset S of C is called a region.
Remark. If S is a region and we have found open subsets U and V such that U ∩ V = ∅ and
U ∪ V = S then either U = S, V = ∅ or U = ∅, V = S.
8
We will soon see that any open disc is connected and that C is connected. In the meantime, we
have the following.
Theorem 1.16. Let S be a region and f : S → C a function. Then f is locally constant if and
only if f is constant.
Proof We have already remarked that, if f is constant, then it is clearly locally constant.
Conversely, suppose f is locally constant. Fix a ∈ S, put b = f (a) and define
U = {z ∈ S : f (z) = b},
V = {z ∈ S : f (z) 6= b}
Then V is open as f is continuous and U is open as f is locally constant (check the details!). But
U ∪ V = S and U ∩ V = ∅ so, since S is connected, either U = ∅ or V = ∅. But we have a ∈ U so
V = ∅ and U = S, i.e. f (z) = b for all z ∈ S.
Putting this together with Theorem 1.9, we get:
Corollary 1.17. Let S be a region and f : S → C a holomorphic function with f ′ (z) = 0 for all
z ∈ S. Then f is constant.
We now have what will turn out to be a (very) closely related notion:
Definition 1.18. Let S be an open subset of C.
(i) Two points w, z ∈ S are path-connected in S if there exists a path γ with γ ∗ ⊆S, initial point
w and finial point z.
(ii) S is path-connected if every pair of points w, z ∈ S is path-connected in S.
Proposition 1.19. Let S be an open subset of C. Then S is connected if and only if it is pathconnected.
Proof Suppose first S is connected and pick w ∈ S. Then put
U = {z ∈ S : z, w are path-connected in S},
V = {z ∈ S : z, w are not path-connected in S}.
Now for z ∈ S there exists δ > 0 such that D(z; δ)⊆S, since S is open. Then, for z ′ ∈ D(z; δ), the
straight-line path [z, z ′ ] is contained in S. Hence w, z are path-connected in S if and only if w, z ′
are path-connected in S, since we can just join the line segment [z, z ′ ] to any path from w to z or
z ′ . This shows that U, V are open. But U ∪ V = S and U ∩ V = ∅ so, since S is connected and
w ∈ U 6= ∅, we have V = ∅ and U = S.
Conversely, suppose S is path-connected but we have non-empty open sets U, V with S = U ∪ V
and U ∩ V = ∅. We pick u ∈ U and v ∈ V and let γ : [a, b] → S be a path joining u and v. At
least one continuous piece of γ has one end in U and the other in V so, restricting to this piece if
necessary, we may assume that γ is continuous. Define a function h : [a, b] → R by
(
0
if γ(t) ∈ U,
h(t) =
1
if γ(t) ∈ V.
We will show that h is continuous, giving a contradiction by the Intermediate Value Theorem.
9
Let c ∈ [a, b] and suppose γ(c) ∈ U (the argument for γ(t) ∈ V is identical). Since U is open, there
exists ε > 0 such that D(γ(c); ε)⊆U and, since γ is continuous, there exists δ > 0 such that
|γ(c) − γ(d)| < ε,
for |c − d| < δ.
Hence, for |c − d| < δ, γ(d) ∈ U also so |h(c) − h(d)| = 0 and h is continuous at c.
N.B The notions of connectedness and path-connectedness for general subset of C are not equivalent. There are pathological sets like
A=
∞
[
1
n=1
1
,
+
i
∪ [0, 1] ∪ {i},
n n
which is connected but not path-connected (there is no path from i to any other point of A).
Example 1.20. An open subset S of C is convex if, for all w, z ∈ S, the line segment [w, z]⊆S.
A convex open subset is clearly path-connected so connected, i,e a region. In particular, the open
discs D(a; r) and C are regions.
The notion of connectedness allows us to distinguish topologically between open discs and disjoint
unions of them. Now we seek a notion which will allow us to distinguish topologically between
open discs and annuli. Informally, the difference is that every simple closed path in a disc can be
continuously shrunk to a point but this is not true in an annulus. We will not be very rigorous in
our approach here.
Let γ, η be any two paths in an open set S. After reparametrizing, if necessary, we may assume
that they are defined on the same interval [a, b] of R.
Definition 1.21. With notation as above, we say that γ is homotopic to η if there is a continuous
function
ψ : [a, b] × [c, d] → U
Defined on a rectangle [a, b] × [0, 1], such that
ψ(t, 0) = γ(t)
and
ψ(t, 1) = η(t),
for t ∈ [a, b].
For each s ∈ [0, 1], the function
ψs (t) = ψ(t, s),
for t ∈ [a, b]
is a continuous curve; the family of curves ψs is a deformation of the path γ to the path η. If γ
and η are closed paths, then we will assume that all the ψs are closed also.
Definition 1.22. Fairly rigorous: A connected open set S is called simply connected if every closed
path in S is homotopic to some point w ∈ S (i.e. homotopic to the constant path γ(t) = w, for
t ∈ [a, b].)
Less rigorous but acceptable: A connected open set S is simply connected if every closed path in S
can be continuously shrunk to a point.
Very informally, simply connected regions have no “holes” in them.
Examples 1.23.
(i) Any convex region is simply connected, in particular any disc D(a; r) or C.
(ii) No open annulus is simply connected, though the proof is non-trivial.
(iii) The set C− := C \ (−∞, 0] is simply connected.
10
2
Integration
In this chapter, we will introduce the notion of integration along a path, for continuous complex
functions and give some important results in the theory: the estimation theorem, which will be
especially important when doing contour integrations next semester; the fundamental theorem of
calculus; and the integration of power series.
2.1
Complex integration
We will assume a familiarity with the definitions (though these are not so important) and simple
properties of real integration of continuous functions. We will actually need a slight generalization
of this:
Let [a, b] be a closed interval in R and let h : [a, b] → R be a piecewise continuous function, i.e. h
is continuous except possibly at a finite number of points a = a0 < a1 < · · · < an = b. Then we
define the integral:
Z b
n−1
X Z ak+1
h(t)dt =
h(t)dt.
a
ak
k=0
Now let g : [a, b] → C be a piecewise continuous function. As usual, we can write the function
g(t) = Re g(t) + iIm g(t), for t ∈ [a, b]. Then we define the complex integral:
Z
Z
b
g(t)dt :=
a
b
Re g(t)dt + i
Z
b
Im g(t)dt.
a
a
The following properties of the complex integral follow straightforwardly from the same properties
of the real integral:
• Linearity: for f, g piecewise continuous functions on [a, b] and λ, µ ∈ C,
Z
Z
b
(λf + µg)(t)dt = λ
a
b
f (t)dt + µ
Z
b
g(t)dt.
a
a
• Additivity: for f : [a, b] → C and a < c < b,
Z
b
f (t)dt =
a
Z
c
f (t)dt +
Z
b
f (t)dt,
and
a
c
a
Z
b
f (t)dt = −
Z
a
f (t)dt.
b
• Substitution: If φ : [a, b] → [c, d] is continuous and f : [c, d] → C then
Z
d
f (s)ds =
Z
b
f (φ(t))φ′ (t)dt.
a
c
• Estimation: If f : [a, b] → C then
Z b
Z b
|f (t)|dt.
f (t)dt ≤
a
a
11
(2.1)
Rb
To prove this, put I = a f (t)dt. Let θ = argI, 0 ≤ θ < 2π, and λ = e−iθ so that |λ| = 1 and
Rb
λ a f (t)dt is real and positive. Then
Z b
Z b
Z b
f (t)dt = λ
λf (t)dt
f (t)dt =
a
=
Z
a
2.2
a
a
b
Z
Re(λf (t))dt ≤
b
a
|λf (t)|dt =
Z
b
a
|f (t)|dt.
Integration along a path
Definition 2.2. Let γ : [a, b] → C be a path and a = a0 < a1 < · · · < an = b such that γ|[ak ,ak+1 ]
is smooth, for 0 ≤ k ≤ n − 1. Let f : γ ∗ → C be continuous so that (f ◦ γ)γ ′ : [a, b] → C is a
piecewise continuous function and hence integrable. Then we define
Z b
Z
f (γ(t)) γ ′ (t)dt.
f (z)dz :=
a
γ
The motivation for this definition is the formal substitution of γ(t) for z (and γ ′ (t)dt for dz).
For example, to compute the integral of f (z) = |z| along the straight√line path [0, 1 + i]: we have
γ(t) = t + it, with t ∈ [0, 1], so γ ′ (t) = 1 + i and f (γ(t)) = |t + it| = t 2; then
" √
#1
√
Z 1
Z
t2 2(1 + i)
2
|t + it|(1 + i)dt =
|z|dz =
=
(1 + i).
2
2
0
[0,1+i]
0
Example 2.3. This example is very important. We consider f (z) = z n , for n ∈ Z, and γ = γ(0; r);
then γ(t) = reit , for t ∈ [0, 2π], so γ ′ (t) = ireit and f (γ(t)) = (reit )n = r n eint . Then
Z 2π
Z 2π
Z
r n eint ireit dt = ir n+1
z n dz =
ei(n+1)t dt
0
γ(0;r)
0
(
0
for n 6= −1;
=
2πi for n = −1.
We continue with some straightforward (but important!) properties. Suppose γ : [a, b] → C is a
path, that f, g : γ ∗ → C are continuous and that λ, µ ∈ C.
R
R
R
• Linearity: γ (λf + µg)(z)dz = λ γ f (z)dz + µ γ g(z)dz.
R
R
• Inversion: −γ f (z)dz = − γ f (z)dz.
• Additivity: if a < c < b put γ1 = γ|[a,c] and γ2 = γ|[c,b] then
Z
Z
Z
f (z)dz.
f (z)dz +
f (z)dz =
γ2
γ1
γ
• Reparametrization: suppose ψ : [e
a, eb] → [a, b] is a continuously differentiable bijection and
put γ
e : γ ◦ ψ. Then
Z
Z
f (z)dz.
f (z)dz =
γ
γ
e
12
Proof By additivity, we may (and do) assume γ (so γ
e also) is smooth. For t ∈ [e
a, eb], we
′
′
′
have γ
e (t) = γ (ψ(t)) ψ (t), by the chain rule. Then, making the substitution s = ψ(t),
Z eb
Z eb
Z
f (γ(ψ(t))) γ ′ (ψ(t))ψ(t)dt
f (e
γ (t)) γ ′ (t)dt =
f (z)dz =
a
e
e
a
γ
e
=
Z
b
′
f (γ(s))γ (s)ds =
a
Z
f (z)dz.
γ
Proposition 2.4 (Fundamental Estimate). Suppose that γ : [a, b] → C is a path and that f : γ ∗ →
C is continuous. Then
Z
Z b
′ f (z)dz ≤
γ (t) dt.
|f
(γ(t))|
γ
a
Z b
Z
Z b
′
|f (γ(t))| γ ′ (t) dt, by (2.1).
f (γ(t))γ (t)dt ≤
Proof f (z)dz = a
a
γ
Corollary 2.5. In the same situation, put M = supz∈γ ∗ |f (z)| and l =
γ. Then
Z
f (z)dz ≤ M l.
Rb
a
|γ ′ (t)|dt, the length of
γ
Note that this definition of length gives the expected value for line segments and circular arcs.
Also, as γ ∗ is compact and f is continuous on γ ∗ , Proposition 1.6 says that M = supz∈γ ∗ |f (z)| is
finite.
Proof Since M = supz∈γ ∗ |f (z)|, we have |f (γ(t))| ≤ M , for all t ∈ [a, b]. Hence
Z
Z b
Z b
Z b
′ ′ ′ f (z)dz ≤
γ (t) dt ≤
γ (t) dt = M
γ (t) dt = M l.
|f
(γ(t))|
M
γ
a
a
a
Example 2.6. Let f (z) = 1/(z 2 + 1) and let γR : [0, π] → C be the semicircular arc γ(t) = Reit ,
with R > 1. Then l(γR ) = πR so, by Corollary 2.5,
Z
1
πR
f (z)dz ≤ sup 2 2it
+ 1
0≤t≤π R e
γR
Now R2 e2it + 1 ≥ |R2 e2it | − 1 = R2 − 1, by the reverse triangle inequality. Hence
Z
Rπ
f (z)dz ≤ 2
.
R −1
γR
Note also that, as R → ∞,
Hence
π/R
0
Rπ
=
= 0.
→
2
2
R −1
1 − 1/R
1−0
Z
lim R→∞
γR
f (z)dz = 0.
13
Warning When using the fundamental estimate, do not make the mistake of writing
Z
Z
f (z)dz ≤
|f (z)|dz,
γ
γ
as this may be false. For example, if f (z) = 1/z and γ = γ(0; 1) then, from Example 2.3, we have
Z
Z 2π
Z
2π
f (z)dz = |2πi| = 2π
|e−it |ieit dt = eit 0 = 0.
|f (z)|dz =
but
γ
0
γ
Theorem 2.7 (Fundamental Theorem of calculus). Suppose that γ : [a, b] → C is a path, that F
is a holomorphic function on an open set containing γ ∗ and that F ′ (z) is continuous at each point
of γ ∗ . Then
Z
γ
F ′ (z)dz = F (γ(b)) − F (γ(a)) .
In particular, if γ as above is closed then
R
γ
F ′ (z)dz = 0.
Proof By additivity, we may (and do) assume that γ is smooth. Then
Z
F (z)dz =
γ
Z
b
F ′ (γ(t))γ ′ (t)dt =
a
=
Z
Z
b
a
b
a
(F ◦ γ)′ (t)dt
′
Re (F ◦ γ) (t)dt + i
Z
b
a
Im (F ◦ γ)′ (t)dt
= [Re(F ◦ γ)(t) + iIm(F ◦ γ)(t)]ba
= F (γ(b)) − F (γ(a)) ,
where the penultimate equality is by applying the fundamental theorem of calculus for real functions
to Re(F ◦ γ) and Im(F ◦ γ).
Example . If f (z) = cos z and γ : [0, π] → C is the semicircular arc γ(t) =
sin′ z = cos z,
Z
cos z dz = sin − π2 − sin π2 = −2.
π it
2e
then, since
γ
On the other hand, if we tried to do this without the fundamental theorem, we would get
Z π
Z
cos π2 eit i π2 eit dt,
cos z dz =
γ
0
which looks horrible!
Remark
. If we are given a function f which is continuous on the image some closed path γ,
R
and γ f (z)dz 6= 0 then, by the Fundamental Theorem of Calculus, there can be no holomorphic
function F on γ ∗ such that F ′ = f . For example, we have see in Example 2.3 that
Z
1
dz = 2πi 6= 0.
z
γ(0;1)
Hence there is no holomorphic function F : C → C such that F ′ (z) = z1 , for all z ∈ C.
14
We recall the definition of uniform convergence. Let S be a subset of C, let fn : S → C be functions,
for n = 1, 2, . . ., and let f : S → C be another function. Then fn converges uniformly to f on S if,
for all ε > 0, there exists N such that, for all n ≥ N ,
sup |fn (z) − f (z)| < ε,
z∈S
i.e. limn→∞ (supz∈S |fn (z) − f (z)|) = 0.
Recall also that the uniform limit of continuous functions is continuous so, in particular, integrable.
Theorem 2.8. Let γ : [a, b] → C be a path and, for n = 1, 2, . . ., let fn : γ ∗ → C be a continuous
function. Suppose that fn converges uniformly o f : γ ∗ → C on γ ∗ . Then
Z
Z
f (z)dz.
fn (z)dz =
lim
n→∞ γ
γ
R
R
Proof Write I = γ f (z)dz and In = γ fn (z)dz, so we need to show that In → I as n → ∞; that
is, we need to show that, for any ε > 0, there is an N such that, for all n ≥ N , |In − I| < ε.
So let ε > 0 and put l = l(γ), the length of γ. By uniform convergence, there is an n such that, for
all n ≥ N , supz∈γ ∗ |fn (z) − f (z)| < ε/l. Then, for n ≥ N ,
Z
|In − I| = fn (z) − f (z)dz ≤ sup |fn (z) − f (z)| l(γ) by Corollary 2.5
γ
<
z∈γ ∗
ε
ll =
ε,
as required.
In particular, we can apply Theorem 2.8 together with the Weierstrass M
P-test: Suppose that γ is
a path and that u0 , u1 , . . . are continuous functions on γ ∗ . Put fn (z) = kn=1 uk (z), so that fn is
a continuous function o γ ∗ and, by additivity,
Z
n Z
X
fn (z)dz =
uk (z)dz.
γ
k=0 γ
Suppose f : γ ∗ → C is such that fn (z) converges to f (z), for all z ∈ γ ∗ .
Suppose further that there exist real constants Mk such that
(i) |uk (z)| ≤ Mk , for all z ∈ γ ∗ and k = 0, 1, . . .;
P
(ii)
Mk converges.
Then, by the Weierstrass M -test, fn converges uniformly to f on γ ∗ so, by Theorem 2.8,
Z
Z
∞ Z
n Z
X
X
uk (z)dz.
uk (z)dz =
fn (z)dz = lim
f (z)dz = lim
γ
n→∞ γ
n→∞
k=0
γ
k=0
γ
In other words, where we can apply the Weierstrass M -test, we can exchange the order of integrals
and infinite sums:
Z X
∞
∞ Z
X
uk (z)dz =
uk (z)dz.
γ k=0
k=0 γ
15
k−3 /k! and
Examples. (i) Let f (z) = ez /z 3 and γ = γ(0; r), with
P r > 0. We put uk (z) = z
k−3
∗
Mk = r /k!, so |uk (z)| ≤ Mk , for all z ∈ γ , and
Mk converges, by the ratio test:
r k−2 /(k + 1)!
r
Mk+1
=
=
→ 0 as k → ∞.
k−3
Mk
r /k!
k+1
(In fact, the sum converges to er /r 3 .) Then
Z X
∞ Z
∞
X
z k−3
2πi
z k−3
dz =
dz =
= πi.
f (z)dz =
k!
2!
γ k!
γ
γ
Z
k=0
k=0
(ii) Consider the function f (z) =
1
z−a
integrated around γ = γ(0; 1).
Suppose first |a| < 1; then, if z ∈ γ ∗ , |a/z| < 1 so
∞
X
ak
1/z
=
.
f (z) =
1 − a/z
z k+1
k=0
P
Putting Mk = |a|k , we have that
Mk converges (either by the ratio test again, or by
noticing that the limit is (1 − |a|)−1 ) so
Z
γ(0;1)
1
dz =
z−a
Z X
∞
∞ Z
X
ak
ak
dz
=
dz = 2πi.
k+1
z k+1
γ z
γ
k=0
k=0
Now suppose |a| > 1; then, if z ∈ γ ∗ , |z/a| < 1 so
f (z) =
∞
X zk
−1/a
= −
.
1 − z/a
ak+1
k=0
Putting Mk = 1/|a|k+1 , we have that
Z
γ(0;1)
1
dz = −
z−a
P
Mk converges to (|a| − 1)−1 so
Z X
∞
∞ Z
X
zk
zk
dz
=
−
dz = 0.
k+1
ak+1
γ a
γ
k=0
16
k=0
3
Cauchy’s Theorem
3.1
The proof of Cauchy’s Theorem
The aim of this section is to prove some deep theorems about the integral of holomorphic functions,
especially along closed paths. The prototype for these is the following Lemma:
Lemma 3.1 (Goursat’s Lemma). Suppose f is holomorphic inside and on a triangular path γ.
Then
Z
f (z)dz = 0.
γ
The proof is technically difficult, but the idea is the following: We split our triangle into many
smaller triangles, and use the fact that f is holomorphic to show that the integral around each
of these small triangles is so small that, even once we have added up the integrals around all the
small triangles, we get that the total integral is very small. Splitting up into smaller and smaller
triangles, we can show that the integral is as small as we like.
Proof Without loss of generality, we assume the path goes anticlockwise around γ.
Write γ = γ0 , the boundary of the closed triangle ∆0
shown:
Put I =
Z
γ0∗
f (z)dz. We will show that, for any ε > 0,
γ
we have |I| ≤ ε; since this will be true for any ε > 0, we
deduce that |I| = 0 so I = 0, as required. So we fix ε > 0.
∆0
First, we split the triangle into four similar triangles by
joining the midpoints of the sides of ∆0 . Let η1 , . . . , η4
denote the (anti-clockwise) paths around these 4 triangles:
Then
I =
Z
γ
f (z)dz =
4 Z
X
i=1
η1∗
γ0∗
η3∗
f (z)dz,
η2∗
ηi
η4∗
∆0
because the integrals along the lines inside the triangle cancel, since they are in opposite directions.
Taking absolute values and applying the triangle inequality, we get
4 Z
X
f (z)dz .
|I| ≤
i=1
ηi
Z
Hence, for at least one value of i (namely, the one for which f (z)dz is maximal),
ηi
Z
|I| ≤ 4 f (z)dz .
ηi
17
We relabel ηi as γ1 . Notice that the length l(γ1 ) is half the length L = l(γ) of the original path.
Now we repeat the process with this smaller triangle. In this way, we get a sequence γ0 , γ1 , γ2 , . . .
of (anti-clockwise) triangular paths such that:
(i) γ0 = γ;
(ii) for each n, ∆n+1 ⊆∆n , where ∆n is the closed triangle with boundary γn∗ ;
(iii) the length l(γn ) = 2−n L;
Z
n
f (z)dz , for n ≥ 0.
(iv) |I| ≤ 4 γn
We now need to know the following:
from:
T∞
n=0 ∆n
6= ∅. Since the triangles ∆n are compact, this follows
Lemma. For n ≥ 0, let Sn be a compact subset of C, with Sn+1 ⊆ Sn . Then
T∞
n=0 Sn
6= ∅.
Proof For each n ≥ 0, pick any zn ∈ Sn . Then the sequence {zn } is bounded (since it is
contained in the bounded set S0 ) so, by the Bolzano-Weierstrass Theorem, it has a convergent
subsequence {znk }, with limit
T w. This w is then a limit point of each set Sn so, since Sn is
closed, w ∈ Sn . Hence w ∈ ∞
n=0 Sn .
T
Now choose any w ∈ ∞
n=0 ∆n . Since f is holomorphic at w, there is an r > 0 such that, for
z ∈ D(w; r) with z =
6 w,
f (z) − f (w)
ε
′
− f (w) < 2 .
z−w
L
We multiply through by (z − w) to get
f (z) − f (w) − (z − w)f ′ (w) <
(∗)
ε
|z − w|,
L2
for all z ∈ D(w; r).
Now choose N large enough so that ∆n ⊆D(w; r). Then, for any z ∈ ∆N , we have |z − w| < l(γN )
∗ ⊆∆ , we get
since w ∈ ∆N also. Putting this together with (∗), and since γN
N
sup f (z) − f (w) − (z − w)f ′ (w) ≤
(†)
∗
z∈γN
ε
ε
l(γN ) = N .
L2
2 L
Now we use the trick of adding and subtracting something to the integrand in our integral around
γN :
Z
Z
Z
′
f (z)dz =
f (z) − f (w) − (z − w)f (w) dz +
f (w) + (z − w)f ′ (w) dz.
(‡)
γN
γN
γN
Now the second integral on the RHS is 0 by the Fundamental Theorem of Calculus 2.7, since the
integrand
d
(z − w)2 ′
′
f (w) + (z − w)f (w) =
f (w)
f (w)z +
dz
2
and the path γN is closed.
18
Finally, we apply the estimation of Corollary 2.5 on (‡), and use (†) to get
Z
Z
′
f (z)dz = f (z) − f (w) − (z − w)f (w) dz γN
γN
≤ sup f (z) − f (w) − (z − w)f ′ (w) l(γN )
∗
z∈γN
≤
Then property (iv) gives
ε
L
2N L 2N
Z
|I| ≤ 4 N
γN
=
ε
.
4N
ε
f (z)dz ≤ 4N N = ε,
4
which completes the proof of Goursat’s Lemma.
Theorem 3.2 (Antiderivative Theorem). Let S be a convex region and let f : S → C be a holomorphic function. Then there exists a holomorphic function F : S → C such that F ′ = f .
Proof Choose a point a ∈ S. We define a function F : S → C by
Z
f (w)dw,
F (z) =
[a,z]
where [a, z] is the straight line segment from a to z.
Now fix z ∈ S. Since S is open, there is an r > 0 such that D(z; r)⊆S, so that z + h ∈ S whenever
|h| < r. We need to show that F ′ (z) = f (z), that is
F (z + h) − F (z)
lim − f (z) = 0.
h→0
h
Since S is convex, for any h with |h| < r, the triangle with vertices a, z, z + h is contained in S.
Then
Z
Z
Z
f (w)dw = 0,
f (w)dw +
f (w)dw +
[a,z]
[z+h,a]
[z,z+h]
by Goursat’s Lemma 3.1, since we are integrating around a triangle. Hence
Z
Z
Z
f (w)dw.
f (w)dw =
f (w)dw −
F (z + h) − F (z) =
[a,z]
[a,z+h]
Finally, since
Z
[z,z+h]
f (z)dw = f (z)h (recall that we have fixed z so f (z) is a constant in this
[z,z+h]
integral with respect to w) we get, using the estimation of Corollary 2.5,
Z
F (z + h) − F (z)
1
=
f
(w)dw
−
f
(z)h
−
f
(z)
h
|h| [z,z+h]
Z
1 =
(f (w) − f (z)) dw
|h| [z,z+h]
≤
=
1
|h| sup |f (z) − f (w)|
|h| w∈[z,z+h]
sup
w∈[z,z+h]
by continuity of f at z.
|f (z) − f (w)| → 0 as h → 0,
19
Putting this together with the Fundamental Theorem of Calculus, we get an improvement on
Goursat’s Lemma:
Corollary 3.3. Let S be a convex region, let f : S → C be a holomorphic function, and let γ be a
closed path in S. Then
Z
f (z)dz = 0.
γ
Proof By the Antiderivative Theorem 3.2, there is a function F : S → C such that F ′ = f . Then,
since γ is closed
Z
Z
F ′ (z)dz = 0,
f (z)dz =
γ
γ
by the Fundamental Theorem of Calculus 2.7.
Finally, we can prove a first version of Cauchy’s Theorem:
Theorem 3.4 (Cauchy’s Theorem – version I). Suppose f is a function which is holomorphic on
and inside a simple closed path γ. Then
Z
f (z)dz = 0.
γ
Remarks. (i) We are assuming a very deep and difficult result. This is called the Jordan Curve
Theorem, which says that every simple closed path has an inside (which is open and bounded)
and an outside (which is open and unbounded). We will mostly be looking at paths which
are contours (joins of straight line segments and circular arcs), for which there is a fairly
elementary proof in Priestley’s book.
(ii) Note that we can also describe Cauchy’s Theorem another way: suppose γ1 , γ2 are simple
paths with the same initial and final points, and with no intersection. Suppose f is holomorphic on γ1∗ and γ2∗ and between them. Then
Z
Z
f (z)dz.
f (z)dz =
γ2
γ1
This follows from Cauchy’s Theorem by considering the simple closed path which is the join
of γ1 and −γ2 .
Proof of Theorem 3.4 The proof goes in two steps: First we prove the Theorem when γ is a
polygonal path – that is, it is made up of straight line segments. Then we consider the general case
and approximate our path by a polygonal path.
So suppose first that
Z γ is a polygonal path. Then we can triangulate it: that is, we cut it up into
f (z)dz is the sum of the integrals around the triangles (the integrals on the
triangles. Then
γ
interior lines cancel, as in the proof of Goursat’s
Lemma 3.1) so, since the integral around each
Z
f (z)dz = 0, as required.
triangle is 0 (by Goursat’s Lemma), we get
γ
Now let γ be any simple closed path and let S be an open set containing γ ∗ and its interior, and
such that f is holomorphic in S. Note that there is such a set: if we write Γ for the union of γ ∗
20
and its interior, then f is holomorphic in
SΓ so, for all z ∈ Γ, there is an open neighbourhood Nz of
z on which f is holomorphic; then S = z∈Γ Nz is an open set with the required properties.
Now we need to use a second result, whose proof can also be found in Priestley’s book:
Fact (Covering Theorem) We can cover γ ∗ by a finite number of overlapping open discs D1 , . . . , DN
contained in S.
We pick points γ(t1 ) ∈ D1 ∩ D2 , γ(t2 ) ∈ D2 ∩ D3 , etc. and join these with straight lines. Since
each disc Di is convex and γ(ti ), γ(ti+1 ) ∈ Di , Corollary 3.3 shows that
Z
Z
f (z)dz.
f (z)dz =
[γ(ti ),γ(ti+1 )]
γ|[ti ,ti+1 ]
[Recall that γ|[ti ,ti+1 ] is the restriction of the path γ, while [γ(ti ), γ(ti+1 )] is the straight line segment.]
Then
Z
f (z)dz =
γ
XZ
i
XZ
f (z)dz =
γ|[ti ,ti+1 ]
i
f (z)dz = 0,
[γ(ti ),γ(ti+1 )]
since the final integral is the integral around the closed polygonal path joining γ(t1 ), . . . , γ(tN ). Cauchy’s Theorem can then be applied to give the following, which describes how integrals of
holomorphic functions do not change when we deform the path around which we are integrating.
We say that a closed path γ is positively oriented if, as t increases, γ(t) ends up moving anticlockwise
around any point inside γ.
Theorem 3.5 (Deformation Theorem). Suppose γ1 , γ2 are positively oriented simple closed paths
and γ2∗ is inside γ1∗ . Suppose f is holomorphic on γ1 and γ2 and between them. Then
Z
Z
f (z)dz.
f (z)dz =
γ2
γ1
Proof We add two lines to the paths, as shown:
Then
Z
Z
Z
Z
f (z)dz
f (z)dz +
f (z)dz =
f (z)dz −
γ1
η1
γ2
η2
γ1∗
γ2∗
= 0,
η2∗
by Cauchy’s Theorem 3.4.
η1∗
Example. Let γ be any positively oriented simple closed path with 0 inside. Then, for some r > 0,
its image γ ∗ is inside the image of the circular path γ(0; r)∗ . Since 1z is holomorphic in C \ {0}, it
is certainly holomorphic on, and between, γ(0; r) and γ. Hence, by the Deformation Theorem 3.5,
Z
Z
1
1
dz =
dz = 2πi.
γ(0;r) z
γ z
21
There is also a second version of the Deformation Theorem, which says the following: Suppose S
is open, f : S → C is holomorphic and γ1 , γ2 are two closed paths in S which are homotopic in S.
Then
Z
Z
f (z)dz.
f (z)dz =
γ2
γ1
As a Corollary of this, we get a second version of Cauchy’s Theorem:
Theorem 3.6 (Cauchy’s Theorem II). Let S be a simply connected region, let f : S → C be a
holomorphic function, and let γ be a simple closed path with γ ∗ ⊆S. Then
Z
f (z)dz = 0.
γ
Example. Let S = {z ∈ C : a < |z| < b}, an annulus in C, which is a region (Exercise). If we put
1
f (z) = , then f : S → C is holomorphic. If a < r < b then γ(0; r) is a simple closed path in S.
z
But
Z
1
dz = 2πi 6= 0.
γ(0;r) z
Hence, by Cauchy’s Theorem 3.6, S cannot be simply-connected.
3.2
Consequences of Cauchy’s Theorem
In this section, we will see several important consequences of Cauchy’s Theorem.
Theorem 3.7 (Cauchy’s Integral Formulae). Let γ be a positively oriented simple closed path and
let f be a function which is holomorphic in and inside γ ∗ . If a is inside γ ∗ then:
Z
f (z)
1
dz;
(i) f (a) =
2πi γ z − a
(ii) for n = 1, 2, . . ., the nth derivative of f at a exists and
Z
n!
f (z) n+1
(n)
f (a) =
dz.
2πi γ z − a
Proof (i) Since a is inside γ ∗ and the inside I(γ) of γ ∗ is open, there is an r > 0 such that
D(a; r) ⊆ I(γ). Since f (z)/(z − a) is holomorphic on I(γ) \ {a}, the Deformation Theorem 3.5 tells
us that
Z
Z
f (z)
f (z)
dz =
dz,
γ(a;ε) z − a
γ z−a
for any 0 < ε < r. Similarly, we have
Z
Z
1
1
dz =
dz = 2πi,
γ(a;ε) z − a
γ z−a
as in Example 2.3. Then
Z
1 Z
1
f
(z)
f
(z)
−
f
(a)
= dz
−
f
(a)
dz
2πi
z
−
a
2πi
z
−
a
γ
γ(a;ε)
≤
1
2π
=
sup
z∈γ(a;ε)∗
sup
z∈γ(a;ε)∗
22
|f (z) − f (a)|
2πε
|z − a|
|f (z) − f (a)|,
which tends to 0 as ε → ∞. Hence the LHS is zero, as required.
(ii) We proceed by induction on n, the base case being n = 0 which is (i). So we assume the result
is true for n = k. The proof is then an exercise in the (inductive) definition of nth derivative and
the use of the Fundamental Estimate, rather similar to (i) but with nastier details – see Priestley’s
book.
Corollary 3.8. Suppose f is holomorphic in an open set S. Then f has derivatives of all orders
in S.
So this says that, as soon as f is differentiable once, it is infinitely differentiable. Note here the
contrast with the theory of real functions: the function f : R → R given by f (x) = x|x| is once
differentiable but not twice differentiable.
Proof Since S is open, for a ∈ S there is an r > 0 such that D(a; r)⊆S. Then the path γ =
γ(a; r/2) is a simple closed path in S, the function f is holomorphic on and inside γ, and a is inside
γ. Then Cauchy’s Integral Formulae 3.7 tell us that f (n) (a) exists, for all n ≥ 0.
Cauchy’s integral formulae can be used directly to compute some integrals:
Z
z2
Examples. (i) Compute
dz.
2
γ(i,1) z + 1
To do this we write
z 2 /(z + i)
z2
=
,
z2 + 1
z−i
since the numerator is then holomorphic on and inside γ(0; 1). Then we apply Cauchy’s
Integral Formulae 3.7(i) to get
2 Z
Z
z2
z 2 /(z + i)
z
−1
dz =
dz = 2πi
= 2πi
= −π.
2
z−i
z + i z=i
2i
γ(i,1) z + 1
γ(i,1)
(ii) Compute
Z
γ(0,1)
ez
dz.
z3
For this we apply Cauchy’s Integral Formulae 3.7(ii), with f (z) = ez (which is holomorphic
everywhere) and a = 0 (since the z 3 in the denominator is (z − 0)3 ) and n = 2 to get
Z
2πi d2 z
ez
dz =
(e )
= πi [ez ]z=0 = πi.
3
2! dz
γ(0,1) z
z=0
(iii) Compute
Z
γ(0,1)
Re(z)
dz.
z − 12
Here we need to be a little more cunning, since Re(z) is nowhere holomorphic, so we cannot
use Cauchy’s Integral Formulae with it. However, we are only interested in the function on the
path γ(0; 1) around which we are integrating, we can replace Re(z) with any other function
which takes the same values as Re(z) on γ(0; 1)∗ . Here, we can write
Re(z) =
z + z −1
z+z
=
,
2
2
23
since z = z −1 for z ∈ γ(0; 1)∗ . Notice that, by doing this, we have introduced an extra point
(namely 0) inside γ(0; 1) where our function is not defined.
So now we are integrating
(z + z −1 )/2
z2 + 1
=
.
z − 12
2z(z − 21 )
Now there are two approaches, one involving cutting our path, the other involving partial
fractions:
(a) We split our path into two closed paths γ1 , γ2 , as shown.
Notice that, as usual, the integrals along
the internal lines cancel, and that there is
only one point where the function is not
defined inside each of γ1 , γ2 .
γ1∗
γ2∗
γ∗
1
2
0
Then we apply Cauchy’s Integral Formulae 3.7(i) to get
Z
γ(0,1)
Z
Z
(z 2 + 1)/(2(z − 21 ))
(z 2 + 1)/2z
dz
dz +
z
(z − 21 )
γ1
γ1
"
#
2
z2 + 1
z +1
= 2πi
+ 2πi
2z
2(z − 12 )
z= 1
Re(z)
dz =
z − 12
z=0
= 2πi[−1] + 2πi
(b) We write
5
4
2
=
1
2 πi.
5
1 1
z2 + 1
4
−
+
=
,
2 z z − 21
2z(z − 21 )
where we have used partial fractions to simplify our function. Finally, we can apply
Cauchy’s Integral Formulae 3.7(i) to get
Z
γ(0,1)
Re(z)
dz =
z − 21
Z
γ(0,1)
1
dz −
2
Z
γ(0,1)
1
dz +
z
Z
γ(0,1)
5
4
z−
1 dz
2
= 0−2πi+2πi
as before.
5
4
=
1
2 πi,
Now we have all the tools to prove Taylor’s Theorem, which says that any function holomorphic
in an open disc can be written as a power series about the centre of the disc. This is a Theorem
which you have already been using in Calculus over the last two years.
Theorem 3.9 (Taylor’s Theorem). Suppose a function f is holomorphic in an open disc D(a; R).
Then there exist constants cn ∈ C such that
f (z) =
∞
X
n=0
cn (z − a)n ,
24
for all z ∈ D(a; R).
Moreover, for any 0 < r < R, we have
Z
1
f (w)
f (n) (a)
cn =
dw
=
.
2πi γ(a;r) (w − a)n+1
n!
Proof Let z ∈ D(a; R) and r be such that |z − a| < r < R, and put γ = γ(a; r); note then that z
is inside γ so, by Cauchy’s Integral Formula 3.7(i) we get
Z
f (w)
1
dw.
(∗)
f (z) =
2πi γ w − z
z−a ∗
Now, for w ∈ γ , we have w−a < 1 so (using the binomial expansion)
1
w−z
=
1
w−a
1
1−
z−a
w−a
!
∞
X
(z − a)n
=
.
(w − a)n+1
n=0
Substituting this in (∗), we get
1
f (z) =
2πi
(†)
Z X
∞
(z − a)n
f (w)dw.
(w − a)n+1
γ
n=0
We would like to exchange the sum with the integral but, to do this, we must first know that
we have uniform convergence. We do this (as in the examples following Theorem 2.8) using the
Weierstrass M -test.
Since f is continuous on the compact set γ ∗ , Proposition 1.6 implies that f is bounded on γ ∗ , by
M say: that is, |f (z)| ≤ M for all z ∈ γ ∗ . Now, for n ≥ 0, put
M |z − a| n
.
Mn =
r
r
Then:
n
(z − a)n
≤ |z − a| M = Mn ;
(i) for z ∈
f
(w)
(w − a)n+1
r n+1
P∞
M
|z − a| −1
(ii)
.
1−
n=0 Mn converges to
r
r
γ ∗,
Hence, by the Weierstrass M -test, the sum in the RHS of (†) converges uniformly in γ so, by
Theorem 2.8, we may interchange the sum and integral in (†) and we have
Z
∞ X
f (w)
1
dw
(z − a)n ,
f (z) =
n+1
2πi
(w
−
a)
γ
n=0
as required.
As an application of Taylor’s Theorem, we get the following amazing result:
25
Theorem 3.10 (Liouville’s Theorem). Suppose f is holomorphic in C and bounded, that is, there
is an M ∈ R such that |f (z)| ≤ M , for all z ∈ C. Then f is constant.
So the only bounded functions which are differentiable everywhere in C are the constant functions.
Again, notice the contrast with real function, where the sine function sin : R → R is everywhere
differentiable, bounded (by 1) but not constant. In case you are wondering what goes wrong with
the complex sine function sin : C → C, remember that sin(iz) = i sinh(z) so, since sinh(x) → ∞ as
x → ∞, the complex sine function is unbounded on the imaginary axis.
Proof of Theorem 3.10 By Taylor’s Theorem 3.9, we can write f (z) =
Z
1
f (z)
cn =
dz,
for any R > 0.
2πi γ(0;R) z n+1
P∞
n=0 cn z
n,
where
Now we apply Corollary 2.5 to get
cn
f (z) 1
M
M
1
l(γ(0; R)) sup n+1 ≤
2πR n+1 = n .
≤
2π
z
2π
R
R
|z|=R
This is true for any R > 0. But, for n > 0, we have RMn → 0 as R → ∞, so |cn | = 0 and cn = 0.
Substituting this into the power series expansion for f , we get f (z) = c0 , which is constant.
Finally, as an application of Liouville’s Theorem, we get:
Corollary 3.11 (Fundamental Theorem of Algebra). Suppose p(z) is a non-constant polynomial
with coefficients in C. Then there is a ξ ∈ C such that p(ξ) = 0.
So any polynomial with coefficients in C has a root in C and, by an easy induction on the degree,
in fact factorizes completely:
p(z) = z n + an−1 z n−1 + · · · + a1 z + a0 = (z − ξ1 ) · · · (z − ξn ),
for some ξ1 , . . . , ξn ∈ C.
Proof of Corollary 3.11 For contradiction, suppose there is no ξ ∈ C such that p(ξ) = 0. Then,
since p(z) is holomorphic in C and never 0, the function
f (z) =
1
p(z)
is holomorphic in C (by the quotient rule). We will show that f is bounded and then, by Liouville’s
Theorem 3.10, f must be constant; but then p is also constant, which contradicts the hypotheses.
To show that f is bounded, note first that |p(z)| → ∞ as |z| → ∞. (This is not hard to prove, so it
is left as an Exercise.) Hence there exists R > 0 such that |p(z)| > 1, for all z ∈ C with |z| > R, so
|f (z)| < 1,
for all z ∈ C \ D(0; R).
On the other hand, f is continuous on the compact set D(0; R), so f is bounded on D(0; R), by
Proposition 1.6: that is, there is an M such that |z| ≤ M , for all z ∈ D(0; R). But then
|f (z) < max{1, M },
so f is bounded, as required.
for all z ∈ C,
26
3.3
Zeros of holomorphic functions
In the final part of this section, we look at the zeros of holomorphic functions. Recall that D ′ (a; r) =
D(a; r) \ {a} is the punctured open disc of centre a and radius r.
Definition 3.12. Suppose f is holomorphic in D(a; r).
(i) a is a zero of f if f (a) = 0.
(ii) a is an isolated zero of f if f (a) = 0 and there exists ε > 0 such that D ′ (a; ε) contains no
zeros of f .
So (immediately from the definitions) a zero of f is isolated if and only if it is not a limit point of
zeros of f .
Theorem 3.13 (Identity Theorem). Suppose S is a region and f : S → C is holomorphic. Let
Z(f ) be the set of zeros of f in S. If Z(f ) has a limit point in S, then f ≡ 0 in S.
So, in the situation of the Theorem, if f is not the zero function then all the zeros of f in S are
isolated.
Proof P
First let a ∈ S and r > 0 be such that D(a; r)⊆S. By Taylor’s Theorem 3.9, we can write
n
f (z) = ∞
n=0 cn (z − a) in D(a; r). There are two possibilities:
(i) All coefficients cn = 0. In this case f ≡ 0 in D(a; r).
(ii) There is a least n ≥ 0 such that cn 6= 0. Putting g(z) =
f (z) = (z − a)n g(z),
P∞
m=0 cm+n (z
− a)m , we have
for z ∈ D(a; r),
while g is holomorphic in D(a; r) and g(a) = cn 6= 0. But then the continuity of g implies
that there is an ε > 0 such that g is non-zero in D(a; ε). Hence f is non-zero in the punctured
disc D ′ (a; ε) and a is not a limit point of Z(f ).
Now let E be the set of limit points of Z(f ) in S.
• If z ∈ E then there exists a sequence {zn } of elements of Z(f ) which converges to z. But
then, by continuity, f (z) = limn→∞ f (zn ) = 0, so z ∈ Z(f ). Hence E⊆Z(f ).
• If a ∈ E then we cannot be in case (ii) above so we must be in case (i); hence f ≡ 0 in some
D(a; r) and D(a; r)⊆E. So E is open.
• If a ∈ S \ E then we are in case (ii) above so f is non-zero in D ′ (a; ε) and hence D ′ (a :
ε)⊆S \ Z(f )⊆S \ E.
Hence S = E ∪ (S \ E), with both pieces open. Since S is connected, we have either E = ∅, in
which case Z(f ) has no limit points in S, or E = S, in which case f ≡ 0 on S.
Corollary 3.14 (Uniquenes Theorem). Suppose S is a region and f, g : S → C are holomorphic.
Let I(f, g) = {z ∈ S : f (z) = g(z)}. If I(f, g) has a limit point in S, then f ≡ g in S.
27
Proof Apply the Identity Theorem 3.13 to f − g.
(i) Suppose f is holomorphic in D(0; 1) and
f n1 = sin n1 ,
for n = 2, 3, . . .
The set n1 : n = 2, 1, . . . has a limit point 0 ∈ D(0; 1) so, by the Uniqueness Theorem 3.14,
f (z) = sin(z) for all z ∈ D(0; 1).
Example.
(ii) Suppose f is holomorphic in D(0; 1) and f (x) = f (−x) for x ∈ R, |x| < 1. Put g(z) =
f (z) − f (−z), which is holomorphic in D(0; 1). Then g(x) = 0 for x ∈ R ∩ D(0; 1), and this
set certainly has limit points in D(0; 1). Hence, by the Identity Theorem 3.13, g(z) = 0 for
all z ∈ D(0; 1), i.e. f (z) = f (−z) for all z ∈ D(0; 1).
(iii) Suppose f is holomorphic in D(1; 1) and
for n = 1, 2, . . . .
f n1 = 0,
The set n1 : n = 2, 1, . . . has a limit point 0, but 0 6∈ D(1; 1) so we cannot apply the Identity
Theorem. Indeed, the function
f (z) = sin πz
is holomorphic in D(1; 1) and satisfies the conditions, but is not the zero function.
28
4
Singularities and Residues
4.1
Laurent Expansions
When f is a function holomorphic in a disc, we know from Taylor’s Theorem 3.9 that it has a power
series expansion in the disc. However, for functions like
f (z) =
1
,
sin z
which is holomorphic in the punctured disc D ′ (0; 1), we cannot hope for this. Instead, we get a
more general Laurent series expansion:
Theorem 4.1. Let 0 ≤ r < R ≤ ∞ and A = {z ∈ C : r < |z − a| < R}, an annulus with centre a.
Let f : A → C be a holomorphic function. Then f has an expansion
∞
X
f (z) =
n=−∞
where
cn (z − a)n ,
1
=
2πi
cn
Z
γ
for z ∈ A,
f (w)
dw,
(w − a)n+1
with γ = γ(a; s) for any r < s < R.
The expansion given by the Theorem is called a Laurent expansion.
Proof Let z ∈ A and choose P, Q with
γ2∗
r < P < |z − a| < Q < S.
γ1∗
γ(a; Q)∗
Let γ1 , γ2 be the contours shown in the
picture. Then
γ(a; P )∗
a
f (z) = f (z) + 0
Z
Z
f (w)
f (w)
1
1
dz +
dz
=
2πi γ1 w − z
2πi γ2 w − z
[by Theorem 3.7(i)] [by Theorem 3.4]
=
1
2πi
Z
γ(a;Q)
f (w)
1
dz −
w−z
2πi
Z
γ(a;P )
f (w)
dz.
w−z
As in the proof of Taylor’s Theorem 3.9, we have
∞
X (z − a)n



f (w)

n+1

(w
−
a)

n=0
f (w)
=

w−z
∞

X

(w − a)m



f (w)
−
(z − a)m+1
m=0
29
(†)
for w ∈ γ(a; Q)∗ ,
for w ∈ γ(a; P )∗ ,
and these sums converge uniformly on γ(a; Q)∗ and γ(a; P )∗ respectively. Hence, substituting
in (†), we can exchange the sum and the integral to get
!
!
Z
Z
∞
∞
X
X
1
f (w)
f
(w)
1
f (z) =
dw (z −a)n +
dw (z −a)−m−1 .
2πi γ(a;Q) (w − a)n+1
2πi γ(a;P ) (w − a)−m
n=0
m=0
Finally, put m = −(n + 1) in the second sum and use the Deformation Theorem 3.5 to replace
γ(a; Q) and γ(a; P ) by the path γ in the statement of the Theorem.
A similar (but easier) proof shows that the coefficients in the Laurent expansion are unique. Hence,
if we can find a Laurent expansion for a function f then it is the Laurent expansion. Note that,
from the Autumn semester analysis, we can do this by multiply together known series expansions.
Examples.
(i) f (z) =
1
is holomorphic in
z(1 − z)
(a) A1 = D ′ (0; 1) = {z ∈ C : 0 < |z| < 1}. Here we have
∞
∞
X
1
1 X n
1
zn.
z =
+
=
+
f (z) =
z 1−z
z
n=0
(b) A2 = {z ∈ C : |z| > 1}. Here we have
1 1
f (z) =
−
z z
1
1−
1
z
!
=
n=−1
∞
1 1X 1
−
=
z z
zm
m=0
−2
X
n=−∞
−z n .
(c) A′1 = D ′ (1; 1) = {z ∈ C : 0 < |z − 1| < 1. Here we have
∞
∞
X
1
1
1 X
m
m
(−1)n (z − 1)n .
f (z) = −
(−1) (z − 1) =
= −
z − 1 1 + (z − 1)
z − 1 m=0
n=−1
1
is holomorphic in C except at z = kπ, for k ∈ Z. In particular, it is
sin z
holomorphic in A = D ′ (0; π). In A, we have
(ii) cosec (z) =
z3 z5
sin(z) = z −
+
− ···
3! 2 5! 4
z
z
6
+
+ O(z ) .
= z 1−
3!
5!
Hence
1
cosec (z) =
sin(z)
−1
z2 z4
6
−
+ O(z )
1−
3!
5!
!
2
2
2
4
z
z
z4
z
1
1+
−
+ O(z 6 ) +
−
+ O(z 6 ) + O(z 6 )
z
3!
5!
3!
5!
1
z4
z2 z4
6
−
+
+)(z )
1+
z
3!
5!
(3!)2
7z 3
1 z
+ +
+ O(z 5 ).
z 6 360
1
z
=
=
=
30
cos z
is also holomorphic in A = D ′ (0; π). We have just found an expression for
sin z
cosec (z), while we also have the expansion
(iii) cot(z) =
cos(z) = 1 −
z2 z4
+
+ O(z 6 ).
2!
4!
Multiplying together these two expression, we get
1 z
7z 3
z2 z4
6
5
+
+ O(z )
+ +
+ O(z )
cot(z) =
1−
2!
4!
z 6 360
1 z z
z3
z3
7z 3
=
− + +
−
+
+ O(z 5 )
z 2 6 24 12 360
1 z
z3
=
− +
+ O(z 5 ).
z 3 45
4.2
Singularities
P
n
From the examples above we see that, in a Laurent expansion f (z) = ∞
n=−∞ cn (z −a) , sometimes
we have non-zero terms cn for n right down to −∞, other times maybe only down to some finite
negative number. We will see that these two situations have rather different implications for the
function f .
Definition 4.2. Let f be a complex function.
(i) The point a ∈ C is a regular point of f if f is holomorphic at a.
(ii) The point a ∈ C is a singularity of f if it is not a regular point but it is a limit point of
regular points. A singularity a of f is called
(a) an isolated singularity if f is holomorphic in some punctured disc D ′ (a; r);
(b) an non-isolated essential singularity otherwise.
Non-isolated essential singularities are rather nasty but isolated singularities are somewhat nicer:
for example, we know from Theorem 4.1 that an isolated singularity a of f has a (unique) Laurent
series expansion about a.
Definition 4.3 (Classification of isolated singularities). Suppose a is an isolated singularity of f ,
so f has a unique Laurent series expansion
f (z) =
∞
X
n=−∞
cn (z − a)n ,
for z in some punctured disc D ′ (a; r). The point a is called
• a removable singularity if cn = 0 for all n < 0, so that f (z) =
P∞
n=0 cn (z
− a)n ;
• aPpole of order m (where m ≥ 1) if c−m 6= 0 but cn = 0 for all n < −m, so that f (z) =
∞
n
n=−m cn (z − a) ;
• an isolated essential singularity otherwise.
31
Poles of order 1, 2, 3, . . . are called simple poles, double poles, triple poles, etc.
Examples.
(ii)
1 3
z−i
(i)
has a triple pole at i.
1−cos z
z2
is holomorphic except at z = 0, where it has an isolated singularity. The Laurent
expansion about 0 is
1
1 z2
z2 z4
+
·
·
·
=
−
+ ···
1
−
1
−
z2
2!
4!
2 24
so the singularity at z = 0 is removable.
(iii) We saw that
1 z
7z 3
+ +
+ O(z 5 ),
z 6 360
so cosec (z) has a simple pole at 0. Since
cosec (z) =
for z ∈ D ′ (0, π)
cosec (z − kπ) = (−1)k cosec (z),
the singularities of cosec (z) at kπ (for k ∈ Z) are simple too.
(iv) For z ∈ C \ {0}, we have
∞
∞
1 2n+1
X
X
(−1)n −(2n+1)
1
n z
(−1)
=
z
.
=
sin
z
(2n + 1)!
(2n + 1)!
n=0
n=0
1
z
has an isolated singularity at 0.
1
(v) The function cosec 1z has singularities at kπ
, for each k ∈ Z \ {0}, each of which is a simple
pole. But is also has a singularity at 0, which is then a non-isolated essential singularity since
there are singularities arbitrarily close to it.
Hence sin
Poles and zeros
It is clear that there is a relationship between zeros of a function f and singularities of the function
1/f , since 1/f (z) is undefined whenever f (z) = 0. Suppose that f is holomorphic in an open disc
D(a; r) and f (a) = 0, but f 6≡ 0 in D(a; r). Then f has a Taylor expansion
f (z) = =
∞
X
n=m
cn (z − a)n ,
for z ∈ D(a; r),
for some integer m ≥ 1 with cm 6= 0. In this case, we say that f has a zero of order m at a. As for
poles, zeros of order 1, 2, . . . are called simple, double,. . .
We note that f has a zero of order m at a if and only if we can write
f (z) = (z − a)m g(z),
for z ∈ D(a; r),
where g is a function holomorphic in D(a; r) for which g(a) 6= 0 – in terms of the Taylor expansion
above, the function g is given by
g(z) =
∞
X
n=0
cn+m (z − a)n ,
32
for z ∈ D(a; r).
We also note that, since we have the formula
cn =
f (n) (a)
n!
for the coefficients in the Taylor expansion, f has a pole of order m at a if and only if
f (a) = 0,
f ′ (a) = 0,
...
f (m−1) (a) = 0,
f (m) (a) 6= 0.
This is a much more useful way to find the order of a zero than to write out the Taylor expansion.
Example. The function f (z) = z sin(z) has zeros at z = nπ, for n ∈ Z. Then
f ′ (nπ) = [sin(z) + z cos(z)]z=nπ = (−1)n nπ,
which is non-zero for n 6= 0. Hence f has simple zeros at z = nπ, for n ∈ Z \ {0}. On the other
hand, at z = 0 we have f ′ (0) = 0, while
f ′′ (0) = [2 cos(z) − z sin(z)]z=0 = 2 6= 0,
so f has a double zero at z = 0.
The relationship between poles and zeros is given by the following result:
Proposition 4.4. (i) Suppose f is holomorphic in some open disc D(a; r). Then f has a zero
of order m at a if and only if 1/f has a pole of order m at a.
(ii) Suppose f has a pole of order m at a.
(a) If g is holomorphic in D(a; r) then the function f g has
• a pole of order m at a if g(a) 6= 0;
• a pole of order m − n at a if g has a zero at a of order n < m;
• a removable singularity at a if g has a zero at a of order n ≥ m.
(b) If g has a pole of order n a a then f g has a pole of order m + n at a.
(iii) Suppose f, g are holomorphic in D(a; r) and g has a zero of order m at a. The function f /g
has
• a pole of order m at a if f (a) 6= 0;
• a pole of order m − n at a if f has a zero at a of order n < m;
• a removable singularity at a if f has a zero at a of order n ≥ m.
(i) We saw that z sin(z) has simple zeros at z = nπ, for n ∈ Z\{0}, and a double zero
1
at z = 0. Hence, by Proposition 4.4(i),
has simple poles at z = nπ, for n ∈ Z \ {0},
z sin(z)
and a double pole at z = 0.
Examples.
(ii) Consider
F (z) =
(z − 1)2 cos2 πz
.
(2z − 1)(z − i)4 sin3 πz
The denominator function g(z) = (2z − 1)(z − i)4 sin3 πz has
33
• a simple zero at 12 ;
• a zero of order 4 at i;
• triple zeros at k, for k ∈ Z.
The numerator function f (z) = (z − 1)2 cos2 πz has
• a double zero at 1;
• double zeros at
2k+1
2 ,
for k ∈ Z.
Hence, by Proposition 4.4(iii), the function F = f /g has
• a quadruple pole at i;
• a triple pole at k, for k ∈ Z \ {1};
• a simple pole at 1;
• a removable singularity at 12 .
Behaviour near an isolated singularity
Suppose f is holomorphic in some punctured disc D ′ (a; r), with an isolated singularity at z = a.
The behaviour of the function near a actually determines the type of singularity we have:
Removable singularity
P
′
Suppose f has Laurent expansion f (z) = ∞
n=0 , for z ∈ D (a; r). Then, (re)defining f (a) = c0 , we
see that f is holomorphic in D(a; r) – we have removed the singularity. In particular, f is bounded
close to a.
Pole
If f has a pole at a then 1/f has a zero at a. Hence 1/f (z) → 0 as z → a and |f (z)| → ∞ as
z → a.
Essential singularity
The Casorati-Weierstrass Theorem says: Let w be any complex number. Then there is a
sequence of complex numbers {an } converging to a such that f (an ) → w... so f is about as far
from being continuous at a is is possible.
Even more spectacularly, Picard’s Theorem says: there exists w0 ∈ C such that, for all ε > 0
and w ∈ C \ {w0 }, there is a z ∈ D ′ (a, ε) with f (z) = w... so, in any punctured disc around a, the
function f takes all but at most one value of C.
4.3
Residues
Definition 4.5. Suppose f is holomorphic in a punctured open disc D ′ (a; r) and has a pole at
a. TheP
residue of f at a is the (unique) coefficient c−1 of (z − a)−1 in the Laurent expansion
f (z) = n cn (z − a)n , for z ∈ D ′ (a; r). We write
res{f (z); a} = c−1 .
34
Theorem 4.6 (Cauchy’s Residue Theorem). Suppose f is holomorphic on and inside a positively
oriented simple closed path γ, except for (a finite number of ) poles at a1 , . . . , am inside γ. Then
Z
f (z)dz = 2πi
γ
m
X
k=1
res{f (z); ak }.
Proof We choose r > 0 such that D(ak ; 2r) is inside γ, for each pole ak , and such that D(ak ; 2r)
contains no pole other than ak .
We deform our contour to the contour γ ′ shown:
By the Deformation Theorem 3.5,
(‡)
Z
Z
m Z
X
f (z)dz =
f (z)dz =
γ
a1
f (z)dz.
γ ′∗
k=1 γ(ak ;r)
γ′
a2
′ (a ; 2r), we have Laurent expanNow, for z ∈
k
PD
n
sion f (z) = ∞
n=−m cn (z − ak ) . Then, using uniform convergence as in the proof of Laurent’s Theorem 4.1, we have
Z
f (z)dz =
Z
∞
X
am
γ(ak ;r) n=−m
γ(ak ;r)
=
∞
X
n=−m
cn
Z
γ∗
cn (z − ak )n dz
γ(ak ;r)
(z − ak )n dz = 2πic−1 .
Since c−1 = res{f (z); ak }, we get the result by summing the integrals in (‡).
Now, if we want to integrate a function f which is holomorphic on and inside a simple closed path
γ, except for a finite number of poles, we just need to compute the residues at each pole. One
way would be to write out the Laurent expansion, but this is not such a useful method as Laurent
expansions can be difficult to compute, so we need other methods.
Classification of poles
Suppose a function f is holomorphic in D ′ (a; r) with a pole of order m at a. The pole is
• simple if m = 1;
• multiple if m > 1;
• Type I (overt) if the function is written as
f (z) =
where g is holomorphic in D(a; r);
35
g(z)
,
(z − a)m
• Type II (covert) otherwise – in this case we can always write
h(z)
k(z)
f (z) =
where h, k are holomorphic in D(a; r), h(a) 6= 0 and k has a zero of order m at a.
The distinction between type I and type II poles is useful but rather artificial as it depends on how
we have written our function f – indeed, we can often convert type II poles into type I poles.
Examples.
1
has simple poles of type II at z = ±i.
z2 + 1
1
(which is the same function) has simple poles of type I at z = ±i.
(z − i)(z + i)
1
cosec 2 z =
has double poles of type II at z = nπ, for n ∈ Z.
sin2 z
The residue at a simple pole
We have the following useful lemma:
Lemma 4.7. Suppose f has a simple pole at a. Then
res{f (z); a} = lim (z − a)f (z).
z→a
Proof We have Laurent expansion
f (z) =
∞
X
n=−1
Then
(z − a)f (z) =
cn (z − a)n ,
∞
X
n=−1
for z in some D ′ (a; r).
cn (z − a)n+1 → c−1
as z → a,
since (z − a)n+1 → 0 as z → a, for n > −1.
From this, we immediately get:
Type I If f (z) =
g(z)
, with g holomorphic in D(a; r) and g(a) 6= 0, then
(z − a)
res{f (z); a} = g(a).
Type II If f (z) =
h(z)
, where h, k are holomorphic in D(a; r), h(a) 6= 0 and k has a simple zero
k(z)
at a, then
res{f (z); a} =
36
h(a)
.
k′ (a)
Proof Using Lemma 4.7, the fact that k(a) = 0, and the definition of derivative, we have
h(z)
k(z)
z−a
= lim h(z)
z→a
k(z) − k(a)
h(a)
=
k′ (a)
res{f (z); a} =
lim (z − a)
z→a
The residue at a multiple pole
Suppose f is holomorphic in D ′ (a; r), with a pole of order m > 1 at a.
Type I If f (z) =
g(z)
then
(z − a)m
res{f (z); a} =
g(m−1) (a)
.
(m − 1)!
Proof By Cauchy’s Integral Formula 3.7(ii), we have
g
(m−1)
(m − 1)!
(a) =
2πi
Z
γ(a; 21 r)
(m − 1)!
g(z)
dz =
m
(z − a)
2πi
Z
γ(a; 12 r)
dz = (m − 1)! res{f (z); a},
by Cauchy’s Residue Theorem 4.6.
Type II There is no easy way to compute the residue at a multiple pole of type II. Either convert
it to a type I pole or else compute the Laurent expansion explicitly.
z2
has simple poles at 1 (type I) and at ±i (type
(z − 1)(z 2 + 1)
z 2 /(z 2 + 1)
II). We compute the residues using the methods above. Firstly, at 1 we have f (z) =
so
z−1
2 z
1
res{f (z); 1} =
= .
2
z + 1 z=1
2
Examples. (i) The function f (z) =
z 2 /(z − 1)
at z = ±i, we have
z2 + 1
2
z /(z − 1)
1
1−i
−1
=
=
res{f (z); i} =
=
2z
(i
−
1)2i
2(1
+
i)
4
z=i
Now writing f (z) =
and
res{f (z); −i} =
z 2 /(z − 1)
2z
=
z=−i
37
1
1+i
−1
=
=
.
(−i − 1)2(−i)
2(1 − i)
4
(ii) The function g(z) =
eiz
has a pole of order 4 at z = 0 (type I). Then
z4
1
1 d3 iz
i
e
= i3 ei0 = − .
res{g(z; 0)} =
3
3! dz
6
6
z=0
Alternatively, we could do this by writing out the Laurent expansion for eiz .
π cot(πz)
has a triple pole at z = 0 (type II) and simple poles at z = n,
z2
π cos(πz)/z 4
so that
for n ∈ Z \ {0} (also type II). For n 6= 0, we write h(z) =
sin(πz)
(iii) The function h(z) =
res{h(z); n} =
π cos(πz)/z 4
π cos(πz)
=
z=n
1
.
n2
At z = 0 we must use the Laurent expansion. We saw that
cot(z) =
1 z
z3
− −
+ O(z 5 )
z 3 45
so we get
h(z) =
π
π(πz) π(πz)3
π
π2 π4 z
3
−
−
+
O(z
)
=
−
−
+ O(z 3 ).
z3
3z 2
45z 2
z3
3z
45
and
res{h(z); 0} = −
4.4
π2
.
3
Applications to real integrals
We can apply complex analysis, in particular Cauchy’s Residue Formula, to compute real integrals
for which the methods of real analysis do not suffice. This section is devoted to examples of this.
Example. Evaluate I =
Z
∞
x4
0
1
dx.
+1
Solution The idea is to integrate the complex function f (z) =
contour γ shown:
1
around a semicircular
z4 + 1
γ is made up of the straight line segment γ1 = [−R, R],
and γ2 , the semicircular arc with centre 0 and radius
R joining R to −R (where R > 1).
Now the function f has simple poles when z 4 + 1 = 0,
that is at e±iπ/4 , e±i3π/4 . Since only eiπ/4 and ei3π/4 are
inside our contour, Cauchy’s Residue Theorem gives
(∗)
Z
γ1
f (z)dz +
Z
γ2
iR
ei3π/4
−R
γ2∗
eiπ/4
γ1∗
n
o
n
o
f (z)dz = 2πi res f (z); eiπ/4 + res f (z); ei3π/4 .
38
R
First we find the residues, noting that f has simple poles of type II.
o
n
1
i
1 −i3π/4
1
1
1
√
√
res f (z); eiπ/4 =
−
.
=
e
=
=
−
4z 3 z=eiπ/4
4
4
4ei3π/4
2
2
o
n
1
i
1 −iπ/4
1
1
1
i3π/4
√ −√
.
=
res f (z); e
= e
=
=
4z 3 z=ei3π/4
4
4
4ei9π/4
2
2
Now we look at the terms on the LHS of (∗). For the first term, we can take the path to be
γ1 : [−R, R] → C given by γ1 (t) = t; then γ1′ (t) = 1 and
Z R
Z
Z R
1
1
f (z)dz =
dt = 2
dt,
4
4
γ1
−R t + 1
0 t +1
since f (t) = f (−t). In particular, letting R → ∞ we get
Z ∞
Z
1
f (z)dz = 2
lim
dt = 2I.
4
R→∞ γ1
t +1
0
Finally, we estimate the integral around γ2 , noting that it has length l(γ2 ) = πR:
Z
1 ≤ πR .
f (z)dz ≤ πR sup 4
∗
R4 − 1
z∈γ2 z + 1
γ2
Again, letting R → ∞ we get
π/R3
πR
0
=
= 0.
→
4
4
R −1
1 − 1/R
1
Hence
lim
Z
R→∞ γ2
f (z)dz = 0.
Now we substitute everything back into (∗) and let R → ∞. We get
√
π 2
i
1
1
i
1
1
√ −√
=
2I + 0 = 2πi
−√ − √
+
4
4
2
2
2
2
2
Hence
I =
Example. Evaluate J =
Z
∞
−∞
x2
Z
∞
0
√
π 2
1
dx =
.
x4 + 1
4
cos x
dx.
+x+1
cos z
around the semicircle γ from the
+z+1
previous example but unfortunately cos z is not bounded as |z| = R → ∞: indeed,
Solution We could try to integrate the function
z2
cos(iR) = cosh R → ∞
as R → ∞.
Instead, we consider the function
g(z) =
cos z
sin z
eiz
= 2
+i 2
2
z +z+1
z +z+1
z +z+1
39
and we will take real parts later. The reason is that (as in question 7 on the problem sheets)
iz e = eRe(iz) = e−Im(z) ,
which is bounded (by e0 = 1) for z in the upper half-plane (that is, Im(z) ≥ 0).
The function g is holomorphic except for simple poles where the denominator z 2 + z + 1 is zero,
that is at
√
1
3
z = − ±i
= e±i2π/3 .
2
2
We integrate g around
the contour γ from the previous example. Only the simple pole of type II
√
3
1
i2π/3
at e
= − 2 + i 2 is inside the contour, where we have
i2π/3
res{g(z); e
} =
ei z
2z + 1
z=ei2π/3
1
e− 2 i e−
√
=
i 3
√
3
2
.
Then Cauchy’s Residue Theorem gives us
√ !
1
3
Z
Z
√3
e− 2 i e− 2
2π
√
g(z)dz = 2πi
g(z)dz +
(†)
= √ cos( 21 ) + i sin( 21 ) e− 2 .
i 3
3
γ2
γ1
Now we turn to the terms on the LHS. For the first one, we have γ1 : [−R, R] → ∞ so γ1 (x) = x
and
Z R
Z
Z R
Z R
eix
cos x
cos x
g(z)dz =
dx =
dx + i
dx.
2
2
2
γ1
−R x + x + 1
−R x + x + 1
−R x + x + 1
For γ2 , we estimate as in the previous example:
iz Z
e 1
≤ πR 2
→0
g(z)dz
≤
πR
sup
2
R −R−1
z∈γ1∗ |z + z + 1|
γ2
as R → ∞.
Substituting into (†) and letting R → ∞, we get
Z ∞
Z ∞
− √3
cos x
cos x
2π
1
1
√
cos(
)
+
i
sin(
)
e 2
dx
+
i
dx
=
2
2
2
2
3
−∞ x + x + 1
−∞ x + x + 1
and, comparing real parts, we get
Z
J =
∞
−∞
Example. Evaluate K =
Z
0
∞
√
cos x
2π
1 − 23
√
cos(
)e
dx
=
.
2
x2 + x + 1
3
sin x
dx.
x3 + x
eiz
, which has simple poles (type II) at 0, ±i.
z3 + z
We cannot integrate around the same semicircular contour as previously, because there is a singularity at 0,
Γ∗R
so we make a small indentation there: our path is
the join of the semi-circular arc ΓR , the line segment
[−R, −ε], the semicircular arc Γε , and the line segment
i
Γ∗ε
[ε, R]. The only pole inside the contour is at i, where
we have
−ε ε
Solution We consider the function h(z) =
−R
40
R
res{h(z); i} =
eiz
3z 2 + 1
=
z=i
1
e−1
= − .
−2
2e
Then Cauchy’s Residue Theorem gives us
(‡)
Z
Z
Z
Z
1
h(z)dz = 2πi −
h(z)dz +
h(z)dz +
h(z)dz +
2e
[ε,R]
Γε
[−R,−ε]
ΓR
= −
πi
.
e
We will (eventually) let ε → 0 and R → ∞ in (‡). Three of the terms in are easy to deal with: we
have
Z
iz
1
≤ πR sup |e |
≤ πR 3
→ 0 as R → ∞,
h(z)dz
∗ |z 3 + z|
R −R
z∈ΓR
ΓR
while on [ε, R] we have γ(x) = x, γ : [ε, R] → C, so
Z
h(z)dz =
Z
R
ε
[ε,R]
eix
dx =
x3 + x
Z
R
ε
cos x
dx + i
x3 + x
Z
R
ε
sin x
dx
x3 + x
and on [−R, −e] we have γ(x) = −x, γ : [R, ε] → C, so
Z
h(z)dz =
[−R,−ε]
Z
ε
R
ei(−x)
(−1)dx = −
(−x)3 + (−x)
Z
R
ε
cos x
dx + i
x3 + x
Z
R
ε
sin x
dx.
x3 + x
Finally, we must deal with the integral around Γε . For this, we need the following:
Lemma 4.8. Suppose f is holomorphic in D ′ (a; r) and has a simple pole at a. For 0 < ε < r and
θ1 < θ2 let γε : [θ1 , θ2 ] → C be the path
γε (t) = a + εeit .
Then
lim
Z
ε→0 γe
f (z)dz = i(θ2 − θ1 ) res{f (z); a}.
Note that Cauchy’s Residue Theorem is given by the special case θ1 = 0, θ2 = 0. However, this
Lemma is only valid for simple poles. For multiple poles, other methods must be found.
Proof Put b = res{f (z); a} so, by Lemma 4.7,
lim f (z)(z − a) = b.
z→a
Let g(z) = (z − a)f (z) − b so that, for any η > 0, there is a 0 < δ < r such that
|g(z)| <
η
,
θ2 − θ1
for all z ∈ D ′ (a; δ).
Let ε > 0 be such that ε < δ; we notice that, for z = γε (t), we have
γε′ (t) = iεeit = i(z − a).
41
Then
Z
Z
=
f
(z)dz
−
ib(θ
−
θ
)
2
1
γε
Z
= θ2
θ1
θ2
θ1
f (γε (t))γε′ (t)
g(γε (t))dt
− ib dt
≤ (θ2 − θ1 ) sup |g(ζ)| ≤ η.
z∈γε∗
Since this is true for any η > 0, the LHS tends to 0 as ε → 0.
Now we return to the example. We have
res{h(z); 0} =
eiz
2
3z + 1
=
z=0
e0
= 1.
1
Since the semicircular arc Γε is negatively oriented, Lemma 4.8 give us
Z
lim
h(z)dz = −iπres{h(z); 0} = −iπi.
ε→0 Γε
Substituting everything back into (‡) we get
Z
h(z)dz +
ΓR
Z
h(z)dz + 2i
Γε
Z
R
ε
sin x
πi
dx = − .
x3 + x
e
Letting ε → 0 and R → ∞, we get
0 − iπ + 2iK = −
which rearranges to give
K =
Example. Evaluate L =
Z
0
∞
x4
Z
∞
0
πi
,
e
sin x
π(e − 1)
dx =
.
x3 + x
2e
log x
dx.
+ x2 + 1
Solution For this, we will use the complex logarithm function, which we saw on question 9 of the
problem sheets
log(z) = log |z| + i arg(z).
We note that the argument arg(z) is only defined modulo 2π so we need to choose which argument
we want. In question 9, we chose the principal value of the argument
−π < arg(z) ≤ π,
but recall that our function was not defined on the negative real axis; that is, it was only defined
for −π < arg(z) < π. We showed that this function was then holomorphic on C− = C \ (−∞, 0],
d
log(z) = 1z .
and dz
42
Note that it was certainly necessary for us to remove the negative real axis: for x > 0, the limit
limy→0+ arg(−x + iy) = π, while limy→0− arg(−x + iy) = −π, so the choice of argument we have
made is not even continuous at −x; then our logarithm function would not be continuous at −x,
let alone holomorphic.
In fact, we could define a logarithm function by cutting out any line from 0 to ∞. For φ ∈ R,
define
Cφ = {z ∈ C : z 6= 0 and φ is not an argument of z}.
[So Cπ = C− = C−π .] Then we can define log(z) on Cφ by
log(z) = log |z| + i arg(z),
with φ − 2π < arg(z) < φ.
All these different possibilities are called branches of the complex logarithm function. [In fact, it is
possible to define a single logarithm, but this is not defined on C but rather on a Riemann surface.]
In the example, we will use the branch on C3π/2 , so that −π/2 < arg(z) < 3π/2; in particular, this
branch of log(z) is holomorphic on all of the (closed) upper half-plane except 0.
We will integrate
f (z) =
z4
log(z)
+ z2 + 1
around the semicircular contour with an indentation at 0 used in the previous example. Then f
has poles whenever the denominator is 0, that is z 4 + z 2 + 1 = 0. There are (at least) three ways
to find the roots of this polynomial:
(i) We notice that z 6 − 1 = (z 2 − 1)(z 4 + z 2 + 1) so that the roots of z 4 + z 2 + 1 are the complex
6th roots of 1 which are not also square roots. The 6th roots of 1 are
1, e2πi/6 , e4πi/6 , e6πi/6 = −1, e8πi/6 , e10πi/6 ,
of which ±1 are square roots of 1. Simplifying (and recalling that we require the argument
to lie between −π/2 and 3π/2), we get that the roots of z 4 + z 2 + 1 are
e−πi/3 , eπi/3 , e2πi/3 , e4πi/3 ,
or, in Cartesian form,
√
±1 ± i 3
.
2
(ii) We notice that we can factorize,
z 4 + z 2 + 1 = (z 2 + z + 1)(z 2 − z + 1)
and solve the two quadratics (using the quadratic formula) to get the roots in Cartesian form
as above.
(iii) We notice that, since there are only even powers, we really have a polynomial in z 2 . So we
substitute y = z 2 to get the quadratic polynomial y 2 + y + 1, whose roots (using the quadratic
formula) are
√
1±i 3
= e±πi/3 ,
2
43
where we have computed the modulus and argument to write them in polar form. Recalling
that y = z 2 , we now need to solve
z 2 = e±πi/3 .
Writing z = reiθ , we have r 2 ei2θ = e±πi/3 and, comparing modulus and argument, we get the
same roots as before (in polar form).
The function f has just two simple poles of type II inside the contour. At z = eiπ/3 =
have residue
log(eiπ/3 )
log(z)
iπ/3
=
res{f (z); e
} =
.
4z 3 + 2z z=eiπ/3
4eiπ + 2eiπ/3
√
1+i 3
2 ,
we
Since |eiπ/3 | = 1, while arg(eiπ/3 ) = π/3 and eiπ = −1, we get
iπ/3
res{f (z); e
√
√
iπ/3
(−3 − i 3)
iπ
π( 3 − 3i)
√
√
√
.
} =
=
=
36
−4 + (1 + i 3)
3(−3 + i 3) (−3 − i 3)
Similarly, at z = ei2π/3 =
√
−1+i 3
,
2
we get
i2π/3
res{f (z); e
√
π(2 3 + 6i)
} =
.
36
[This is an Exercise for you!] In particular, the sum of the residues inside the contour is
√
√
√
π 3
π
π( 3 − 3i) π(2 3 + 6i)
+
=
+i ,
36
36
12
12
and Cauchy’s Residue Theorem gives
Z
Z
Z
f (z)dz +
f (z)dz +
(§)
[−R,−ε]
ΓR
√
π2
π2 3
f (z)dz = − + i
f (z)dz +
.
6
6
[ε,R]
Γε
Z
The integrals along the straight line segments are easy to deal with:
Z R
Z ∞
Z
log(x)
log(x)
f (z)dz =
dx →
dx = L
4 + x2 + 1
4 + x2 + 1
x
x
ε
[ε,R]
0
as ε → 0, R → ∞. For x ∈ [ε, R], note that arg(−x) = π so log(−x) = log(x) + iπ. Then
Z
f (z)dz =
[−R,−ε]
Z
ε
R
log(−x)
(−1)dx
4
(−x) + (−x)2 + 1
=
Z
R
ε
log(x) + iπ
dx
x4 + x2 + 1
→ L + iπ
as ε → 0, R → ∞.
Z
0
∞
1
dx
x4 + x2 + 1
Now we need to estimate the integrals around the semicircular arcs. To do this, we will need to
know what happens, for example, to log(R)/R3 , as R → ∞. For this, we use the following Lemma:
Lemma 4.9. For any k > 0,
(i) xk e−x → 0 as x → ∞;
44
(ii) x−k log(x) → 0 as x → ∞;
(iii) xk log(x) → 0 as x → 0.
Proof (i) We fix some integer n > k. Then, for x > 0,
ex = 1 + x + · · · +
But then
0 < xk e−x =
xn
xn
+ ··· >
.
n!
n!
xk
n!
xk
<
= n−k → 0
ex
xn /n!
x
as x → ∞,
since n − k > 0.
(ii) For x > 0 we can write x = ey/k (that is, y = k log(x)), so that y → ∞ as x → ∞. Then
y
x−k log(x) = e−y → 0
as x → ∞,
k
by applying (i).
(iii) Since x−1 > 0 whenever x > 0, we can replace x by x−1 in (ii) to get
−k
x−1
log x−1 → 0
as x−1 → ∞,
that is,
xk log(x) → 0
as x → 0.
Now we return to the example and to the estimates of the integrals around ΓR and Γε . For large
R, we have
Z
(log(R) + π)
R−3 log(R) + πR−3
0
| log(z)|
≤ πR sup
≤
πR
=
π
→
= 0
f
(z)dz
4 + z 2 + 1|
4 − R2 − 1
−2 − R−4
∗
|z
R
1
−
R
1
z∈ΓR
ΓR
as R → ∞, where we are using Lemma 4.9(ii). Similarly, for small ε we have
Z
(log(ε) + π)
ε log(ε) + πε
0
| log(z)|
≤ πε
= π
→
= 0
f (z)dz ≤ πε sup 4
2
2
4
2
4
1−ε −ε
1−ε −ε
1
z∈Γ∗ε |z + z + 1|
Γε
as ε → 0, where we are using Lemma 4.9(iii).
Putting this all together in (§) and letting ε → 0 and R → ∞, we get
√
Z ∞
π2
π2 3
1
dx + 0 + L = − + i
.
0 + L + iπ
x4 + x2 + 1
6
6
0
Comparing real parts, and dividing by 2, we get
Z ∞
π2
log x
dx = − .
L =
4
2
x +x +1
12
0
Note that we also get another integral for free: comparing imaginary parts, and dividing by π, we
have
√
Z ∞
π 3
1
dx =
.
x4 + x2 + 1
6
0
45
4.5
Application to summation of series
We end with an application to summation of series. In first year analysis, we used the Integral Test
to show that certain series converge, like
∞
X
1
,
ns
n=0
whenever s > 1. When s is an even integer, we can use contour integration to evaluate this sum.
(When s is an odd integer, the problem is much more subtle.)
Example. Prove that
∞
X
π2
1
=
.
2
n
6
n=0
Solution We consider the function f (z) =
holomorphic except for:
π cot(πz)
, which (we saw in an earlier example) is
z2
• a triple pole (type II) at z = 0, with residue −π 2 /3;
• simple poles (type II) at z = n, for n ∈ Z \ {0}, with residue 1/n2 .
N + 21 (1 + i)
We integrate f around the square contour γN ,
with centre 0 and side of length 2N + 1:
γN∗
Note that we avoid having any poles actually
on the contour.
From Cauchy’s Residue Theorem, we get
(∗)
Z
We will show that

γN

N
N
X
 X
1
1
π2 
π2
 = 2πi 2
f (z)dz = 2πi 
−
−

n2
3
n2
3
n=0
n=−N
n6=0
lim
Z
N →∞ γN
for then, letting N → ∞ in (∗), we get
!
.
f (z)dz = 0,
N
X
1
π2
,
=
n2
6
n=0
as required. So we estimate our integral, noting that the square contour γN has length l(γN ) =
∗ . We get
4(2N + 1) < 8N , while |z| ≥ N + 12 > N , for all z ∈ γN
Z
< 8N sup π| cot(πz)| < 8π sup | cot(πz)|.
(†)
f
(z)dz
∗
|z 2 |
N z∈γN∗
z∈γN
γN
46
∗ . This is given by the following Lemma:
In order to finish, we need a bound for | cot(πz)|, for z ∈ γN
Lemma 4.10. With notation as above, there is a constant C such that, for all n ≥ 1 and for all
∗ ,
z ∈ γN
| cot(πz)| ≤ C.
Plugging this into (†), we get
Z
γN
as required.
8πC
f (z)dz <
→ 0
N
as N → ∞,
Proof of Lemma 4.10 On the vertical sides, we have z = ±(N + 21 ) + iy, so
cos(πz) = cos(±(N + 12 )π) cos(iπy) − sin(±(N + 12 )π) sin(iπy) = ± sin(iπy).
Similarly, sin(πz) = ± cos(iπy), so
| cot(πz)| = | tan(iπy)| = | tanh(πy)| ≤ 1.
On the horizontal sides, we have z = x ± i(N
inequalities, we get
cos(πz) =
| cot(πz)| = sin(πz) + 12 ). Using the fact that |eiπx | = 1 and the triangle
iπz
e + e−iπz eiπz − e−iπz eiπx e∓(N + 12 ) + e−iπx e±(N + 12 ) = eiπx e∓(N + 12 ) − e−iπx e±(N + 12 ) iπx ∓(N + 12 ) −iπx ±(N + 12 ) e
e
+
e
e
≤ 1 1
eiπx e∓(N + 2 ) − e−iπx e±(N + 2 ) =
1
1
1
1
e(N + 2 ) + e−(N + 2 )
e(N + 2 ) − e−(N + 2 )
= coth (N + 21 )π
≤ coth 23 π ,
as coth(t) is decreasing for t ≥ 0. Hence we can take C = 32 π , since this is greater than 1.
47