Complex Analysis
Fall 2004
Professor L. Marcoux
CHRIS ALMOST
Contents
1 Introduction
1.1 Motivation . . . . . .
1.2 Definitions . . . . . .
1.3 Algebraic properties
1.4 Polar coordinates . .
1.5 DeMoivre’s formula .
1.6 nth root functions . .
1.7 Embed C in M2 (R) .
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3
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4
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2 Elementary functions
2.1 The complex exponential function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Sine and cosine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 The complex logarithm function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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4
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3 Properties of C
3.1 C is complete . . . .
3.2 Riemann sphere . .
3.3 Continuous curves
3.4 Connnectedness . .
3.5 Compactness . . . .
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4 Analyticity
8
4.1 Complex differentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
4.2 The Cauchy-Riemann equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
4.3 The Möbius Tranformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
5 Integration
6 Cauchy’s Theorem
6.1 Cauchy’s integral theorem
6.2 Cauchy’s integral formula
6.3 Harmonic Functions . . .
6.4 Winding Number . . . . .
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12
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2
CONTENTS
7 Series
21
7.1 Uniform Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
7.2 Integration of Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
7.3 Differentiation of Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
8 Power Series
24
8.1 Radius of Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
9 Taylor Series
27
9.1 Uniqueness Theorem for Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
10 Laurent Series
31
11 Residues
35
11.1 Applications of residues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
11.2 Evaluation of Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
INTRODUCTION
1
1.1
3
Introduction
Motivation
In real analysis, we observe that a distinct advantage of the real numbers over the rational numbers is that R
is complete. This is the propositionerty that allows us to define things like suprema and infima, and in turn
allows us to prove theorems such as the Mean/Intermediate value theorems. One way in which the real numbers
are deficient is that they are not algebraically closed. (e.g. x 2 + 1 = 0 has no roots in R.) Our goal is study
a new number system, the set of complex numbers, which contains R and is algebraically closed. As we shall
see, whereas notions of differentiability of real valued functions lead to examples which exhibit pathologies, the
theory of “analytic” complex valued functions is much richer and the examples are much better behaved.
1.2
Definitions
Denote by C the set R2 equipped with the following operations.
(a, b) + (c, d) = (a + b, c + d) and (a, b)(c, d) = (ac − bd, ad + bc)
It is not hard to see that C is a commutative ring. Here (1, 0) is the multiplicative identity. We shall identify the
real numbers with the subset R × {0} of C. We write i for the ordered pair (0, 1), and refer to iR as the imaginary
axis of C, and R as the real axis of C. Note that i 2 = (−1, 0) = −1. We tend to use a, b for real numbers and w, z
for complex numbers.
1.1 Exercise. C is a field
1.3
Algebraic properties
Given z = a + bi, the number z = a − bi is called the complex conjugate of z. The map z → z is an involution on
an algebra.
zz = a2 + b2 =: |z|2 . This is the Euclidean length of the vector z in C. This is also called the norm or modulus
zz
−1
of z. Note that if 0 6= z if and only if |z| 6= 0, in which case |z|
= |z|z 2 = aa−bi
2 = 1, whence z
2 +b 2 . We define the real
and imaginary parts of z to be ℜz = a and ℑz = b, so that z = ℜz + iℑz. Then functions z → ℜz and z → ℑz are
real-linear (but not complex linear).
1.4
Polar coordinates
z = a + bi. Define r = |z| and θ by r cos θ = a and r sin θ = b. We call θ the argument of z and write θ = arg z
(not defined if z = 0. Of course, sin and cos are periodic, and so the argument is uniquely defined once we
choose some interval [λ, λ + 2π) for some λ ∈ R. This is called the branch of the argument. This representation
is useful for multiplying complex numbers. The product of complex numbers is equal to the complex number
who’s modulus is the product of the moduli and who’s argument is the sum of the arguments (modulo 2π). For
example, z → iz is the map that takes a complex number and rotates it counterclockwise π/2. Using induction,
we can prove:
1.5
DeMoivre’s formula
If z = r(cos θ + i sin θ ) and if n ∈ N then z n = r n (cos(nθ ) + i sin(nθ )).
Now consider the inverse problem, given w ∈ C, n ∈ N, find z ∈ C so that z n = w. We see clearly that if w = 0
1
then z = 0 works, and if w = ρ(cos φ + i sin φ) then z = φ n (cos(φ/n) + i sin(φ/n)) will work. But there are
more solutions, in fact replacing φ/n with (φ + 2kπ)/n will give us a solution zk for any k ∈ Z, and the solutions
are distinct for 0 ≤ k < n.
4
1.6
COMPLEX ANALYSIS
nth root functions
The situation becomes more delicate if we consider the behaviour of functions, f (z) = z n and their “inverses”.
Indeed, except at the point 0, the function f (z) = z n is an n to 1 function (i.e. it maps n different numbers to
the same number), and maps each section S of the disk of angle 2π/n onto the whole disk. As such, we cannot
1
talk of an inverse function. Can we at least find a continuous function g such that g n (z) = z? Try g(z) = z n . Fix
any branch of the arg function, say [0, 2π). Consider the behaviour of g as z traverses the unit circle T. With
z0 = 1, we have arg(z0 ) = 0. Suppose that g(z0 ) = 1. Then as z → z0 in the first quadrant, g(z) → g(z0 ) = 1, so
arg g(z) → arg g(z0 ) = 0. In the forth quadrant, as z → z0 , the argument of g(z) is approaching 2π/n 6= 0. Thus
g is not continuous at 1.
In 1851, Riemann understood this and propositionosed the following “solution”. We extend the domain of
1
the function g(z) = z n as follows. For 0 6= w, we treat each w k = r(cos(θ + 2kπ) + i sin(θ + 2kπ)) as being
different numbers for 0 ≤ k < n. One way of doing this is considering C × {0, 1, . . . , n − 1}, each of which sheet is
“cut” along [0, ∞). We identify the lower part of the cut on sheet k with the upper part of the cut on sheet k + 1
1
mod n. The result is a n-sheeted surface, [R], called the Riemann surface associated with g(z) = z n . The point 0
is called a branch point because there is only one copy of it.
F /
As a result, we get the following commutative diagram C
[R]
AA
AA
AA
AA π
f
C
where π : (z, k) → z is projection, f is as above, and F (z) = F (r(cos(θ + 2kπ/n) + i sin(θ + 2kπ/n)) =
(r n (cos(nθ ) + i sin(nθ )), k). Notice that where f was n to 1, F : C → [R] is 1 to 1 and continuous.
1.7
Embed C in M2 (R)
Suppose we fix w = a + bi. Consider the map Mw : C → C : z 7→ wz. We can think of Mw as a map from R2
to itself. Mw : (u, v) 7→ (au − bv, av + bu). Thus Mw corresponds to a matrix (with repsect to the standard basis
{(1, 0), (0, 1)} for R2 :
a −b
Mw ↔
b
a
This correspondence establishes an algebra isomorphism between C and a subalgebra of M2 (R).
2
2.1
Elementary functions
The complex exponential function
Recall the Taylor expansions of cos(x), sin(x), e x ,
sin(x) = x −
cos(x) = 1 −
x3
3!
x2
2!
+
+
ex = 1 + x +
x5
5!
x4
4!
x2
2!
−
−
+
x7
7!
x6
6!
x3
3!
+ −···
+ −···
+
x4
4!
+ ···
ELEMENTARY
5
FUNCTIONS
Now e x > 0 and (e x )0 = e x , and so x 7→ e x is strictly increasing on R. As such, we can talk about the inverse of
e x , which we call the (natural) logarithm, log(x), which satisfies:
elog(x) = x for x > 0 and log(e x ) = x for all x ∈ R
We would like to make sense of ez , cos(z), sin(z) and log(z) for complex numbers z. However, we do not
know anything about convergence of complex series! Informally, if the series were to converge, and if we could
reorder the terms at will, then we would expect for all y ∈ R that
e i y = cos( y) + i sin( y)
We sometimes write cis( y) for cos( y) + i sin( y).
Therefore we choose to define our “extended” exponential to satisfy this expression. Namely, for x, y ∈ R
ex = 1 + x +
x2
2
+
x3
6
+
x4
24
+ ···
e i y = cos( y) + i sin( y)
Notice that both definitions agree where they overlap at zero.
If the complex-exponential is to behave like the real-exponential, we would also expect that exp(x + y i) =
exp(x) exp(i y), and so we define the complex- exponential to achieve that.
2.1 Definition. For z = x + i y ∈ C, we define exp(z) = e x cis( y)
2.2 Definition. A function f : C → C is said to be periodic with period ω if f (z + ω) = f (z) ∀ z ∈ C
2.3 Proposition.
1. exp(z + w) = exp(z) exp(w)
2. exp(z) 6= 0
3. | exp(x + i y)| = e x
4. exp(·) is periodic, and each period has the form 2πni for some n ∈ Z.
5. exp(z) = 1 if and only if z ∈ (2πi)Z
PROOF: Exercise.
ƒ
We point out in passing that as y goes from 0 to 2π, exp(i y) revolves around the unit circle T in a counterclockwise direction.
2.2
Sine and cosine
When y ∈ R, we can use our definition of exp to solve for cos and sin.
cos( y) =
exp(i y) + exp(−i y)
2
sin( y) =
exp(i y) − exp(−i y)
2i
This observation leads to:
2.4 Definition. For z ∈ C, we define
cos(z) =
exp(iz) + exp(−iz)
2
sin(z) =
exp(iz) − exp(−iz)
2i
2.5 Proposition.
1. sin (z) + cos (z) = 1
2. sin(z + w) = sin(z) cos(w) + sin(w) cos(z)
3. cos(z + w) = cos(z) cos(w) − sin(z) sin(w)
2
PROOF: Exercise.
2
ƒ
6
2.3
COMPLEX ANALYSIS
The complex logarithm function
The discussion of Riemann surfaces for the power functions applies to the log function as well. z 7→ exp(z) is
periodic with (minimal) period 2πi. In particular, if we fix y0 ∈ R and set
B y0 = {x + y i | x ∈ R, y0 ≤ y < y0 + 2π}
It is clear that exp(B y0 ) = C \ {0}. In particular, horizontal lines are mapped onto rays emmanating from the
origin, and vertical line segments get mapped to circles centred at the origin. Not only this, but the restricted
map is one to one and onto from this strip to C \ {0}.
2.6 Definition. We define log : C \ {0} → C with range B y0 (that is, y0 ≤ ℑ(log z) < y0 + 2π) via log z =
log |z| + i arg(z), where arg(z) takes values in B y0 .
The Riemann surface for the log function looks like the the Riemann surface for z 7→ z n , except there are
infinitely many sheets and zero is removed.
3
Properties of C
3.1 C is complete
3.1 Theorem. A sequence (zn )∞
n=1 in C converges if and only if it is Cauchy.
"
PROOF: Suppose that (zn )∞
n=1 converges to z ∈ C. Then given " > 0, there is N such that for all n ≥ N , |zn −z| < 2 .
But then for n, m ≥ N ,
|zn − zm | ≤ |zn − z| + |z − zm | < "
and so (zn )∞
n=1 is Cauchy.
∞
Now suppose that (zn )∞
n=1 is Cauchy. Let "0 = 1. Since (zn )n=1 is Cauchy, we can find N0 such that for all
n, m ≥ N0 , |zn − zm | < "0 . In particular, for n ≥ N0 , |zn − zN0 | < 1, which implies that |zn | < |zN0 | + 1 and therefore
(zn )∞
n=1 is bounded by max{|z1 |, . . . , |zN0 −1 |, |zN0 | + 1}.
∞
By the Bolzano-Weierstraß theorem we can find a subsequence (znk )∞
k=1 of (zn )n=1 which converges to some
∞
z ∈ C. We claim that (zn )n=1 converges to z.
Let " > 0. We can find K such that for all k ≥ K, |znk − z| < 2" . We can also find N such that for all n, m ≥ N ,
|zn − zm | < 2" . Thus if n, k ≥ max{N , K} then nk ≥ k and
|zn − z| ≤ |zn − znk | + |znk − z| < "
and so (zn )∞
n=1 converges.
3.2
ƒ
Riemann sphere
We can represent the elements of C as the points on a sphere as follows. Let Σ be a sphere of radius 1 tangent
to the complex plane at the point 0. The diameter of Σ containing 0 intersects Σ at a point which we will call
N , the north pole. Given z ∈ C, let σ(z) denote the intersection of Σ with the line segment from N to z. This
establishes a bijection between C and Σ \ {N }. (σ is actually a homeomorphism.) We think of N as the “point at
infinity” for C, and we denote it simply ∞.
The map C ∪ {∞} → Σ : z 7→ σ(z) : ∞ 7→ N is called the stereographic projection of the extended complex
plane onto Σ.
3.2 Theorem. (Bolzano-Weierstraß) Every sequence in C has a limit point in the extended complex plane, C ∪
{∞}.
PROPERTIES
3.3
OF
7
C
Continuous curves
Let x : [a, b] → R and y : [a, b] → R be two continuous functions. We define a continuous curve C =
{(x(t), y(t)) | t ∈ [a, b]} ⊆ R2 , or with z(t) = x(t) + i y(t), we think of C as lying in the complex plane. The
point z(a) (resp. z(b)) is called the initial (resp. final) point of C . This assigns a direction to the curve (from the
initial to the final point, duh). C is closed if the initial and final points are the same, otherwise it is called an arc.
3.3 Definition. A set E ⊆ C is called arcwise connected if for any two (distinct) points in E there exists an arc
between them that lies entirely in E.
We say that a set G ⊆ C is a domain if it is open and arcwise connnected. A closed domain is a set of the form
G (the closure of G), where G is a domain. A region is a set of the form G ∪ H, where G is a domain and H ⊆ ∂ G.
It is possible for a region to be a domain or a closed domain.
3.4 Definition. Let C = {z(t) = x(t) + i y(t) | t ∈ [a, b)}, where x and y are continuous on [a, b). Suppose that
t 1 6= t 2 =⇒ z(t 1 ) 6= z(t 2 ). Then C is called a Jordan curve if lim t→b− x(t) and lim t→b− y(t) both exist. If
lim x(t) = x(a) and lim− y(t) = y(a)
t→b−
t→b
then C is called a closed Jordan curve.
3.5 Theorem. (Jordan Curve Theorem) Let C be a closed Jordan curve. Then C \ C is the union of two domains
with C as their common boundry. The first domain G1 is bounded and called the interior of C , while the second
domain G2 is not bounded and called the exterior of C . We shall assume that the positive orientation of a closed
Jordan curve keeps the interior on the “left”.
3.4
Connnectedness
3.6 Definition. A domain G is said to be simply connected if, given any closed Jordan curve C ⊆ G, the interior
of C is contained in G. Otherwise, G is said to be multiply connected.
In the extended complex plane, a domain G is simply connected if given any Jordan curve in G, either the
interior or the exterior is contained in G. This means that C ∪ {∞} \ D is simply connected in the extended
complex plane, but this set (minus ∞) is not simply connected in the normal complex plane.
3.7 Definition. Let C0 , . . . , Cn be n+1 closed Jordan curves such that no two of them intersect and that C1 , . . . , Cn
lie in the interior of C0 . Suppose further that that the interiors of the curves are pairwise disjoint. Then the interior
of C0 minus the closures of the interiors of C1 , . . . , Cn is a domain G. We say that G is (n + 1)-connected. We say
that G is simply connected if n = 0.
For example, the annulus is 2-connected and the disc is simply connected.
3.8 Definition. Let G ⊆ C be a domain and suppose that f : G → C is a function. If z0 ∈ G then we say that f
has a limit L at z0 and we write limz→z0 f (z) = L if ∀ " > 0, there exists δ > 0 so that 0 < |z − z0 | < δ implies that
| f (z) − L| < ". If f (z0 ) = limz→z0 f (z) then we say that f is continuous at z0 . If f is continuous at all points in G
then f is said to be continuous on G.
Notice that with f : G → C as above, we can define two functions u : R2 → R and v : R2 → R such that
u(x, y) = ℜ f (x + i y) and v(x, y) = ℑ f (x + i y), so that f (x + i y) = u(x, y) + iv(x, y). It follows that f is
continuous at z0 = x 0 + i y0 if and only if u and v are continuous at (x 0 , y0 ).
3.9 Example. f (z) = z n is continuous on C. (Prove this.)
8
COMPLEX ANALYSIS
If G̃ is a region which contains some of the boundry points of the domain G, and if z0 ∈ G̃ and z0 is on the
boundry of G, then f : G̃ → C is said to be continuous at z0 if f (z0 ) = limz→z0 ,z∈G̃ f (z). In a similar way, we can
define the notion of continuity of a function defined on a curve C .
3.10 Definition. Let K ⊆ C be a set and f : K → C be a function. We say that f is uniformly continuous on K if
∀ " > 0 there exists δ > 0 so that |z − w| < δ implies that | f (z) − f (w)| < ". (The key point is that one choice of
δ works for all pairs z, w.)
We wish to establish conditions on a set K which will ensure that every continuous function on K is in fact
uniformly continuous.
3.5
Compactness
3.11 Definition. Let K ⊆ C. We say that a collection {Gα }α∈Λ of sets is an open cover of K if each Gα is open
and K ⊆ ∪α∈Λ Gα . A finite subcover of K is a finite subset {Gα1 , . . . , Gαn } such that K ⊆ ∪nk=1 Gαk . We say that K is
compact if every open cover of K admits a finite subcover.
3.12 Theorem. (Heine-Borel) Suppose that K ⊆ C is closed and bounded. Then K is compact.
PROOF: Suppose otherwise. Let {Gα }α∈Λ be an open cover of K that does not admit a finite subcover. Since K
is bounded, we can find M > 0 such that K ⊂ R0 = {x + i y | x, y ∈ [−M , M ]}. We can partition R0 into four
closed subsquares R0i of equal area. At least one of the K ∩ R0i ’s cannot be covered by finitely many of the Gα ’s.
Let R1 be one of those squares. Continue this process to get a sequence R0 ⊇ R1 ⊇ R2 ⊇ · · · such that K ∩ R i
cannot be covered by finitely many of the Gα ’s. Choose zm ∈ K ∩ R m . Then this is a bounded sequence, so by
the Bolzano-Weierstraß theorem, there is a subsequence (zmn )∞
n=1 that converges to some z0 ∈ C. K is closed,
so z0 ∈ K. Since K ⊆ ∪α∈Λ Gα , there exists α0 ∈ Λ so that z ∈ Gα0 . Since Gα0 is open, there exists some δ > 0
so that the ball around z0 of radius δ is contained in Gα0 . Moreover, z0 ∩∞
m=1 R m and diamR m → 0. Thus for m
sufficiently large, z, w ∈ R m implies that |z − w| < δ, which implies that R m ⊆ Gα0 . This contradiction shows that
K is compact.
ƒ
3.13 Theorem. If K ⊆ C is compact and f : K → C is continuous, then f is uniformly continuous on K.
PROOF: Exercise.
4
ƒ
Analyticity
The definition of the derivative of a complex function at a point w is the same in form as that of the derivatives
for real-valued functions. It is therefore quite surprising how different the two theories turn out to be.
4.1 Definition. Let G ⊆ C be a domain and let w ∈ G. Suppose that f : G → C is a function. We say that f is
differentiable at w if
f (z) − f (w)
f (w + h) − f (w)
= lim
f 0 (w) = lim
z→w
h→0
h
z−w
exists as a complex number. We say that f is analytic in G if f is differentiable at each point w ∈ G. f is said
to be analytic at w if f is analytic in Vδ (w) for some δ > 0. If f is analytic on the entire plane, we say that f is
entire.
4.2 Example.
1. Let k ≥ 1 and f (z) = z k . Then for all w ∈ C
f 0 (w) = lim
z→w
as you would expect. Thus f is entire.
f (z) − f (w)
z−w
= lim
z→w
zk − wk
z−w
= kw k−1
ANALYTICITY
9
2. For z = x + i y ∈ C, define g(z) = z = x − i y. Then
lim
g(w + h) − g(w)
h
h→0
= lim
h→0
h
h
which does not exist. Thus complex conjugation is not differentiable at any point.
3. Exercise: neither ℜ(·) nor ℑ(·) are differentiable at any point in C.
Since the proofs of the algebraic propositionerties of the (real) derivative only depend on the definition of
the derivative, and this is the same for complex-valued functions, the proofs carry over to this setting.
1. [λ f ]0 (z) = λ( f 0 (z)) for all λ ∈ C
2. [ f + g]0 (z) = f 0 (z) + g 0 (z)
3. [ f g]0 (z) = f 0 (z)g(z) + f (z)g 0 (z)
4. [ f /g]0 (z) =
f 0 (z)g(z)− f (z)g 0 (z)
(g(z))2
when g(z) 6= 0
5. [ f ◦ g]0 (z) = f 0 (g(z))g 0 (z) if g 0 (z), f 0 (g(z)) exist
4.1
Complex differentials
Let w = f (z) be a complex function. For ∆z = ∆x + i∆ y set ∆w = f (z + ∆z) − f (z). Observe that f is
differentiable at z if and only if
∆w
f 0 (z) = lim
∆z→0 ∆z
exists. The differential of the function w = f (z) is dw := f 0 (z)dz, where dz = ∆z. As in the real case, this gives
us the suggestive equation f 0 (z) = dw
.
dz
In CALCULUS 3, we saw that a function u : R2 → R is differentiable at a point (x, y) if and only if there exists
A ∈ L(R2 , R) (say [A] = [a, b] with respect to the standard basis) such that
lim
|u(x + ∆x, y + ∆ y) − u(x, y) − A(∆x, ∆ y)|
k(∆x, ∆ y)k2
(∆x,∆ y)→(0,0)
=0
If we write ∆u for u(x ∆ x, y + ∆ y) − u(x, y) this becomes
∆u = A(∆x, ∆ y) = ηk(∆x, ∆ y)k2
where η → 0 as k(∆x, ∆ y)k2 → 0. When u is differentiable at (x, y), recall from mulitvariable calculus that
a=
∂u
∂x
and b =
∂u
∂y
We saw before that if f : C → C is a complex function, then f (x +i y) = u(x, y)+iv(x, y) and f is continuous
if and only if u and v are continuous. Consider fR (z) = ℜ(z). Then u(x, y) = x and v(x, y) = 0. The partials exist
and are 1, 0 respectively, and the second partials are continuous, so u, v are differentiable. There is a problem
here. It appears as though one cannot choose the real and imaginary parts of a complex differentiable function
arbitrarily.
10
COMPLEX ANALYSIS
4.2
The Cauchy-Riemann equations
4.3 Theorem. Let G ⊆ C be a domain and f : G → C a function. Let z = x + i y ∈ G. Then f is differentiable at
z if and only if
1. u, v are differentiable at (x, y)
2. u, v satisfy the Cauchy-Riemann equations
∂u
∂x
=
∂v
∂y
and
∂u
∂y
=−
∂v
∂x
PROOF: See text.
ƒ
4.4 Example. Let G ⊆ C be a domain and C ⊆ G be a Jordan curve. Let z(t), t ∈ [a, b] be a parametrization
of C and fix t 0 ∈ (a, b). If C has a tangent line τ at z0 = z(t 0 ), then ∆z = z(t 0 + ∆t) − z(t 0 ) has a “limiting
direction” in the sense that ϕ = lim∆t→0 arg ∆z exists. Since ∆t → 0 implies that ∆z → 0, we may say that
ϕ lim∆z→0 arg ∆z. Now suppose that f : G → C is continuous and f 0 (z0 ) exists and is not zero. Then w(t) =
f (z(t)) defines a new curve Γ in C. Let ∆w = f ◦z(t 0 +∆t)− f ◦z(t 0 ) = f (z0 +∆z)− f (z0 ). Since f is continuous
lim∆z→0 ∆w = 0. Since f 0 (z0 ) 6= 0, f 0 (z0 ) = lim∆z→0 ∆w
. In particular,
∆z
arg f 0 (z0 ) = lim arg
∆z→0
∆w
∆z
= lim arg ∆w − arg ∆z
∆z→0
Equivalently, lim∆z→0 arg ∆w = arg f 0 (z0 ) + lim∆z→0 arg ∆z, and so Γ has a tangent line T at w0 = f (z0 ) and the
angle that T makes with R is arg f 0 (z0 ) + ϕ.
4.5 Example. Suppose that C1 and C2 are two curves in G which pass through z0 . Suppose that the tangent
line Ci at z0 meets R at an angle of ϕi . We define the angle between the curves to be ϕ2 − ϕ1 . From above, if
f 0 (z0 ) 6= 0, then the angle between Γ1 = f C1 ) and Γ2 = f (C2 ) at w0 = f (z0 ) is
θ2 − θ1 = (ϕ2 + arg f 0 (z0 )) − (ϕ1 + arg f 0 (z0 )) = ϕ2 = ϕ1
Thus f preserves angle between curves passing through z0 when f 0 (z0 ) 6= 0.
4.6 Definition. Let G ⊆ C be a domain, f : G → C a continuous function, and z0 ∈ G. If f preserves angles
between curves passing through z0 , we say that f is conformal at z0 . If f preserves the magnitude of the angle
between curves, we say that f is isogonal at z0 .
Thus if f : G → C is continuous on a domain G, then f is conformal at each z0 ∈ G where 0 6= f 0 (z0 ) exists.
It can be shown that if f is analytic in G and 0 = f 0 (z0 ) then f is not conformal at z0 .
|∆w|
, it follows that | f 0 (z0 )| = lim∆z→0 |∆z| , and so we can think of | f 0 (z0 )| as the local
Since f 0 (z0 ) = lim∆z→0 ∆w
∆z
“stretching” factor of f at z0 .
4.3
The Möbius Tranformation
Suppose that a, b, c, d ∈ C and |c| + |d| 6= 0. The map
f (z) =
az + b
cz + d
is called a Möbius transformation (or a fractional linear transformation).
INTEGRATION
11
1. If c = 0 then f is linear and f 0 (z) =
2. If ad − bc = 0 then either
a
d
for all z ∈ C
(a) d = 0, whence c 6= 0 and so b = 0. Therefore f is just a constant function (undefined at z = 0)
(b) d =
6 0. In this case, (az + b)d = (ad)z + (bd) = (bc)z + bd = b(cz + d), and so f is constant again.
These are the so-called trivial cases.
3. Suppose now that c 6= 0 and that ad − bc 6= 0. If z 6= − dc then f 0 (z) =
limz→− d az + b = −ad+bc
c
c
hand, limz→∞ f (z) = ac .
ad−bc
(cz+d)2
6= 0. Thus f is conformal at z.
6= 0, while the limit on the bottom is 0, and so limz→− d f (z) = ∞. On the other
c
If we think of f (− dc ) = ∞ and f (∞) = ac then a routine calculation shows that f is analytic on the extended
complex plane and f is a bijection with inverse (another Möbius transformation)
f −1 (z) =
5
1
dz − b
ad − bc −cz + a
Integration
5.1 Definition. A curve C with parametric equation z(t) = x(t)+i y(t), t ∈ [a, b] is called smooth if z 0 (t) exists,
z(t) 6= 0 for all t ∈ [a, b], and z 0 (t) is continuous in t.
Let G be a domain and f : G → C be a function. Suppose that C ⊆ G is a smooth curve. Define the integral
of f in exactly the way you would expect. (I’m serious. This should be the fifth time you’ve seen this, so I’m not
going to explain it again.)
5.2 Proposition. The arclength λC of a smooth curve C is independent of parametrization.
PROOF: Suppose that C = {z1 (t) | y ∈ [a, b]} = {z2 (t) | y ∈ [c, d]}. We are assuming that there exists a
differentiable function f : [a, b] → [c, d] so that f (a) = c, f (b) = d and f 0 (t) 6= 0 for any t ∈ [a, b]. Then
λC =
d
Z
|z20 (s)|ds
c
=
b
Z
|z20 ( f
(s))|d f (s) =
a
b
Z
|z20 ( f
0
(s))| f (s)ds =
b
Z
a
|z20 ( f
0
(s)) f (s)|ds =
a
so λC is independent of the parametrization.
b
Z
|z10 (s)|ds
a
ƒ
Next we investigate some familiar rules of integration from multivariable calculus. If C = {z(t) | t ∈ [a, b]}
is a curve, then we denote by {z((b + a) − t) | t ∈ [a, b]} by C − . Note that the graphs of these curves coincide,
they are merely traversed opposite directions.
5.3 Theorem. Let f be a continuous function on a smooth curve C . Then
Z
Z
f (z)dz = −
f (z)dz
C−
C
5.4 Theorem. Let f , g be continuous functions on a smooth curve C . Let a, b ∈ C. Then
Z
Z
Z
(a f + b g)(z)dz = a
C
f (z)dz + b
C
g(z)dz
C
12
COMPLEX ANALYSIS
5.5 Theorem. Let f : C → C be a continuous function on a piecewise smooth curve C . Suppose that | f (z)| ≤ M
for all z ∈ C . Then
Z
f (z)dz ≤ M λC
C
Fix a smooth curve C ⊆ C. Let
AC = { f : C → C | f is continuous}
It is clear that AC is a vectors space. Moreover, it is a ring under pointwise multiplication. Such objects are called
algebras. Let
Z
f (z)dz
T : AC → C : f 7→
C
Then T is a linear functional. The set of all linear functionals is referred to as the algebraic dual, sometimes
denoted A#
C . We can define a norm on AC , k f k∞ = sup{| f (z)| : z ∈ C }. Then T is bounded with respect to this
norm. And so on. . .
6
Cauchy’s Theorem
R
If n 6= −1 and C is a closed smooth curve in C (with 0 6∈ C if n < 0) then C z n dz = 0. In particular, if n ≥ 0
R
and C ∈ C is smooth then C p(z)dz = 0 for all polynomials p. As we shall now see, this is just one example of
a much more general phenomenon.
6.1
Cauchy’s integral theorem
6.1 Definition. A triangular contour C is a union C1 ∪ C2 ∪ C3 of curves where
1. the final point of Ci is the initial point of Ci+1
2. Ci is given by {zi (t) | t ∈ [ai , bi ]} where zk (t) is linear in t
Note that C is automatically piecewise smooth.
6.2 Lemma. Suppose that G ⊆ C is a simply connected domain, ∆ ⊆ G is a triangular contour, and f : G → C
is an analytic function. Then
Z
f (z)dz = 0
∆
The fact that G is simply connected implies that the interior of ∆ is contained in G , which is crucial to our
argument.
PROOF: Suppose that ∆ = ABC. Bisect each side of ∆, say P bisects AB, Q bisects BC, and R bisects CA. By
connecting P to Q, Q to R, and R to P, we obtain 4 congruent triangular contours, ∆1 , ∆2 , ∆3 , ∆4 , each of which
λ
has perimeter (i.e. arclength) equal to half that of ∆ (i.e. λ∆i = 2∆ ). Using the orientation inherited from ∆, we
CAUCHY’S THEOREM
13
can complete the direction on ∆1 , ∆2 , ∆3 so that they again form triangular contours. Then
Z
Z
f (z)dz =
∆
f (z)dz
ABC
=
Z
f (z)dz −
Z
P BQ
=
Z
f (z)dz +
Z
PQ
f (z)dz +
Z
∆1
APR
f (z)dz +
Z
∆2
Orient ∆4 in such a way that −
R
PRQ
f (z)dz −
f (z)dz =
R
∆4
f (z)dz −
Z
f (z)dz
RQ
f (z)dz
PRQ
f (z)dz =
4
X
i=1
Z
f (z)dz
∆i
f (z)dz|. We want to show that M = 0. Now M ≤
∆
f (z)dz −
RQC
Z
f (z)dz. Then
∆
R
f (z)dz +
Z
PR
∆3
Z
Let M = |
satisfies
Z
P4
i=1 |
R
∆i
f (z)dz|, thus a least one of the ∆i ’s
Z
M
f (z)dz ≥
∆
4
i
(1)
Choose one of them and call it ∆ . We can iterate the procedure to obtain a nested sequence of triangular
contours {∆(n) }∞
n=1 such that
Z
M
f (z)dz ≥ n , n ≥ 1
4
∆(n)
UsingSan argument similar to the principal of nested rectangles, we can show that there exists a unique point
∞
z0 ∈ n=1 ∆(n) ∩ G . f is analytic on G , so f 0 (z0 ) exists and
f (z) − f (z0 )
0
− f (z0 ) = 0
lim
z→z0 z − z0
Equivalently,
f (z) − f (z0 ) − f 0 (z0 )(z − z0 ) = "(z)(z − z0 )
where limz→z0 "(z) = 0. Moreover,
Z
f (z) − f (z0 ) − f 0 (z0 )(z − z0 )dz =
∆(n)
Z
f (z)dz − ( f (z0 ) − z0 f 0 (z0 ))
∆(n)
=
Z
Z
∆(n)
1 dz − f 0 (z0 )
Z
z dz
∆(n)
f (z)dz
∆(n)
Thus
M
4n
≤|
R
∆(n)
f (z)dz| = |
R
∆(n)
"(z)(z − z0 )dz|. Let η > 0. Since limz→z0 "(z) = 0, we can find δ > 0 such that
0 < |z − z0 | < δ implies that |"(z)| < η. Moreover, since the arclength of ∆(n) goes to zero as n → ∞, we can find
N0 > 0 so that n ≥ N0 implies that λ∆(n) < δ. So if n ≥ N0 , then for z ∈ ∆(n) , |z − z0 | < λ∆(n) , and so
Z
ηλ2∆
M λ∆
(n)
≤
"(z)(z
−
z
)dz
λ
=
≤
η
0
∆
4n ∆(n)
2n
4n
Thus M ≤ ηλ2∆ , so M = 0 since η was arbitrary. Hence
R
∆
f (z)dz = 0.
ƒ
14
COMPLEX ANALYSIS
6.3 Definition. A polygonal closed curve is just a closed curve L = L1 ∪ · · · ∪ L n , where the final point of L k is the
initial point of L k+1 and each L k is a linear curve (i.e. a line segment).
R If G ⊆ C is a simply connected domain, f : G → C is analytic, and L ⊆ G is a closed polygonal curve then
f (z)dz can be reduced to a sum of integrals over triangular contours (A.I. Markushevich). Because of this fact,
L
we have
6.4 Corollary. Suppose that G ⊆ C is a simply connected domain. If f : G → C is analytic and L ⊆ G is a closed
polygonal curve then
Z
f (z)dz = 0
L
Our next goal is to approximate smooth curves by polygonal curves. First we need the following lemma,
proved on assignment 3.
6.5 Lemma. Let G ( C be a domain and C ⊆ G be a continuous curve. Let ρ = dist(C , ∂ G ). Then
1. ρ > 0
2. if D = {w ∈ C | |w − z| <
ρ
2
for some z ∈ C } then D is a bounded domain and C ⊆ D ⊆ D ⊆ G
6.6 Theorem. Let G ⊆ C be a domain and C ⊆ G be a piecewise smooth curve. Suppose that f : G → C is
continuous. Given " > 0, there exists a polygonal curve L so that
1. L ⊆ G
2. L is inscribed in C (the vertices of L lie in C and the initial and final points of L are the same as the intial
and final points of C , respectively)
R
R
3. C f (z)dz − L f (z)dz < "
PROOF: By the above lemma we can find a bounded domain D such that C ⊆ D ⊆ D ⊆ G , and ρ = dist(C , ∂ D) >
0. Since f is continuous in G and D is compact, f is uniformly continuous on D.
Let " > 0 and let λC denote the arclength of C . We can find δ > 0 such that if z, w ∈ D and |z − w| < δ then
| f (z) − f (w)| < 2λ " +1 . Let γ = min{δ, ρ}. Choose a partition Π = {z0 , z1 , . . . , zm } of C where the arclength λCi
C
of the subcurve Ci from zi−1 to zi of C is less than γ for all 1 ≤ i ≤ m. Let L be the polygonal curve obtained
from Π by joining zi−1 to zi for 1 ≤ i ≤ m. Then it will satisfy the conclusions of the theorem.
1. Suppose z ∈ L. Then z ∈ zi−1 zi for some i. Since zi ∈ C and since
|z − zi | ≤ |zi−1 − zi | < γ < ρ
it follows that z ∈ D. Thus L ⊆ D.
2. This is clear.
3. Consider S =
Pm
k=1
f (zk )∆zk which approximates the value of
∆zk = zk − zk−1 =
R
C
Z
1dz
Ck
f (z)dz. Now
CAUCHY’S THEOREM
and so S =
15
Pm R
k=1
Ck
f (zk )dz. Hence
Z
n Z
X
=
f
(z)
−
f
(z
)dz
f
(z)dz
−
S
k
k=1 C
C
k
Z
n X
≤
f (z) − f (zk )dz C
k=1
k
≤
n
X
λC k
k=1
=
since λCk < γ implies that | f (z) − f (zk )| <
"
2λC +1
"
2λC + 1
"
n
X
2λC + 1
k=1
λC k <
"
2
for all z ∈ Ck . Similarily, we have
Z
"
f (z)dz − S ≤
L
2
and so, by the triangle inequality,
Z
Z
f (z)dz −
f (z)dz < "
C
L
ƒ
As an imediate consequence of the above results, we get
6.7 Theorem. (Cauchy’s Integral Theorem) If G ⊆ C is a simply connected domain, f : G → C is analytic and
C ⊆ G is a piecewise smooth closed curve, then
Z
f (z)dz = 0
C
PROOF: Let " > 0. By the above theorem, there exists a closed polygonal contour L inscribed in C such that
Z
Z
f (z)dz < "
f (z)dz −
L
C
By a corollary above,
R
L
f (z)dz = 0, so |
R
C
f (z)dz| < ". Since " > 0 was arbitrary, the result follows.
ƒ
6.8 Corollary. Let G ⊆ C be a simply connected domain and f : G → C be analytic. Suppose that w1 , w2 ∈ G
and let C1 , C2 be two piecewise smooth curves in G with initial point w1 and final point w2 . Then
Z
Z
f (z)dz =
C1
f (z)dz
C2
PROOF: Let C2− be the curve C2 traversed in the reverse direction. Then C = C1 ∪ C2− is a closed curve in G . By
Cauchy’s Integral Theorem,
Z
Z
Z
Z
Z
0=
f (z)dz =
C
f (z)dz +
C1
f (z)dz =
C2−
f (z)dz −
C1
f (z)dz
C2
ƒ
16
COMPLEX ANALYSIS
Suppose that C0 , . . . , Cn are n + 1 piecewise smooth Jordan curves such that
1. Ci ∪ int(Ci ) ⊆ int(C0 ) for i = 1, . . . , n
T
2. Ci ∪ int(Ci ) C j ∪ int(C j ) = ∅ for 1 ≤ i < j ≤ n
Tn
Then D := int(C0 )∩( i=1 ext(Ci )) is an (n+1)-connected domain. Suppose that f is a function which is analytic
on D. We claim that
Z
Z
Z
f (z)dz =
f (z)dz + · · · +
C0
C1
f (z)dz
Cn
Indeed, it can be seen that with these assumptions, we can find n+1 non-intersecting arcs γ1 , . . . , γn which divide
D into two closed regions D1 and D2 bounded by two closed piecewise smooth Jordan curves Γ1 and Γ2 . Now f
is analytic on D and therefore it is analytic on each of D1 and D2 . But Di = Γi ∪ int(Γi ), i = 1, 2, and since Γ1
and Γ2 are closed Jordan curves, their interiors are simply connected. So we can find simply connected domains
Gi ⊇ Di , i = 1, 2, so that f is analytic. Therefore
Z
Z
f (z)dz
f (z)dz =
0=
Γ1
Γ1
by Cauchy’s Integral Theorem. Hence
0=
Z
f (z)dz +
C0
n
X
Z
f (z)dz
Ck−
k=1
In particular, when there are only two curves C0 and C1 with int(C1 ) ⊆ int(C
R 0 ), then if CR0 , C1 are piecewise
smooth Jordan curves and f is analytic on C0 ∪ C1 ∪ (int(C0 ) ∩ ext(C1 )) then C f (z)dz = C f (z)dz.
0
1
6.9 Example.
1. Suppose that C is a piecewise smooth Jordan curve and that 0 ∈ ext(C ). Let f (z) = 1z .
Then f (z) is analytic everywhere except 0, so by assignment 3 we can find a simply
R connected domain G
containing C ∪ int(C ) so that f is analytic on G . By Cauchy’s Integral Theorem, C dz
= 0.
z
2. Now suppose that 0 ∈ int(C ). Since int(C ) is open we can find some r > 0 such that V2r (0) ⊆ int(C ).
Let C = {r e i t | 0 ≤ t ≤ 2π}. Since our function is analytic on C ∪ C1 ∪ (int(C ) ∩ ext(C1 )), we see that
R
R 2πi
R dz 1 R dz
= C z . Now for z ∈ C1 , dz = izd t, and so C dz
= 0 idz = 2πi.
z
C z
1
1
3. More generally, suppose that C is any R
piecewise Rsmooth closed Jordan curve and z0 ∈ int(C ). Given the
dz
substitution w = z − z0 , d w = dz, then C z−z
= C dw
= 2πi.
w
0
6.2
Cauchy’s integral formula
A key step in proved Cauchy’s Integral Formula is provided by the next result, which acts like a complex version
of the fundemental theorem of calculus
6.10 Theorem. (Local Primitive Theorem) Suppose
G ⊆ C is a simply connected domain, z0 ∈ G , and f : G → C
Rz
is analytic (see remark below). Then F (z) = z f (w)dw, taken along any piecewise smooth curve from z0 to z,
0
defines a single-valued analytic function in G and F 0 (z) = f (z) for all z ∈ G .
R
PROOF: Let z ∈ G . The fact that C f (w)dw = 0 for all piecewise smooth closed curves in G implies that
R
R
f (w)dw = C f (w)d w for all piecewise smooth curves C1 and C2 from z0 to z. Let R > 0 so that |h| < R
C1
2
implies that z + h ∈ G . Let LR be the line segment from z to z + h. Then
Z z+h
Zz
Z z+h
F (z + h) − F (z) =
f (w)dw −
z0
f (w)dw =
z0
f (w)dw
z
CAUCHY’S THEOREM
Now f (z) =
f (z)
h
17
R z+h
z
dw =
1
h
R z+h
z
f (z)d w. Thus
F (z + h) − F (z)
h
− f (z) =
1
h
!
z+h
Z
f (w) − f (z)dw
z
Since f is analytic on G , it is continuous at z. Thus, given " > 0 we can find 0 < δ < R so that |w − z| < δ implies
R z+h
| f (w) − f (z)| < ". Since z f (w) − f (z)dw is independent of the curve from z to z + h, shall assume that we
are integrating along the line segment from z to z + h. If |h| < δ and w lies on the line segment from z to z + h,
then |w − z| < δ, so we have
!
Z z+h
F (z + h) − F (z)
1
1
f
(w)
−
f
(z)dw
−
f
(z)
=
"|h| = "
≤
h
|h|
h
z
Thus F 0 (z) = limh→0
F (z+h)−F (z)
h
= f (z).
ƒ
Remark.
1. Our proof of Cauchy’s Integral
Theorem used the fact that f was analytic on a simply connected
R
domain G only where we needed ∆ f (z)dz = 0 for all triangular contours ∆. If we only know that f is
R
continuous on G but we assume that ∆ f (z)dz = 0 for all such contours, then the proofs of the lemmas
leading to Cauchy’s Integral Theorem, and the theorem itself, did not require analycity. Thus we may also
replace the assumption
that f is analytic in the Local Primitive Theorem with the assumption that f is
R
continuous and ∆ f (z)dz = 0 for all triangular contours ∆. As we shall see below, this will imply that f is
analytic on G .
2. We refer to F above as the antiderivative or indefinite integral of f on G.
6.11 Proposition. Let G ⊆ C be a domain and suppose that F1 , F2 : G → C are analytic with F10 (z) = F20 (z) for
all z ∈ G . Then there exists k ∈ C such that F1 (z) = F2 (z) + k for all z ∈ G .
PROOF: Let F (z) = F1 (z) − F2 (z). Then F is analytic in G and we can write F (x + i y) = u(x, y) + iv(x, y). Then
0 = F 0 (z) =
∂u
∂x
+i
∂v
∂x
=
∂v
∂y
−i
∂u
∂y
So ∂∂ ux = ∂∂ uy = 0 = ∂∂ xv = ∂∂ vy . From mulitvariable calculus, it follows that there are k1 , k2 ∈ R such that u(x, y) =
k1 and v(x, y) = k2 . Thus F (z) = k1 + ik2 for all z ∈ G .
ƒ
We are now ready to prove one of the most important results in the course.
6.12 Theorem. (Cauchy’s Integral Formula) Let G ⊆ C be a domain. Suppose that C is a piecewise smooth
Jordan curve with int(C ) ⊆ G . If z0 ∈ int(C ) and f : G → C is analytic then
Z
1
f (z)
f (z0 ) =
dz
2πi C z − z0
f (z)
PROOF: If z0 lies inside C then z−z is analytic on G , except for z0 itself. Let γR denote the circle of radius R
0
centered at z0 , where R > 0 is chosen sufficiently small that γR ⊆ intC . By an above corollary,
Z
Z
f (z)
f (z)
dz =
dz
z − z0
z − z0
C
γ
R
18
Now
COMPLEX ANALYSIS
R
1
dz
γR z−z0
= 2πi, so there exists δ > 0 such that for any R with 0 < R < δ then
Z
Z
Z
Z
f
(z)
1
f
(z)
−
f
(z
)
f
(z)
0
dz − 2πi f (z0 ) = dz − f (z0 )
dz = dz γ z − z0
γ z − z0
z
−
z
z
−
z
0
0
γ
γ
R
R
R
R
Since f is continuous at z0 , given " > 0 we can choose 0 < δ < δ such that |w − z0 | < δ0 implies that
| f (w) − f (z0 )| < ". Then if 0 < R < δ0
Z
Z
f (z)
f (z) − f (z0 ) "
dz − 2πi f (z0 ) = dz ≤ 2πR = 2π"
γ z − z0
γ
R
z − z0
0
R
Since
f (z)
dz
γR z−z0
R
R
is independent of 0 < R < δ , it (and thus
0
f (z)
dz)
C z−z0
R
is equal to 2πi f (z0 ).
ƒ
6.13 Theorem. Let G ⊆ C be a domain and suppose f : G → C is analytic. Let C be any closed piecewise
smooth Jordan curve in G and z0 ∈ int(C ) ⊆ G . For any n ∈ N,
Z
f (z)
n!
(n)
dz
f (z0 ) =
2πi C (z − z0 )n+1
PROOF: By induction on n. When n = 0 we have Cauch’y Integral Formula, proved above. Suppose n > 1 and the
theorem holds for n − 1 in all possible cases. We must show that
Z
f (n−1) (z0 + h) − f (n−1)
n!
f (z)
→
dz
h
2πi C (z − z0 )n+1
as h → 0. Take d > 0 so small that Vδ (z0 ) ⊆ int(C ) and take γ r , a circle of radius r > 0, such that γ r ⊆ Vδ (z0 ).
By a theorem,
Z
Z
f (z)
n!
f (z)
n!
dz =
dz
n+1
2πi C (z − z0 )
2πi γ (z − z0 )n+1
r
So we must check that
f (n−1) (z0 + h) − f (n−1)
h
→
n!
2πi
Z
C
f (z)
(z − z0 )n+1
dz
as h → 0. Since h → 0, we may assume that |h| < r so that z0 + h is inside γ r . Now by the induction hypothesis
Z
1 (n − 1)!
f (z)
f (z)
f (n−1) (z0 + h) − f (n−1)
=
−
dz
n
h
h 2πi
(z
−
z
−
h)
(z
− z0 )n
0
γr
Z
(n − 1)!
(z − z0 )n−1 + (z − z0 )n−2 (z − z0 − h) + · · · + (z − z0 − h)n−1
=
f (z)h
2hπi
(z − z0 )n (z − z0 − h)n
γr
Z
(n − 1)!
(z − z0 )n + (z − z0 )n−1 (z − z0 − h) + · · · + (z − z0 )(z − z0 − h)n−1
=
f (z)
2πi
(z − z0 )n+1 (z − z0 − h)n
γ
r
R
R
f (z)n(z−z0 −h)n
f (z)
(n−1)!
n!
Now 2πi
dz = 2πi γ (z−z )n+1 (z−z
n dz Subtracting these and applying the triangle inequality gives
γ r (z−z0 )n+1
0
0 −h)
r
that the difference is less than or equal to
Z n
n−1
n−1
(n − 1)!
− n(z − z0 − h)n (z − z0 ) + (z − z0 ) (z − z0 − h) + · · · + (z − z0 )(z − z0 − h)
max | f |
dz
γr
2πi
(z − z0 )n+1 (z − z0 − h)n
γ
r
which goes to 0 as h → 0. Thus the theorem is proved by mathematical induction.
ƒ
CAUCHY’S THEOREM
19
6.14 Theorem. Let G be a simply connected domain and f : G → C continuous and such that
for all piecewise smooth closed Jordan curves in G . Then f is analytic.
R
C
f (z)dz = 0
Rz
PROOF: Pick any z0 ∈ G . For z ∈ G , let g(z) = z f (w)dw, taken along any path γ from z0 to z. This is
0
R
independent of the path chosen because C f = 0 for all closed C . We know g is analytic on G and g 0 (z) = f (z).
By the above theorem, g 00 exists, so f 0 exists over G .
ƒ
Pn
P∞
6.15 Example. (Weierstraß M-test) Let f n : A → C and let sn = k=1 f k . We say that k=1 f k converges
uniformly
P∞
when sn converges uniformly. This happens if k f n kA ≤ M
Pn∞for all n, for some constant Mn , and k=1 Mn converges.
In particular, if f n is continuous for each n, then so is k=1 f k .
6.16 Proposition. If f n , f : G → C, where G is simply connected and f n is analytic and f n → f uniformly on
every compact subset K of G , then f is analytic.
PROOF: G is covered by the closed discs inside it, and each disk is compact. To get f continuous on G it suffices
to show that f is continuous on every closed disc in G . But f n → f uniformly on such discs.
Since each f n is
R
continuous, f is continuous. To see that f is analytic use Morera. We must prove that C f (z)dz = 0 for all
R
R
R
closed curves in G . We know that C f n (z)dz = 0 for all curves C , by Cauchy’s Theorem. Since C f n → C f by
R
uniform convergence on compact sets, we have C f = 0.
ƒ
P∞
6.17 Example. nz = ez log n is analytic on the whole plane. Let G = {z | ℜz > 1}. show that the series n=1 n1z
Pk
converges to an analytic function on G . It is enough to show that sk := n=1 n1z converges uniformly on compact
subsets of G . If K ⊆ G is compact,
there is " > 0 such that ℜz P
≥ 1 + " for all z ∈ K. Then for all z ∈ K,
P∞ then
∞
1
|nz | ≥ n1+" . From CALCULUS 2, k=1 n1+"
converges, so by the M-test, n=1 n1z converges.
In multivariable calculus, Green’s Theorem tells us that if G ⊆ R2 is a simply connected domain and if P, Q
are continuously differentiable on some domain G1 ⊇ G then
∂Q
ZZ
G
∂x
−
∂P
∂y
dA =
Z
P dx +Qd y
∂G
We can identify G with the subset GC = {x + i y | (x, y) ∈ G } of C. Let C = ∂ G . If f : GC → C is analytic, we
can write f (x + i y) = u(x, y) + i v(x, y) and
Z
f (z)dz =
Z
udx − v d y + i
C
C
Z
v dx +ud y
C
Suppose that u and v are continuously differentiable. Then by Green’s Theorem
Z
C
udx − v d y =
ZZ
−
G
∂v
∂x
−
∂u
∂y
dA = 0 and
Z
C
v dx +ud y =
ZZ
G
∂u
∂x
−
∂v
∂y
dA = 0
R
by the Cauchy-Riemann equations. Hence C f (z)dz = 0. The issue is that we have to assume that u and v are
continuously differentiable. In our original proof we needed only that u and v are differentiable and satisfy the
Cauchy-Riemann equations. Since we concluded that f 0 is differentiable, we are able to conclude that u and v
are continuously differentiable.
20
COMPLEX ANALYSIS
6.3
Harmonic Functions
6.18 Definition. Let G ⊆ R2 be a domain and u : G → R be a function. If u has continuous second partial
derivatives on G and they satisfy the Laplace equation
∂ 2u
∂ x2
+
∂ 2u
∂ y2
=0
everywhere in G , then u is said to be harmonic. Suppose u, v are hormonic in G and further suppose that they
satisfy the Cauchy-Riemann equations
∂u
∂x
=
∂v
∂y
and
∂u
=−
∂y
∂v
∂x
at every point in G . Then u and v are said to be conjugate harmonic functions.
Our interest in harmonic functions stems from the following two theorems
6.19 Theorem. Let G ⊂ C be a domain and f (x + i y) = u(x, y) + iv(x, y) be a complex-valued function on G .
Then f is analytic on G if and only if u and v are conjugate harmonic functions.
PROOF: Suppose that f is analytic. Then u and v are differentiable and satisfy the Cauchy-Riemann equations.
f 0 is analytic and
∂u
∂v
∂u
∂u
∂v
∂v
f0=
+i
=
−i
=
+i
∂x
∂x
∂x
∂y
∂y
∂x
Thus all of these functions are differentiable and satisfy the Cauchy-Riemann equations, which is to say
∂ 2u
∂
∂u
∂
∂u
∂ 2u
=
=
−
=
−
∂x ∂x
∂y
∂y
∂ x2
∂ y2
so u is harmonic, and similarily for v. Now since f 00 is analytic, all of the second partials are continuous as well,
so u and v are harmonic conjugates.
Conversely, suppose that u and v are harmonic conjugates. Then u and v have continuous second partials, so
in particular they have continuous partials. This in turn implies that u and v are differentiable. Since u and v
satisfy the Cauchy-Riemann equations it follows that f is analytic (by Green’s Theorem).
ƒ
6.20 Theorem. Let G ⊂ R2 be a simply connected domain and u : G → R be a harmonic function. Then there
exists an analytic function f : GC → C such that u = ℜ f .
PROOF: Consider g =
so in particular
∂u
∂x
∂ 2u
∂ x2
∂u
∂x
− i ∂∂ uy . We claim that g is analytic. Since u is harmonic it has continuous second partials,
and − ∂∂ uy are differentiable. Furthermore,
=−
∂ 2u
∂ y2
by Laplace’s equation, and
∂
∂y
∂u
∂x
=
∂
∂x
∂u
∂y
=−
∂
∂x
−
∂u
∂y
so g satisfies the Cauchy-Riemann equations. Hence g is analytic. By the Local Primitive theorem there exists an
analytic function f˜ = ũ + i ṽ so that f˜0 = g. But
∂ f˜
∂x
Thus
∂ ũ
∂x
=
∂u
.
∂x
=
∂ ũ
∂x
+i
∂ ṽ
∂y
=g=
Also, by the Cauchy-Riemann equations,
f := f˜ − k = u + i v is analytic and ℜ f = u.
∂ ũ
∂y
∂u
∂x
−i
∂v
∂y
= − ∂∂ xṽ =
∂u
.
∂y
Thus ũ − u = k is constant. That is,
Remark. The function f above is completely determined (up to a constant).
ƒ
SERIES
6.4
21
Winding Number
6.21 Definition. Let C be a piecewise smooth closed curve in C. For z0 6∈ C , we define the index or winding
number of C about z0 to be
Z
1
1
dz
IndC (z0 ) =
2πi C z − z0
6.22 Theorem. The function z0 7→ IndC (z0 ) is a continuous, integer valued function on C \ C , so that IndC (·) is
constant on the connected components of C \ C . On the unbounded component of C \ C , IndC (·) = 0.
PROOF: See text, or algebraic topology course notes.
7
ƒ
Series
P∞
Let n=0 zn be an infinite series where each zn ∈ C, n ≥ 1. As in the real case, we sk =
to sk as the kth partial sum of the series.
Pk
n=n zn ,
k ≥ 1 and refer
P∞
7.1 Definition. If limk→∞ sk ∈ C then we say that the series converges to s := limk→∞ sk and write s = n=n zn .
Otherwise, we say that the series diverges. If limk→∞ sk = ∞, we say the series propositionerly diverges (note
that such a series will converge on the Riemann sphere). If limk→∞ sk does not exist, we say that the series is
oscillatory.
Since R ⊆ C, every real series is a complex series.
P∞
P∞
7.2 Definition. As in the real case, we say that a series n=0 zn converges
absolutely if n=0 |zn | converges. If
P
P
∞
∞
n=0 zn converges but is not absolutely convergent then we say that
n=0 zn is conditionally convergent.
The following are easy adaptations of the corollaryresponding results for real series.
P∞
7.3 Lemma. Let n=0 zn be a complex series
P∞
1. If n=0 zn converges then limn→∞ zn = 0.
P∞
P∞
2. If n=0 |zn | converges then n=0 zn converges.
P∞
P∞
P∞
3. If n=0 zn = a and n=0 w n = b then n=0 (zn + w n ) = a + b.
P∞
P∞
4. If n=0 zn is absolutely convergent and if σ : N → N is a bijection then n=0 zσ(n) is absolutely convergent,
with the same limit.
P∞
P∞
P∞ P∞
5. If n=0 zn = a and n=0 w n = b are absolutely convergent series then k=0 j=0 z j w k− j converges absolutely to a b.
P∞
P∞
7.4 Lemma. (The Comparison Test) If |zn | ≤ rn , n ≥ 1 and if n=0 rn converges then n=0 zn is absolutely
convergent.
7.1
Uniform Convergence
P∞
7.5 Definition. Let n=0 f n be a series, each of whose terms is a function defined on a subset E ⊆ C. For each
Pk
n ≥ 1, let skP= n=0 f n be the kth partial sum of the series. Suppose that s(z) = limk→∞ sk (z) exists for all
∞
z ∈ E. Then n=0 f n is said to be uniformly convergent on E if ∀ " > 0, there exists N > 0 so that k ≥ N implies
|s(z) − sk (z)| < " for all z ∈ E.
22
COMPLEX ANALYSIS
P∞
7.6 Theorem. If the
P∞series n=0 f n is uniformly convergent on a set E and if each f n is continuous at some fixed
point z0 ∈ E, then n=0 f n is also continuous at z0 .
PROOF: Assignment 5.
ƒ
P∞
7.7
P∞Theorem. (Weierstraß
P∞ M-test) Let n=0 f n be a series and suppose | f n (z)| ≤ an for all z ∈ E and n ≥ 1, where
n=0 an < ∞. Then
n=0 f n converges uniformly (and pointwise absolutely) on E.
P∞
PROOF: That n=0 | f n (z)| converges follows from the comparison
P∞test. To see that the series converges uniformly
on E, let " > 0 and choose N > 0 so that n ≥ N implies that k=n+1 ak < ". Then for any z ∈ E, if we denote
P∞
s(z) = n=0 f n (z), we have
|s(z) −
n
X
∞
X
f k (z)| = |
k=0
∞
X
f k (z)| ≤
k=n+1
| f k (z)| ≤
k=n+1
∞
X
ak < "
k=n+1
Thus the series converges uniformly on E.
ƒ
7.8 Example. The series f1 (z) = z and f n (z) = z n − z n−1 for n > 1 does not converge uniformly on D, even
though it converges pointwise for each point in D. Fix 0 < r < 1 and suppose z ∈ rD. Then for n ≥ 1 we have
| f n (z)| ≤ r n−1 (r + 1). Thus
∞
∞
X
X
r +1
| f n (z)| ≤
r n−1 (r + 1) =
<∞
1−r
n=1
n=0
P∞
By the last theorem, n=1 f n converges uniformly (and pointwise absolutely) on rD. Thus the series converges
uniformly on any compact subset of D. This is refered to as UCC (Uniform Convergence on Compacta).
7.2
Integration of Series
Clearly, if C ⊆ C is a piecewise smooth curve and if f1 , . . . , f n : C → C are continuous, then
Z
n
X
f k (z) dz =
C k=0
n
X
k=0
Z
f k (z) dz
C
We extend this to infinite series as follows.
∞
7.9 Theorem. Let C ⊆ C be a piecewise smooth
P∞ curve. Suppose that ( f n )n=0 is a sequence of functions from C
to C, each of which is continuous on C . If n=0 f n converges uniformly on C , then
Z
∞
X
C n=0
f n (z) dz =
∞
X
n=0
Z
f n (z) dz
C
P∞
PROOF: P
From assignment 5, we see that n=0 f n is continuous on C , and therefore it is integrable there. Let
n
sn (z) = k=0 f k (z) for z ∈ C (sn is clearly continuous). Let " > 0 and choose N > 0 so that n ≥ N implies
∞
n
X
X
"
f k (z) −
f k (z) <
k=0
λC + 1
k=0
SERIES
23
for all z ∈ C . Then for n ≥ N ,
Z
Z
Z n
∞
X
X
f k (z) dz −
f k (z) dz = C k=0
C
C k=0
≤
This says that
7.3
R P∞
C
k=0 f k (z) dz
=
P∞ R
k=0
C
"
∞
X
f k (z) −
k=0
λC + 1
n
X
k=0
!
f k (z)
dz λC < "
f k (z) dz.
ƒ
Differentiation of Series
P∞
P∞
7.10 Lemma. Suppose that n=0 f n is aP
uniformly convergent series with sum s(z) = n=0 f n (z), z ∈ E ⊆ C. Let
∞
ϕ : E → C be a bounded function. Then n=0 ϕ(z) f n (z) is uniformly convergent with sum ϕ(z)s(z) for z ∈ E.
PROOF: Trivial.
ƒ
7.11 Theorem. P
(Weierstraß) Let G ⊆ C be a domain. For each n ≥ 1, suppose that f n : G → C is analytic.
∞
Suppose s(z) = n=0 f n (z) exists for all z ∈ G and that the series is uniformly convergent on compact subsets of
G (i.e. the series is UCC on G ). Then s(z) is analytic on G . Moreover, the series is infinitely differentiable term
by term, so that
∞
X
s(k) (z) =
f n(k) (z)
n=0
for all k ≥ 1, for all z ∈ G , and the convergence of each differentiated series is also uniform on compact subsets
of G .
PROOF: Let z0 ∈ G and choose r > 0 small enough so that Vr (z0 ) ⊆ G . Since Vr (z0 ) is closed and bounded
P∞
(i.e. compact) the series n=0 f n(k) (z) converges uniformly on Vr (z0 ). In particular, it converges uniformly on
γ r := ∂ Vr (z0 ). If we let ϕk (z) =
1
k!
,
2πi (z−z0 )k+1
then for z ∈ γ r , |ϕk (z)| ≤
k!
s(z)
2πi (z − z0
)k+1
∞
X
k!
=
n=0
k!
,
2πr k+1
a fixed constant. By the last lemma,
f n (z)
2πi (z − z0 )k+1
converges uniformly on γ r . By the last section, we may integrate term by term, so for each k ≥ 0
Z
Z
∞
X
k!
s(z)
k!
f n (z)
dz =
dz
k+1
2πi γ (z − z0 )k+1
2πi
γ r (z − z0 )
n=0
r
When k = 0
1
2πi
Z
γr
s(z)
(z − z0 )
dz =
∞
X
1
n=0
f n (z)
Z
2πi
γr
(z − z0 )
dz =
∞
X
f n (z0 ) = s(z0 )
n=0
Thus s(·) satisfies Cauchy’s Integral formula. It now follows from the proof of Professor Zorzitto’s “infinite
differentiability theorem” that s(·) is analytic and thus s(·) is infinitely differentiable. (Note: analyticity of the
function in the statement of that theorem was only required to establish that the original function satisfied
Cauchy’s Integral formula.) With this in mind, we have for k ≥ 1
s(k) (z0 ) =
∞
X
n=0
f n(k) (z0 )
24
COMPLEX ANALYSIS
It remains to show that this series converges uniformly on compact sets. Let L ⊆ G be compact and " > 0. Fix
z0 ∈ L and choose r = r(z0 ) > 0 so that V2r (z0 ) ⊆ G . Then γ r (z0 ) ⊆ V2r (z0 ) ⊆ G . Moreover, V r (z0 ) ⊆ G . Since the
2
original series converges uniformly on compact sets and since γ r (z0 ) is compact we can find some N = N (z0 , ")
so that n ≥ N implies |s(w) − sn (w)| < " for all w ∈ γ r (z0 ). Then n ≥ N implies for any z ∈ V r (z0 )
2
k! Z
2πi γ
s(w)
r (z0 )
(w − z)
dw −
k+1
n
X
k!
j=0
2πi
Z
γ r (z0 )
since " > 0 is arbitrary, this shows that
compactness yields the result.
8
k! Z
f j (w)
dw = 2πi γ
(w − z)k+1
P∞
(k)
n=0 f n (z)
r (z0 )
s(w) − sn (w)
"
k!
2πr € Šk+1
dw
≤
k+1
r
2π
(w − z)
2
converges uniformly to s(k) (z) for z ∈ V r (z0 ). Applying
2
ƒ
Power Series
We now consider a special class of series P
of functions known
P∞ as power series. Let a ∈ C. Suppose that f n (z) =
∞
cn (z − a)n where cn ∈ C and n ≥ 0. Then n=0 f n (z) = n=0 cn (z − a)n is called a power series at a. The region
of convergence of the series is the set of all points where the series converges.
8.1 Example.
1. Let a = 0 and cn = nn for n ≥ 1 and c0 = 1. The series 1 + z + 4z 2 + 27z 3 + · · · converges
only at the point z = 0. Indeed, if z 6= 0 then there is n0 such that n ≥ n0 implies that |nn z n | ≥ 1. Therefore
the terms of the series do not tend to zero, so the series cannot converge.
2. Let a = 0, c0 = 1 and cn =
the comparison test.
1
nn
for n ≥ 1. The series 1 + z +
z2
4
+
z3
27
+ · · · converges for every point of C by
P∞
8.2 Lemma. Let a ∈ C and n=0 cn (z − a)n be a power series. Suppose the series converges at some point z0 6= a.
If 0 < |z1 − a| < |z0 − a|, then the series converges absolutely at z1 .
PROOF: Exercise. This is a trivial consequence of the Comparison Test.
ƒ
P∞
8.3 Proposition. If n=0 cn (z − a)n converges
P∞ for some z0n6= a but diverges for some z1 6= a then there is some
R
>
0
such
that
|z
−
a|
<
R
implies
that
n=0 cn (z − a) converges absolutely and |z − a| > R implies that
P∞
n
n=0 cn (z − a) diverges.
8.1
Radius of Convergence
P∞
PROOF: Let R = sup{r > 0 | n=0 cn (z − a)n converges absolutely for all |z − a| < r}.
P∞By the lemma above,
|z0 − a| ≤ R ≤ |z1 − a|. Clearly the series converges for |z − a| < R. If |z2 − a| > R and n=0 cn (z2 − a)n were to
converge then by the lemma above we would have R > |z2 − a|, a contradiction.
ƒ
P∞
We refer to the R above as the radius of convergence of the power series n=0 cn (z − a)n and the circle
γR = {z ∈ C : |z − a| = R} as the circle of convergence.
Whether of not the series converges for points on the
P∞
circle depends upon the series itself. If the series n=0 cn (z − a)n converges for all z ∈ C we say that R = ∞. If
the series diverges for all z 6= a then we say that R = 0.
P∞
8.4 Example. Consider the power series n=0 z n . This is a geometric series, as such, if |z| < 1 then the series
1
converges to 1−z
. If |z| ≥ 1 then the series diverges.
POWER SERIES
25
Note however that the series does not converge uniformly on D. Indeed, suppose otherwise. Let " > 0 and
choose N > 0 such that n ≥ N implies
n
1
X
−
z k < " for all z ∈ D
1 − z k=0 But z ∈ D implies
1
z∈D 1−z
whereas limz→1
|z| ≤ r and n ≥ N0 ,
n
n
X
X
|z k | ≤ n + 1
zk ≤
k=0 k=0
= ∞. On the other hand, if 0 < r < 1 we can choose N0 > 0 such that
P
N0 +1
r n < ". If
∞
∞
n
∞
n
X
1
X
X
X
X
k
k
k
k
≤
|z|
≤
rk < "
z
z
=
−
z
1 − z k=0 k=0 k=0 k=n+1
k=N +1
0
Thus
P∞
k=0 z
k
converges uniformly on all disks rD, and hence on all compact subsets of D.
It turns out that this behavior is typical power series.
P∞
8.5 Theorem. If n=0 cn (z − a)n has radius of convergence R > 0 then the series converges uniformly on every
compact subset of VR (a).
PROOF: Assignment.
ƒ
P∞
8.6 Corollary. If n=0 cn (z − a)n has radius of convergence R > 0 then the sum s(z) is continuous for all |z − a| <
R.
In fact, much more is true. Notice that for each n ≥ 0 then f n (z) = cn (z − a)n is analytic on C.
P∞
s(z) is
8.7 Theorem. Let s(z) = n=0 cn (z − a)n be a power series with radius of convergence
P∞R > 0. Thenn−1
0
and
analytic on VR (a). Moreover, the series can be differentiated term by term to get s (z) = n=1 ncn (z − a)
this series has the same radius of convergence R.
PROOF: As in the previous result, if 0 < r < R then by Theorem 8.5 the series converges uniformly on Vr (a) and
hence on
every compact subset of VR (a). By Weierstraß’s Theorem we find that s(z) is analytic on VR (a) and
P∞
s0 (z) = n=1 ncn (z − a)n−1 converges uniformly on compact subsets of VR (a).
If R0 is the radius of convergence of s0 (z), it follows that R0 ≥ R. Suppose that R0 > R. Let R < |z0 − a| < R0
and let C be the line segment connecting a to z0 . Then C is a compact subset of VR0 (a), and so the series s0 (z)
converges uniformly on C . We may integrate s0 (z) term by term along C to yield
Z ∞
∞ Z
∞
X
X
X
n−1
ncn (z − a) dz =
ncn (z − a)n−1 dz =
cn (z0 − a)n
C n=1
Thus this series converges and hence so does
n=1
P∞
C
n=0 cn (z0
n=1
− a)n , contradicting that R < |z0 − a|. Hence R0 = R. ƒ
We can now attack the problem of finding the radius of convergence of a given power series.
P∞
1
8.8 Theorem. (Cauchy-Hadamard) If n=0 cn (z − a)n is a power series and if Q := lim(|cn | n ) then the radius of
convergence of the series is R = Q1 (R = 0 if Q = ∞ and R = ∞ if Q = 0).
26
COMPLEX ANALYSIS
PROOF: We consider 3 cases.
1
Case 1: Q = ∞. In this case, the sequence P
(|cn | n )∞
n=0 is unbounded. We wish to show that the series converges only
∞
at z = a. Suppose otherwise, that n=0 cn (z0 −a)n converges for some z0 6= a. Then limn→∞ cn (z0 −a)n = 0,
and so in particular there exists an M > 1 such that
|cn (z0 − a)n | < M ∀ n ≥ 0
But then
1
1
n
|cn | ≤
Mn
|z0 − a|
≤
M
|z0 − a|
a contradiction. Hence the series converges only if z = a, so R = 0.
P∞
1
n
n
n=0 cn (z − a) for all z ∈ C. Since |cn | ≥ 0 for all n ≥ 0 and
Case 2: Q = 0. In this case we wish to prove that
1
1
limn→∞ |cn | n = 0, we must have limn→∞ |cn | n = 0. Fix z0 ∈ C and choose " > 0 such that "|z0 − a| < 21 .
1
Choose N = N (") so that |cn | n < " for all n ≥ N . Then if n ≥ N ,
|cn (z0 − a) | ≤ "|z0 − a|
n
n
<
n
1
2
P∞
By the comparison test, it follows that n=0 cn (z − a)n converges absolutely. Since z0 ∈ C was arbitrary,
R = ∞.
P∞
Case 3: 0 < Q < ∞. Our goal now is to show that if |z0 − a| < Q1 then n=0 cn (z − a)n converges, whereas if
P∞
1
|z0 − a| > Q1 then n=0 cn (z − a)n diverges. We know that Q is the largest limit point of (|cn | n )∞
n=0 . Thus,
given " > 0, we can find N > 0 such that n ≥ N implies that
1
|cn | n < Q + "
If |z0 − a| <
1
Q
then Q|z0 − a| < 1, so let
"=
1 − Q|z0 − a|
2|z0 − a|
Then for appropositionriate N > 0 we have that if n ≥ N then
1
|cn | n < Q +
=
1
|cn | n |z0 − a| <
Again by the comparison test,
Conversely, for |z0 − a| >
1
Q
P∞
n=0 cn (z
1 − Q|z0 − a|
2|z0 − a|
1 + Q|z0 − a|
2|z0 − a|
1 + Q|z0 − a|
2
=: q < 1
− a)n converges.
and any " > 0 we can find infinitely many n’s such that
1
|cn | n > Q − "
If |z0 − a| >
1
Q
then Q|z0 − a| > 1, so let
"=
Q|z0 − a| − 1
|z0 − a|
>0
TAYLOR SERIES
27
Then we have
1
|cn | n > Q −
for infinitely many n’s. Since the terms of
Q|z0 − a| − 1
|z0 − a|
P∞
n=0 cn (z
=
1
|z0 − a|
− a)n do not go to zero, the series diverges.
ƒ
Remark. As previously noted, the behavior
of convergence depends
P∞ nupon the coefficients
P∞ of a series on the
Pcircle
n
∞
of the series. For example, let f1 (z) = n=0 z n , f2 (z) = 1 + n=1 zn , and f3 (z) = 1 + n=1 nz 2 . Then they all have
radius of convergence R = 1. On the other hand, f1 diverges for every point on the circle of convergence, as the
terms of the series don’t go to zero. f2 diverges at z = 1 but converges at z = −1, and in fact converges for all
points on the circle of convergence except for z = 1. Finally, f3 converges absolutely at every point on the circle
of convergence.
9
Taylor Series
P∞
Let f (z) = n=0 cn (z − a)n be a power series with radius of convergence R > 0. As we saw in the last section, we
may differentiate f and get a power series with the same radius of convergence. Hence for any k ≥ 0
f (k) =
∞
X
(k + n)k ck+n (z − a)n
n=0
By setting z = a we see that
f (k) (a)
ck =
k!
If a power series is related to an analytic function f by the equation above, we say that the series is a Taylor
series for f at the point a. We refer to the cn ’s as the Taylor coefficients. It is clear that if f is given by a power
series with radius of convergence R > 0 then it is it’s own Taylor series.
With f as above and 0 < r < R, let γ r be the circle of radius r centred at a (traversed once in the positive
direction). By the Cauchy Integral formula,
f
(n)
(a) =
Z
n!
2πi
γr
f (z)
(z − a)n+1
dz
Combining this with the above equation gives us
cn =
1
2πi
f (z)
Z
γr
(z − a)n+1
dz
Suppose that | f (z)| ≤ M for all |z − a| < R. Then we obtain Cauchy’s Inequalities
|cn | ≤
1
2π
(2πr)
M
r n+1
=
M
rn
This is true for each r < R, and therefore |cn | ≤ RMn .
that a power series with radius of convergence R > 0 converges to an analytic function s(z) =
P∞We have seen
n
c
(z
−
a)
in
the disk |z − a| < R. The following result may be considered a converse to this.
n=0 n
28
COMPLEX ANALYSIS
9.1 Theorem. Suppose that
P∞f is analytic on VR (a). Then f admits a Taylor series expansion in VR (a), which is to
say that the power series n=1 cn (z − a)n with coefficients
f (n) (a)
cn =
=
n!
f (z)
Z
1
2πi
(z − a)n+1
γ
dz
converges to f (z) for all z ∈ VR (a).
PROOF: Let x ∈ VR (a) and with 0 < |w − a| < r < R, let γ r denote the circle of radius r centred at a. Then
w ∈ intγ r . By Cauchy’s Integral formula
f (w) =
f (z)
Z
1
2πi
z−w
γr
dz
To obtain a power series from this, first observe that
1
z−w
=
1
(z − a) − (w − a)
=
1
1
z−a 1−
w−a
z−a
Notice that α := w−a
< 1 for all z ∈ γ r . Hence
z−a
1
z−w
It follows that
1
=
1
1
z−a 1−
f (z)
2πi z − w
w−a
z−a
=
∞
1 X
=
2πi
∞
X
(w − a)n
n=0
(z − a)n+1
f (z)
(z − a)n+1
n=0
(w − a)n
P∞ 1 M αn
Since α < 1, n=0 2πi
< ∞ and so the series above is uniformly convergent on γ r by the Weiestraß M-test.
r
Thus we can integrate along γ r term by term. This gives
f (w) =
=
=
Z
1
2πi
γr
∞
X
1
2πi
n=0
∞
X
f (z)
z−w
Z
γr
dz
f (z)
(z − a)n+1
dz (w − a)n
cn (w − a)n
n=0
Thus the series converges to f on VR (a).
ƒ
9.2 Definition. If a function f is analytic at a point z0 ∈ C (which is to say that f is differentiable in some
neighbourhood Vδ (z0 )) then we say that z0 is a regular point of f . Otherwise we say that z0 is a singular point
(or singularity) of f .
P∞
9.3 Theorem. Suppose that f has a Taylor expansion f (z) = n=0 cn (z − a)n at the point a, with radius of
convergence R > 0. If f is defined on γR (a) then f has at least one singularity there.
TAYLOR SERIES
29
PROOF: Suppose that f has no singularities on γR (a). For each a ∈ γR (a) we can find an open disk Vδz (z) on
which f is analytic. Since γR (a) is compact, we can find finitely many z1 , . . . , zn ∈ γR (a) such that
γR (a) ⊆
n
[
Vδz (zk ) =: G
k
k=1
Let ρ := dist(γR (a), ∂ G ) > 0. Then f is analytic inside γR+ ρ (a) and so has a convergent Taylor series inside
2
γR+ ρ (a). But the coefficients of this series coincide with the coefficients of the original series since they (effec2
tively) depend only on a. This contradicts the fact that the radius of convergence of the series is R.
ƒ
Consider the following situation with real series. Given the function f (x) =
around the origin
1
= 1 − x2 + x4 − x6 + · · ·
1 + x2
1
,
1+x 2
we can expand f as a series
which diverges for |x| ≥ 1 and converges for |x| < 1. The behavior of f (x) at x = ±1 is not exceptional in any
1
way, and yet the series representation behaves badly. By considering g(z) = 1+z
2 we see that z = ±i are singular
points of g, and therefore by the above theorems we find that the power series will diverge |z| > 1, hence for
|x| > 1.
9.1
Uniqueness Theorem for Power Series
We now turn our attention P
to the uniqueness of the coefficients in a Taylor expansion. Suppose that f (z) =
P
∞
∞
n
n
n=0 cn (z − a) and g(z) =
n=0 dn (z − a) have radii of convergence R f , R g > 0. Suppose furthermore that for
some 0 < δ < min(R f , R g ) we know that f (z) = g(z) for all |z − a| < δ. But then for all n ≥ 0
cn =
f (n) (a)
n!
=
g (n) (a)
n!
= dn
9.4 Theorem. (Uniqueness Theorem P
for Power Series) Let E ⊆ CPbe an infinite set and suppose a ∈ C is an
∞
∞
accumulation point of E. Let f (z) = n=0 cn (z − a)n and g(z) = n=0 dn (z − a)n and suppose that these series
converge on VR (a) for some R > 0. Suppose also that f (e) = g(e) for all e ∈ E. Then cn = dn for all n ≥ 0.
PROOF: Now f and g are analytic and therefore continuous on VR (a). Let (zn )∞
n=1 be a sequence of points in E
which converge to a. Since f and g are continuous at a
c0 = f (a) = lim f (zn ) = lim g(zn ) = g(a) = d0
n→∞
P∞
n→∞
P∞
Let f1 (z) = n=0 cn+1 (z − a)n and g1 (z) = n=0 dn+1 (z − a)n and notice that the radius of convergence of f1 and
g1 is the same as for f and g. Notice further that
f1 (zn ) =
f (zn ) − c0
zn − a
=
g(zn ) − d0
zn − a
= g1 (zn )
for all n ≥ 0. From the argument above we get that c1 = d1 . Continuing by induction, we get cn = dn for all
n ≥ 0.
ƒ
9.5 Theorem. (Uniqueness Theorem for Analytic Functions) Suppose that G ⊆ C is a domain and that E ⊆ G is
an infinite set with accumulation point a ∈ G . Suppose f , g : G → C are analytic and that f (e) = g(e) for all
e ∈ E. Then f (z) = g(z) for all z ∈ G .
30
COMPLEX ANALYSIS
PROOF: Let w 6= a be any point in G . Choose a continuous path C = {z(t) | t ∈ [0, 1]} such that z(0) = a and
z(1) = w. If G = C let r = 1, otherwise let r = dist(C , ∂ G ) > 0.
Since z(t) is continuous and therefore uniformly continuous on [0, 1], we can find δ > 0 such that |t 1 −t 2 | < δ
implies |z(t 1 )−z(t 2 )| < r. Let Π = {0 = t 0 < t 1 < · · · < t n = 1} be a partition of norm at most δ. Since |t i −t i−1 | <
δ, the centres of the disks Vr (z(t i )) and Vr (z(t i−1 )) are less than r apart, and so Vr (z(t i )) ∩ Vr (z(t i−1 )) 6= ∅ for
i = 1, . . . , n. Let (en )∞
in E converging to a.
n=1 be a sequence
P∞
P∞Since f and g are analytic on Vr (a), they admit
Taylor series expansions f (z) = n=0 cn (z − a)n and g(z) = n=0 dn (z − a)n for |z − a| < r. Since f (en ) = g(en )
for all n ≥ 1, the uniqueness theorem for power series give us that f (z) = g(z) for all z ∈ Vr (a) = Vr (z(t 0 )).
Choose any point z ∈ Vr (z(t 0 )) ∩ Vr (z(t 1 )). Then z is an accumulation point of the set E1 = Vr (z(t 0 )) ∩
Vr (z(t 1 )), on which f (e) = g(e) for all e ∈ E1 . By the same reasoning as above, f (u) = g(u) for all u ∈ Vr (z(t 1 )).
Continue this process to find that f (u) = g(u) for all u ∈ Vr (z(t k )) for all k = 0, . . . , n. But then f (w) =
f (z(t n )) = g(z(t n )) = g(w). Since w ∈ G \ {a} was arbitrary and since both f and g are continuous at a, we
conclude that f = g on G .
ƒ
€1Š
9.6 Example.
1. Suppose that f is entire and f n = 0 for all n ≥ N then f = 0.
2. Suppose f : V2 (0) → C is analytic and f (z) = 0 for all z ∈ [0, 1]. Then f = 0.
3. In constrast, let f : R → R be defined by
¨
0
x ≤π
f (x) =
cos(x) + 1 x > π
Then f is differentiable on R, f = 0 on (−∞, π], but f 6= 0. Thus the structure of analytic complex
functions is stronger than that of real differentiable functions.
9.7 Definition. A zero of a function f is a point z0 ∈ C such that f (z0 ) = 0.
If GPis a domain and f : G → C is analytic and a ∈ G is a zero of f , then f admits a Taylor expansion
∞
f (z) = n=0 cn (z − a)n which converges in the largest disk centred at a which is contained in G . Since f (a) = 0,
c0 = 0. Unless f ≡ 0 on G , there exists some m0 ∈ N so that cm0 6= 0. In particular, if k is the smallest such index,
then we say that a is a zero of order k for f . If k = 1 we also say that a is a simple zero.
As an easy corollary of the uniqueness theorem, every zero of a non-zero analytic function f on a domain G
is isolated.
The following result is useful in proving the Maximum Modulus Principle.
9.8 Theorem. Let G ⊆ C be a domain and suppose that there is M ≥ 0 such that | f (z)| = M for all z ∈ G . Then
f is constant on G .
PROOF: This was on the midterm.
ƒ
9.9 Theorem. (Maximum Modulus Principle) Let G ⊆ C be a domain and f : G → C be a non-constant analytic
function. Then | f | does not attain its maximum in G .
PROOF: Suppose otherwise, say z0 ∈ G is such that M := | f (z0 )| ≥ | f (z)| for all z ∈ G . Now M > 0, for otherwise
f is (constantly) zero on G . Choose δ > 0 so that f is analytic on Vδ (z0 ) ⊆ G . Let 0 < r < δ. By Cauchy’s
integral formula,
Z
1
f (z)
f (z0 ) =
dz
2πi γ z − z0
r
In fact, with z = z0 + r e , dz = i r e dθ , so
iθ
iθ
f (z0 ) =
1
2π
Z
2π
0
f (z0 + r e iθ )dθ
LAURENT SERIES
Thus
31
Z 2π
Z 2π
1 Z 2π
1
1
iθ
iθ
M = | f (z0 )| = f (z0 + r e )dθ ≤
| f (z0 + r e )|dθ ≤
M dθ = M
2π 0
2π 0
2π 0
Hence all of the inequalities are equalities. We claim that | f (z0 + r e iθ )| = M , and this is obvious. The choice
of r was arbitrary, so | f (z)| = M for all z ∈ Vδ (z0 ). Therefore f is constant on the disk, and by the uniqueness
theorem f is constant on G . This contradiction shows that f does not attain its maximum on G .
ƒ
9.10 Corollary. (Minimum Modulus Principle) Let G ⊆ C be a domain and f : G → C be a non-constant analytic
function that does not vanish on G . Then | f | does not attain its minimum in G .
PROOF:
1
f
is analytic on G .
ƒ
9.11 Corollary. Let G ⊆ C be a bounded domain and f : G → C be an analytic function in G that is continuous
on G (resp. and non-vanishing on G ). Then ∂ G contains a point a so that | f (a)| = max{| f (z)| | z ∈ G } (resp.
| f (a)| = min{| f (z)| | z ∈ G }).
PROOF: Since f is constant on G and G is compact, | f | attains its max (min) at some point a ∈ G . If a ∈ G then
f is constant so we may choose any point on the boundary. Otherwise we are done.
ƒ
9.12 Corollary. Let G ⊆ C be a bounded domain and f : G → C be an analytic function in G that is continuous
on G and non-vanishing on G . Suppose that | f ||∂ G is constant. Then f is constant on G .
PROOF: f attains both its max and min on the boundary, hence they are equal and f is a constant function.
10
ƒ
Laurent Series
We now look at the case of series with negative exponents.
P0
1
10.1 Theorem. Given a series n=−∞ cn (z − a)n , let ` = lim |cn | n . Then there are three possibilities:
1. ` = 0 and the series convergec for all a 6= z ∈ C ∪ {∞}
2. 0 < ` < ∞ and the series converges outside of the circle |z − a| = ` and diverges inside this circle.
3. ` = ∞ and the series diverges for all z ∈ C
PROOF: If we set w =
1
.
`
1
,
z−a
then the above series becomes c0 +c−1 w+c−2 w 2 +· · · and this has radius of convergence
The points z = a, z = ∞ are carried to w = ∞, w = 0, while the points outside the circle |z − a| = ` are carried
to points inside the circle |w| < 1` and visa versa. This observation proves the theorem.
ƒ
Clearly the map w =
0
1
z−a
maps every closed bounded region D outside the circle |z − a| = ` into a closed
0
bounded region D inside the circle |w| < 1` . Since the series in w converges uniformly in D , it follows that
P0
n
n=−∞ cn (z − a) converges uniformly in D. Since each term of the series is analytic outside of |z − a| = `, the
sum s(z) of the series is an analytic function at all points |z − a| > `. If ` = 0, the circle |z − a| = ` is just the point
a, in which case s(·) is analytic for all z 6= a.
P∞
P∞
10.2 Definition. A Larent series is a formal series n=−∞ cn (z − a)n which is the sum of n=0 cn (z − a)n , the
P−1
regular part of the series, and n=−∞ cn (z − a)n , the principal part. By definition, the series converges if and only
if both the regular and principal parts converge.
32
COMPLEX ANALYSIS
The regular part of the series converges inside the disk of radius
1
R=
1
limn≥0 |cn | n
while the principal part of the series converges outside of the circle of radius
1
r = lim |cn | n
Thus the Larent series converges absolutely and uniformly on every compact set in the annulus A = {z ∈ C | r <
|z − a| < R}, provided r < R. Moreover, the sum s(z) is analytic on A.
P∞
10.3 Theorem. Let s(z) = n=−∞ cn (z − a)n be a Laurent series with annulus of convergence r < |z − a| < R. If
r < ρ < R and γρ (a) is the circle centred at a of radius ρ then for each n ∈ Z
cn =
Z
1
2πi
s(z)
γρ
(z − a)n+1
dz
PROOF: The series converges on the compact set γρ (a) and hence the same is true of each of the series
1
s(z)
2πi (z − a)n+1
1
for each n ∈ Z, since 2πi(z−a)
n+1 =
1
2πρ n+1
∞
X
1
=
2πi
ck (z − a)k−n−1
k=−∞
for all z ∈ γρ (a). (We can multiply a uniformly convergent series by
a bounded function and get another uniformly convergent series.) Since the series converges uniformly we can
integrate term by term,
Z
Z
∞
X
1
s(z)
1
dz =
ck
(z − a)k−n−1 dz
2πi γ (z − a)n+1
2πi
γ
k=−∞
ρ
As for computing
R
γρ
ρ
(z − a)k−n−1 dz, let z = a + ρe iθ as θ runs over [0, 2π]. Then dz = ρie iθ dθ and
Z
2π
(ρe )
iθ k−n−1
ρie dθ =
Z
0
Hence
R
γρ
(z − a)k−n−1 dz = δk,n and so
2π
¨
ρ
iθ
k−n i(k−n)θ
e
idθ =
0
1
2πi
s(z)
dz
γρ (z−a)n+1
R
= cn .
0
2πi
if k 6= n
if k = n
ƒ
10.4 Theorem. Let A = {z ∈ C | P
r < |z − a| < R} and suppose f : A → C is analytic. Then f has a Laurent
∞
expansion at a. That is, the series n=−∞ cn (z − a)n where
cn =
1
2πi
Z
γρ (a)
f (z)
(z − a)n+1
dz
converges to f (z) for all z ∈ A.
PROOF: Let w ∈ A, and choose r 0 > r, R0 < R so that 0 < r < r 0 < |w − a| < R0 < R. Choose ρ so that γρ (w) ⊆
{z ∈ C | r 0 < |z − a| < R0 }. Now the function
f (z)
z−w
is analytic on {z ∈ C | r 0 < |z − a| < R0 } ∩ {z ∈ C | |z − w| > ρ}.
LAURENT SERIES
33
By Cauchy’s integral formula
Z
1
2πi
f (z)
γR0 (a)
z−w
dz =
f (w) =
We’ll show that
1
2πi
f (z)
dz
γR0 (a) z−w
R
(resp.
series exapansion. If z ∈ γR0 (a) then
1
2πi
2πi
2πi
f (z)
z−w
γR0 (a)
f (z)
dz)
γ r 0 (a) w−z
R
z−w
If z ∈ γ r 0 (a) then
w−z
z−w
γ r 0 (a)
Z
1
1
1
f (z)
Z
1
=
=
dz +
1
2πi
1
2πi
f (z)
Z
γρ (w)
Z
γ r 0 (a)
z−w
f (z)
w−z
dz
dz
yields the non-negative (resp negative) powers of the
∞
X
(w − a)n
n=0
(z − a)n+1
∞
X
(w − a)n
n=0
dz +
(z − a)n+1
=
∞
X
(w − a)−n
n=1
(z − a)−n+1
The uniform convergence of the (geometric) series on γR0 and γ r 0 respectively means that we can integrate term
by term, giving the result.
ƒ
Suppose that | f (z)| ≤ M for all z ∈ A. Then Cauchy’s inequalities become
|cn | ≤
1
2πi
2πρ
M
ρ n+1
=
M
ρ n+1
for all n ∈ Z and r < ρ < R.
Let z0 ∈ C and suppose that for some δ > 0, f is analytic on Vδ (z0 ) \ {z0 }. Then z0 is called an isolated
singular point of f . Consider the Laurent expansion of f
f (z) =
∞
X
cn (z − z0 )n
n=−∞
valid in Vδ (z0 ) \ {z0 }. Then are 3 possibilities
1. If cn = 0 for all n < 0 then z0 is called a removable singularity.
2. There exists M > 0 so that cn = 0 for n < −M . In this case z0 is called a pole. The order of the pole is that
1
largest power of z−z
which appears in this expansion.
0
3. If there are infinitely many m < 0 such that cm 6= 0 then z0 is called an essential singularity of f .
P∞
If z0 is a removable singularity, then f (z) = n=0 cn (z − z0 )n , which clearly converges at z0 . If we define
g(z) = f (z) and g(z0 ) = c0 then g is analytic on the whole disk Vδ (z0 ).
If z0 is a pole of order m for f , then
(z − z0 )m f (z) = c−m + c−m+1 (z − z0 ) + c−m+2 (z − z0 )2 + · · ·
is an ordinary power series with non-zero constant term c−m . Thus z0 is a removable singularity of (z − z0 )m f (z).
Verify that limz→z0 f (z) = ∞. We now study the relationship between zeroes and poles.
10.5 Theorem. Let z0 be a zero of order m of a function 0 6= f analytic at z0 . Then
neighbourhood of z0 , with a pole of order m at z0 .
1
f
is analytic in a punctured
34
COMPLEX ANALYSIS
PROOF: We can write f (z) =
non-zero
so z0 is a
P∞
k
k=m ck (z − z0 ) with cm 6= 0, and so f (z)
−m
(z−z0 )
1
at z0 . Thus f (z)
= ϕ(z)
. Now ϕ1 has a power series expansion
pole of order m of 1f .
= (z − z0 )ϕ(z) where ϕ is analytic and
at z0 with non-zero constant coefficient,
ƒ
The converse is also true, and the proof is similar.
10.6 Theorem. Let f be analytic in a punctured neighbourhood of z0 , where z0 is a pole of order m of f . If we
set f (z1 ) := 0 then 1f is analytic at z0 with a zero of order m.
0
PROOF: Exercise.
ƒ
10.7 Corollary. Suppose that f is analytic and non-vanishing in a punctured neighbourhood of z0 and has an
essential singularity at z0 . Then 1f has an essential singularity at z0 .
10.8 Theorem. (Casorati-Weierstraß) Let f be analytic in a punctured neighbourhood of z0 and suppose that z0
is an essential singularity of f . If A ∈ C ∪ {∞} then there is a sequence zn → z0 and f (zn ) → A.
This is a special case of Picard’s Theorem: If f has an essential singularity at z0 and δ > 0 is small enough that
f is analytic on Vδ (z0 )\{z0 } then there is w0 ∈ C such that for all w0 6= w ∈ C, the equation f (z) = w has infinitely
many solutions inside of Vδ (z0 ) \ {z0 }. We will not prove Picard’s theorem, but will prove the Casorati-Weierstraß
theorem.
PROOF: There are two cases.
A = ∞: Suppose to the contrary that there is δ > 0 such that | f (z)| < M for all 0 < |z − z0 | < δ. By the Cauchy
Inequalities,
|cn | ≤
M
ρn
∀ n∈Z∀0<ρ<δ
where the cn are the coefficients in the Laurent expansion of f at z0 . Now if n < 0 then we can let ρ → 0
to show that cn = 0. That is, the Laurent expansion of f at z0 has no negative powers, so z0 is a removable
singular point of f , a contradiction. Thus | f | is unbounded on every Vδ (z0 ) \ {z0 }, from which a sequence
converging to z0 such that the image under f diverges to ∞ may be found.
A ∈ C: If every punctured neighbourhood of z0 contains a point z so that f (z) = A, we are done. Otherwise,
1
suppose that δ > 0 is chosen so that s ∈ Vδ (z0 ) \ {z0 } implies that f (z) 6= A. Then ϕ(z) = f (z)−A
has an
essential singularity at z0 (by the corollary above). By the first case, we may find a sequence zn → z0 such
that ϕ(zn ) → ∞, which is to say that f (zn ) → A.
ƒ
1
10.9 Example. Let f (z) = e z = 1 +
1
z
+
1
2z 2
+
1
6z 3
+ · · · , which has an essential singularity at z = 0. For A = ∞
let zn = 1n . Then f (zn ) = e n → ∞ = A. For A = 0 let zn = − 1n . Then f (zn ) = e−n → 0 = A. For A ∈ C \ {0}, solving
1
e z = A yields z =
1
.
log A
Let log0 denote the branch of the complex logarithm so that arg z lies in the interval
[0, 2π). Then z0 =
1
log0 A
and f (z0 ) = e z0 = elog0 A = A. In fact zn :=
1
for all n ≥ 0. Note that f (z) = 0 has no solution in any Vδ (0) \ {0}.
1
log0 A+2πni
is such that zn → 0 and f (zn ) = A
RESIDUES
11
35
Residues
11.1 Definition. Let z0 be an isolated singular point of an analytic function f . The residue of f at z0 is the
coefficient c−1 in the Laurent expansion of f at z0 , denoted Res( f ; z0 ).
If f has a pole or an essential singularity at z0 then Res( f ; z0 ) may or may not be zero. If z0 is a removable
singularity of f then Res( f ; z0 ) = 0.
11.2 Theorem. (Residue Theorem) Suppose that f is analytic inside and on a piecewise smooth closed Jordan
curve C , except for isolated singular points z1 , . . . , zN lying inside C . Then
Z
1
2πi
f (z)dz =
C
N
X
Res( f ; zk )
k=1
PROOF: Let γ1 , . . . , γN be circles centred at z1 , . . . , zN , respectively, and small enough to be contained inside C
and not to intersect each other. By (a corollary to) Cauchy’s Theorem,
Z
f (z)dz =
C
N
X
Z
f (z)dz
γk
k=1
Suppose that the Laurent series for f at zk is
∞
X
f (z) =
cn(k) (z − a)n for k = 1, . . . , N
n=−∞
Integrating term by term (as the series converges uniformly along γk ), we obtain
Z
f (z)dz =
∞
X
Z
γk n=−∞
γk
∞
X
=
=
n=−∞
∞
X
cn(k) (z − a)n dz
cn(k)
Z
(z − a)n dz
γk
cn(k) 2πiδn,−1
n=−∞
= 2πiRes( f ; zk )
Therefore
1
2πi
R
C
f (z)dz =
PN
k=1 Res( f
11.3 Example. Evaluate the integral
pole of order n. Indeed
; zk ).
ez
dz.
|z|=2 (z−1)n
R
f (z) =
and so
ez
dz
|z|=2 (z−1)n
R
ƒ
= 2πiRes( f ; 1) =
ez
(z − 1)n
2πie
.
(n−1)!
=
The only singularity of f (z) =
eez−1
(z − 1)n
=
∞
X
e(z − 1)i−n
i=0
i!
ez
(z−1)n
is at z = 1 and it is a
36
COMPLEX ANALYSIS
Next we show how to calculate the residues at a pole without making explicit use of the Laurent expansions.
c−1
First suppose that z0 is a simple pole of f . Then f (z) = (z−z
+ c0 + c1 (z −z0 )+ c2 (z −z0 )2 +· · · in some punctured
0)
neighbourhood of z0 . Thus Res( f ; z0 ) = c−1 = limz→z0 (z − z0 ) f (z).
In particular, if f (z) =
g(z)
h(z)
where g(z0 ) 6= 0 and z0 is a simple zero of h then z0 is a simple pole of f , so
Res( f ; z0 ) = lim (z − z0 )
z→z0
g(z)
h(z)
g(z)
= lim
z→z0 h(z)−h(z0 )
z−z0
=
g(z0 )
h0 (z0 )
Suppose that z0 is a pole of order m > 1 for f . Then f (z) = c−m (z − z0 )−m + · · · + c0 + c1 (z − z0 ) + · · · , and so
(z − z0 )m f (z) = c−m + c−m+1 (z − z0 ) + · · · + c−1 (z − z0 )m−1 . But then [(z − z0 )m f (z)](m−1) = (m − 1)!c−1 + · · · , so
taking limits gives
1
lim [(z − z0 )m f (z)](m−1)
Res( f ; z0 ) = c−1 =
(m − 1)! z→z0
In the example above,
en
(z−1)n
Res
11.1
has a pole of order n at 1, so
en
(z − 1)
;1
n
=
1
lim (z − 1)n
(n − 1)! z→1
(n−1)
en
(z
− 1)n
=
e
(n − 1)!
Applications of residues
11.4 Definition. The logarithmic residue of an analytic function f at a point a is Res( f 0 / f ; a).
Suppose that a is a zero of order m of f . Then f (z) = cm (z − a)m + · · · in a neighbourhood of a, and so
f (z) = mcm (z − a)m−1 + · · · in that neighbourhood. From this we get
0
f 0 (z)
f (z)
=
=
=
mcm (z − a)m−1 + · · ·
cm (z − a)m + · · ·
m
z−a
m
z−a
cm +
m+1
c
(z
m m+1
− a) + · · ·
!
cm + cm+1 (z − a) + · · ·
ϕ(z)
where ϕ(z) is analytic on some neighbourhood of a and ϕ(a) = 1. From this it follows that
Res( f 0 / f ; a) = lim(z − a)
z→a
m
z−a
ϕ(z) = m
the order of the zero at a. Similarily, if b is a pole of order n of f then Res( f 0 / f ; b) = −n.
11.5 Theorem. Let C be a piecewise smooth closed Jordan curve and suppose that f is analytic on and inside
C , expect at finitly many poles b1 , . . . , bn inside C . Suppose also that f has zeros a1 , . . . , am inside C . then
1
2πi
Z
C
f 0 (z)
f (z)
dz =
m
X
k=1
αk −
n
X
βj
j=0
where αk (resp. β j ) is the order of the zero at ak (resp. the order of the pole at b j ).
RESIDUES
37
PROOF: The function f 0 / f only has poles where f has a pole or a zero, so
1
2πi
Z
C
f 0 (z)
f (z)
dz =
s
X
Res( f 0 / f ; p r )
r=1
where p r is the r th pole. The result follows from the calculations above.
ƒ
11.6 Theorem. (Rouché) Suppose that f and g are analytic inside and on a piecewise smooth closed Jordan
curve C and that | f (z)| > |g(z)| at every point of C . Then f and f + g have the same number of zeroes (counted
with multiplicity) inside C .
PROOF: Let t ∈ [0, 1]. Since | f (z)| > |g(z)| for all z ∈ C , ( f + t g)(z) 6= 0 for all z ∈ C . Let
Z
1
( f + t g)0 (z)
ϕ(t) =
dz
2πi C ( f + t g)(z)
By Theorem 11.5, ϕ(t) is the number of zeroes of f + t g inside C counted with multiplicity. Since ϕ is integer
valued, if we can show that ϕ is continuous on [0, 1] then it must be constant. In particular, ϕ(0) = ϕ(1) implies
that the number of zeroes of f is equal to the number of zeroes of f + g inside C (counted with multiplicity). To
see that ϕ is continuous, fix t ∈ [0, 1]. For s ∈ [0, 1],
Z
f 0 + t g0
1 f 0 + sg 0 |ϕ(t) − ϕ(s)| =
−
2π C f + t g
f + sg Z
(g 0 f − f 0 g)(s − t) 1 =
2π C ( f + t g)( f + sg) 0
(g f − f 0 g)(s − t) 1
≤
λC |t − s| max C
2π
( f + t g)( f + sg) We can find some M1 > 0 so that | f + t g| ≥ M1 on C , since f + t g is continuous and non-vanishing on the
compact set C . Also, g 0 f − f 0 g is continuous on the compact set C , so it is bounded by some M2 > 1. If
M
|s − t| < 21 M1 then
2
|( f + s g)(z)| ≥ |( f + t g)(z)| − |(s − t)g(z)| ≥
Thus |s − t| <
1 M1
2 M2
1
2
M1 > 0
implies that
|ϕ(t) − ϕ(s)| ≤
1
2π
‚
λC
2M2
M12
Œ
|t − s|
so ϕ is Lipshitz and hence continuous.
ƒ
11.7 Example. How many zeros does the function h(z) = z 8 −4z 5 +z 2 −1 have in the unit circle? Let f (z) = −4z 5
and g(z) = z 8 + z 2 − 1. Then for |z| = 1
| f (z)| = | − 4z 5 | = 4 > 3 = |z 8 | + |z 2 | + | − 1| ≥ |g(z)|
By Rouché’s Theorem, f and h = f + g have the same number of zeroes (including multiplicities) inside the the
curve |z| = 1, which is to say on the unit disc. f has 5 zeroes there, so h does as well.
38
COMPLEX ANALYSIS
11.8 Example. Show that h(z) = 2 + z 2 − e iz has precisely one simple zero in the upper open half plane. Let
f (z) = 2 + z 2 and g(z) = −e iz . Let CR = ΓR ∪ [−R, R], where ΓR = {Re iθ | θ ∈ [0, π]}. Then CR is a closed,
piecewise smooth Jordan curve and f , g are entire. For z ∈ [−R, R] we have
| f (z)| = |2 + z 2 | ≥ 2 > 1 = |e iz | = |g(z)|
For z ∈ ΓR with R >
p
3 we have
| f (z)| = |2 + z 2 | ≥ |z|2 − 2 = R2 − 2 > 1 ≥ |e−ℑz | = |e iz | = |g(z)|
By Rouché’s Theorem, f and h = f + g have the same number of zeroes inside of CR . Since R was arbitrary, f
and h have the same number of zeroes on the whole upper open half plane. f has one zero there, so h does as
well.
11.2
Evaluation of Improper Integrals
Techniques from complex analysis may be used to evaluate difficult looking improper real integrals with ease.
We will end this section with two examples – see the textbook for many more.
R∞
11.9 Example. Evaluate −∞ (x 2 d+ax 2 )3 , where a > 0.
Let f (z) =
z0 = ai. Now
1
(z 2 +a2 )3
and CR = ΓR ∪ [−R, R] (as above). Then f has only one singular point inside CR , namely
1
Res( f ; ai) = lim
z→ai
Therefore
R
CR
2!
f (z)dz = 2πiRes( f ; ai) =
(z − ai)3
3π
.
8a5
(2)
1
(z 2 + a2 )3
1 (−3)(−4) 3
=
5
2 (z + ai) z=ai 16a5 i
=
On the other hand,
Z
Z
dz
1
πR
≤
f
(z)dz
=
≤
λ
sup
ΓR
(R2 − a2 )3
2
2
3
2
2
3
Γ (z + a ) Γ
(z
+
a
)
|x+i y|=R
R
R
R
So limR→∞
ΓR
y>0
f (z)dz = 0 and hence
Z
∞
dx
(x 2 + a2 )3
−∞
11.10 Example. Evaluate
R∞
0
sin x
x
Z
= lim
R→∞
R
f (z)dz = lim
R→∞
−R
Z
f (z)dz =
CR
3π
8a5
d x.
e iz
.
z
Consider f (z) =
Then ℑ f (z) = sinz z when z ∈ R \ {0}. Let CR,r = ΓR ∪ [−R, −r] ∪ Γ r ∪ [r, R], oriented in
the anticlockwise direction. Then f is analytic on and inside CR,r for all R > r > 0. By Cauchy’s Integral Theorem
R
f (z)dz = 0. Taking limits as R → ∞ and r → 0 we get
C
R,r
0=
Z
0
−∞
Consider
e iz
ΓR z
R
ei x
x
d x + lim
r→0
Z
Γr
e iz
z
dz +
Z
∞
0
ei x
x
d x + lim
R→∞
Z
ΓR
e iz
z
dz
dz. Notice that ou usual estimate yields that this integral is less than or equal to π, which is of
RESIDUES
39
little help to us. In stead we use integration by parts:
Z
Z
e iz
de iz
dz =
dz
z
iz
ΓR
ΓR
Z
e iz R
e iz
=
− 2 dz
−
iz −R
iz
ΓR
Z
e iR + e−iR 1
e iz
=
+
− 2 dz −→ 0 as R → ∞
iR
i Γ
z
R
R
iz
− ez 2 dz| ≤ πR R12 .
R iz
As for lim r→0 Γ ez dz, notice that
since |
ΓR
r
e iz
z
=
1
z
+i+
i2z
2!
+
i3z2
3!
+ ··· =
1
z
+ ϕ(z)
where ϕ is analytic on a neighbourhood of zero. Hence ϕ is continuous on a neighbourhood of zero, and so
bounded there, say by M > 0. Then
Z
Z
Z
e iz
dz
ϕ(z)dz = −πi
lim
dz = lim
+ lim
r→0
r→0
r→0
z
z
Γ
Γ
Γ
r
R
r
R
r
since Γ ϕ(z)dz ≤ λΓr M → 0 as r → 0 and lim r→0 Γ dz
= lim r→0 r e1iθ ir e iθ dθ = −πi. Taking imaginary parts,
r z
r
we get
Z0
Z∞
sin x
sin x
dx +
dx = π
x
x
−∞
0
R∞
And since these integrals are equal, 0 sinx x d x = π2 .