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Title: complex analysis
Description: This is the note of complex analysis in university of Manchester.
Description: This is the note of complex analysis in university of Manchester.
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MATH20101
Complex Analysis
Charles Walkden
2nd December, 2016
MATH20101 Complex Analysis
Contents
Contents
0 Preliminaries
2
1 Introduction
4
2 Limits and differentiation in the complex plane and the Cauchy-Riemann
equations
10
3 Power series and elementary analytic functions
21
4 Complex integration and Cauchy’s Theorem
36
5 Cauchy’s Integral Formula and Taylor’s Theorem
56
6 Laurent series and singularities
64
7 Cauchy’s Residue Theorem
72
8 Solutions to Part 1
95
9 Solutions to Part 2
99
10 Solutions to Part 3
107
11 Solutions to Part 4
116
12 Solutions to Part 5
122
13 Solutions to Part 6
125
14 Solutions to Part 7
130
1
MATH20101 Complex Analysis
0
...
Preliminaries
§0
...
Charles Walkden, Room 2
...
walkden@manchester
...
uk
...
If you want to see me at another time then please
email me first to arrange a mutually convenient time
...
2
§0
...
1
Course structure
Lectures
There will be approximately 21 lectures in total
...
The course webpage is available
via Blackboard or directly at www
...
manchester
...
uk/~cwalkden/complex-analysis
...
The lectures will be recorded via the University’s ‘Lecture Capture’ system
...
§0
...
2
Exercises
The lecture notes also contain the exercises (at the end of each section)
...
The lecture notes also contain the solutions to the exercises
...
§0
...
3
Tutorials and support classes
The tutorial classes start in Week 2
...
You will be assigned to a class
...
If you go to a class other
than the one you’ve been assigned to then you will normally be recorded as being absent
...
Each week I will prepare a worksheet
...
I will often break the exercises down
into easier, more manageable, subquestions; the idea is that then everyone in the class can
make progress on them within the class
...
) You still
need to work on the remaining exercises (and try past exams) in your own time!
I will not put the worksheets on the course webpage
...
2
MATH20101 Complex Analysis
§0
...
4
0
...
These recap topics
where, from my experience of teaching the course before, there is often some confusion and
a second explanation (in addition to that given in the lecture notes/lectures) may be useful
...
§0
...
5
Coursework
The coursework for this year will be a 1 hour closed-book multiple choice test taking place on
Thursday 3rd November (week 6, reading week) between 2pm and 3pm in Simon Theatres
E and B
...
There is no negative marking
...
2
...
The exam for MATH20101 consists
of Part A (examining the Real Analysis part of the course) and Part B (examining the
Complex Analysis part of the course)
...
You must answer 5 of these questions, with at least 2 from each part
...
In terms of what is examinable: anything that I cover (including proofs) in the lectures
can be regarded as being examinable (unless I explicitly say otherwise in the lectures)
...
§0
...
If you do want a text to refer to then the most suitable is
I
...
Stewart and D
...
Tall, Complex Analysis, Cambridge University Press, 1983
...
)
The best book (in my opinion) on complex analysis is
L
...
Ahlfors, Complex Analysis, McGraw-Hill, 1979
although it is probably too advanced for this course
...
3
MATH20101 Complex Analysis
1
...
Introduction
§1
...
For example, if f : R → R is defined by f (x) = 3x2 + 2x then you already
know that f ′ (x) = 6x + 2 and that f (x) dx = x3 + x2 + c
...
In this course, we will look at what it means for
functions defined on the complex plane to be differentiable or integrable and look at ways
in which one can integrate complex-valued functions
...
This theorem gives a new method for calculating real integrals that would be difficult
or impossible just using techniques that you know from real analysis
...
(1
...
1)
2 + a2 )(x2 + b2 )
−∞ (x
If you try calculating this using techniques that you know (integration by substitution,
integration by parts, etc) then you will quickly hit an impasse
...
1
...
2
Recap on complex numbers
A complex number is an expression of the form x + iy where x, y ∈ R
...
) We denote the set of complex numbers by C
...
2
...
If a + ib, c + id ∈ C then we can add and multiply them as follows
(a + ib) + (c + id) = (a + c) + i(b + d)
(a + ib)(c + id) = ac + iad + ibc + i2 bd = (ac − bd) + i(ad + bc)
...
2 b2
2
2
a + ib
a + ib a − ib
a −i
a +b
a +b
a + b2
We shall often denote a complex number by the letters z or w
...
We call x the real part of z and write x = Re(z)
...
(Note: the imaginary part of x + iy is y, and not iy
...
4
MATH20101 Complex Analysis
1
...
2
...
Here z = x + iy
...
In the complex plane, the set of real numbers corresponds to the x-axis (which we will often
call the real axis) and the set of imaginary numbers corresponds to the y-axis (which we
will often call the imaginary axis)
...
¯
Let z = x + iy, x, y ∈ R
...
(If z is real then this is just the usual absolute value
...
z
Here are some basic properties of |z|:
Proposition 1
...
1
Let z, w ∈ C
...
Remark
...
The
inequality ||z| − |w|| ≤ |z − w| is often called the reverse triangle inequality
...
Parts (i), (ii) and (iii) follow easily from the definition of |z|
...
6)
...
Introduction
x2 + y 2 = |z|
...
5(iv)
¯
≤ |z|2 + |w|2 + 2|z w|
¯
= |z|2 + |w|2 + 2|z||w|
¯
= |z|2 + |w|2 + 2|z||w|
= (|z| + |w|)2
...
If we plot the point z in the complex plane then |z| denotes the length of the
vector joining the origin 0 to the point z
...
2
...
The angle θ in Figure 1
...
2 is
z
|z|
θ
Figure 1
...
2: The modulus |z| and argument arg z of z
...
We have that tan θ = y/x
...
However, there is a unique value of θ such that −π < θ ≤ π; this is called the principal
value of arg z
...
Let z ∈ C
...
First write z = x + iy
and draw z in the complex plane; see Figure 1
...
3
...
We call (r, θ) the polar co-ordinates of z
and write z = r(cos θ + i sin θ)
...
There is a short video recapping complex numbers on the course webpage (Video
1)
...
Introduction
z
r sin θ
r
θ
r cos θ
Figure 1
...
3: If z has polar coordinates (r, θ) then the real part of z is r cos θ and the
imaginary part of z is r sin θ, and conversely
...
Exercises for Part 1
Exercises for Part 1
The following exercises are provided for you to revise complex numbers
...
1
Write the following expressions in the form x + iy, x, y ∈ R:
(i) (3 + 4i)2 ; (ii)
2 + 3i
1 − 5i
1−i
1
; (iii)
; (iv)
− i + 2; (v)
...
2
Express
(1 − i)23
√
( 3 − i)13
in the form reiθ , r > 0, −π ≤ θ < π
...
Exercise 1
...
Exercise 1
...
Show that (i) Re(z ± w) = Re(z) ± Re(w), (ii) Im(z ± w) = Im(z) ± Im(w)
...
Exercise 1
...
Show that (i) z ± w = z ± w, (ii) zw = z w, (iii)
z + z = 2 Re(z), (v) z − z = 2i Im(z)
...
6
Let z, w ∈ C
...
Exercise 1
...
Exercise 1
...
Compute the product zw
...
8
MATH20101 Complex Analysis
1
...
(iii) Use De Moivre’s Theorem to derive formulæ for cos 3θ, sin 3θ, cos 4θ, sin 4θ in terms
of cos θ and sin θ
...
9
Let w0 be a complex number such that |w0 | = r and arg w0 = θ
...
Exercise 1
...
Give an example to show that,
in general, Arg(z1 z2 ) = Arg(z1 ) + Arg(z2 ) (c
...
Exercise 1
...
Exercise 1
...
1
...
e
...
(There will be a prize for anyone who can do this integral by hand in under 2
pages using such methods!)
9
MATH20101 Complex Analysis
2
...
Limits and differentiation in the complex plane and the
Cauchy-Riemann equations
§2
...
For example, the only connected subsets of R
are intervals, whereas there are far more complicated connected subsets in C (‘connected’
has a rigorous meaning, but for now you can assume that a subset is connected if it ‘looks’
connected: i
...
any two points in the subset can be joined by a line that does not leave the
subset)
...
Remark
...
Thus A ⊂ B
means that A is a subset (indeed, possibly equal to) B
...
Let z0 ∈ C and let ε > 0
...
We
call Bε (z0 ) the ε-neighbourhood of z0
...
Let D ⊂ C
...
Definition
...
Note that a set is closed precisely when the complement is open
...
In our setting, one can often decide whether a set is open or not by looking at it and
thinking carefully
...
) For example, any open disc {z ∈ C | |z − z0 | < r} is an open set;
see Figure 2
...
1
We will also need the notion of a polygonal arc in C
...
We denote the
straight line from z0 to z1 by [z0 , z1 ]
...
, zr ∈ C
...
, [zr−1 , zr ] a polygonal arc joining z0 to zr
...
We will only be interested in ‘nice’ open
sets called domains
...
Let D ⊂ C be a non-empty set
...
Differentiation, the Cauchy-Riemann equations
r
z0
z
Figure 2
...
1: The open disc Br (z0 ) with centre z0 and radius r > 0
...
(i) D is open;
(ii) given any two point z1 , z2 ∈ D, there exists a polygonal arc contained in D that joins
z0 to z1
...
(i) The open disc {z ∈ C | |z − z0 | < r} centred at z0 ∈ C and of radius r is a domain
...
(iii) A half-plane such as {z ∈ C | Re(z) > a} is a domain
...
(v) The set D = {z ∈ C | Im(z) = 0}, corresponding to the complex plane with the real
axis deleted, is not a domain
...
See Figure 2
...
2 for examples of domains
...
2
Limits of complex sequences
Let zn ∈ C be a sequence of complex numbers
...
(Equivalently, in the
language of MATH10242 Sequences and Series, we say that zn → z if |zn − z| is a null
sequence
...
2
...
Then zn converges if and only if xn and
yn converge
...
Differentiation, the Cauchy-Riemann equations
D
D
(i)
D
(ii)
Figure 2
...
2: In (i), D is a domain
...
Proof
...
Then
|xn − x| ≤
|xn − x|2 + |yn − y|2 = |zn − z| → 0
as n → ∞
...
A similar argument show that yn → y
...
Then
|zn − z| =
|xn − x|2 + |yn − y|2 → 0
so that zn → z
...
3
✷
Complex functions and continuity
Let D ⊂ C, D = ∅
...
Write z = x + iy
...
Example
...
Then f (x + iy) = (x + iy)2 = x2 − y 2 + 2ixy
...
Example
...
Then f (x + iy) = x − iy
...
¯
Video
...
Why is it impractical to draw the graph of a complex function f : D → C? There is a short
video recapping what is meant by a function and answering this question on the course
webpage (see Video 2)
...
Let z0 ∈ D
...
That is, f (z) → ℓ as z → z0 means that if z is very close (but not equal to) z0 then f (z)
is very close to ℓ
...
12
MATH20101 Complex Analysis
2
...
Let f : C → C be defined by f (z) = 1 if z = 0 and f (0) = 0
...
Here limz→0 f (z) = f (0)
...
Definition
...
We say that f is
continuous at z0 ∈ D if
lim f (z) = f (z0 )
...
Continuity obeys the same rules as in MATH20101 Real Analysis
...
Then
f (z) + g(z), f (z)g(z), cf (z) (c ∈ C)
are all continuous at z0 , as is f (z)/g(z) provided that g(z0 ) = 0
...
4
Differentiable functions
Let us first consider how one differentiates real valued functions defined on R
...
Let (a, b) ⊂ R be an open interval and
letf : (a, b) → R be a function
...
The idea is that f ′ (x0 ) is the slope of the
graph of f at the point x0
...
We then say that f is differentiable at x0 if this limit exists, and define the
derivative of f at x0 to be the value of this limit
...
Let (a, b) ⊂ R be an interval and let x0 ∈ (a, b)
...
4
...
We call f ′ (x0 ) the derivative of f at x0
...
Remark
...
The definition of the derivative in (2
...
1) requires the
limit to exist from both the left and the right and for the value of these limits to be the
same
...
For
example, consider f (x) = |x| defined on R
...
x→0+
x→0− x
x−0
lim
13
MATH20101 Complex Analysis
2
...
) Thus the left-handed and right-handed derivatives are not equal, so f is not
differentiable at the origin
...
)
Remark
...
If f was defined on the closed
interval [a, b] then we could only consider right-handed derivatives at a (and left-handed
derivatives at b)
...
Definition
...
Let z0 ∈ D
...
4
...
(Note that in (2
...
2) we are allowing z to converge to z0 from any direction
...
If f is differentiable at every point z0 ∈ D then we say
that f is differentiable on D
...
Sometimes we use the notation
df
(z0 )
dz
to denote the derivative of f at z0
...
For example (as we shall see) there are
many functions that are differentiable when restricted to the real axis but that are not
differentiable as a function defined on C
...
Definition
...
Then we say that
f is holomorphic on D
...
The higher derivatives are defined similarly, and we denote them by
f ′′ (z0 ), f ′′′ (z0 ),
...
Example
...
Let z0 ∈ C be any point
...
z→z0 z − z0
z→z0
z→z0
z − z0
z − z0
Hence f ′ (z0 ) = 2z0 for all z0 ∈ C
...
All of the standard rules of differentiable functions continue to hold in the complex case:
14
MATH20101 Complex Analysis
2
...
4
...
Let c ∈ C
...
Proof
...
(Usually the only modification needed is to replace the absolute value
| · | defined on R with the modulus | · | defined on C
...
Proposition 2
...
2
Suppose that f is differentiable at z0
...
Proof
...
e
...
Note that
lim f (z) − f (z0 ) = lim
z→z0
z→z0
as required
...
5
f (z) − f (z0 )
(z − z0 ) = f ′ (z0 ) × 0 = 0,
z − z0
✷
The Cauchy-Riemann equations
Throughout, let D be a domain
...
Let f : D → C be a complex valued
function
...
Example
...
Then
f (z) = z 3 = (x + iy)3 = x3 − 3xy 2 + i(3x2 y − y 3 ) = u(x, y) + iv(x, y)
where u(x, y) = x3 − 3xy 2 and v(x, y) = 3x2 y − y 3
...
To state them, we need to recall the notion of a partial derivative
...
Suppose that g(x, y) is a real-valued function depending on two co-ordinates
x, y
...
For brevity (and provided there is no confusion), we leave out the
(x, y) and write
∂g ∂g
,
...
Differentiation, the Cauchy-Riemann equations
Thus, to calculate ∂g/∂x we treat y as a constant and differentiate with respect to x, and
to calculate ∂g/∂y we treat x as a constant and differentiate with respect to y
...
5
...
Suppose that f is differentiable at
z0 = x0 + iy0
...
∂x
∂y
∂y
∂x
(2
...
1)
Remark
...
5
...
Proof
...
4
...
The trick is to calculate f ′ (z0 ) in two different ways: by
looking at points that converge to z0 horizontally, and by looking at points that converge
to z0 vertically
...
Then as h → 0 we have z0 + h → z0
...
=
∂x
∂x
f ′ (z0 ) =
lim
(2
...
2)
Now take k to be real and consider z0 + ik = x0 + i(y0 + k)
...
Hence
f (z0 + ik) − f (z0 )
ik
u(x0 , y0 + k) + iv(x0 , y0 + k) − u(x0 , y0 ) − iv(x0 , y0 )
= lim
k→0
ik
u(x0 , y0 + k) − u(x0 , y0 )
v(x0 , y0 + k) − v(x0 , y0 )
= lim
+i
k→0
ik
ik
∂v
∂u
(x0 , y0 ),
= −i (x0 , y0 ) +
∂y
∂y
f ′ (z0 ) =
lim
k→0
(2
...
3)
recalling that 1/i = −i
...
5
...
5
...
✷
16
MATH20101 Complex Analysis
2
...
We can use the Cauchy-Riemann equations to examine whether the function
f (z) = z might be differentiable on C
...
Hence f (z) = u(x, y) + iv(x, y) with u(x, y) = x and v(x, y) = −y
...
∂x
∂y
∂x
∂y
Hence there are no points at which
∂v
∂u
=
∂x
∂y
so that f (z) = z is not differentiable at any point in C
...
Notice however that f (z) = z is continuous at every point in C
...
Such functions also exist in real analysis, but they are much harder to write down
and considerably harder to study (one of the simplest is known as Weierstrass’ function
w(x) = ∞ 2−nα cos 2πbn x where α ∈ (0, 1), b ≥ 2; such functions are still of interest in
n=0
current research)
...
One could ask whether the
converse is true: if the Cauchy-Riemann equations are satisfied at the point z0 then is f
differentiable at z0 ? The answer is no, as the following example shows
...
Note that if we write f (x + iy) = u(x, y) + iv(x, y) then u(x, y) = f (x + iy) and v(x, y) = 0
...
Clearly, ∂v/∂x, ∂v/∂y are all equal to zero at the origin
...
However,
f is not continuous at the origin; this is because h + ih → 0 as h → 0 but 1 = f (h + ih) →
f (0) as h → 0
...
The problem with the above example is that in the definition of differentiability (2
...
2)
we need to let z tend to z0 in an arbitrary way
...
Hence we need
some extra hypotheses on u, v at z0 ; the correct hypotheses are to assume the continuity
of the partial derivatives
...
5
...
Let
z0 = x0 + iy0 ∈ D
...
Then f is differentiable at z0
...
Differentiation, the Cauchy-Riemann equations
Proof
...
Lemma 2
...
3
Suppose that ∂w/∂x, ∂w/∂y exist at (x0 , y0 ) and ∂w/∂x is continuous at (x0 , y0 )
...
Now consider z = z0 + h + ik
...
Using the Cauchy-Riemann equations we can write the above expression as
f (z) − f (z0 ) = (h + ik)
= (z − z0 )
∂u
∂v
+i
∂x
∂x
∂u
∂v
+i
∂x
∂x
+ hε1 + kη1 + ihε2 + ikη2
+ρ
where ρ = hε1 + kη1 + ihε2 + ikη2
...
To see this, note that
|ρ|
|h||ε1 | + |k||η1 | + |h||ε2 | + |k||η2 |
ρ
√
=√
≤
≤ |ε1 | + |η1 | + |ε2 | + |η2 |
z − z0
h2 + k2
h2 + k2
which tends to zero as h, k → 0
...
There is a video recapping the Cauchy Riemann Theorem and its converse on the
course webpage (see Video 3)
...
Exercises for Part 2
Exercises for Part 2
Exercise 2
...
(i) {z ∈ C | Im(z) > 0},
(ii) {z ∈ C | Re(z) > 0, |z| < 2},
(iii) {z ∈ C | |z| ≤ 6}
...
2
Using the definition in (2
...
2), differentiate the following complex functions from first principles:
(i) f (z) = z 2 + z; (ii) f (z) = 1/z (z = 0); (iii) f (z) = z 3 − z 2
...
3
(i) In each of the following cases, write f (z) in the form u(x, y)+iv(x, y) where z = x+iy
and u, v are real-valued functions
...
z
(ii) Show that u and v satisfy the Cauchy-Riemann equations everywhere for (a), and for
all z = 0 in (b)
...
Using the Cauchy-Riemann
equations, decide whether there are any points in C at which f is differentiable
...
4
Verify the Cauchy-Riemann equations for the functions u, v given by
(i) u(x, y) = x3 − 3xy 2 , v(x, y) = 3x2 y − y 3 ;
(ii)
u(x, y) =
4xy 3 − 4x3 y
x4 − 6x2 y 2 + y 4
, v(x, y) =
(x2 + y 2 )4
(x2 + y 2 )4
where x2 + y 2 = 0
...
Exercise 2
...
(i) Show from the definition (2
...
2) that f is not differentiable at the origin
...
Exercises for Part 2
(ii) Show however that the Cauchy-Riemann equations are satisfied at the origin
...
5
...
6
Suppose that f (z) = u(x, y) + iv(x, y) is holomorphic
...
(Functions
which satisfy Laplace’s equation are called harmonic functions
...
7
For f (z) = z 3 calculate u, v so that f (z) = u(x, y) + iv(x, y) (where z = x + iy)
...
Exercise 2
...
Suppose we know that u(x, y) = x5 − 10x3 y 2 + 5xy 4
...
(The Cauchy Riemann equations have the following remarkable implication: suppose
f (z) = u(x, y) + iv(x, y) is differentiable and that we know a formula for u, then we
can recover v (up to a constant); similarly, if we know v then we can recover u (up to a
constant)
...
)
Exercise 2
...
Find all values of k for which u is the real part of a holomorphic
function f : C → C
...
10
Show that if f : C → C is holomorphic and f has a constant real part then f is constant
...
11
Show that the only holomorphic function f : C → C of the form f (x + iy) = u(x) + iv(y)
is given by f (z) = λz + a for some λ ∈ R and a ∈ C
...
12
Suppose that f (z) = u(x, y) + iv(x, y), f : C → C, is a holomorphic function and that
2u(x, y) + v(x, y) = 5 for all z = x + iy ∈ C
...
20
MATH20101 Complex Analysis
3
...
Power series and elementary analytic functions
§3
...
Recall that if sn ∈ C then we say that sn converges to s ∈ C if for
all ε > 0 there exists N ∈ N such that |s − sn | < ε for all n ≥ N
...
We say that the series ∞ zk converges if the sequence of partial sums
k=0
sn = n zk converges
...
A series which does not converge is called divergent
...
One can show (see Exercise 3
...
is convergent if, and only if, both
We will need a stronger property than just convergence
...
Let zn ∈ C
...
Lemma 3
...
1
Suppose that
∞
n=0 zn
∞
n=0 zn
is absolutely convergent if the real series
is absolutely convergent
...
∞
Proof
...
Let zn = xn + iyn
...
Hence by the comparison test, the real series
n=0 xn and
are absolutely convergent
...
By the above remark,
∞
n=0 zn
is convergent
...
It is easy to give an example of a series which is convergent but not absolutely
convergent
...
Recall from MATH10242
Sequences and Series that ∞ (−1)n /n is convergent but ∞ |(−1)n /n| = ∞ 1/n is
n=0
n=0
n=0
divergent
...
Indeed, two series which converge absolutely may be multiplied in a
similar way to two finite sums
...
(We remark that Proposition 3
...
2 is not true, in general, if one of the infinite series converges but is not absolutely
convergent
...
1
...
Suppose that
3
...
Then
=
n=0
n=0
where cn = a0 bn + a1 bn−1 + a2 bn−2 + · · · + an b0 and
∞
cn
n=0
∞
n=0 cn
is absolutely convergent
...
Omitted
...
The same tests continue to hold for complex series and we state them
below as propositions
...
1
...
Suppose that
|zn+1 |
= ℓ
...
If ℓ > 1 then
(3
...
1)
∞
n=0 zn
diverges
...
If ℓ = 1 in (3
...
1) then we can say nothing: the series may converge absolutely,
it may converge but not absolutely converge, or it may diverge
...
1
...
Suppose that
lim |zn |1/n = ℓ
...
1
...
If ℓ > 1 then
∞
n=0 zn
diverges
...
Again, if ℓ = 1 in (3
...
2) then we can say nothing: the series may converge
absolutely, it may converge but not absolutely converge, or it may diverge
...
Consider the series
∞
in
...
We can use the ratio test to show that this series converges absolutely
...
zn
2
i
2
2
Hence limn→∞ |zn+1 /zn | = 1/2 < 1 and so by the ratio test the series converges absolutely
...
To
see this, note that
in 1/n
1 1/n
|zn |1/n = n
= 1/2
...
22
MATH20101 Complex Analysis
§3
...
Power series, analytic functions
Power series and the radius of convergence
Definition
...
∞
n=0 an (z
− z0 )n where an ∈ C, z ∈ C is called a power
By changing variables to z ′ = z − z0 we need only consider power series at 0, i
...
power
series of the form
∞
an z n
...
n=0
(We allow R = ∞ if no finite supremum exists
...
2
...
Then
n=0
(i)
∞
n
n=0 an z
converges absolutely for |z| < R;
(ii)
∞
n
n=0 an z
diverges for |z| > R
...
We cannot say what happens in the case when |z| = R: the power series may
converge, it may converge but not absolutely converge, or it may diverge
...
Let z ∈ C be such that |z| < R
...
As ∞ an z1 converges, it follows that an z1 → 0 as n → ∞
...
It follows that |a z n | is a bounded sequence; that is, there
Hence |an z1
n 1
n
exists K > 0 such that |an z1 | < K for all n
...
As |z| < |z1 |, we have that
q < 1
...
z1
Hence by the comparison test, ∞ |an z n | converges (noting that ∞ Kq n = K/(1−q))
...
n=0
n
Now suppose that ∞ an z2 diverges
...
Hence
n=0
∞
n
n=0 an z diverges
...
✷
Definition
...
2
...
We call the set {z ∈ C | |z| < R} the disc of convergence
...
Proposition 3
...
2
Let ∞ an z n be a power series
...
R n→∞ |an |
23
MATH20101 Complex Analysis
(ii) If limn→∞ |an |1/n exists then
3
...
R n→∞
(Here we interpret 1/0 as ∞ and 1/∞ as 0
...
If the limit in (i) exists then the limit in (ii) exists and they give the same
answer for the radius of convergence
...
Remark
...
The reason is to make the formulæ in Proposition 3
...
2 resemble the ratio test and
the root test (Propositions 3
...
3 and 3
...
4, respectively) for the convergence of infinite
series
...
Consider
∞
zn
...
In this case
n
1
an+1
=
→1=
an
n+1
R
as n → ∞
...
Example
...
2n
Here an = 1/2n
...
2
...
Alternatively, we could use Proposition 3
...
2(ii) and see that
1
= lim |an |1/n = lim
n→∞
R n→∞
1
2n
1/n
=
1
2
so that again R = 2
...
2
...
We prove (i)
...
e
...
lim
n→∞ an
Then
|an+1 z n+1 |
lim
→ ℓ|z|
...
Hence the radius of convergence R = 1/ℓ
...
Suppose that |an |1/n → ℓ as n → ∞
...
Hence the radius of convergence R = 1/ℓ
...
Power series, analytic functions
Remark
...
2
...
However, there is a formula for the radius of convergence R that works for any power series
∞
n
n=0 an z
...
For each n, consider supk≥n xk
...
Recall that any decreasing sequence of reals converges
...
We denote the limit by lim supn→∞ xn
...
(One can show that if limn→∞ xn = ℓ then lim supn→∞ xn =
ℓ
...
R
n→∞
§3
...
This suggests that a power series
∞
f (z) =
an z n
(3
...
1)
nan z n−1
...
3
...
There are two
steps to this: (i) we have to show that if (3
...
1) converges for |z| < R then (3
...
2) converges
for |z| < R, and (ii) that f (z) is differentiable for |z| < R and the derivative is given by
(3
...
2)
...
3
...
Then g(z) =
n=0
for |z| < R
...
Let |z| < R and choose r such that |z| < r < R
...
Hence the summands must be bounded, so there exists K > 0 such that
|an r n | < K for all n ≥ 0
...
Then
K
z n−1 n−1
r
< n q n−1
...
Hence by the comparison test,
But n=1 nq
converges
...
n=0
|nan z n−1 | = n|an |
25
∞
n−1 |
n=0 |nan z
✷
MATH20101 Complex Analysis
3
...
3
...
Then f (z) is holomorphic on the disc
n=0
of convergence {z ∈ C | |z| < R} and f ′ (z) = ∞ nan z n−1
...
By Lemma 3
...
1 we know that this converges for
Proof
...
We have to show that if |z0 | < R then f (z) is differentiable at z0 and, moreover, the
derivative is equal to g(z0 ), i
...
we have to show that if |z0 | < R then
f ′ (z0 ) := lim
z→z0
f (z) − f (z0 )
= g(z0 )
z − z0
or equivalently
lim
z→z0
f (z) − f (z0 )
− g(z0 )
z − z0
= 0
...
Let ε > 0
...
Then, as in the proof of Lemma 3
...
1,
is absolutely convergent
...
4
Since |z0 | < r, provided z is close enough to z0 so that |z| < r then we have that
|Σ2,N (z)| ≤
∞
n=N +1
ε
2n|an |r n−1 <
...
3
...
This is a polynomial in z and so is a continuous function
...
Hence, as z → z0 , we have that Σ1,N (z) → 0
...
3
...
2
26
MATH20101 Complex Analysis
3
...
3
...
3
...
≤
2 2
As ε is arbitrary, it follows that f ′ (z0 ) = g(z)
...
If f (z) = ∞ an z n conn=0
verges for |z| < R then we can differentiate it as many times as we like within the disc of
convergence
...
3
...
Then all of the higher derivatives
n=0
f ′ , f ′′ , f ′′′ ,
...
of f exist for z within the disc of convergence
...
(n − k)!
Proof
...
✷
Instead of using a power series at the origin, by replacing z by z − z0 we can consider
a power series at the point z0
...
)
Suppose that f (z) = ∞ an z n has disc of convergence |z| < R
...
That is, the power series g(z) converges for all z inside the disc with
centre z0 and radius R
...
g (z) =
(n − k)!
n=k
§3
...
4
...
Here we study the (complex) exponential function
...
The exponential function is defined to be the power series
∞
exp z =
n=0
zn
...
2
...
Hence this series has radius convergence ∞, and so converges absolutely
for all z ∈ C
...
Power series, analytic functions
By Theorem 3
...
2 we may differentiate term-by-term to obtain
∞
∞
∞
d
z n−1
z n−1
zn
n
exp z =
=
=
= exp z,
dz
n!
(n − 1)! n=0 n!
n=1
n=1
which we already knew to be true in the real-valued case
...
This is also true in the
complex-valued case, and the proof involves a neat trick
...
4
...
Then f is constant on D
...
This is well-known in the real case: a function with zero derivative must be
constant
...
(See
Stewart and Tall, p
...
)
Proposition 3
...
2
Let z1 , z2 ∈ C
...
Proof
...
Then
f ′ (z) = exp(z) exp(c − z) − exp(z) exp(c − z) = 0
by the product rule
...
4
...
Hence exp(z) exp(c − z) = exp(c)
...
✷
Remark
...
4
...
Hence exp z = 0 for any z ∈ C
...
)
Finally, we want to connect the real number e to the complex exponential function
...
Then, iterating Proposition 3
...
2 inductively, we
obtain
en = exp(1)n = exp(1 + · · · + 1) = exp n
...
Thus the notation ez = exp z does not conflict with the usual
definition of ex when z is real
...
In particular, if
we write z = x + iy then Proposition 3
...
2 tells us that
ex+iy = ex eiy
...
Hence to understand complex exponentials
we need to understand expressions of the form eiy
...
4
...
Power series, analytic functions
Trigonometric functions
Define
cos z =
∞
(−1)n
n=0
z 2n
, sin z =
(2n)!
∞
(−1)n
n=0
z 2n+1
...
2
...
Substituting z = −z we see that cos is an even function and that sin is an odd function,
i
...
cos(−z) = cos z, sin(−z) = − sin z
...
By Theorem 3
...
2 we can differentiate term-by-term to see that
d
d
cos z = − sin z,
sin z = cos z
...
Replacing z by −z we see that e−iz = cos z − i sin z
...
2
2i
Squaring the above expressions and adding them gives cos2 z + sin2 z = 1
...
Carrying on in the same way, one
can prove the addition formulæ cos(z1 + z2 ) = cos z1 cos z2 − sin z1 sin z2 , etc, for complex
z1 , z2 , and all the other usual trigonometric identities
...
4
...
2
2
Differentiating these we see that
d
d
cosh z = sinh z,
sinh z = cosh z
...
We also have the relations
cos iz = cosh z, sin iz = i sinh z;
these follow from Exercise 3
...
29
MATH20101 Complex Analysis
§3
...
4
3
...
Let f : C → C
...
Clearly if p ∈ C is a period and n ∈ Z is any integer then np is also a period
...
Hence 2πi is a period for exp, as is 2nπi for any integer n
...
11 we shall see
that these are the only periods for exp
...
§3
...
5
The logarithmic function
In real analysis, the (natural) logarithm is the inverse function to the exponential function
...
(Throughout we will write ln to denote the (real) natural
logarithm
...
Let z ∈ C, z = 0, and consider the equation
exp w = z
...
4
...
4
...
4
...
Each of these values is a called
a logarithm of z, and we denote any of these values by log z
...
We want to find a formula for log z
...
4
...
Then
z = exp w = exp(x + iy) = ex (cos y + i sin y)
...
4
...
4
...
Note that both x and
|z| are real numbers
...
By taking the argument of both sides of (3
...
2) we
see that y = arg z
...
Definition
...
Then a complex logarithm of z is
log z = ln |z| + i arg z
where arg z is any argument of z
...
e
...
We denote the principal logarithm by
Log z:
Log z = ln |z| + i Arg z
...
Dealing with multivalued functions is tricky
...
30
MATH20101 Complex Analysis
3
...
4
...
Definition
...
See Figure 3
...
1
...
4
...
Proof
...
✷
Having seen that the principal logarithm is continuous, we can go on to show that it is
differentiable
...
4
...
dz
z
Proof
...
Then z = exp w
...
Then by Proposition 3
...
3
Log is continuous on the cut plane so we have that k → 0 as h → 0
...
z
✷
31
MATH20101 Complex Analysis
3
...
For
b, z ∈ C with b = 0 we define the principal value of bz to be
bz = exp(z Log b)
and the subsiduary values to be exp(z log b)
...
There is a video on the course webpage recapping properties of the complex
logarithm and complex powers (see Video 4)
...
Exercises for Part 3
Exercises for Part 3
Exercise 3
...
Show that
n=0 zn is convergent if, and only if, both
∞
n=0 Im(zn ) are convergent
...
2
Find the radii of convergence of the following power series:
(i)
∞
n=1
2n z n
, (ii)
n
∞
n=1
∞
zn
, (iii)
n!
n!z n , (iv)
∞
n=1
n=1
np z n (p ∈ N)
...
3
Consider the power series
∞
an z n
n=0
where an = 1/2n if n is even and an = 1/3n if n is odd
...
2
...
Show
by using the comparison test that this power series converges for |z| < 2
...
4
(i) By multiplying two series together, show using Proposition 3
...
2 that for |z| < 1, we
have
∞
1
nz n−1 =
...
1
...
n!
n!
n!
n=0
n=0
n=0
Exercise 3
...
Use the formula 1/R = limn→∞ |an |1/n
to find the radii of convergence of the following power series in terms of R:
(i)
∞
n3 an z n , (ii)
n=0
∞
n=0
an z 3n , (iii)
∞
a3 z n
...
6
Show that for z, w ∈ C we have
(i) cos z =
eiz − e−iz
eiz + e−iz
, (ii) sin z =
...
Exercises for Part 3
(iii) sin(z + w) = sin z cos w + cos z sin w,
(iv) cos(z + w) = cos z cos w − sin z sin w
...
7
Derive formulæ for the real and imaginary parts of the following complex functions and
check that they satisfy the Cauchy-Riemann equations:
(i) sin z, (ii) cos z, (iii) sinh z, (iv) cosh z
...
8
For each of the complex functions exp, cos, sin, cosh, sinh find the set of points on which it
assumes (i) real values, and (ii) purely imaginary values
...
9
We know that the only real numbers x ∈ R for which sin x = 0 are x = nπ, n ∈ Z
...
e
...
Also show that if cos z = 0, z ∈ C then z = (n + 1/2)π, n ∈ Z
...
10
Find the zeros of the following functions
(i) 1 + ez , (ii) 1 + i − ez
...
11
(i) Recall that a complex number p ∈ C is called a period of f : C → C if f (z + p) = f (z)
for all z ∈ C
...
(ii) We know that p = 2nπi, n ∈ Z, are periods of exp z
...
Exercise 3
...
Here is one instance of where they can differ
...
Show that
Log z1 z2 = Log z1 + Log z2 + 2nπi
...
Give an explicit example of two
complex numbers z1 , z2 for which Log z1 z2 = Log z1 + Log z2
...
13
Calculate the principal value of ii and the subsiduary values
...
14
(i) Let α ∈ C and suppose that α is not a non-negative integer
...
= 1+
n!
n=1
f (z) = 1 + αz +
34
MATH20101 Complex Analysis
3
...
)
Show that the this power series has radius of convergence 1
...
1+z
(iii) By considering the derivative of the function g(z) =
(1 + z)α for |z| < 1
...
Integration and Cauchy’s Theorem
4
...
1
Introduction
Consider the real integral
b
f (x) dx
...
That is, we think of starting at the
point a and moving along the real axis to b, integrating f as we go
...
How might we define
z1
f (z) dz?
z0
We want to start at z0 , move through the complex plane to z1 , integrating f as we go
...
Suppose γ is a path
from z0 to z1 (we shall make precise what we mean by a path below, but intuitively just
think of it as a continuous curve starting at z0 and ending at z1 )
...
γ
A priori this looks like it will depend on the path γ
...
§4
...
Definition
...
Remark
...
We say that the path γ
starts at γ(a) and ends at γ(b)
...
Note that a path is a function
...
e
...
However, we should
regard this set of points as having an orientation: a path starts at one end-point and ends
at the other
...
Note that the same path can have different parametrisations
...
We shall
see later (Proposition 4
...
1) that when we calculate an integral along a path then it is
independent of the choice of parametrisation
...
Integration and Cauchy’s Theorem
As an example of a path, let z0 , z1 ∈ C
...
(4
...
1)
Then γ(0) = z0 , γ(1) = z1 and the image of γ is the straight line joining z0 to z1
...
See Figure 4
...
1
...
2
...
We sometimes denote this path by [z0 , z1 ]
...
Let γ : [a, b] → C be a path
...
e
...
Example
...
(4
...
2)
This is the path that describes the circle in C with centre 0 and radius 1, starting and ending
at the point 1, travelling around the circle in an anticlockwise direction
...
2
...
Definition
...
(By differentiable at a we mean that the one-sided derivative exists, similarly
at b
...
We can use integrals to define the lengths of paths:
Definition
...
Then the length of γ is defined to be
b
length(γ) =
a
|γ ′ (t)| dt
...
It is straightforward to check from (4
...
1) that
length([z0 , z1 ]) = |z1 − z0 |
...
2
...
37
MATH20101 Complex Analysis
4
...
2
...
Note that it starts at 1 and travels
anticlockwise around the unit circle
...
One could make
the latter a path by constructing a suitable reparametrisation, but in practice this makes
things complicated; in particular the joins may not be smooth
...
Definition
...
, γn where the end-point
of γr coincides with the start point of γr+1 , 1 ≤ r ≤ n − 1
...
If the end-point of γn coincides with the start point of γ1 then we call γ a closed contour
...
A contour looks
like a smooth path but with finitely many corners
...
Let 0 < ε < R
...
Then γ = γ1 + γ2 + γ3 + γ4 is a closed contour (see Figure 4
...
3)
...
The length of a contour γ = γ1 + · · · + γn is defined to be
length(γ) = length(γ1 ) + · · · + length(γn )
...
Then we can
consider the reverse of this path, where we start at γ(b) and, travelling backwards along γ,
end at γ(a)
...
38
MATH20101 Complex Analysis
4
...
2
...
Definition
...
Define −γ : [a, b] → C to be the path
−γ(t) = γ(a + b − t)
...
Video
...
(See Video 5
...
3
Contour integration
Let f : D → C be a complex functions defined on a domain D
...
Definition
...
3
...
f (z) dz
...
Strictly speaking we should write f (γ(t))γ ′ (t) = u(t)+iv(t) where u, v : [a, b] →
b
b
R and define γ f to be a u(t) dt + i a v(t) dt
...
Take f (z) = z 2 and γ(t) = t2 + it, 0 ≤ t ≤ 1
...
Hence
1
f (z) dz =
1
f (γ(t))γ ′ (t) dt =
0
0
γ
1
=
0
=
=
2t5 − 4t3 dt + i
1 6
t − t4
3
−2
2
+i
...
Integration and Cauchy’s Theorem
The following proposition shows that the definition (4
...
1) is independent of the choice of
parametrisation of the path
...
3
...
Let φ : [c, d] → [a, b] be an increasing smooth bijection
...
Moreover,
f
f=
γ
γ◦φ
for any continuous function f
...
It is clear that both γ and γ ◦ φ have the same image
...
Note that
d
f
=
f (γ(φ(t)))(γφ)′ (t) dt
c
γ◦φ
d
=
f (γ(φ(t)))γ ′ (φ(t))φ′ (t) dt by the chain rule
c
b
=
f (γ(t))γ ′ (t) dt by the change of variables formula
...
If φ in Proposition 4
...
1 is a decreasing smooth bijection then γφ has the same
image as φ but the path traverses this in the opposite direction, i
...
γφ is a parametrisation
of −γ
...
Now suppose that γ = γ1 + · · · + γn is a contour in D
...
γn
The following basic properties of contour integration follow easily from this definition
...
3
...
Suppose that γ, γ1 , γ2 are contours in D
...
Then
(i)
f;
f+
f=
γ2
γ1
γ1 +γ2
(ii)
γ
γ
(iii)
f;
cf = c
γ
γ
40
g;
f+
(f + g) =
γ
MATH20101 Complex Analysis
4
...
γ
Recall from real calculus (or, indeed, from A-level or high school) that one way to
calculate the integral of f is to find an anti-derivative, i
...
find a function F such that
b
F ′ = f
...
We have an analogue of this in for the complex integral
...
Definition
...
We say that a function F : D → C
is an anti-derivative of f on D if F ′ = f
...
3
...
Then
γ
f = F (z1 ) − F (z0 )
...
3
...
It is sufficient to prove the theorem for smooth paths
...
Let w(t) = f (γ(t))γ ′ (t) and let W (t) = F (γ(t))
...
Write w(t) = u(t) + iv(t) and W (t) = U (t) + iV (t) so that U ′ = u, V ′ = v
...
✷
Remark
...
3
...
Example
...
Then
F (z) = z 3 /3 is an anti-derivative for f and
2 2
1 3 1 3 (1 + i)3
= − + i
...
Integration and Cauchy’s Theorem
Remark
...
e
...
However, possessing an antiderivative is a very strong hypothesis on f (see the following remarks)
...
In real analysis, any sufficiently nice function f has an anti-derivative: we define
x
f (t) dt
...
In complex analysis, however, the existence of an anti-derivative in on domain
D is a very strong hypothesis
...
Does this function have an anti-derivative on D? The natural candidate would be Log z
...
So Log z is not an anti-derivative of f (z) = 1/z
...
Let γ(t) = eit , 0 ≤ t ≤ 2π denote the unit circle in C described anticlockwise
...
Let f (z) = 1/z
...
Thus to evaluate γ f we need to
use the definition of the contour integral given in (4
...
1)
...
eit
If f had an anti-derivative on a domain that contains γ then, by the Fundamental Theorem
of Contour Integration, we would have that γ f = 0
...
In general, looking for an anti-derivative is not the best way of calculating complex
integrals
...
One such technique that
applies in the case when γ is a closed contour is Cauchy’s Theorem
...
§4
...
Firstly
b
b
f (x) dx ≤
a
a
|f (x)| dx
(4
...
1)
and secondly, if |f (x)| ≤ M then
b
a
f (x) dx ≤ M (b − a)
...
4
...
4
...
4
...
Both of these results have analogies in the context of complex analysis
...
Here is the complex analogue of (4
...
1)
...
Integration and Cauchy’s Theorem
|f |
f
a
a
b
b
Figure 4
...
1: If f [a, b] → R is negative on some subset of [a, b] then the area underneath
that part of the graph is negative
...
M
f
a
b
Figure 4
...
2: The graph of f is contained inside the rectangle of width b − a and height
M
...
Lemma 4
...
1
Let u, v : [a, b] → R be continuous functions
...
Write
a
|u(t) + iv(t)| dt
...
a
Then
X 2 + Y 2 = (X − iY )(X + iY )
b
(X − iY )(u(t) + iv(t)) dt
=
a
b
b
Xu(t) + Y v(t) dt + i
=
a
a
43
Xv(t) − Y u(t) dt
...
4
...
Integration and Cauchy’s Theorem
However, X 2 + Y 2 is real, hence the imaginary part of the above expression must be zero,
i
...
b
a
Xv(t) − Y u(t) dt = 0
so that
b
X2 + Y 2 =
Xu(t) + Y v(t) dt
...
4
...
4
...
Recalling
that for any complex number z we have that Re(z) ≤ |z|, we have that
Xu(t) + Y v(t) ≤ |(X − iY )(u(t) + iv(t))|
= |X − iY ||u(t) + iv(t)|
X 2 + Y 2 |u(t) + iv(t)|
...
a
|u(t) + iv(t)| dt
✷
We can now prove the following important result—the complex analogue of (4
...
2)—
which we will use many times in the remainder of the course
...
4
...
Suppose that |f (z)| ≤ M for all
z on γ
...
Remark
...
Proof
...
4
...
✷
44
MATH20101 Complex Analysis
4
...
Let f (z) = 1/(z 2 + z + 1) and let γ(t) = 5eit , 0 ≤ t ≤ 2π, be the circle of
radius 5 centred at 0
...
First note that if z is a point on γ then |z| = 5
...
Thus for z on γ we have that
|f (z)| =
z2
1
1
≤
...
Thus, by the Estimation Lemma,
γ
§4
...
19
Winding numbers and Cauchy’s Theorem
Suppose that f : D → C
...
3
...
We say that γ : [a, b] → D is closed if it begins and ends at the same point, i
...
if
z0 = γ(a) = γ(b) = z1
...
3
...
5
...
What happens if we do not know if f has an anti-derivative?
In this case, Cauchy’s Theorem gives conditions under which (4
...
1) continues to hold
...
We will give
one version expressed in terms of winding numbers
...
Imagine you have a
piece of string
...
Now trace around the closed path γ with the other end of the piece of string
...
This number (counted positively for anti-clockwise turns and negatively
for clockwise turns) is the winding number of γ at z0
...
5
...
In examples, it is easy to calculate winding numbers by eye and this is how we shall
always do it
...
Let us first consider the case when the closed path γ does not pass
through the origin 0
...
45
MATH20101 Complex Analysis
4
...
5
...
In (ii), w(γ2 , z0 ) = −1 and
w(γ2 , z1 ) = 0 as γ2 winds clockwise around z0
...
Proposition 4
...
1
Let γ be a path in C \ {0}
...
Any other choice of parametrisation with
a continuous choice of argument differs from this argument by a constant integer multiple
of 2π
...
For example, consider
γ(t) =
eit , 0 ≤ t ≤ π
ei(t+2π) , π < t ≤ 2π
...
Here
arg γ(t) =
t, 0 ≤ t ≤ π
t + 2π, π < t ≤ 2π
...
However, we can find a parametrisation of γ for which the
argument is continuous, for example
γ(t) = eit , 0 ≤ t ≤ 2π
and note that arg γ(t) = t, 0 ≤ t ≤ 2π, is continuous
...
We can reinterpret the winding number w(γ, 0) of γ
around 0 as the multiple of 2π giving the total change in argument along γ
...
5
...
Then
w(γ, 0) =
1
2πi
γ
1
dz
...
Let γ(t) = e4πit , 0 ≤ t ≤ 1
...
We can check this using
46
MATH20101 Complex Analysis
4
...
5
...
=
0
Example
...
In this case, γ winds around the origin once, but
clockwise
...
Again, we can check this using Proposition 4
...
2 as follows:
1
2πi
γ
1
dz =
z
1
2πi
2π
=
0
2π
0
1
e−it
(−i)e−it dt
−1
dt = −1
...
5
...
Intuitively this is clear: let γ : [a, b] → C \ {0} be a closed
path that does not pass through the origin
...
Then (and we put
quotes around the following to indicate that this is not a valid proof)
“
γ
1
dz =
z
b
a
1 ′
γ (t)
γ(t)
= [log(γ(t))]b
a
= (ln |γ(b)| + i arg γ(b)) − (ln |γ(a)| + i arg γ(a))
= i (arg γ(b) − arg γ(a))
= 2πiw(γ, 0)”
...
This is because
Log(z) is not continuous on C \ {0} and so cannot be differentiable
...
More generally, one can define a logarithm continuously on a cut plane
where one removes any ray from C
...
)
For each α ∈ [−π, π) define the cut plane at angle α to be
Cα = C \ {reiα | r > 0},
i
...
the complex plane with the ray inclined at angle α from the positive x-axis removed
...
(The case α = π, m = 0 corresponds to
the usual principal value of the argument
...
In general, γ will not lie
in one cut plane
...
, γn defined on [t0 , t1 ],
...
Along each γr we will choose a value of the
47
MATH20101 Complex Analysis
4
...
Hence
γr
1
dz = log γ(tr ) − log(γ(tr−1 ))
z
= log |γ(tr )| − log |γ(tr−1 )| + i argαr (γ(tr )) − argαr (γ(tr−1 ))
...
The real parts cancel
...
e
...
✷
More generally, we have the following formula for the winding number around z0 for a
closed path that does not pass through z0
...
5
...
Then
w(γ, z0 ) =
1
2πi
γ
1
dz
...
This is just a change-of-origin argument
...
Consider the path γ1 (t) = γ(t) − z0 ; this is γ translated by z0
...
Now
1
2πi
γ
1
dz =
z − z0
1
2πi
b
a
1
γ ′ (t) dt
γ(t) − z0
b
1
1 ′
′
γ (t) dt as γ ′ (t) = γ1 (t)
2πi a γ1 (t) 1
1
1
dz
=
2πi γ1 z
= w(γ1 , 0)
...
Proposition 4
...
4
(i) Let γ1 , γ2 be closed paths that do not pass through z0
...
48
MATH20101 Complex Analysis
4
...
Then
w(−γ, z0 ) = −w(γ, z0 )
...
We prove (i)
...
3
...
w(γ1 + γ2 , z0 ) =
γ2
1
dz
z − z0
We prove (ii)
...
3
...
w(−γ, z0 ) =
✷
We can now state Cauchy’s Theorem
...
5
...
e
...
Then γ f = 0
...
The strength of Cauchy’s Theorem is that we do not need to know if f has
an anti-derivative on D
...
See Theorem 4
...
3 and the
remarks following it
...
See Figure 4
...
2 for examples of the hypotheses of Cauchy’s Theorem
...
There are many proofs of Cauchy’s Theorem; here we give one based on Green’s
Theorem (see MATH10121 Calculus and Vectors)
...
Green’s theorem states the following: suppose that γ is a piecewise smooth closed
contour bounding a region Γ, g, h are C 1 functions defined on an open set containing Γ,
then
∂h ∂g
−
dx dy
...
5
...
Note that
dz = dx + i dy
...
Integration and Cauchy’s Theorem
γ
γ
γ
D
D
D
(i)
(ii)
(iii)
Figure 4
...
2: In (i) and (ii), γ has winding number zero around every point outside D, so
the hypotheses of Cauchy’s Theorem (Theorem 4
...
5 hold
...
=
γ
u dx − v dy + i
=
Γ
−
∂u
∂v
−
∂x ∂y
v dx + u dy
γ
dx dy +
Γ
∂u ∂v
−
∂x ∂y
dx dy
= 0
as, by the Cauchy-Riemann equations, ∂u/∂x = ∂v/∂y and ∂u/∂y = −∂v/∂x hold everywhere on Γ
...
In many ways, this proof is cheating: Green’s Theorem is a deep theorem and
not easy to prove
...
(The idea is to build D up from smaller pieces, often starting with the case when
D is a triangle; see Stewart and Tall, p
...
)
Another reason for why the above proof is cheating is that Green’s theorem requires
the partial derivatives in (4
...
2) to be continuous
...
e
...
In fact, as we shall see, the existence of the derivative on a
domain forces the derivative (and so the partial derivatives) to be continuous (indeed, if
the derivative exists on a domain then the function is differentiable infinitely many times)
...
There are many variants of Cauchy’s Theorem
...
Our first variant deals with simply connected domains
...
(For example, in Figure 4
...
2(i) the domain
D is simply connected; however the domains D in Figures 4
...
2(ii) and (iii) are not simply
connected as they have holes in them
...
A domain D is simply connected if for all closed contours γ in D and for all
z ∈ D, we have w(γ, z) = 0
...
Integration and Cauchy’s Theorem
Theorem 4
...
6 (Cauchy’s Theorem for simply connected domains)
Suppose that D is a simply connected domain and f is a holomorphic function on D
...
More generally, we can ask about integrating around several closed contours
...
5
...
Let γ1 ,
...
Suppose that
w(γ1 , z) + · · · + w(γn , z) = 0 for all z ∈ D
...
γn
Remark
...
5
...
Consider the example in Figure 4
...
3
...
If z is
in the ‘hole’ in D then w(γ1 , z) = 1, w(γ2 , z) = −1 so that w(γ1 , z) + w(γ2 , z) = 0
...
γ1
γ2
D
Figure 4
...
3: An example of closed contours that satisfy the hypotheses in the Generalised
Cauchy Theorem (Theorem 4
...
7)
...
5
...
Suppose that γr starts and ends at zj ∈ D, 1 ≤ j ≤ n
...
, σn in D which join z0 to z1 ,
...
(See Figure 4
...
4
...
We see
that
γ = σ1 + γ1 − σ1 + · · · + σn + γn − σn
is a closed contour that starts and ends at z0
...
Then, using Proposition 4
...
4,
w(γ, z) = w(σ1 + γ1 − σ1 + · · · + σn + γn − σn , z)
51
MATH20101 Complex Analysis
4
...
5
...
n
=
j=1
n
w(σj + γj − σj , z)
w(γj , z)
=
j=1
= 0
...
Hence
n
n
j=1
as
−σj
f =−
σj
σj
f
f+
f+
−σj
γj
f
...
There is a video on the course webpage that summarises the main facts about
complex integration, the Fundamental Theorem of Contour Integration, Cauchy’s Theorem
and the Generalised Cauchy Theorem
...
)
52
MATH20101 Complex Analysis
4
...
1
Draw the following paths:
(i) γ(t) = e−it , 0 ≤ t ≤ π,
(ii) γ(t) = 1 + i + 2eit , 0 ≤ t ≤ 2π,
(iii) γ(t) = t + i cosh t, −1 ≤ t ≤ 1,
(iv) γ(t) = cosh t + i sinh t, −1 ≤ t ≤ 1
...
2
Find the values of
γ
x − y + ix2 dz
where z = x + iy and γ is:
(i) the straight line joining 0 to 1 + i;
(ii) the imaginary axis from 0 to i;
(iii) the line parallel to the real axis from i to 1 + i
...
3
Let
γ1 (t) = 2 + 2eit , 0 ≤ t ≤ 2π,
γ2 (t) = i + e−it , 0 ≤ t ≤ π/2
...
From the definition
γ
f=
b
′
a f (γ(t))γ (t) dt,
(i)
γ1
dz
, (ii)
z−2
calculate
γ2
dz
...
4
Evaluate γ |z|2 dz where γ is the circle |z − 1| = 1 described anticlockwise
...
5
For each of the following functions find an anti-derivative and calculate the integral along
any smooth path from 0 to i:
(i) f : C → C, f (z) = z 2 sin z;
(ii) f : C → C, f (z) = zeiz
...
Exercises for Part 4
Exercise 4
...
What does this tell you about possibility of the existence of an anti-derivative for f (z) =
|z|2 ?
Exercise 4
...
Figure 4
...
1: See Exercise 4
...
Exercise 4
...
3
...
Let −γ denote the reversed path of γ
...
γ
Exercise 4
...
Let γ be a smooth path in D starting at z0 and ending
at z1
...
γ
Exercise 4
...
Exercises for Part 4
1
γ2 (t) = 1 + eit , 0 ≤ t ≤ 2π,
2
γ(t) = 2eit , 0 ≤ t ≤ 2π
...
Use the Generalised Cauchy Theorem to deduce that
f dz
...
11
Let γ1 denote the unit circle centred at 0, radius 1, described anti-clockwise
...
Show that γ1 f = 2πi
...
6
...
Use the
Generalised Cauchy Theorem on an appropriate domain to calculate γ2 f
...
6
...
55
MATH20101 Complex Analysis
5
...
Cauchy’s Integral Formula and Taylor’s Theorem
§5
...
(This partly explains why
complex analysis is so much easier than real analysis
...
Then
C 1 ⊃ C 2 ⊃ · · · and we think of a function that is C r for a large r as being ‘nice’
...
e
...
Integration, however, works the other way:
the indefinite integral of a C r function is C r+1
...
In terms of complex analysis, this distinction into C r functions does not have
any meaning: as we shall see, if a function is differentiable once then it is differentiable
infinitely many times!)
Theorem 5
...
1 (Cauchy’s Integral Formula for a circle)
Suppose that f is holomorphic on the disc {z ∈ C | |z − z0 | < R}
...
Then for |w − z0 | < r we have that
f (w) =
1
2πi
Cr
f (z)
dz
...
1
...
Equation (5
...
1) has the following remarkable corollary: if we know the value
of the function f along the closed path Cr then we know the values of the function at all
points inside the disc Cr
...
Remark
...
1
...
This is not necessary, and a more general version of
Cauchy’s Integral Formula holds provided that f is holomorphic on a simply connected
domain D and we replace Cr by an appropriate simple closed loop
...
)
Proof
...
Consider the function
g(z) =
f (z) − f (w)
...
Define the circle
Sε to be the circle centred at w and of radius ε > 0
...
Then, provided ε > 0 is sufficiently small, both Cr and Sε lie inside D
...
Cauchy’s Integral Formula
We apply the Generalised Cauchy Theorem (Theorem 4
...
7) to the contours Sε and
−Cr
...
Then either |z − z0 | > R or z = w
...
In the second case, if z = w then
w(Sε , z) = 1 and w(−Cr , z) = −1
...
Noting that −Cr g = − Cr g we have that, by the Generalised Cauchy Theorem
(Theorem 4
...
7),
g(z) dz
...
1
...
As |f ′ (w)| is finite, it
follows that g(z) is bounded for z sufficiently close to w, i
...
there exist δ > 0 and M > 0
such that if 0 < |w − z| < δ then |g(z)| < M
...
4
...
Cr
g(z) dz ≤ M 2πε,
By (5
...
2) it follows that
and since we can take ε > 0 to be arbitrarily small, it follows that
g(z) dz = 0
...
1
...
1
...
Hence
f (w) =
§5
...
z−w
✷
Taylor series
The integral formula allows us to express a differentiable function as a power series (the
Taylor series expansion)
...
3
...
Theorem 5
...
1 (Taylor’s Theorem)
Suppose that f is holomorphic in the domain D
...
Cauchy’s Integral Formula
f has a Taylor series expansion given by
∞
f (z) =
n=0
f (n) (z0 )
(z − z0 )n
...
(z − z0 )n+1
Remark
...
For example, if
2
f (x) =
e−1/x , x = 0
0, x = 0
then f is differentiable arbitrarily many times
...
As f = 0 near
0, it follows that f is not equal to its Taylor series
...
If, for each z0 ∈ D, a function f : D → C is equal to its Taylor series at
z0 on some open disc then we say that f is analytic
...
2
...
)
Proof of Theorem 5
...
1
...
1−w
Put w = h/(z − z0 )
...
Hence
1
1
h
hm
hm+1
1
=
=
...
Then
Cauchy’s Integral formula, together with the above identity, gives
f (z0 + h)
1
=
2πi
Cr
f (z)
dz
z − (z0 + h)
58
MATH20101 Complex Analysis
=
1
2πi
1
h
hm
+
+ ··· +
z − z0 (z − z0 )2
(z − z0 )m+1
f (z)
Cr
+
m
5
...
=
n=0
where
an =
and
Am =
1
2πi
1
2πi
Cr
Cr
f (z)
dz
(z − z0 )n+1
f (z)hm+1
dz
...
As f is differentiable on Cr , it is bounded
...
By the reverse triangle inequality, using the facts that |h| < r = |z − z0 | for z on Cr , we
have that
|z − z0 − h| ≥ ||z − z0 | − |h|| = r − |h|
...
4
...
Since |h| < r, this tends to zero as m → ∞
...
However, the integral is unchanged if we vary r in
the whole range 0 < r < R
...
Finally, we put h = z − z0
...
From Theorem 3
...
2 we know that a power series
can be differentiated term-by-term as many times as we please and that
an =
f (n) (z0 )
...
3
Applications of Cauchy’s Integral Formula
Cauchy’s Integral Formula has many applications; here we give just three
...
3
...
Cauchy’s Integral Formula
Cauchy’s Estimate
As a consequence of the formula for the nth derivative of f in terms of an integral given in
Taylor’s Theorem, we have the following estimate
...
3
...
If 0 < r < R and |f (z)| ≤ M for
all z such that |z − z0 | = r then, for all n ≥ 0,
|f (n) (z0 )| ≤
M n!
...
By Theorem 5
...
1 we know that
f (n) (z0 ) =
n!
2πi
Cr
f (z)
dz
...
4
...
rn
✷
§5
...
2
Liouville’s Theorem
Theorem 5
...
2 (Liouville’s Theorem)
Suppose that f is holomorphic and bounded on the whole of C
...
Remark
...
Remark
...
It is easy to think of functions
f : R → R that are differentiable and bounded but not constant
...
)
Proof
...
Let z0 ∈ C
...
By Cauchy’s Estimate (Lemma 5
...
1), we have for 0 < r < R
|f ′ (z0 )| ≤
M
...
Hence
we can let r → ∞
...
Hence f is a constant
...
3
...
Cauchy’s Integral Formula
The Fundamental Theorem of Algebra
Consider the equation x − n = 0 where n ∈ N
...
If, however, we consider x + n = 0, n ∈ N, then we need to introduce
negative integers to be able to solve this equation
...
Continuing this theme, one can see that one needs to introduce
surds (to solve x2 − 2 = 0) and complex numbers (to solve x2 + 1 = 0)
...
Theorem 5
...
3 (The Fundamental Theorem of Algebra)
Let p(z) = z n + an−1 z n−1 + · · · + a1 z + a0 be a polynomial of degree n ≥ 1 with coefficients
aj ∈ C
...
Corollary 5
...
4
Let p(z) = z n + an−1 z n−1 + · · · + a1 z + a0 be a polynomial of degree n ≥ 1 with coefficients
aj ∈ C
...
j=1
Proof of Theorem 5
...
3
...
e
...
If p(z) = 0 for all z ∈ C then 1/p(z) is holomorphic for all z ∈ C
...
For z = 0
p(z)
a1
an−1
a0
+ · · · + n−1 + n → 1
=1+
n
z
z
z
z
as |z| → ∞
...
zn
2
Re-arranging this implies that for |z| > K we have that
2
2
1
≤ n ≤ n
...
We shall show that this bound continues to hold for |z| ≤ K
...
Let Cr (t) = z0 + reit , 0 ≤ t ≤ 2π, denote the circular path with centre z0 and radius r
...
Hence, for such an r, if z is any point on Cr then |z| > K
...
By Cauchy’s Estimate (Lemma 5
...
1) it follows that
2
1
≤ n
...
Cauchy’s Integral Formula
Hence |1/p(z)| ≤ 2/K n for all z ∈ C, so that p is a bounded holomorphic function on C
...
3
...
✷
Proof of Corollary 5
...
4
...
By
Theorem 5
...
3 we can find α1 ∈ C such that p(α1 ) = 0
...
The proof then follows by induction
on n
...
Exercises for Part 5
Exercises for Part 5
Exercise 5
...
Exercise 5
...
What is the radius of convergence?
Exercise 5
...
Exercise 5
...
Prove that f is a polynomial function
of degree at most k
...
)
Exercise 5
...
)
Let f (z) = |z + 1|2
...
(i) Show that f is not holomorphic on any domain that contains γ
...
)
(ii) Find a function g that is holomorphic on some domain that contains γ and such that
f (z) = g(z) at all points on the unit circle γ
...
) (Hint:
recall that if w ∈ C then |w|2 = ww
...
63
MATH20101 Complex Analysis
6
...
Laurent series and singularities
§6
...
e
...
1
...
The idea of Laurent series is to generalise (6
...
1) to allow negative powers of
(z − z0 )
...
§6
...
A Laurent series is a series of the form
∞
n=−∞
an (z − z0 )n
...
2
...
2
...
We define
(6
...
1) to mean
∞
n=1
a−n (z − z0 )−n +
∞
n=0
an (z − z0 )n = Σ− + Σ+
...
2
...
Now Σ+ converges for |z − z0 | < R2 for some R2 ≥ 0, where R2 is the radius of
convergence of Σ+
...
This has a radius of convergence
−1
−1
equal to, say, R1 ≥ 0
...
In other words,
Σ− converges when |z − z0 | > R1
...
2
...
See Figure 6
...
1
...
(Compare this with Taylor’s Theorem:
if f is holomorphic on a disc then it can be expressed as a Taylor series
...
64
MATH20101 Complex Analysis
6
...
2
...
Theorem 6
...
1 (Laurent’s theorem)
Suppose that f is holomorphic on the annulus {z ∈ C | R1 < |z − z0 | < R2 }, where
0 ≤ R1 < R2 ≤ ∞
...
2
...
Then
1
f (z)
an =
dz
2πi Cr (z − z0 )n+1
for n ∈ Z
...
Note that in this case we cannot conclude that an = f (n) (z0 )/n! as we do not
know that f is differentiable at z0 (indeed, it may not even be defined at z0 )
...
The proof is similar to the proof of Taylor’s Theorem and can be found in
Stewart and Tall’s book
...
2
...
We call
−1
n=−∞
an (z − z0 )n
the principal part of the Laurent series
...
Example
...
Recall that ez =
e1/z = ∞ z −n /n! for all z = 0
...
Hence
1
1
1
z2
zn
+ ··· +
+ +2+z+
+ ··· +
+ ···
...
Laurent series, singularities
where
1
1
for n ≥ 1, a0 = 2, a−n =
for n ≥ 1
...
e
...
an =
Example
...
Now 1/z is already a Laurent series at 0 (the only non-zero coefficient is a−1 = 1)
...
Now, by summing a geometric progression, we have that 1/(1 − z) = ∞ z n and this
n=0
power series converges for |z| < 1
...
Example
...
First note that we can write
f (z) =
∞
1
−1
zn
=
=−
z−1
1−z
n=0
(6
...
3)
(summing a geometric progression) and that this is valid for |z| < 1
...
This converges for |z −1 | < 1, i
...
|z| > 1
...
2
...
Similarly, we can write
1
−1
1
=
=−
z−2
2−z
2
1
1−
z
2
1
=−
2
∞
n=0
z
2
n
(6
...
5)
∞
n
by noting that
n=0 (z/2) = 1/(1 − z/2) is the sum of a geometric progression with
common ratio z/2
...
e
...
66
MATH20101 Complex Analysis
6
...
2
...
This converges when |(2/z)−1 | < 1, i
...
when |z| > 2
...
2
...
2
...
Using (6
...
4) and (6
...
5) we can expand
f (z) =
∞
∞
1
1
z
+
n
z
2 n=0 2
n=1
= ··· +
n
1 1
z
zn
1
+ · · · + + + 2 + · · · + n+1 + · · ·
zn
z 2 2
2
and this is valid on the annulus {z ∈ C | 1 < |z| < 2}
...
2
...
2
...
§6
...
A singularity of a function f (z) is a point z0 at which f (z) is not differentiable
...
Here is a common way for a singularity to occur: if f is not defined at z0 then
it cannot be differentiable at z0
...
If f (z) = 1/z then f is not defined at the origin (we are not allowed to divide
by 0)
...
Suppose that f has a singularity at z0
...
If there exists a punctured disc 0 < |z − z0 | < R such that f is differentiable
on this punctured disc then we say that z0 is an isolated singularity of f
...
In the above example, 0 is an isolated singularity of f (z) = 1/z
...
Laurent series, singularities
In this course we will only be interested in isolated singularities
...
Then f is holomorphic on an annulus of the form {z ∈ C | 0 <
|z − z0 | < R}
...
Consider the principal part of the Laurent series
∞
n=1
bn (z − z0 )−n
...
3
...
§6
...
1
Removable singularities
Suppose that f has an isolated singularity at z0 and that the principal part of the Laurent
series (6
...
1) has no terms in it
...
The radius of convergence of this power series is at least R, and so f (z) extends to a function
that is differentiable at z0
...
Let
sin z
, z = 0
...
However, we know
that
z2 z4
sin z
=1−
+
− ···
z
3!
5!
for z = 0
...
Then f (z) is differentiable for all z ∈ C
...
f (z) =
§6
...
2
Poles
Suppose that f has an isolated singularity at z0 and that the principal part of the Laurent
series (6
...
1) has finitely many terms in it
...
In this case, we say that f has a pole of order m at z0
...
68
MATH20101 Complex Analysis
6
...
Let
sin z
, z = 0
...
We can write
f (z) =
sin z
1
1
1
11
+ z − z3 + · · ·
...
We will often consider functions f : D → C defined on a domain D that are differentiable
except at finitely many points in D and f has either removable singularities or poles at
these points
...
Let D be a domain
...
§6
...
3
Isolated essential singularities
Suppose that f has an isolated singularity at z0 and that the principal part of the Laurent
series (6
...
1) has infinitely many terms in it
...
Isolated essential singularities are difficult to deal with and we will not consider them
in this course
...
Let f (z) = sin 1/z, z = 0
...
3
z 3!z
5!z 5
Hence f has an isolated essential singularity at z = 0
...
Exercises for Part 6
Exercises for Part 6
Exercise 6
...
Exercise 6
...
Exercise 6
...
(Hint: introduce a new variable w = z − 1
...
4
Let f (z) = (z − 1)−2
...
Exercise 6
...
2+1
+1
z + 16
z + 2z
z +z−1
Exercise 6
...
z2
MATH20101 Complex Analysis
6
...
7
Let D be a domain and let z0 ∈ D
...
Show that f has a removable singularity at z0
...
Cauchy’s Residue Theorem
7
...
1
Introduction
One of the more remarkable applications of integration in the complex plane in general,
and Cauchy’s Theorem in particular, is that it gives a method for calculating real integrals
that, up until now, would have been difficult or even impossible (assuming that you only
had the tools of 1st year calculus or A-level mathematics to hand)
...
However, in only very few examples were you able
n=0
to say what the limit actually is! Using complex analysis, it becomes very easy to evaluate
infinite series such as ∞ 1/n4 = π 4 /90
...
2
Zeros and poles of holomorphic functions
Recall that a function f has a singularity at z0 if f is not differentiable at z0
...
In the examples we have seen so far
f (z) has a pole at z0 because we have been able to write f (z) = p(z)/q(z) and q(z0 ) = 0
(so that f is not even defined at z0 )
...
Definition
...
We will only be interested in isolated zeros
...
More formally, we have the following definition
...
A function f defined on a domain D has an isolated zero at z0 if f (z0 ) = 0
and there exists ε > 0 such that f (z) = 0 for all z such that 0 < |z − z0 | < ε
...
By Taylor’s
Theorem (Theorem 5
...
1), we can expand f as a Taylor series in some neighbourhood
around z0
...
2
...
Definition
...
We say that z0 is a simple zero if it is a zero of order 1
...
(i) Let f (z) = z 2
...
(ii) Let f (z) = z(z + 2i)3
...
(iii) Let f (z) = z 2 + 4
...
72
MATH20101 Complex Analysis
7
...
The coefficients an in the Taylor expansion are given by an = f (n) (z0 )/n!
...
In partiocular, if f (z0 ) = 0 but f ′ (z0 ) = 0 then z0 is a simple zero
...
(i) Let f (z) = sin z
...
Note that f ′ (kπ) =
cos kπ = (−1)k = 0
...
(ii) Let f (z) = 1 − cos z
...
Now f ′ (z) = sin z and
f ′ (2kπ) = 0, but f ′′ (2kπ) = cos 2kπ = 1 = 0
...
Lemma 7
...
1
Suppose that the holomorphic function f has a zero of order m at z0
...
Proof
...
2
...
Take g(z) =
centred on z0 and g(z0 ) = am = 0
...
∞
n=0
an+m (z − z0 )n
Then g is holomorphic on an open disc
✷
We can now link poles of a function f (z) = p(z)/q(z) with zeros of the function q
...
2
...
Then f has a pole of order m at z0
...
By Lemma 7
...
1, we can write q(z) = (z − z0 )m r(z) where r is holomorphic and
r(z0 ) = 0
...
Then g(z) is holomorphic at z0 , and so we can expand
it as a Taylor series at z0 as
g(z) =
∞
n=0
an (z − z0 )n
and this expression is valid in some disc |z − z0 | < R, for some R > 0
...
m
m−1
(z − z0 )
(z − z0 )
(z − z0 )m−2
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MATH20101 Complex Analysis
7
...
Hence f has a pole of order m at z0
...
(i) Let
f (z) =
sin z
...
This is because sin z = 0 when z = 3 and
(z − 3)2 has a zero of order 2 at z = 3
...
sin z
Then f has a simple pole at kπ for each k ∈ Z
...
f (z) =
§7
...
Definition
...
Suppose that on {z ∈ C | 0 < |z − z0 | < R} ⊂ D, f has Laurent
expansion
f (z) =
∞
n=0
n
an (z − z0 ) +
∞
n=1
bn (z − z0 )−n
...
That is, the residue of f at the isolated singularity z0 is the coefficient of (z − z0 )−1 in the
Laurent expansion
...
By Laurent’s Theorem (Theorem 6
...
1) we have the alternative expression
1
Res(f, z0 ) =
f (z) dz
2πi Cr
where Cr (t) = z0 + reit , 0 ≤ t ≤ 2π is a circular anticlockwise path around z0 in the annulus
of convergence
...
Cauchy’s Residue Theorem relies on using Cauchy’s Theorem in just the right way
...
We make the
following definition
...
A closed contour γ is said to be a simple closed loop if, for every point z not
on γ, the winding number is either w(γ, z) = 0 or w(γ, z) = 1
...
Thus a simple closed loop is a loop that goes round anticlockwise in a loop once, and
without intersecting itself; see Figure 7
...
1
...
We can now state the main result of this section
...
Cauchy’s Residue Theorem
γ2
γ3
γ1
Figure 7
...
1: Here γ1 is a simple closed loop
...
Theorem 7
...
1 (Cauchy’s Residue Theorem)
Let D be a domain containing a simple closed loop γ and the points inside γ
...
, zn inside γ
...
f (z) dz = 2πi
γ
j=1
Remark
...
A very common mistake is to either miss a 2πi out,
or put one in by mistake
...
§7
...
In some cases, ad hoc manipulations have to be used to calculate the Laurent series, but
there are many cases where one can calculate them more systematically
...
We say that a pole of order 1
is a simple pole
...
If we can write f (z) = p(z)/q(z) where p and q are differentiable and p(z) = 0
when q(z) = 0 then the poles of f occur at the zeros of q
...
It is easy to calculate the residue at a simple pole
...
4
...
z→z0
75
MATH20101 Complex Analysis
7
...
q (z0 )
Proof
...
Hence
(z − z0 )f (z) = b1 +
∞
n=0
an (z − z0 )n+1
so that limz→z0 (z − z0 )f (z) = b1
...
By part (i) and the fact that
q(z0 ) = 0, the residue is
lim
z→z0
(z − z0 )p(z)
= lim
z→z0
q(z)
p(z)
q(z)−q(z0 )
z−z0
=
p(z0 )
...
For example, let
cos πz
...
4
...
Hence
f (z) =
Res(f, 1) =
1
cos π
=
...
4
...
Lemma 7
...
2
Suppose that f has a pole of order m at z0
...
(m − 1)! dz m−1
Proof
...
Hence
m
m−1
(z − z0 ) f (z) = bm + (z − z0 )bm−1 + · · · + (z − z0 )
Differentiating this m − 1 times gives
dm−1
(z − z0 )m f (z) = (m − 1)!b1 +
dz m−1
∞
n=0
n=0
an (z − z0 )m+n
...
(n + 1)!
Dividing by (m − 1)! and letting z → z0 gives the result
...
Cauchy’s Residue Theorem
Example
...
This has a pole of order 3 at z = 1
...
Hence
1 d2
6
6
(z − 1)3 f (z) = (z + 1) → × 2 = 6
2! dz 2
2!
2!
as z → 1
...
Let us check this by calculating the Laurent series
...
Then z = w + 1 and we can write
z+1
z−1
3
=
=
=
=
(w + 2)3
w3
w3 + 6w2 + 12w + 8
w3
8
12
6
+ 2 + +1
3
w
w
w
8
12
6
+
+
+ 1
...
In other cases, one has to manipulate the formula for f to calculate the residue
...
Let
f (z) =
1
z 2 sin z
...
Hence the singularities are at
z = 0, kπ
...
Recalling the power series for sin z we can write
f (z) =
1
z 2 sin z
=
z2
=
=
=
1
z3
z−
+ ···
6
1
z2
+ ···
1−
z3
6
z2
1
+ ···
1+
z3
6
1
1
+
+ ···
3
z
6z
−1
where we have omitted higher order terms
...
) Hence
Res(f, 0) = 1/6
...
Cauchy’s Residue Theorem
For the poles at kπ, k = 0, we could change variables to w = z − kπ and calculate the
Laurent series
...
4
...
First note that we can write
f (z) =
p(z)
q(z)
where p(z) = 1 and q(z) = z 2 sin z
...
Hence
Res(f, kπ) =
p(kπ)
(−1)k
=
q ′ (kπ)
(kπ)2
as q ′ (z) = 2z sin z + z 2 cos z so that q ′ (kπ) = (kπ)2 cos kπ = (−1)k (kπ)2
...
There is a short video on the course webpage that summarises poles and residues
...
)
§7
...
5
...
Thus C2 is the circle of radius 2 centred at 0 described
anticlockwise, and C4 is the circle of radius 4 centred at 0 described anticlockwise
...
Consider the function
3
f (z) =
z−1
Then f has a pole at z = 1 and no other poles
...
As the pole at z = 1 lies inside C2 , by Cauchy’s Residue Theorem we
have that
f dz = 2πi Res(f, 1) = 6πi
...
C4
See Figure 7
...
1
...
z 2 + (i − 3)z − 3i
Then f has a pole when the denominator has a zero
...
Thus f has
simple poles z = 3 and z = −i
...
4
...
3+i
3+i
78
MATH20101 Complex Analysis
7
...
5
...
See Figure 7
...
2
...
The pole z = −i is inside C2 but the pole z = 3 is outside
...
5
...
10
5
f dz
...
f dz = 2πi (Res(f, −i) + Res(f, 3)) = 2πi
1
−1
+
3+i 3+i
= 0
...
e
...
5
...
First we need to make precise what (7
...
1) means
...
We then define
equal to this limit
...
5
...
Cauchy’s Residue Theorem
C4
C2
4
3
2
−i
Figure 7
...
2: The function f (z) = 1/(z 2 + (i − 3)z − 3i) has simple poles at z = −i and
z = 3
...
(7
...
3)
However, there are many functions f for which (7
...
3) exists but (7
...
2) does not
...
Then
R
R
x dx =
f (x) dx =
−R
−R
1 2
x
2
R
=
x=−R
R2 R2
−
=0
2
2
and so converges to 0 as R → ∞
...
5
...
However
B
B
x dx =
f (x) dx =
−A
−A
1 2
x
2
B
=
x=−A
B 2 A2
−
2
2
does not converge if we first let B tend to ∞ and then let A tend to ∞
...
5
...
Lemma 7
...
1
Suppose that f : R → C is a continuous function and there exist constants K > 0, C > 0
and r > 1 such that for |x| ≥ K we have
|f (x)| ≤
C
...
5
...
5
...
80
(7
...
4)
∞
−∞ f (x) dx,
exists and is equal
MATH20101 Complex Analysis
7
...
We will show how to use Cauchy’s Residue Theorem to evaluate
∞
(x2
−∞
1
dx
+ 1)(x2 + 4)
(7
...
5)
(the fact that 1 and 4 are squares will make the calculations notationally easier, but this is
not essential to the method)
...
To see this, first recall from (4
...
1) that [−R, R] denotes the straight line
path from −R to R and that this has parametrisation σ(t) = t, −R ≤ t ≤ R
...
−R
−R
[−R,R]
R
f (σ(t))σ ′ (t) dt =
Note that there exists a constant C > 0 such that
(x2
1
C
≤ 4
...
5
...
5
...
e
...
(x2 + 1)(x2 + 4)
Let
f (z) =
(z 2
1
+ 1)(z 2 + 4)
(note that we have introduced a complex variable)
...
This has parametrisation t, −R ≤ t ≤ R
...
5
...
[−R,R]
To use Cauchy’s Residue Theorem, we need a closed contour
...
5
...
−R
R
Figure 7
...
3: The ‘D-shaped’ contour ΓR
...
81
MATH20101 Complex Analysis
7
...
To use Cauchy’s Residue Theorem, we need to know
the poles and residues of f (z)
...
2 + 4)
(z + 1)(z
(z − i)(z + i)(z − 2i)(z + 2i)
Hence f (z) has simple poles at z = +i, −i, +2i, −2i
...
Now by
Lemma 7
...
1,
Res(f, i) = lim(z − i)f (z)
z→i
1
z→i (z + i)(z − 2i)(z + 2i)
1
=
6i
= lim
and
Res(f, 2i) =
lim (z − 2i)f (z)
z→2i
1
z→2i (z − i)(z + i)(z + 2i)
−1
...
6i 12i
6
If we can show that
f (z) dz = 0
lim
R→∞ SR
(7
...
6)
then we will have that
∞
−∞
(x2
1
dx = lim
R→∞
+ 1)(x2 + 4)
f (z) dz =
[−R,R]
π
...
5
...
We shall use the Estimation
Lemma
...
Note that |z| = R
...
≤
(z 2 + 1)(z 2 + 4)
(R2 − 1)(R2 − 4)
Hence, by the Estimation Lemma,
f (z) dz
SR
1
length(SR )
− 1)(R2 − 4)
πR
=
(R2 − 1)(R2 − 4)
→ 0
≤
(R2
as R → ∞, which is what we wanted to check
...
Cauchy’s Residue Theorem
Remark
...
5
...
(ii) Construct a ‘D-shaped’ contour ΓR as in Figure 7
...
3
...
(iv) Use Cauchy’s Residue Theorem to write down
ΓR
f (z) dz
...
Use the
Estimation Lemma to conclude that the integral over SR converges to 0 as R → ∞
...
Remark
...
You should always check that your answer makes sense
...
2 + 4)
+ 1)(x
6
This is obviously wrong: the left-hand side is a real number, whereas the (incorrect) righthand side is imaginary
...
§7
...
3
Trigonometric integrals
We can use Cauchy’s Residue Theorem to calculate integrals of the form
2π
Q(cos t, sin t) dt
(7
...
7)
0
where Q is some function
...
)
The first step is to turn (7
...
7) into a complex integral
...
Then
cos t =
z − z −1
z + z −1
, sin t =
...
Finally, note that
dz = ieit dt so that
dz
...
Q
Q(cos t, sin t) dt =
2
2i
iz
C1
0
83
MATH20101 Complex Analysis
7
...
Instead of stating a general theorem, we shall compute some examples to illustrate the
method
...
We shall how to compute
2π
(cos3 t + sin2 t) dt
...
Then
2π
(cos3 t + sin2 t) dt
0
C1
C1
C1
+
z − z −1
2i
2
dz
iz
z 3 3z 3z −1
z −3 z 2 1 z −2
+
+
+
−
+ −
8
8
8
8
4
2
4
=
=
3
z + z −1
2
=
z 2 3 3z −2 z −4 z z −1 z −3
+ +
+
− +
−
8
8
8
8
4
2
4
1
i
dz
iz
dz
The integrand has a pole of order 4 at z = 0 with residue 1/2i, and no other poles
...
We shall compute
1
= π
...
0
Again, substituting z = eit we have that
2π
cos t sin t dt
0
=
C1
=
C1
=
C1
1
dz
(z + z −1 )(z − z −1 )
4i
iz
1 2
dz
(z − z −2 )
4i
iz
1
−1
z − 3 dz
...
There are no other poles
...
0
84
MATH20101 Complex Analysis
7
...
The above illustrates a more general method
...
In this case, as t varies from 0 to π then z describes the
unit circle in C with centre 0 and radius 1 described anti-clockwise
...
5
...
Then cot πz has a pole whenever sin πz = 0, i
...
whenever z = n, n ∈ Z
...
Hence cot πz has a simple pole at z = n
...
4
...
π cos πn
π
This suggests a method for summing infinite series of the form ∞ an
...
Consider the function f (z) cot πz
...
For example, we will
∞
2
...
First of all, we need to choose a good contour
to integrate around
...
Secondly, f (z) may have poles of its own and
these will need to be taken into account
...
)
Instead of choosing a D-shaped contour, here we use a square contour
...
5
...
This is a square with each side having length 2N + 1
...
)
Lemma 7
...
2
There is a bound, independent of N , on cot πz where z ∈ CN , i
...
there exists M > 0 such
that for all N and all z ∈ CN , we have | cot πz| ≤ M
...
Consider the square CN
...
Consider first the horizontal sides
...
Then |y| ≥ 1/2
...
Cauchy’s Residue Theorem
−(N + 1) −N
−1
N
0 1
N +1
Figure 7
...
4: The square contour CN
...
Consider now the vertical sides
...
Hence
| cot πz| =
=
=
=
=
≤
eiπz + e−iπz
eiπz − e−iπz
e2πiz + 1
e2πiz − 1
eiπ−2πy + 1
eiπ−2πy − 1
−e−2πiy + 1
−e−2πiy − 1
1 − e−2πy
1 + e2πy
1
...
Cauchy’s Residue Theorem
Hence | cot πz| ≤ max{1, coth π/2} for all z ∈ CN
...
Very similar
techniques and slight modifications to the argument work for many other examples
...
Let f (z) = 1/z 2 and consider the function
n=0
f (z) cot πz =
cot πz
cos πz
= 2
...
These occur when z 2 sin πz = 0, i
...
when z = n, n ∈ Z
...
Let us calculate the residue when n = 0
...
4
...
Then
Res
cot πz
,n
z2
cos πn
+ 2n sin πn
=
πn2 cos πn
1
...
There are (at least) three ways to work out the residue
here, and for completeness we’ll discuss them all
...
(We used the expansion (1−x)−1 = 1+x+x2 +· · ·
...
Alternatively, as another method for calculating the residue at 0, we can use the following power series expansion for cot z:
cot z =
z3
2z 5
1 z
− −
−
− ···
...
Finally, as a third method of calculating the residue at 0, one could use Lemma 7
...
2
...
Note that each side of the square
has length 2N + 1
...
87
MATH20101 Complex Analysis
7
...
By Cauchy’s Residue
Theorem we have that
N
cot πz
,n
z2
Res
2πi
n=−N
=
CN
cot πz
dz
...
5
...
Also
note that |1/z 2 | ≤ 1/N 2 for z on CN
...
Hence
N
N →∞
cot πz
,n
z2
Res
lim
n=−N
= 0
...
5
...
5
...
2
πn
3
n=1
∞
π2
1
...
6
Proof of Cauchy’s Residue Theorem
Let us first recall the statement of Cauchy’s Residue Theorem:
Theorem 7
...
1 (Cauchy’s Residue Theorem)
Let D be a domain containing a simple closed loop γ and the points inside γ
...
, zn inside γ
...
f (z) dz = 2πi
γ
j=1
88
MATH20101 Complex Analysis
7
...
The proof is a simple application of the Generalised Cauchy Theorem (Theorem 4
...
7)
...
, n, we can find circles
Sj (t) = zj + εj eit , 0 ≤ t ≤ 2π
centred at zj and of radii εj , each described once anticlockwise, such that Sj and the points
inside Sj lie in D and such that Sj contains no singularity other than zj (see Figure 7
...
1)
...
6
...
The circles S1 , S2 , S3 (centred on
z1 , z2 , z3 , respectively) have been chosen so that they lie inside γ and do not intersect each
other
...
, zn }
...
, Sn
satisfy the hypotheses of the Generalised Cauchy Theorem (Theorem 4
...
7) with respect
to D ′ : i
...
their winding numbers sum to zero for every point not in D′
...
Hence the hypotheses of the Generalised Cauchy Theorem hold for points z not in D
...
e
...
Since each pole zj lies inside γ, we have that
w(−γ, zj ) = −w(γ, zj ) = −1
...
w(Sk , zj ) =
89
MATH20101 Complex Analysis
7
...
Hence, by the Generalised Cauchy Theorem,
f+
S1
−γ
f + ··· +
f = 0
...
2
...
Sj
Hence
f
γ
=
S1
f + ··· +
f
Sn
= 2πi (Res(f, z1 ) + · · · + Res(f, zn )) ,
concluding the proof
...
Exercises for Part 7
Exercises for Part 7
Remark
...
Exercise
7
...
4
...
2, 7
...
5
...
Exercises 7
...
5,
7
...
7 can be done after §7
...
2
...
8, 7
...
5
...
Exercises 7
...
10 can be done after §7
...
3
...
11 (which is hard) relies on ideas in §7
...
2
...
1, 7
...
4, 7
...
6, 7
...
8, 7
...
Exercise 7
...
Exercise 7
...
Use Cauchy’s
Residue Theorem to evaluate the integrals
(i)
C4
1
dz, (ii)
2 − 5z + 6
z
1
dz, (iii)
2 − 5z + 6
z
C5/2
C2
eaz
dz (a ∈ R)
...
3
Recall that Laurent’s Theorem (Theorem 6
...
1) says the following: Suppose that f is
holomorphic on the annulus {z ∈ C | R1 < |z − z0 | < R2 }
...
The coefficients are given by
an =
1
2πi
Cr
f (z)
dz
(z − z0 )n+1
and Cr (t) = z + reit , 0 ≤ t ≤ 2π, denotes the circular path around z0 of radius r where r
is chosen such that R1 < r < R2
...
Exercises for Part 7
(ii) 1 < |z| < ∞,
(iii) 0 < |z − 1| < 1,
(iv) 1 < |z − 1| < ∞
...
)
In each case, check your answer by directly calculate the Laurent series using the methods described in §6
...
Exercise 7
...
+1
(i) Explain why this integral is equal to its principal value
...
(How would you have
done this without using complex analysis?)
(b)
(i) Now evaluate, using Cauchy’s Residue Theorem, the integral
∞
−∞
e2ix
dx
...
x2 + 1
(Why is it obvious, without having to use complex integration, that one of these
integrals is zero?)
(iii) Why does the ‘D-shaped’ contour used in the lectures for calculating such integrals fail when we try to integrate
∞
−∞
e−2ix
dx?
x2 + 1
By choosing a different contour, explain how one could evaluate this integral
using Cauchy’s Residue Theorem
...
5
Use Cauchy’s Residue Theorem to evaluate the following real integrals:
(i)
∞
−∞
1
dx, (ii)
2 + 1)(x2 + 3)
(x
92
∞
−∞
1
dx
...
Exercises for Part 7
Exercise 7
...
+ 4x + 5
e
Exercise 7
...
Evaluate the integral discussed in §1
...
Exercise 7
...
Hence use Cauchy’s Residue Theorem to evaluate them
...
1 + cos2 t
Exercise 7
...
n=1
Why doesn’t the method work for evaluating ∞ 1/n3 ?
n=1
Exercise 7
...
Consider the function
cot πz
z 2 + a2
Show that this function has poles at z = n, n ∈ Z and z = ±ia
...
Hence show that
∞
π
1
1
=
coth πa − 2
...
11
(The method used in §7
...
3 can be used evaluate other integrals
...
(i) Prove, using Cauchy Residue Theorem, that
C1
ez
dz = 2πi
...
MATH20101 Complex Analysis
7
...
12
(Sometimes, one has to be rather creative in picking the right contour
...
Show that
∞
eaz
π
=
z
sin aπ
−∞ 1 + e
using the following steps
...
(ii) Let f (z) = eaz /(1 + ez )
...
Draw a diagram to illustrate where the poles are
...
(iii) On the diagram from (ii), draw the contour ΓR = γ1,R + γ2,R + γ3,R + γ4,R where:
γ1,R is the horizontal straight line from −R to R,
γ2,R is the vertical straight line from R to R + 2πi,
γ3,R is the horizontal straight line from R + 2πi to −R + 2πi,
γ4,R is the vertical straight line from −R + 2πi to −R
...
(iv) Show, by choosing suitable parametrisations of the paths γ1,R and γ3,R and direct
computation, that γ3 f = −e2πia γ1 f
...
94
f = 0
...
Solutions to Part 1
8
...
1
(i) (3 + 4i)2 = 9 + 24i − 16 = −7 + 24i
...
3 − 4i
3 − 4i 3 + 4i
25
25
25
(iii)
1 − 5i
−8
1
=
+i
...
1+i
(v)
1
= −i
...
2
√
√
First note that |1 − i| = 2 and arg(1 − i) = −π/4 (draw a picture!)
...
Hence
√
√
3 − i = 2e−iπ/6
...
2 e
( 3 − i)13
Note that
5e3iπ/4 + 2e−iπ/6 = 5 cos(3π/4) + 5i sin(3π/4) + 2 cos(−π/6) + 2i sin(−π/6)
√
√
√
1
2
2
3
+ 5i
+2
− 2i
= −5
2
2
√2
√
√2
5 2−2
−5 2 + 2 3
+i
...
3
(i) Write z = x + iy
...
Comparing real and imaginary
parts gives the simultaneous equations x2 − y 2 = −5, 2xy = 12
...
Solving this quadratic equation gives x2 = 4, hence x = ±2
...
Hence z = 2+ 3i, −2− 3i are the solutions
...
Solutions to Part 1
(ii) A bare-hands computation as in (i) will work, but is very lengthy
...
Write
z 2 + 4z + 12 − 6i = (z + 2)2 + 8 − 6i
...
Then (x+ iy)2 = −8+ 6i
...
Hence z = −1 + 3i or z = −3 − 3i
...
4
(i) Let z = a+ib, w = c+id
...
Similarly Re(z − w) = Re(z) − Re(w)
...
Similarly Im(z − w) = Im(z) − Im(w)
...
For example, choose z = i, w = −i
...
However,
Re(zw) = 1 = Re(z) Re(w) = 0 × 0 = 0
...
Solution 1
...
(i) z + w = (a + ib) + (c + id) = (a + c) + i(b + d) = (a+c)−i(b+d) = (a−ib)+(c−id) =
z + w
...
¯ ¯
(ii) zw = (a + ib)(c + id) = (ac − bd) + i(ad + bc) = (ac − bd) − i(ad + bc) = (a − ib)(c −
id) = zw
...
1
a+ib
=
a−ib
a2 +b2
=
a+ib
...
¯
(v) z − z = (x + iy) − (x − iy) = 2iy = 2i Im(z)
...
6
Let z, w ∈ C
...
Hence
|z| − |w| ≤ |z − w|
...
Hence
||z| − |w|| ≤ |z − w|
...
7
(i) Writing z = x + iy we obtain Re(z) = {(x, y) | x > 2}, i
...
a half-plane
...
96
=
a+ib
,
a2 +b2
so
MATH20101 Complex Analysis
8
...
(iv) Write z = x + iy
...
e
...
Multiplying this out (and noting that the ys cancel) gives x > 0, i
...
an open halfplane
...
8
(i) We have
zw = rs ((cos θ cos φ − sin θ sin φ) + i(cos θ sin φ + sin θ cos φ))
= rs (cos(θ + φ) + i sin(θ + φ))
...
(ii) From (i) we have that arg z 2 = 2 arg z
...
Put z =
cos θ + i sin θ so that arg z = θ
...
Hence |z n | = 1
...
(iii) Applying De Moivre’s theorem in the case n = 3 gives
(cos θ + i sin θ)3 = cos3 θ + 3i cos2 θ sin θ − 3 cos θ sin2 θ − i sin3 θ
= cos 3θ + i sin 3θ
...
Similarly,
cos 4θ = sin4 θ − 6 cos2 θ sin2 θ + cos4 θ,
sin 4θ = −4 cos θ sin3 θ + 4 cos3 θ sin θ
...
9
Let w0 = reiθ and suppose that z n = w0
...
Then z n = ρn einφ
...
Thus we have that ρ = r 1/n and we get distinct values of the
argument φ of z when k = 0, 1,
...
Hence
z = r 1/n ei(
θ+2kπ
n
) , k = 0, 1,
...
Solution 1
...
Then Arg(z1 ) = Arg(z2 ) = 3π/4
...
(Draw a picture!) In this case, Arg(z1 z2 ) = Arg(z1 ) + Arg(z2 )
...
)
Solution 1
...
However, there is one technique that
may work (I haven’t tried it), and it’s one that was a favourite of Richard Feynman (Nobel
laureate in physics, safe-cracker, and bongo-player, amongst many other talents)
...
Solutions to Part 1
claimed to have never learned complex analysis but could perform many real integrals using
a trick called ‘differentiation under the integral sign’
...
mit
...
pdf for an account of this, if
you’re interested
...
Solutions to Part 2
9
...
1
(i) This set is open
...
Let z0 ∈ D
...
To do this, write z0 = x0 + iy0 and let ε = y0 /2 > 0
...
Then
y0
|y − y0 | ≤ |x − x0 |2 + |y − y0 |2 = |z − z0 | ≤
...
e
...
Hence z ∈ D
...
1(i)
...
Let D = {z ∈ C | Re(z) > 0, |z| < 2}
...
We have to
find ε > 0 such that Bε (z0 ) ⊂ D
...
1(ii)
...
Let
ε = min
x0 2 − |z0 |
,
2
2
>0
(note that |z0 | < 2 as z0 ∈ D)
...
Then arguing as in (i) we see that
x0
|x − x0 | ≤ |x − x0 |2 + |y − y0 |2 = |z − z0 | ≤
2
so that
x0
x0
− < x − x0 <
2
2
from which it follows that x > x0 /2 > 0, i
...
Re(z) > 0
...
Hence |z| < 2
...
|z| = |z − z0 + z0 | ≤ |z − z0 | + |z0 | ≤
(iii) Let D = {z ∈ C | |z| ≤ 6}
...
If we take the point z0 = 6 on the
real axis, then no matter how small ε > 0 is, there are always points in Bε (z0 ) that
are not in D
...
1(iii)
...
2
(i) For any z0 ∈ C we have
2
z 2 + z − (z0 + z0 )
z→z0
z − z0
(z − z0 )(z + z0 + 1)
= lim
z→z0
z − z0
= lim z + z0 + 1
f ′ (z0 ) =
lim
z→z0
= 2z0 + 1
so that f ′ (z) = 2z + 1
...
Solutions to Part 2
z0
ε
z0
ε
z0
ε
(i)
(ii)
(iii)
Figure 9
...
1
...
(iii) For each z0 ∈ C we have
2
3
(z 3 − z 2 ) − (z0 − z0 )
z→z0
z − z0
2
(z − z0 )(z 2 + z0 z + z0 − z − z0 )
= lim
z→z0
z − z0
2
2
= lim z + z0 z + z0 − z − z0
f ′ (z0 ) =
=
lim
z→z0
2
3z0 −
2z0
so that f ′ (z) = 3z 2 − 2z
...
Solution 2
...
(a) Note that f (z) = (x + iy)2 = x2 + 2ixy − y 2
...
(b) Note that for z = 0
f (x + iy) =
1
x − iy
x
−y
= 2
= 2
+i 2
2
2
x + iy
x +y
x +y
x + y2
so that u(x, y) = x/(x2 + y 2 ), v(x, y) = −y/(x2 + y 2 )
...
Solutions to Part 2
(ii) (a) Here
∂u
∂v ∂u
∂v
= 2x =
,
= −2y = −
∂x
∂y ∂y
∂x
so that the Cauchy-Riemann equations are satisfied
...
∂x
(x + y 2 )2 ∂y
(x + y 2 )2
Hence ∂u/∂x = ∂v/∂y and ∂u/∂y = −∂v/∂x so that the Cauchy-Riemann
equations hold
...
Then for (x, y) = (0, 0) we have
x2 + y 2 , v(x, y) =
∂u
∂v
x
y
∂v
∂u
,
,
= 2
= 2
= 0,
= 0
...
At (x, y) = (0, 0) we have
|h|
∂u
= lim
∂x h→0 h
which does not exist
...
)
Hence f is not differentiable anywhere
...
4
(i) Here
∂v
∂u
= 3x2 − 3y 2 ,
= 3x2 − 3y 2 ,
∂x
∂y
and
∂u
∂v
= −6xy,
= 6xy
∂y
∂x
so that the Cauchy-Riemann equations hold
...
Solutions to Part 2
so that the Cauchy-Riemann equations hold
...
In both cases, the partial derivatives of u and v exist at z0
...
The Cauchy-Riemann equations holds at z0
...
5
...
Hence f = u + iv is differentiable
at z0
...
Solution 2
...
z→0
z
As f (z) = 0 for the function in the question we need to investigate the limit
f ′ (0) = lim
lim
z→0
f (z)
...
Then f (x) = x and
lim
z→0
x
1
f (z)
= lim
=
...
x→0 x − ix
z
1−i
Hence there is no limit of (f (z) − f (0))/z as z → 0
...
Then clearly
∂v
∂v
(0, 0) =
(0, 0)
...
k→0
k→0 k
∂y
k
Hence the Cauchy-Riemann equations are satisfied
...
5
...
To see this, note that for x > 0, y > 0,
√
y
∂u
(x, y) = √
∂x
2 x
so that
lim
(x,y)→(0,0)
does not exist
...
Solutions to Part 2
Solution 2
...
∂x2
∂y
Similarly
∂2v
∂x2
∂ ∂v
∂ ∂u
∂2u
∂2u
=−
=−
=−
∂x ∂x
∂x ∂y
∂x∂y
∂y∂x
2v
∂ ∂u
∂ ∂v
∂
= −
=−
=− 2
∂y ∂x
∂y ∂y
∂y
=
so that
∂2v
∂2v
+ 2 = 0
...
7
Let f (z) = z 3 and write z = x + iy so that
f (x + iy) = (x + iy)3 = x3 − 3xy 2 + i(3x2 y − y 3 )
...
Now
∂2u
∂2u
= 6x,
= −6x
∂x2
∂y 2
and
so that
∂2v
∂2v
= 6y,
= −6y
∂x2
∂y 2
∂2u ∂2u
∂2v
∂2v
+ 2 = 0,
+ 2 = 0
...
Solution 2
...
Then
∂v
∂u
= 5x4 − 30x2 y 2 + 5y 4 =
...
0
...
(Recall that we are looking
for an anti-partial derivative and ∂α(x)/∂y = 0
...
Solutions to Part 2
and integrating with respect to x gives
v(x, y) = 5x4 y − 10x2 y 3 + β(y)
(9
...
2)
for some arbitrary function β(y)
...
0
...
0
...
e
...
The right-hand side depends only on y and the left-hand side depends only on x
...
Hence
v(x, y) = 5x4 y − 10x2 y 3 + y 5 + c
for some constant c ∈ R
...
9
From Exercise 2
...
e
...
Note that
∂2u
= 6x,
∂x2
so that
0=
∂2u
= −2kx
∂y 2
∂2u ∂2u
+ 2 = (6 − 2k)x
...
It remains to show that in the case k = 3, u is the real part of a holomorphic function
...
8
...
∂x
∂y
Hence
v(x, y) = 3x2 y − y 3 + 6y 2 − 12y + α(x)
(9
...
3)
for some arbitrary function α(x) depending only on x
...
0
...
Comparing (9
...
3) and (9
...
4) we
see that
α(x) + 6x2 = β(y) + y 3 − 6y 2 + 12y;
as the left-hand side depends only on x and the right-hand side depends only on y, the
above two expressions must be equal to a constant c ∈ R
...
Note that the partial derivatives for both u and v exist and are continuous at every
point in C and the Cauchy-Riemann equations hold at every point in C, it follows from
the converse of the Cauchy-Riemann Theorem that f (x + iy) = u(x, y) + iv(x, y) is a
holomorphic function on C
...
Solutions to Part 2
Solution 2
...
Then ∂u/∂x = 0
...
Integrating with respect to y gives
that v(x, y) = α(x) for some function α(x) that depends only on x
...
Hence by the Cauchy-Riemann equations −∂v/∂x = 0
...
Hence
v(x, y) = α(x) = β(y)
...
Hence v(x, y) is constant and it follows that f is constant
...
11
Suppose that f (x + iy) = u(x) + iv(y) where the real part depends only on x and the
imaginary part depends only on y
...
∂x
∂y
By the Cauchy-Riemann equations, u′ (x) = v ′ (y)
...
From u′ (x) = λ we have that u(x) = λx + c1 , for some constant
c1 ∈ R
...
Let
a = c1 + ic2
...
Solution 2
...
Partially differentiating
the latter expression with respect to x gives
2
∂u ∂v
+
=0
∂x ∂x
and using the Cauchy-Riemann equations gives
2
∂u ∂u
−
= 0
...
∂x
∂y
This gives us two simultaneous equations in ∂u/∂x and ∂u/∂y
...
∂x
∂y
From ∂u/∂x = 0 it follows that u(x, y) = α(y), an arbitrary function of y
...
This is only possible if u is
constant
...
Solutions to Part 2
If u is constant then
0=
∂v
∂u
=
∂x
∂y
(so that v depends only on x) and
0=
∂u
∂v
=−
∂y
∂x
(so that v depends only on y)
...
106
MATH20101 Complex Analysis
10
...
Solutions to Part 3
Solution 3
...
Let
n
n
sn =
n
zk , xn =
k=0
Re(zk ), yn =
k=0
Im(zk )
k=0
∞
denote the partial sums of zn , Re(zn ), Im(zn ), respectively
...
k=0
Suppose that ∞ zn is convergent
...
Then there exists N such that if n ≥ N
n=0
we have |s − sn | < ε
...
k=0
k=0
Conversely, suppose that ∞ Re(zk ) and ∞ Im(zk ) exist
...
Choose N1
k=0
k=0
such that if n ≥ N1 then |x − xn | < ε/2
...
Then if n ≥ max{N1 , N2 } we have that
|z − zn | ≤ |x − xn | + |y − yn | < ε
...
Solution 3
...
an z n is given by 1/R =
(i) Here an = 2n /n so that
2n+1 n
2n
1
|an+1 |
=
=
→2=
n
|an |
n+12
n+1
R
as n → ∞
...
(ii) Here an = 1/n! so that
n!
1
1
|an+1 |
=
=
→0=
|an |
(n + 1)!
n+1
R
as n → ∞
...
107
MATH20101 Complex Analysis
10
...
Hence the radius of convergence is R = 0 and the series converges for
z = 0 only
...
Hence the radius of convergence is R = 1
...
3
To see that the expression in Proposition 3
...
2(i) does not converge, note that
n
2
if n is even,
an+1
3n+1
=
3n
an
if n is odd
...
Hence limn→∞ |an+1 /an | does not exist
...
2
...
|an |1/n =
Hence limn→∞ |an+1 /an | does not exist
...
Hence
∞
n=0
an z
n
≤
∞
n=0
zn
≤
2n
∞
z
2
n=0
n
,
which converges provided that |z/2| < 1, i
...
if |z| < 2
...
Solution 3
...
Hence
1
1−z
2
=
1
1−z
1
1−z
=
∞
n=0
zn
∞
zn
...
1
...
Hence
1
1−z
2
=
∞
n=1
108
nz n−1
...
Solutions to Part 3
(ii) Using Proposition 3
...
2 we see that
∞
zn
n!
n=0
where
∞
∞
wn
cn
=
n!
n=0
n=0
n
1
z r wn−r
...
Solution 3
...
Hence the radius of convergence of
this series is also R
...
Then the series becomes ∞ an wn
...
Changing back to z this says
that we get convergence for |z|3 < R and divergence for |z|3 > R, i
...
convergence for
|z| < R1/3 and divergence for |z| > R1/3
...
(iii) In this case
lim |a3 |1/n = lim |an |3/n =
n
n→∞
n→∞
Hence the radius of convergence is
lim |an |1/n
n→∞
3
=
1
...
Solution 3
...
Adding these expressions
gives 2 cos z = eiz + eiz so that cos z = (eiz + e−iz )/2
...
(iii) It is easier to start with the right-hand side:
sin z cos w + cos z sin w
1
1 iz
(e − e−iz )(eiw + e−iw ) + (eiz + e−iz )(eiw − e−iw )
=
4i
4i
1
i(z+w)
−i(z+w)
e
−e
=
2i
= sin(z + w)
...
Solutions to Part 3
(iv) Similarly,
cos z cos w − sin z sin w
1 iz
1
=
(e + e−iz )(eiw + e−iw ) − 2 (eiz − e−iz )(eiw − e−iw )
4
4i
1 i(z+w)
−i(z+w)
=
e
+e
2
= cos(z + w)
...
7
(i) We have
sin z = sin(x + iy)
1
ei(x+iy) − e−i(x+iy)
=
2i
1 ix −y
=
(e e − e−ix ey )
2i
1
=
((e−y cos x − ey cos x) + i(e−y sin x + ey sin x))
2i
e−y sin x + ey sin x
ey cos x − e−y cos x
=
+i
2
2
= sin x cosh y + i cos x sinh y
...
Now
∂v
∂u
= cos x cosh y,
= cos x cosh y,
∂x
∂y
and
∂u
∂v
= sin x sinh y,
= − sin x sinh y,
∂y
∂x
so that the Cauchy-Riemann equations are satisfied
...
Hence the real and imaginary parts of cos z are u(x, y) = cos x cosh y and v(x, y) =
− sin x sinh y, respectively
...
Solutions to Part 3
and
∂u
∂v
= cos x sinh y,
= − cos x sinh y,
∂y
∂x
so that the Cauchy-Riemann equations are satisfied
...
cos z = sin z +
(iii) Here we have that
sinh z = sinh(x + iy)
1 x+iy
=
e
− e−(x+iy)
2
1 x iy
(e e − e−x e−iy )
=
2
1 x
=
((e cos y − e−x cos y) + i(ex sin y + e−x sin y))
2
= sinh x cos y + i cosh x sin y
...
Now
∂u
∂v
= cosh x cos y,
= cosh x cos y,
∂x
∂y
and
∂u
∂v
= − sinh x sin y,
= sinh x sin y,
∂y
∂x
so that the Cauchy-Riemann equations are satisfied
...
)
(iv) Here we have that
cosh z = cosh(x + iy)
1 x+iy
e
+ e−(x+iy)
=
2
1 x iy
=
(e e + e−x e−iy )
2
1 x
((e cos y + e−x cos y) + i(ex sin y − e−x sin y))
=
2
= cosh x cos y + i sinh x sin y
...
Now
∂u
∂v
= sinh x cos y,
= sinh x cos y,
∂x
∂y
111
MATH20101 Complex Analysis
10
...
(Alternatively, using the results of (ii), one can use the fact that cosh z = cos iz to
derive this
...
8
(i) A complex-valued function takes real values if and only its imaginary part equals 0
...
As ex > 0 for all x ∈ R, this is zero if and
only if sin y = 0, i
...
y = kπ, k ∈ Z
...
e
...
2
Similarly, the imaginary part of cos z is − sin x sinh y and this equals zero if and only
if x = kπ, k ∈ Z, or y = 0
...
The imaginary part of sinh z is cosh x sin y and this equals zero if and only if y = kπ,
k ∈ Z (as cosh x > 0 for all x ∈ R)
...
Now ez = ex cos y + iex sin y has zero real part if and only if ex cos y = 0, i
...
if
cos y = 0
...
2
The real part of sin z is sin x cosh y and this equals zero if and only if sin x = 0, i
...
x = kπ, k ∈ Z
...
e
...
2
The real part of sinh z is sinh x cos y and this equals zero if and only if either x = 0
or y = π + kπ, k ∈ Z
...
π
2
+ kπ,
Solution 3
...
Then sin z = 0 if and only if both the real parts and imaginary parts of sin z
are equal to 0
...
We know
cosh y > 0 for all y ∈ R so the first equation gives x = kπ, k ∈ Z
...
e
...
Thus the solutions to sin z = 0 are z = kπ,
k ∈ Z
...
Now cos x cosh y = 0 implies cos x = 0 so that x = (k + 1/2)π, k ∈ Z
...
e
...
Hence the solutions to cos z = 0 are z = (k + 1/2)π, k ∈ Z
...
Solutions to Part 3
Solution 3
...
Then
ez = ex cos y + iex sin y = −1
...
As
ex > 0 for all x ∈ R the second equation gives that sin y = 0, i
...
y = kπ, k ∈ Z
...
When k is even this
gives ex = −1 which has no real solutions
...
e
...
Hence the solutions are z = (2k + 1)πi, k ∈ Z
...
Then
ez = ex cos y + iex sin y = 1 + i
and comparing real and imaginary parts gives
ex cos y = 1, ex sin y = 1
...
e
...
√
and so we have ex = 2; hence x = log 2
...
Hence
√
z = log 2 + i(π/4 + 2kπ), k ∈ Z
...
11
(i) Write z = x + iy
...
Putting z = 0 we get sin p = sin 0 = 0, so that p = kπ, k ∈ Z
...
Continuing inductively, we see that sin(z + nπ) = (−1)n sin z
...
Hence the periods of sin are p = 2πn, n ∈ Z
...
Putting z = 0 gives exp p = exp 0 = 1
...
Then
exp p = ex cos y + iex sin y = 1
and comparing real and imaginary parts gives ex cos y = 1, ex sin y = 0
...
e
...
The first equation
then gives (−1)n ex = 1
...
When n is even
this gives ex = 1, i
...
x = 0
...
Solution 3
...
We choose θ1 , θ2 ∈ (−π, π] to be the principal value
of the arguments of z1 , z2
...
Then
z1 z2 = r1 eiθ1 r2 eiθ2 = r1 r2 ei(θ1 +θ2 )
113
MATH20101 Complex Analysis
10
...
However, θ1 + θ2 ∈ (−2π, 2π] and is not necessarily a
principle value of the argument for z1 z2
...
Thus
Log z1 z2 = ln r1 r2 + i(θ1 + θ2 + 2πn)
= Log z1 + Log z2 + 2πn
for some integer n ∈ {−1, 0, 1}
...
Then |z1 | = |z2 | = 2 and Arg z1 = Arg z2 = 3π/4
...
Hence
Log z1 = Log z2 = ln
√
2+
πi
3πi
, Log z1 z2 = ln 2 −
4
2
so that
Log z1 + Log z2 = ln 2 +
3πi
...
(Any two complex numbers z1 , z2 where the principal values of the arguments of z1 , z2 add
to either more than π or less than −π will also work
...
13
Take b = z = i
...
Hence
π
π
=i
2
2
π
π
log(i) = ln(1) + i
+ 2nπ = i
+ 2nπ
...
2
Solution 3
...
2
...
Note that
lim
n→∞
α(α − 1) · · · (α − n + 1)(α − n)
n!
(n + 1)!
α(α − 1) · · · (α − n + 1)
Hence the power series has radius of convergence R = 1
...
Solutions to Part 3
(ii) Recall from Theorem 3
...
2 that a power series is holomorphic on its disc of convergence
and can be differentiated term by term
...
2!
n!
Multiply this by 1 + z
...
1
...
Clearly the constant
term of (1 + z)f ′ (z) is α
...
(iii) Let g(z) = f (z)/(1 + z)α
...
Hence g′ (z) = 0 on {z ∈ C | |z| < 1}
...
4
...
Noting that g(0) = 1 we have that g(z) = 1
for all z with |z| < 1
...
115
MATH20101 Complex Analysis
11
...
Solutions to Part 4
Solution 4
...
See Figure 11
...
(ii) Here γ(t) = x(t) + iy(t), x(t) = 1 + 2 cos t, y(t) = 1 + 2 sin t, 0 ≤ t ≤ 2π
...
e
...
See Figure 11
...
(iii) Here γ(t) = x(t) + iy(t) where x(t) = t, y(t) = cosh t, −1 ≤ t ≤ 1, i
...
y = cosh x
...
See Figure 11
...
(iv) Here γ(t) = x(t) + iy(t) where x(t) = cosh t, y(t) = sinh t, −1 ≤ t ≤ 1
...
e
...
See Figure 11
...
1+i
(i)
(ii)
(iii)
(iv)
Figure 11
...
1
...
2
Let f (z) = x − y + ix2 where z = x + iy
...
Solutions to Part 4
(i) The straight line from 0 to 1 + i has parametrisation γ(t) = t + it, 0 ≤ t ≤ 1
...
Hence
γ
1
x − y + ix2 =
0
1
=
0
=
=
it2 (1 + i) dt
−t2 + it2 dt
−1 3 i 3
t + t
3
3
i
−1
+
...
Here
γ ′ (t) = i and f (γ(t)) = −t
...
2
=
=
1
0
(iii) The line parallel to the real axis from i to 1 + i has parametrisation γ(t) = t + i,
0 ≤ t ≤ 1
...
Hence
γ
x − y + ix2 =
=
=
1
0
(t − 1 + it2 ) dt
t2
it3
−t+
2
3
−1
i
+
...
3
The path γ1 is the circle of radius 2, centre 2, described anticlockwise
...
(i) Let f (z) = 1/(z − 2)
...
Hence
γ1
dz
=
z−2
2π
0
117
1
2ieit dt = 2πi
...
Solutions to Part 4
(ii) Let f (z) = 1/(z − i)3
...
Hence
γ2
dz
=
(z − i)3
π/2
0
π/2
1
(−ie−it ) dt = −i
e−3it
e2it dt =
0
−i 2it
e
2i
π/2
0
= 1
...
4
The circle |z − 1| = 1 described anticlockwise has parametrisation γ(t) = 1 + eit = 1 +
cos t + i sin t, 0 ≤ t ≤ 2π
...
Note that
|γ(t)|2 = (1 + cos t)2 + sin2 t = 1 + 2 cos t + cos2 t + sin2 t = 2(1 + cos t)
...
=
2π
2 cos t +
0
Solution 4
...
We know that differentiation
for complex functions obeys the same rules (chain rule, product rule, etc) as for real
functions, so we first find an anti-derivative for the real function f (x) = x2 sin x
...
Let F (z) = −z 2 cos z + 2z sin z + 2 cos z
...
Hence, by the Fundamental Theorem of Contour Integration (Theorem 4
...
3), if γ is
any smooth path from 0 to i then
γ
f = F (i) − F (0) = −i2 cos i + 2i sin i + 2 cos i − 2 cos 0
= 3 cosh 1 − 2 sinh 1 − 2
...
Solutions to Part 4
(ii) Again, let us first find an anti-derivative for f (x) = xeix
...
Hence F (z) = −izeiz + eiz is an anti-derivative for f
...
Solution 4
...
Note that |γ1 (t)|2 = t2 , γ ′ (t) = i, |γ2 (t)|2 = t2 + 1, γ ′ (t) = 1
...
3 3
1
0
(ii) Similarly, the contour γ that goes horizontally from 0 to 1 and then vertically from 1
to 1 + i is the sum of the paths
γ3 (t) = t, 0 ≤ t ≤ 1
γ4 (t) = 1 + it, 0 ≤ t ≤ 1
...
Hence
γ
|z|2 dz =
γ3
1
=
|z|2 dz +
0
0
=
=
(1 + t2 )i dt
t2 dt +
1 3
t
3
γ4
1
|z|2 dz
1
+ it +
0
it3
3
1
0
1 4i
+
...
119
MATH20101 Complex Analysis
11
...
2: See Solution 4
...
Solution 4
...
2
...
8
Let γ : [a, b] → D be a parametrisation of the contour γ
...
Note that (−γ)′ (t) = γ ′ (a + b − t) = −(γ ′ (a + b − t)), using the chain
rule for differentiation
...
Solution 4
...
Using the formula for integration by parts
from real analysis, we can write
b
f g′ =
f (γ(t))g′ (γ(t))γ ′ (t) dt
a
γ
b
f (γ(t))
=
a
d
(g(γ(t))) dt
dt
120
MATH20101 Complex Analysis
=
11
...
= f (z1 )g(z1 ) − f (z0 )g(z0 ) −
γ
Solution 4
...
To use the Generalised Cauchy Theorem, we need a domain that
excludes these points
...
(There are lots of choices of D that will work
...
Alternatively, one could take D = C \ {±1}
...
Let γ3 (t) = 2e−it , 0 ≤ t ≤ 2π, i
...
γ3 is γ but described in the
opposite direction
...
If |z| ≥ 3 then w(γ1 , z) = w(γ2 , z) = w(γ3 , z) = 0
...
Similarly if |z + 1| ≤ 1/3 then w(γ1 , z) = 0, w(γ2 , z) = −1, w(γ3 , z) = +1
...
Furthermore, since ±1 ∈ D, the function f is holomorphic on D
...
Solution 4
...
Then
2π
f=
γ1
2π
f (γ(t))γ ′ (t) dt =
0
0
1 it
ie dt = 2πi
...
We apply the Generalised Cauchy Theorem to the contours γ2 , −γ1
...
Note that w(γ1 , 0) = 1 (so that w(−γ1 , 0) = −1) and
w(γ2 , 0) = 1
...
By the Generalised Cauchy Theorem we have that
f = 0
...
f=
−γ1
121
γ1
MATH20101 Complex Analysis
12
...
Solutions to Part 5
Solution 5
...
0
...
2
2(2n)!
n=1
(12
...
2)
As the radius of convergence for (12
...
1) is R = ∞, it follows that the radius of
convergence for (12
...
2) is also R = ∞
...
0
...
)
(ii) Here
∞
1
(−2z)n
= 1 − 2z + 4z 2 − 8z 3 + · · · =
1 + 2z
n=0
(by recognising this as a sum of a geometric progression)
...
2
...
(iii) We have
2
ez =
∞
z 2n
n!
n=0
and the radius of convergence is R = ∞
...
2
Let f (z) = Log(1 + z)
...
By Taylor’s Theorem, we can expand f as a Taylor series at 0 valid on
some disc centred at 0 as
f (z) =
∞
n=0
122
an z n
MATH20101 Complex Analysis
where an =
1 (n)
(0)
...
Solutions to Part 5
Here
f ′ (z) =
1
−1
, f ′′ (z) =
,
...
(1 + z)n
Hence an = (−1)n+1 (n − 1)!/n! = (−1)n+1 /n for n = 0 and a0 = 0
...
Solution 5
...
Let a ∈ C
...
Let q(z) = p(z) − a
...
By
the Fundamental Theorem of Algebra, there exists z0 ∈ C such that q(z0 ) = 0
...
Solution 5
...
Applying the bound on |f (z)| we have that Mr ≤ Kr k
...
r k+m
rm
Since this holds for r arbitrarily large, by letting r → ∞ we see that f (k+m) (0) = 0
...
Solution 5
...
Then
f (z) = |(x + 1) + iy|2 = (x + 1)2 + y 2
...
Hence
∂v
∂u
= 2(x + 1),
=0
∂x
∂y
and
∂u
= 2y,
∂y
∂v
= 0
...
(Note that the Cauchy-Riemann equations
do hold at the point z = −1 and that the partial derivatives are continuous at
x = −1, y = 0, hence by Proposition 2
...
2 f is differentiable at z = −1
...
123
MATH20101 Complex Analysis
12
...
Then z = 1/z
...
Let g(z) = (z + 1)2 /z
...
(iii) Let h(z) = (z + 1)2
...
124
MATH20101 Complex Analysis
13
...
Solutions to Part 6
Solution 6
...
Let f (z) = ∞
m=−∞ am z
(i) Since we are looking for an expansion valid for |z| > 3, we should look at powers of
1/z:
1
z−3
=
=
=
1/z
1 − (3/z)
1
3 32
1 + + 2 + ···
z
z z
3
32
3n
1
+ 2 + 3 + · · · + n+1 + · · ·
...
Hence
1
1
= + 1 + z + z2 + · · ·
z(1 − z)
z
is valid for 0 < |z| < 1
...
m
n
2
m!z
n!z
2!z
z
1
1
1
z
+ ··· +
+ + + z2 + z3
...
2!
4!
(2n)!
For z = 0 we have
cos 1/z = · · · +
1
1
(−1)n
+ ··· +
−
+ 1
...
2
Using partial fractions one can write
1
1
2(z − 1)
=
+
...
Solutions to Part 6
Now
∞
∞
1
1
(−1)n z n ,
(−z)n =
=
=
z+1
1 − (−z) n=0
n=0
(13
...
1)
summing a geometric progression with common ratio −z
...
e
...
We also have that
∞
1
1
1
1
=
=
z+1
z 1 − −1
z
z
n=0
−1
z
∞
n
=
(−1)n+1
n=1
1
,
zn
(13
...
2)
summing a geometric progression with common ratio −1/z
...
e
...
Similarly, we have that
1
−1
=
z−3
3
1
1−
z
3
−1
=
3
∞
n=0
z
3
n
(13
...
3)
and this is valid when |z| < 3
...
Hence when |z| < 1 we have the Laurent expansion
f (z) =
∞
n=0
(−1)n −
1
3n+1
zn
...
zn
z
z 3 3
3 +1
For |z| > 3 we have the Laurent expansion
f (z) =
∞
(−1)n+1 + 3n−1
n=1
1
...
3
Put w = z − 1
...
2
2
1−z+z
1+w+w
1 − w3
Now for |w| < 1 we can expand
1
= 1 + w3 + w6 + w9 + · · ·
...
0
...
Solutions to Part 6
Hence
1
1 − z + z2
1−w
1 − w3
= (1 − w)(1 + w3 + w6 + w9 + · · ·)
=
= 1 − w + w3 − w4 + w6 − w7 + · · ·
= 1 − (z − 1) + (z − 1)3 − (z − 1)4 + (z − 1)6 − (z − 1)7 + · · ·
provided |z − 1| < 1
...
4
Let f (z) = 1/(z − 1)2
...
Hence f has Laurent
series
1
f (z) =
(z − 1)2
valid on the annulus {z ∈ C | 0 < |z − 1|}
...
Therefore
we can apply Taylor’s theorem and expand f as a power series
f (z) = 1 + 2z + 3z 2 + · · · + (n + 1)z n + · · ·
valid on the disc {z ∈ C | |z| < 1}
...
Here we can easily compute that
n=0
f (n) (z) = (−1)n (n + 1)!(z − 1)−n−2 so that f (n) (0) = (n + 1)!
...
Alternatively, use the method given in Exercise 3
...
) As a Taylor series is a particular
case of a Laurent series, we see that f has Laurent series
f (z) = 1 + 2z + 3z 2 + · · · + (n + 1)z n + · · ·
valid on the disc {z ∈ C | |z| < 1}
...
Replacing z by 1/z in the first part of the computation in (ii) above, we see that
1
1−
1 2
z
=1+
2
3
n+1
+
+ ··· + n + ···
z z2
z
provided |1/z| < 1, i
...
provided |z| > 1
...
Solution 6
...
127
MATH20101 Complex Analysis
13
...
Now z 2 +1 = (z−i)(z+i)
so the denominator has zeros at z = ±i and both zeros are simple
...
(ii) The poles occur at the roots of the polynomial z 4 + 16 = 0
...
Then we
have
z 4 = r 4 e4iθ = −16 = 16eiπ
...
We get distinct values of z for k = 0, 1, 2, 3
...
All the poles are simple
...
The roots
of this polynomial are at z = ±i, each with multiplicity 2
...
√
(iv) The poles occur at the roots of z 2 + z − 1, i
...
at z = (−1 ± 5)/2, and both poles
are simple
...
6
(i) Since
sin
∞
1
1
(−1)2m+1
=
z m=0
(2m + 1)!z 2m+1
our function has infinitely many non-zero term in the principal part of its Laurent
series
...
(ii) By Exercise 5
...
2
(2n)!
24
26 3
1
−
z+
z − ···
z 2 · 4!
2 · 6!
so that z −3 sin2 z has a simple pole at z = 0
...
Hence 0 is a
removable singularity
...
Solutions to Part 6
Solution 6
...
Notice from Laurent’s Theorem (Theorem 6
...
1) that
bn =
1
2πi
Cr
f (z)(z − z0 )n−1 dz
where Cr is a circular path that lies on the domain D, centred at z0 of radius r > 0, and
described anticlockwise
...
As r is arbitrary and M is independent of r, we can let
r → 0 and conclude that bn = 0 for all n
...
129
MATH20101 Complex Analysis
14
...
Solutions to Part 7
Solution 7
...
The denominator vanishes whenz = 0, ±1 and these are all simple zeros
...
Then by Lemma 7
...
1(i) we have
1
1
= lim
= 1;
Res(f, 0) = lim z
2)
z→0 1 − z 2
z→0 z(1 − z
1
−1
−1
Res(f, 1) = lim (z − 1)
= lim
=
;
2)
z→1
z→1 z(1 + z)
z(1 − z
2
1
−1
1
= lim
=
...
Both sin z and cos z are differentiable on C, so f (z)
is differentiable except when the denominator is 0
...
e
...
These poles are simple (as
(n + 1/2)π is a simple zero of cos z)
...
4
...
− sin(n + 1/2)π
(iii) Let f (z) = (sin z)/z 2
...
By considering the Taylor expansion of sin z around 0 we have that
1
z3 z5
+
− ···
z−
z2
3!
5!
1
z
z2
=
− +
− ···
...
sin z
z2
=
(iv) Let f (z) = z/(1 + z 4 )
...
e
...
To solve this equation, we work in polar coordinates
...
Then
z 4 = −1 implies that r 4 e4iθ = eiπ
...
Hence the four
quartic roots of −1 are:
eiπ/4 , e3iπ/4 , e−iπ/4 , e−3iπ/4
...
Hence by Lemma 7
...
1(ii) we have that
2
3
Res(f, z0 ) = z0 /4z0 = 1/4z0 so that
−i
1
1
=
=
Res(f, eiπ/4 ) =
4i
4
4eiπ/2
i
1
1
=
Res(f, e3iπ/4 ) =
=
3π/2
−4i
4
4e
i
1
1
=
Res(f, e−iπ/4 ) =
=
−iπ/2
−4i
4
4e
1
−i
1
−3iπ/4
Res(f, e
) =
=
...
Solutions to Part 7
(v) Let f (z) = (z + 1)2 /(z 2 + 1)2
...
e
...
Note that we can write
f (z) =
(z + 1)2
...
By Lemma 7
...
2 (with m = 2) we
have that
d
(z − i)2 f (z)
dz
d (z + 1)2
lim
z→i dz (z + i)2
2(z + i)2 (z + 1) − 2(z + 1)2 (z − i)
lim
z→i
(z + i)4
2(2i)2 (i + 1) − 2(i + 1)2 (2i)
(2i)4
−i
...
=
2
Res(f, −i) =
lim
Solution 7
...
Note that
1
1
=
z 2 − 5z + 6
(z − 2)(z − 3)
so that f has simple poles at z = 2, z = 3
...
Hence
C4
1
dz = 2πi Res(f, 2) + 2πi Res(f, 3)
...
4
...
Res(f, 3) = lim
z→3 z − 2
z→3 (z − 2)(z − 3)
Res(f, 2) =
Hence
C4
z2
lim
1
dz = 2πi − 2πi = 0
...
Solutions to Part 7
(ii) Here we have the same integrand as in (i) but integrated over the smaller circle C5/2
...
Hence
C5/2
z2
1
dz = 2πi Res(f, 2) = −2πi
...
Note that
eaz
eaz
=
...
Now
eia
eaz
(z − i)eaz
= lim
=
z→i z + i
z→i (z − i)(z + i)
2i
az
az
(z + i)e
e
e−ia
Res(f, −i) = lim
= lim
=−
...
= 2πi
Solution 7
...
By Theorem 6
...
1 the coefficients an are given by
an =
1
2πi
Cr
f (z)
dz
(z − z0 )n+1
where Cr is a circular path described anticlockwise centred at z0 and with radius r, where
r is chosen such that R1 < r < R2
...
Here z0 = 0
...
We have that
an =
1
2πi
Cr
1
f (z)
dz =
n+1
z
2πi
1
Cr
z n+2 (z
− 1)
dz
where Cr is the circular path with centre 0 and radius r ∈ (0, 1), described once
anticlockwise
...
For all n ∈ Z the
integrand has a simple pole at 1
...
For n = −2, −3,
...
Hence, by
Cauchy’s Residue Theorem, an = 0 for n = −2, −3,
...
Solutions to Part 7
inside Cr
...
4
...
Here
Res
1
z n+2 (z
− 1)
1
dn+1
1
z n+2 n+2
z→0 (n + 1)! dz n+1
z
(z − 1)
dn+1
1
1
= lim
n+1
z→0 (n + 1)! dz
z−1
1
1
= lim
(−1)n (n + 1)!
= −1
...
Hence f has
Laurent series
1
f (z) = − − 1 − z − z 2 − z 3 − · · ·
z
valid on the annulus {z ∈ C | 0 < |z| < 1}
...
(ii) We calculate the Laurent series of f valid on the annulus {z ∈ C | 1 < |z| < ∞}
...
The integrand has, for all n ∈ Z, a simple pole at z = 1 and, for
n ≥ −1, a pole of order n + 2 at 0
...
We have already calculated, for n ≥ −1, the residue of the pole at 0
...
The residue of the pole at 1 is given by
Res
1
z n+2 (z
− 1)
,1
= lim (z − 1)
z→1
1
z n+2 (z
− 1)
= 1
...
n+2 (z − 1)
z
133
MATH20101 Complex Analysis
14
...
To check this directly, first observe that
1−
1
z
−1
=1+
1
1
+
+ ···
z z2
provided that |z| > 1, by summing the geometric progression
...
(iii) We calculate the Laurent series of f valid on the annulus {z ∈ C | 0 < |z − 1| < 1}
...
The integrand has, for all n ∈ Z, a simple pole
at 0 and, for n ≥ −1, a pole of order n + 2 at 1
...
Hence, by Cauchy’s Residue Theorem, an = 0 for n = −2, −3, −4,
...
Using Lemma 7
...
2 we have that
Res
1
,1
z(z − 1)n+2
dn+1
1
1
(z − 1)n+2
n+1
z→1 (n + 1)! dz
z(z − 1)n+2
dn+1 1
1
= lim
z→1 (n + 1)! dz n+1
z
1
1
(n + 1)! n+2 (−1)n+1 = (−1)n+1
...
z−1
To check this directly, it is convenient to change variables and let w = z − 1
...
Solutions to Part 7
where we have used the fact that
1
1
=
= 1 − w + w2 − w3 + · · · ,
1+w
1 − (−w)
summing the geometric progression
...
z−1
valid on the annulus {z ∈ C | 0 < |z − 1| < 1}
...
Hence z0 = 1 and an is given by
an =
1
2πi
Cr
1
f (z)
dz =
(z − 1)n+1
2πi
Cr
1
dz
z(z − 1)n+2
where Cr is the circular path with centre 1 and radius r, described once anticlockwise,
where r is chosen such that r ∈ (1, ∞)
...
Both of these poles lie inside
Cr
...
The residue of the pole at 0 is given by
Res
1
,0
z(z − 1)n+2
= lim z
z→0
1
= (−1)n+2 = (−1)n
...
When
n ≥ −1 we have that
an = Res
1
, 0 + Res
z(z − 1)n+2
1
,1
z(z − 1)n+2
= (−1)n + (−1)n+1 = 0
...
we have that
an = Res
1
,1
z(z − 1)n+2
= (−1)n
...
To see this directly we again change variables and let w = z − 1
...
Solutions to Part 7
for |w| > 1, by summing the geometric progression
...
Solution 7
...
Hence 1/(x2 + 1) ≤ 1/x2
...
5
...
(ii) Let R > 1
...
Let f (z) = 1/(z 2 + 1)
...
Only the pole at z = i lies inside ΓR (assuming
that R > 1)
...
z→i (z + i)(z − i)
z→i z + i
2i
Res(f, i) = lim
By Cauchy’s Residue Theorem,
f = 2πi Res(f, i) = 2πi
f=
f+
ΓR
SR
[−R,R]
1
= π
...
On SR we have
that
|z 2 + 1| ≥ |z 2 | − 1 = R2 − 1
...
Hence
f ≤
R2
∞
−∞
x2
πR
1
length(SR ) = 2
→0
−1
R −1
1
dx = lim
R→∞
+1
f = π
...
)
(b)
(i) Let f (z) = e2iz /(z 2 + 1)
...
2+1
x
|x + 1|
x
By Lemma 7
...
1, the integral is equal to its principal value
...
Solutions to Part 7
Note that
e2iz
e2iz
=
+1
(z − i)(z + i)
f (z) =
z2
so that f has simple poles at z = ±i
...
Only the pole at z = i lies inside this contour
...
1
...
Res(f, i) = lim
z→i z + i
z→i (z − i)(z + i)
2i
i
−R
R
−i
Figure 14
...
By Cauchy’s Residue Theorem,
f = 2πi Res(f, i) = 2πi
f=
f+
ΓR
SR
[−R,R]
e−2
= πe−2
...
On SR we have
that
|z 2 + 1| ≥ |z 2 | − 1 = |z|2 − 1 = R2 − 1
...
Then 0 ≤ y ≤ R, so −R ≤ −y ≤ 0
|e2iz | = |e2i(x+iy) | = |e−2y+2ix | = e−2y ≤ 1
...
Hence by the Estimation Lemma,
SR
as R → ∞
...
[−R,R]
(ii) Taking real and imaginary parts in the above we see that
∞
−∞
cos 2x
dx = πe−2 ,
x2 + 1
137
∞
−∞
sin 2x
dx = 0
...
0
...
Solutions to Part 7
That the latter integral is equal to zero is obvious and we do not need to use
complex integration to see this
...
(iii) Now consider f (z) = e−2iz /(z 2 + 1)
...
Then, with SR as the semi-circle
defined above, we would have to bound |f (z)| on SR in order to use the Estimation Lemma
...
We still have the bound 1/|z 2 + 1| ≤ 1/(R2 − 1)
...
′
Instead, we use a ‘D-shaped’ contour with the ‘negative’ semi-circle SR described
by
Re−it , 0 ≤ t ≤ π
...
Consider
the contour Γ′ which starts at −R travels along the real axis to R, and then
R
′
follows the negative semi-circle SR lying in the bottom half of the plane
...
However, −Γ′ is a simple closed loop and,
R
R
moreover,
Γ′
R
f =−
f
...
2
...
The only
pole inside Γ′ occurs at z = −i
...
z→−i z − i
z→−i (z + i)(z − i)
2i
Res(f, −i) = lim
Hence
−Γ′
R
f = 2πi Res(f, −i) = −πe−2
...
Solutions to Part 7
i
-R
R
-i
(i)
i
-R
R
-i
(ii)
Figure 14
...
Note that Γ′ is not
R
R
R
a simple closed loop but that −Γ′ is
...
Hence |f (z)| ≤ 1/(R2 − 1) for z on Γ′
...
Hence
f ≤
1
πR
′
length(SR ) = 2
→0
R2 − 1
R −1
R
−
−R
f (x) dx −
f=
′
SR
−Γ′
R
f = −πe−2
and letting R → ∞ gives that
∞
−∞
e−2ix
dx = πe−2
...
5
We will use the same notation as above: SR denotes the positive semicircle with centre 0
radius R, ΓR denotes the contour [−R, R] + SR
...
Solutions to Part 7
(i) Let f (z) = 1/(z 2 + 1)(z 2 + 3)
...
Hence by Lemma 7
...
1 the integral converges and equals its principal value
...
Suppose R > 3
...
Now
z−i
(z 2 + 1)(z 2 + 3)
1
lim
z→i (z + i)(z 2 + 3)
1
2i(−1 + 3)
1
4i
√
z−i 3
lim
√
2
2
z→i 3 (z + 1)(z + 3)
1
√
lim
√
2 + 1)(z + i 3)
z→i 3 (z
1
√
(−3 + 1)2i 3
−1
√
...
1− √
3
Now we show that the integral over SR tends to 0 as R → ∞
...
Hence by the Estimation Lemma
SR
f ≤
(R2
1
πR
length(SR ) =
→0
2 − 3)
2 − 1)(R2 − 3)
− 1)(R
(R
as R → ∞
...
140
1
1− √
3
...
Solutions to Part 7
Let f (z) = 1/(z 2 + 4)(z 2 + 7)
...
Hence by Lemma 7
...
1 the integral converges and equals its principal value
...
Suppose R > 7
...
Now
Res(f, 2i) =
=
=
=
√
Res(f, i 7) =
=
=
=
z − 2i
+ 4)(z 2 + 7)
1
lim
z→2i (z + 2i)(z 2 + 7)
1
4i(−4 + 7)
1
12i
√
z−i 7
lim
√
2
2
z→i 7 (z + 4)(z + 7)
1
√
lim
√
2 + 4)(z + i 7)
z→i 7 (z
1
√
(−7 + 4)2i 7
−1
√
...
2
7
Now we show that the integral over SR tends to 0 as R → ∞
...
Hence by the Estimation Lemma
SR
f ≤
(R2
1
πR
length(SR ) =
→0
2 − 7)
2 − 4)(R2 − 7)
− 4)(R
(R
as R → ∞
...
6
Let
f (z) =
eiz
...
MATH20101 Complex Analysis
14
...
Hence by Lemma 7
...
1 the integral −∞ f (x) dx
exists and is equal to its principal value
...
e
...
Both of these poles are
simple
...
Provided R is sufficiently
large, only the pole at −2 + i lies inside ΓR
...
2i
Hence
ΓR
SR
[−R,R]
f = 2πi
f=
f+
e−1−2i
2i
= πe−1 cos 2 − iπe−1 sin 2
...
Then 0 ≤ y ≤ R, so e−y ≤ 1
...
Also, |z 2 + 4z + 5| ≥ |z|2 − 4|z| − 5 = R2 − 4R − 5
...
− 4R + 5
By the Estimation Lemma,
SR
as R → ∞
...
Taking the imaginary part we see that
∞
−∞
x2
sin x
π sin 2
dx = −
...
7
We use the same notation as above: SR denotes the positive semi-circle with centre 0 and
radius R, ΓR denotes the contour [−R, R] + SR
...
(z 2 + a2 )(z 2 + b2 )
Note that |f (x)| ≤ C/|x|3 for some constant C > 0
...
5
...
142
MATH20101 Complex Analysis
14
...
e
...
If R is taken to be larger than b then the poles inside ΓR occur at z = ia, ib
...
2(b2 − a2 )
Res(f, ia) =
lim
Similarly,
(z − ib)zeiz
z→ib (z − ib)(z + ib)(z 2 + a2 )
zeiz
= lim
z→ib (z + ib)(z 2 + a2 )
ibe−b
=
2ib(−b2 + a2 )
−e−b
...
Now if z is a point on SR then |z| > R
...
Also, writing z = x + iy so that 0 ≤ y ≤ R, we have that |eiz | = |ei(x+iy) | = |e−y+ix | =
|e−y | ≤ 1
...
2 − a2 )(R2 − b2 )
(R
By the Estimation Lemma,
SR
f (z) dz ≤
R
πR2
length(SR ) =
(R2 − a2 )(R2 − b2 )
(R2 − a2 )(R2 − b2 )
which tends to zero as R → ∞
...
− a2 )
143
MATH20101 Complex Analysis
14
...
2 )(x2 + b2 )
+a
(b − a2 )
(As a check to see if we have made a mistake, note that the real part is zero
...
+ a2 )(x2 + b2 )
This is obvious as the integrand is an even function, and so must integrate (from −∞ to
∞) to zero
...
8
Denote by C the unit circle C(t) = eit , 0 ≤ t ≤ 2π
...
Then dz = ieit dt = iz dt so that dt = dz/iz and [0, 2π] transforms
to C
...
Hence
2π
2 cos3 t + 3 cos2 t dt =
C
0
(z + z −1 )3 3(z + z −1 )2
+
4
4
dz
...
4
=
=
Hence
2π
2 cos3 t + 3 cos2 t dt
0
=
C
1
i
1
3
3
3
3 2z z 2
+ 3+ 2+
+ +
+
4z 4 4z
4z
2z 4
4
4
dz
...
We can immediately read off the residue at z = 0 as the coefficient of 1/z,
namely 3/2i
...
2i
(ii) As before, substitute z = eit
...
Hence
2π
0
1
dt =
1 + cos2 t
C
1
dz
1
=
−1 )2 /4 iz
1 + (z + z
i
Let
f (z) =
z4
144
4z
...
Solutions to Part 7
This has poles where the denominator vanishes
...
Hence f (z) has poles where
√
√
−6 ± 36 − 4
2
z =
= −3 ± 2 2
...
Hence the poles at z = ±i 3 + 2 2 lie outside C
...
Hence there are poles at z = ±i 3 − 2 2 that lie inside C
...
We have that
√
Res f, i 3 − 2 2
=
=
=
=
√
(z − i 3 − 2 2)4z
lim
√
√
√
√ √
z→i 3−2 2 (z − i 3 − 2 2)(z + i 3 − 2 2)(z 2 − (−3 − 2 2))
4z
lim
√
√
√ √
z→i 3−2 2 (z + i 3 − 2 2)(z 2 − (−3 − 2 2))
√
4i 3 − 2 2
√
√
√
2i 3 − 2 2(−3 + 2 2 + 3 + 2 2)
1
√
2 2
and
√
Res f, −i 3 − 2 2
=
=
=
=
√
(z + i 3 − 2 2)4z
lim √
√
√
√
√
z→−i 3−2 2 (z + i 3 − 2 2)(z − i 3 − 2 2)(z 2 − (−3 − 2 2))
4z
lim √
√
√
√
z→−i 3−2 2 (z − i 3 − 2 2)(z 2 − (−3 − 2 2))
√
−4i 3 − 2 2
√
√
√
−2i 3 − 2 2(−3 + 2 2 + 3 + 2 2)
1
√
...
145
MATH20101 Complex Analysis
14
...
9
Let
1
cos πz
cot πz = 4
z4
z sin πz
Then f has poles when the denominator vanishes, i
...
poles at z = n, n ∈ Z
...
4
...
4 π cos πn
+n
πn4
Res(f, n) =
4n3 sin πn
When z = 0, we use the expansion for cot z:
cot z =
1 z
z3
2z 5
− −
−
− ···
...
Hence Res(f, 0) = −π 3 /45
...
5
...
The poles at z = −N,
...
, N
lie inside CN
...
By Lemma 7
...
2, we have for z on CN
|f (z)| ≤
M
M
≤ 4
...
By the Estimation Lemma,
CN
as N → ∞
...
=
n4
90
n=1
∞
3
n=1 1/n
...
Summing over the residues we get
1/πn
N
n=−N
1
=
πn3
−1
n=−N
1
+
πn3
146
N
n=1
1
+0
πn3
MATH20101 Complex Analysis
14
...
So the residues on the negative integers
cancel with the residues at the positive integers
...
In fact, there is no
known closed formula for ∞ 1/n3
...
wikipedia
...
)
n=1
Solution 7
...
z 2 + a2
As
f (z) =
(z 2
cos πz
+ a2 ) sin πz
this has poles where the denominator vanishes, i
...
poles at z = ±ia and at z = n, n ∈ Z
...
We can calculate
Res(f, ia) =
=
=
=
=
(z − ia) cos πz
(z − ia)(z + ia) sin πz
cos πz
lim
z→ia (z + ia) sin πz
cos iπa
2ia sin iπa
cosh πa
−2a sinh πa
− coth πa
2a
lim
z→ia
using the facts that cos iz = cosh z, sin iz = i sinh z
...
2a
lim
For z = n, we use Lemma 7
...
1(ii) to see that
Res(f, n) =
cos πn
1
=
...
This is because
f (z) does not have a pole at z = 0 and so we have a simple pole at z = 0 for f (z) cot πz
...
Solutions to Part 7
ia
−(N + 1) −N
−1
N
0 1
N +1
−ia
Figure 14
...
, −1, 0, 1,
...
Let CN denote the square contour as described in §7
...
4; see Figure 14
...
If N > |a|
then CN encloses the poles at z = −N,
...
, N and z = ±ia
...
Hence, by the bound on
cot πz from Lemma 7
...
2, and the Estimation Lemma we have that
CN
f ≤
4M (2N + 1)
N 2 − a2
(as length(CN ) = 4(2N + 1)), which tends to zero as N → ∞
...
Solutions to Part 7
and rearranging this gives
∞
n=1
n2
1
1
π
coth πa − 2
...
11
(i) Let f (z) = ez /z
...
Hence f (z) has a
simple pole at z = 0
...
4
...
z→0
z
Noting that 0 lies inside C1 , Cauchy’s Residue Theorem tells us that
C1
ez
dz = 2πi Res(f, 0) = 2πi
...
Then dz = ieit dt = iz dt
...
Hence, noting that z = eit = cos t + i sin t,
ez
dz
z
2πi =
C1
2π
=
0
ecos t+i sin t it
ie dt
eit
2π
= i
ecos t+i sin t dt
0
2π
= i
ecos t ei sin t dt
0
2π
= i
ecos t (cos(sin t) + i sin(sin t)) dt
0
2π
= −
2π
ecos t sin(sin t) dt + i
ecos t cos(sin t) dt
0
0
Comparing real and imaginary parts gives the claimed integrals
...
12
(i) Note that ex < 1 + ex so that 1/(1 + ex ) < 1/ex
...
1 + ex
As a ∈ (0, 1), we have that a − 1 < 0
...
Hence e(a−1)x ≤ C/x2 , provided x > 1
...
5
...
149
MATH20101 Complex Analysis
14
...
Then f is holomorphic except when the denominator vanishes
...
The denominator vanishes precisely when ex+iy = −1
...
The solutions of eiy = −1 are precisely
y = (2k + 1)π, k ∈ Z
...
Write f (z) = p(z)/q(z) with p(z) = eaz , q(z) = 1 + ez
...
So (2k + 1)πi is a simple zero of q
...
From Lemma 7
...
1(ii), the residue at πi is p(πi)/q ′ (πi) = −eaπi
...
4
...
4
...
4: The poles of f (z) = eaz /(1 + ez ) and the contour ΓR
...
Hence, by Cauchy’s Residue Theorem,
ΓR
f = 2πi Res(f, πi) = −2πie−aπi
...
Then
R
f=
−R
γ1,R
150
eat
dt
1 + et
MATH20101 Complex Analysis
14
...
γ1,R
(v) First note that length(γ2,R ) = 2π
...
Then, for z on γR,2 we have
|ea(R+it) |
eaR
≤ R
R+it |
e −1
t∈[0,2π] |1 + e
|f (z)| ≤ sup
(where we have used the reverse triangle inequality to bound the denominator)
...
The case of γR,4 is similar
...
If z is a point on γR,4 then
z = −R + it for some 0 ≤ t ≤ 2π
...
−R+it |
1 − e−R
t∈[0,2π] |1 + e
|f (z)| ≤ sup
Hence, by the Estimation Lemma,
γ4,R
f ≤
2πe−aR
→0
1 − e−R
as R → ∞, as 0 < a < 1
...
dx = −
=
x
2πia
1+e
1−e
sin πa
151
f
Title: complex analysis
Description: This is the note of complex analysis in university of Manchester.
Description: This is the note of complex analysis in university of Manchester.