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Complex Analysis Lecture Notes
Dan Romik
About this document
...
The current 2020 revision (dated June 15, 2021) updates my earlier version
of the notes from 2018
...
M
...
Shakarchi (Princeton University Press, 2003)
...
Acknowledgements
...
Figure 5
on page 27 was created by Jennifer Brown and is used with her permission
...
You too can help me continue to improve these notes by emailing me at
romik@math
...
edu with any comments or corrections you have
...
Created with Mathematica using code by Simon Woods, available at
http://mathematica
...
com/questions/7275/how-can-i-generate-this-domain-coloring-plot
Contents
1 Introduction: why study complex analysis?
1
2 The fundamental theorem of algebra
3
3 Analyticity
7
4 Power series
13
5 Contour integrals
16
6 Cauchy’s theorem
21
7 Consequences of Cauchy’s theorem
26
8 Zeros, poles, and the residue theorem
35
9 Meromorphic functions and the Riemann sphere
38
10 The argument principle
41
11 Applications of Rouché’s theorem
45
12 Simply-connected regions and Cauchy’s theorem
46
13 The logarithm function
50
14 The Euler gamma function
52
15 The Riemann zeta function
59
16 The prime number theorem
71
17 Introduction to asymptotic analysis
79
Problems
92
Suggested topics for course projects
119
References
121
Index
122
1
1
1
INTRODUCTION: WHY STUDY COMPLEX ANALYSIS?
Introduction: why study complex analysis?
These notes are about complex analysis, the area of mathematics that studies
analytic functions of a complex variable and their properties
...
First, it is, in my
humble opinion, one of the most beautiful areas of mathematics
...
e
...
”
The second reason is complex analysis has a large number of applications
(in both the pure math and applied math senses of the word) to things that
seem like they ought to have little to do with complex numbers
...
An important point to keep in
mind is that Cardano’s formula sometimes requires taking operations
in the complex plane as an intermediate step to get to the final answer,
even when the cubic equation being solved has only real roots
...
Using Cardano’s formula, it can be found that the solutions
to the cubic equation
z 3 + 6z 2 + 9z + 3 = 0
are
z1 = 2 cos(2π/9) − 2,
z2 = 2 cos(8π/9) − 2,
z3 = 2 sin(π/18) − 2
...
Here, an ∼ bn is the standard “asymptotic to” relation, defined to mean limn→∞ an /bn = 1
...
2
1
INTRODUCTION: WHY STUDY COMPLEX ANALYSIS?
• Proving many other asymptotic formulas in number theory and combinatorics, e
...
(to name one other of my favorite examples), the HardyRamanujan formula
√
1
p(n) ∼ √ eπ 2n/3 ,
4 3n
where p(n) is the number of integer partitions of n
...
2
(This application is strongly emphasized in older textbooks, and has
been known to result in a mild case of post-traumatic stress disorder
...
• Analyzing alternating current electrical networks by extending Ohm’s
law to electrical impedance
...
• Probability and combinatorics, e
...
, the Cardy-Smirnov formula in percolation theory and the connective constant for self-avoiding walks on
the hexagonal lattice
...
The proofs
make spectacular use of complex analysis (and more specifically, a
part of complex analysis that studies certain special functions known
as modular forms)
...
This is not a mere mathematical convenience or sleight-ofhand, but in fact appears to be a built-in feature of the very equations
describing our physical universe
...
• Conformal maps, which come up in purely geometric applications where
the algebraic or analytic structure of complex numbers seems irrelevant, are in fact deeply tied to complex analysis
...
C
...
See Fig
...
3
2
THE FUNDAMENTAL THEOREM OF ALGEBRA
Figure 1: Print Gallery, a lithograph by M
...
Escher which was discovered to
be based on a mathematical structure related to a complex function z 7→ z α
for a certain complex number α, although it was constructed by Escher
purely using geometric intuition
...
• Complex dynamics, e
...
, the iconic Mandelbrot set
...
2
...
(If you run across some interesting
ones, please let me know!)
In the next section I will begin our journey into the subject by illustrating
a few beautiful ideas and along the way begin to review the concepts from
undergraduate complex analysis
...
This seems like a fitting place to
start our journey into the theory
...
Every nonconstant
polynomial p(z) over the complex numbers has a root
...
[Source: Wikipedia]
The fundamental theorem of algebra is a subtle result that has many
beautiful proofs
...
Let me know if you see
any “algebra”
...
Let
p(z) = an z n + an−1 z n−1 +
...
First, note that it can’t happen as |z| → ∞, since
|p(z)| = |z|n · (|an + an−1 z −1 + an−2 z −2 +
...
Fixing some radius R > 0 for which |z| > R implies
|p(z)| ≥ |a0 |, we therefore have that
m0 := inf |p(z)| = inf |p(z)| = min |p(z)| = |p(z0 )|
z∈C
|z|≤R
|z|≤R
where z0 = arg min |p(z)|, and the minimum exists because p(z) is a continu|z|≤R
ous function on the disc DR (0)
...
We now claim that m0 = 0
...
+ cn (z − z0 )n ,
5
2
THE FUNDAMENTAL THEOREM OF ALGEBRA
where k is the minimal positive index for which cj 6= 0
...
What happens to p(z)? Well, the
expansion gives
p(z0 + reiθ ) = w0 + ck rk eikθ + ck+1 rk+1 ei(k+1)θ +
...
When r is very small, the power r k dominates the other terms r j with k < j ≤
n, i
...
,
p(z0 + reiθ ) = w0 + rk (ck eikθ + ck+1 rei(k+1)θ +
...
To reach a contradiction, it is now enough to
choose θ so that the vector ck r k eikθ “points in the opposite direction” from
w0 , that is, such that
ck rk eikθ
∈ (−∞, 0)
...
It follows that, for
r small enough,
|w0 + ck rk eikθ | < |w0 |
and for r small enough (possibly even smaller than the previous small r )
|p(z0 + reiθ )| = |w0 + ck rk eikθ (1 + g(r, θ))| < |w0 |,
a contradiction
...
Exercise 1
...
Second proof: topological proof
...
If w0 = 0 , we are done
...
Specifically:
1
...
2
...
+ r e
p(re ) = an r e
1+
an
an
n inθ
= an r e (1 + h(r, θ))
iθ
n inθ
6
2
THE FUNDAMENTAL THEOREM OF ALGEBRA
where limr→∞ h(r, θ) = 0 (uniformly in θ )
...
As we gradually increase r from 0 to a very large number, in order to transition from a curve that doesn’t go around the origin to a curve that goes
around the origin n times, there has to be a value of r for which the curve
crosses 0
...
Remark 1
...
It can be made rigorous in a couple of ways — one way we will see a bit
later is using Rouché’s theorem and the argument principle
...
Remark 2
...
That
argument is also “topological” (based on the mean value theorem), although
much more trivial
...
Recall:
Theorem 2 (Liouville’s theorem
...
Assuming this result, if p(z) is a polynomial with no root, then 1/p(z)
is an entire function
...
It follows that 1/p(z) is a constant,
lim|z|→∞ |p(z)|
|z|n
which then has to be 0, which is a contradiction
...
They are all beautiful — the “hocus-pocus” proof certainly
packs a punch, which is why it is a favorite of complex analysis textbooks
— but personally I like the first one best since it is elementary and doesn’t
use Cauchy’s theorem or any of its consequences, or subtle topological concepts
...
” It is a general philosophical principle in analysis
(that has analogies in other areas, such as number theory and graph theory)
that local arguments are easier than global ones
...
3
...
A function f (z) of a complex variable is holomorphic (a
...
a
...
In this case we call f 0 (z) the derivative of f at z
...
Geometrically, this means that to compute f (w), we start
from f (z), and move by a vector that results by taking the displacement vector w − z , rotating it by an angle of θ , and then scaling it by a factor of r
(which corresponds to a magnification if r > 1, a shrinking if 0 < r < 1, or
doing nothing if r = 1)
...
”
1
Note: some people use “analytic” and “holomorphic” with two a priori different definitions that are then proved to be equivalent; I find this needlessly confusing so I may use
these two terms interchangeably
...
A further interpretation of the meaning of analyticity is that analytic functions are conformal mappings where their derivatives don’t vanish
...
If we denote by θ (resp
...
w1 , w2 ), it then
follows that
cos ϕ =
|f 0 (z)|2 hv1 , v2 i
hv1 , v2 i
hw1 , w2 i
= 0
=
= cos θ
...
A function with this property is
said to be conformal at z
...
Thus the theory of analytic functions contains the theory of planar conformal maps as a special (and largely
equivalent) case, although this is by no means obvious from the purely geometric definition of conformality
...
Lemma 1
...
Exercise 2
...
2 on page 10 of [11])
...
2
3
ANALYTICITY
The Cauchy-Riemann equations
In addition to the geometric picture associated with the definition of the
complex derivative, there is yet another quite different but also extremely
useful way to think about analyticity, that provides a bridge between complex analysis and ordinary multivariate calculus
...
Now, if f is analytic at z then
f (z + h) − f (z)
h→0
h
v(x + h + iy) − v(x + iy)
u(x + h + iy) − u(x + iy)
+i
= lim
h→0, h∈R
h
h
∂u
∂v
=
+i
...
∂y
∂y
∂y
∂y
f 0 (z) = lim
Since these limits are equal, by equating their real and imaginary parts we
get a famous system of coupled partial differential equations, the CauchyRiemann equations:
∂u
∂v
=
,
∂x
∂y
∂v
∂u
=−
...
Conversely, we now claim
if f = u + iv is continuously differentiable at z = x + iy (in the sense that
each of u and v is a continuously differentiable function of x, y as defined in
ordinary real analysis) and satisfies the Cauchy-Riemann equations there, f
is analytic at z
...
The assumption implies that f has a differential at z , i
...
, in the notation of vector calculus, denoting f = (u, v), z = (x, y)> , ∆z = (h1 , h2 )> , we
have
!
f (z + ∆z) =
∂u
∂x
∂v
∂x
u(z)
+
v(z)
∂u
∂y
∂v
∂y
h1
h2
+ E(h1 , h2 ),
where E(h1 , h2 ) = o(|∆z|) as |∆z| → 0
...
So, we have shown that (again, in complex analysis notation)
f (z + ∆z) − f (z)
lim
= lim
∆z→0
∆z→0
∆z
∂u E(∆z)
∂u
−i
+
∂x
∂y
∆z
=
∂u
∂u
−i
...
∂y
∂u
∂x
−
With the help of the Cauchy-Riemann equations, we can now prove our
earlier claim that conformality implies analyticity
...
If f = u + iv is conformal at z , continuously differentiable in the
real analysis sense, and satisfies det Jf > 0 (i
...
, f preserves orientation as a
planar map), then f is holomorphic at z
...
In the notation of the proof above, we have as before that
u(z)
f (z + ∆z) =
+
v(z)
∂u
∂x
∂v
∂x
∂u
∂y
∂v
∂y
!
h1
+ E(h1 , h2 ),
h2
where E(h1 , h2 ) = o(|∆z|) as |∆z| → 0
...
So the theorem will follow once we prove the
simple claim about 2 × 2 matrices contained in Lemma 2 below
...
Assume that A =
ANALYTICITY
a b
is a 2 × 2 real
c d
matrix
...
a b
(b) A takes the form A =
for some a, b ∈ R with a2 + b2 > 0
...
(That
sin θ cos θ
is, geometrically A acts by a rotation followed by a scaling
...
Note that both columns of A are nonzero vectors by
the assumption that det A > 0
...
On the other hand, applying the conformality assumption with w1 = (1, 1)> and w2 = (1, −1)> yields that (a + b, c + d) ⊥ (a − b, c − d),
which is easily seen to be equivalent to a2 + c2 = b2 + d2
...
So A is of one of the two forms
a −c
c a
or
a c
...
Exercise 3
...
Another curious consequence of the Cauchy-Riemann equations, which
gives an alternative geometric picture to that of conformality, is that analyticity implies the orthogonality of the level curves of u and of v
...
Since ∇u (resp
...
the level
curve {v = d}, this proves that the level curves {u = c}, {v = d} meet at right
angles whenever they intersect
...
(c) shows the superposition of both families of level curves
...
0
1
...
5
0
...
0
0
0
...
5
-1
-0
...
0
-1
...
0
-0
...
5
1
...
0
-2
-1
(a)
0
1
2
(b)
-1
...
5
0
...
5
1
...
x
x2 +y 2
−
can be removed) the functions u, v are harmonic functions
...
Then
∂ 2u ∂ 2u
∂ ∂u
∂ ∂u
+
=
+
∂x2 ∂y 2
∂x ∂x
∂y ∂y
∂ ∂v
∂ ∂v
∂ 2v
∂ 2v
=
−
=
−
= 0,
∂x ∂y
∂y ∂x
∂x∂y ∂y∂x
i
...
, u satisfies Laplace’s equation
4u = 0,
2
2
∂
∂
where 4 = ∂x
2 + ∂y 2 is the two-dimensional Laplacian operator
...
∂x2 ∂y 2
13
4
POWER SERIES
That is, we have shown that u and v are harmonic functions
...
We will later see that the assumption of twice continuous differentiability
is unnecessary, but proving this requires some subtle complex-analytic ideas
...
This can also be understood geometrically
...
Of course, there are the standard functions that you probably
encountered already in your undergraduate studies: polynomials, rational
functions, ez , the trigonometric functions, etc
...
Of course, there is such a way: power series, which—nonobviously—turn out to be essentially as general a family of functions as one
could hope for
...
These functions are defined wherever the respective series converge
...
n→∞
R is called the radius of convergence of the power series
...
Assume 0 < R < ∞ (the edge cases R = 0 and R = ∞ are left as
an exercise)
...
Let z ∈ DR (0)
...
That implies that for n > N (for some large
enough N as a function of ),
∞
X
n
|an z | <
n=N
∞
X
1
R
n=N
n
+ |z| ,
so the series is dominated by a convergent geometric series, and hence converges
...
Taking a subsequence (ank )∞
k=1 for which |ank | > R −
to exist by the definition of R), we see that
∞
X
n
|an z | ≥
n=0
∞
X
|z|
k=1
nk
1
−
= ∞,
R
so the power series diverges
...
Complete the argument in the extreme cases R = 0, ∞
...
Power series are holomorphic functions in the interior of the
disc of convergence and can be differentiated termwise
...
Denote
f (z) =
SN (z) =
EN (z) =
∞
X
n=0
N
X
an z n = SN (z) + EN (z),
an z n ,
n=0
∞
X
an z n ,
n=N +1
g(z) =
∞
X
n=1
nan z n−1
...
Since n1/n → 1 as n → ∞, it is easy to see that f (z)
and g(z) have the same radius of convergence
...
We
f (z0 +h)−f (z0 )
wish to show that
converges to g(z0 ) as h → 0
...
To bound the second
term, fix some > 0, and note that, if we assume that not only |z0 | < r but
also |z0 + h| < r (an assumption that’s clearly satisfied for h close enough to
0) then
∞
X
EN (z0 + h) − EN (z0 )
≤
|an |
h
n=N +1
(z0 + h)n − z0n
h
P
∞
h n−1 hk (z + h)n−1−k
X
0
k=0
=
|an |
h
n=N +1
≤
∞
X
|an |nrn−1 ,
n=N +1
where we use the algebraic identity
an − bn = (a − b)(an−1 + an−2 b +
...
The last expression in this chain of inequalities is the tail of an absolutely
convergent series, so can be made < be taking N large enough (before
taking the limit as h → 0)
...
Finally, having thus chosen N , we get that
f (z0 + h) − f (z0 )
− g(z0 ) ≤ 0 + + = 2
...
f (z0 +h)−f (z0 )
h
→ g(z0 )
The proof above can be thought of as a special case of the following more
conceptual result: if gn is a sequence of holomorphic functions on a region Ω,
16
5
CONTOUR INTEGRALS
and gn → g uniformly on closed discs in Ω, gn0 → h uniformly on closed discs
on Ω, and h is continuous, then g is holomorphic and g 0 = h on Ω
...
)
Corollary 1
...
Corollary 2
...
n!
In other words g(z) satisfies Taylor’s formula
g(z) =
∞
X
g (n) (z0 )
n=0
5
n!
(z − z0 )n
...
Contour integrals, like many other types of integrals, take as input a function to be integrated and a “thing” (or “place”) over which the function is
integrated
...
We start by developing some terminology to discuss such objects
...
The value γ(a) is called the starting point and γ(b) is called the ending point
...
A “curve” is an equivalence class of
parametrized curves with respect to this equivalence relation
...
(Meta-exercise: think of 2–3 other examples
of this phenomenon
...
More
generally, one can assume them to be rectifiable, but we will not bother to
develop that theory
...
You probably encountered curves and parametrized curves in your earlier
studies of multivariate calculus, where they were used to define the notion
of line integrals of vector and scalar fields
...
The line integral of the first kind of a scalar (usually real-valued)
function u(z) over a curve γ is defined as
n
X
Z
u(z)ds =
γ
lim
max ∆sj →0
u(zj )∆sj
(line integral of the first kind),
j=1
j
where the limit is a limit of Riemann sums with respect to a family of partitions of the interval [a, b] over which the curve γ is defined, as the norm of the
partitions shrinks to 0
...
< tn = b,
the points zj = f (tj ) are their images on the curve γ , and the symbols ∆sj
refer to finite line elements, namely ∆sj = |zj − zj−1 |
...
It is well-known from calculus that line integrals can be expressed in
terms of ordinary (single-variable) Riemann integrals
...
18
5
CONTOUR INTEGRALS
As a further reminder, the basic result known as the fundamental theorem
of calculus for line integrals states that if F = ∇u then
Z
F · ds = u(γ(b)) − u(γ(a))
...
For a function
f = u + iv of a complex variable z and a curve γ , define
Z
γ
Z
f (z) dz = “ (u + iv)(dx + idy)”
γ
Z
Z
=
u dx − v dy + i
v dx + u dy
Z
γ
b
γ
f (γ(t))γ 0 (t) dt
Z
Za
Z
Z
f (z) |dz| = f (z) ds = u ds + i v ds
=
γ
γ
γ
(contour integral),
(arc length integral)
...
e
...
A special case of an arc length integral is the length of the curve, defined
as the integral of the constant function 1:
b
Z
Z
|γ 0 (t)| dt
...
Indeed: if γ1 ∼ γ2 are
representatives of the same equivalence class of parametrized curves, that
is, γ2 (t) = γ1 (I(t)) for some nicely-behaved function, then using a standard
change of variables in single-variable integrals we see that
Z
Z
d
f (γ2 (t))γ20 (t)dt
f (z) dz =
γ2
c
Z
Z
=
d
f (γ1 (I(t)))(γ1 ◦ I)0 (t) dt
c
d
=
f (γ1 (I(t)))γ10 (I(t))I 0 (t) dt
Zc
=
=
a
f (z) dz
...
Show that the integral with respect to arc length similarly does
not depend on the parametrization
...
Contour integrals satisfy
the following properties:
(a) Linearity as an operator on functions:
β
R
γ
R
γ
(αf (z)+βg(z)) dz = α
R
γ
f (z) dz+
g(z) dz
...
γ2
Similarly, if γ2 is the “reverse” contour of γ1 , then
Z
Z
f (z) dz = −
γ2
f (z) dz
...
z∈γ
γ
Exercise 6
...
Contour integrals have their own version of the fundamental theorem of
calculus
...
If
γ is a curve connecting two points w1 , w2 in a region Ω on which a function F
is holomorphic, then
Z
F 0 (z) dz = F (w2 ) − F (w1 )
...
k
...
anti-derivative) of f , that is, a
γ
function F such that F 0 (z)R = f (z) on all of Ω
...
20
5
CONTOUR INTEGRALS
Proof
...
For piecewise smooth curves, this is a trivial extension that is left as an
exercise
...
As an example of this
interplay, the above result has an easy — but important — consequence for
integrals over closed contours
...
If f = F 0 where F is holomorphic on a region Ω (in that case
we say that f has a primitive), γ is a closed curve in Ω, then
I
f (z) dz = 0
...
if f : Ω → C is a continuous function on a region Ω such that
I
f (z) dz = 0
γ
holds for any closed contour in Ω, then f has a primitive
...
Fix some z0 ∈ Ω
...
Define
Z
F (z) =
f (w) dw
...
We now claim that
F is holomorphic and its derivative is equal to f
...
When |h| is
sufficiently small so that the disc D(z, h) is contained in Ω, one can take
γ(z, z + h) as the straight line segment connecting z and z + h
...
Remark 3
...
This is in
fact true, and is called Morera’s theorem (and is an important fact in complex analysis), but we won’t be able to prove it until we’ve proved Cauchy’s
theorem
...
Compute |z|=1 z n dz for n ∈ Z
...
If f is holomorphic on Ω and f 0 ≡ 0 then f is a constant
...
Fix some z0 ∈ Ω
...
Then
Z
f (z) − f (z0 ) =
f 0 (w) dw = 0,
γ(z0 ,z)
hence f (z) ≡ f (z0 ), so f is constant
...
Theorem 6 (Cauchy’s theorem
...
γ
22
6
CAUCHY’S THEOREM
The challenges facing us are: first, to prove Cauchy’s theorem for curves
and regions that are relatively simple (where we do not have to deal with
subtle topological considerations); second, to define what simply-connected
means; third, which will take a bit longer and we won’t do immediately, to
extend the theorem to the most general setting
...
Theorem 7 (Goursat’s theorem)
...
Theorem 8 (Morera’s theorem)
...
Proof of Goursat’s theorem
...
” Namely, try to translate a global statement about
the integral around the triangle to a local statement about behavior near a
specific point inside the triangle, which would become manageable since we
have a good understanding of the local behavior of a holomorphic function
near a point
...
The idea can be made more precise using triangle subdivision
...
...
,jn , 1 ≤ j1 ,
...
order n triangles:
...
23
6
CAUCHY’S THEOREM
(n)
Here, the triangles Tj1 ,
...
,jn−1 into 4 subtriangles whose vertices are the ver(n−1)
tices and/or edge bisectors of Tj1 ,
...
Now, given the way this subdivision was done, it is clear that we have the
equality
I
f (z) dz =
(n−1)
1 ,
...
,jn
∂Tj
jn =1
due to cancellation along the internal edges, and hence
I
f (z) dz =
∂T (0)
4
X
I
f (z) dz
...
,jn =1
(n)
1 ,
...
Now, the crucial
observation is that one of these integrals has to have a modulus that is at
least as big as the average
...
,jn =1
I
I
f (z) dz ≤ 4n
f (z) dz
(n)
∂T (n)
∂Tj ,
...
, jn ) is some n-tuple chosen such that the second inequality holds
...
, jn−1
, k) for some 1 ≤ k ≤ 4 — to make this happen,
H
choose k to be such that ∂T (n)
f (z) dz is greater than (or equal to) the
(j(n−1),k)
average
I
4
1 X
f (z) dz ,
(n)
∂T
4
lently j(n) = (j1
d=1
(j(n−1),d)
which in turn is (by induction) greater than or equal to
4 I
I
I
1 X
−(n−1)
f (z) dz =
f (z) dz ≥ 4
f (z) dz
...
That is, we have
∞
\
n=0
(n)
Tj(n) = {z0 }
24
6
CAUCHY’S THEOREM
for some point z0 ∈ T
...
g
...
Having defined z0 , write f (z) for z near z0 as
f (z) = f (z0 ) + f 0 (z0 )(z − z0 ) + ψ(z)(z − z0 ),
where
ψ(z) =
f (z) − f (z0 )
− f 0 (z0 )
...
Denote by d(n)
(n)
the diameter of Tj(n) and by p(n) its perimeter
...
It follows that
Z
Z
0
(n) f (z) dz = (n) f (z0 ) + f (z0 )(z − z0 ) + ψ(z)(z − z0 ) dz
∂T
∂T
j(n)
Z j(n)
=
ψ(z)(z − z0 ) dz ≤ p(n) d(n) sup |ψ(z)|
(n)
∂T
(n)
z∈T
j(n)
j(n)
H
Finally, combining this with the relationship between
∂T (0)
R
f (z) dz and | ∂T (n) f (z) dz|,
j(n)
we get that
Z
∂T
f (z) dz ≤ p(0) d(0) sup |ψ(z)| −−−→ 0,
n→∞
(0)
(n)
z∈Tj(n)
which finishes the proof
...
But first, let us note a few not-veryamazing consequences
...
Theorem 7 is also true
when we replace the word “triangle” with “rectangle
...
Obvious: a rectangle can be decomposed as the union of two triangles, with the contour integral around the rectangle being the sum of the
integrals around the two triangles due to cancellation of the integrals going
in both directions along the diagonal (see Prop 1(b))
...
If f is holomorphic on a disc D , then f = F 0 for some holomorphic function
F on D
...
The idea is similar to the proof of Proposition 2 above
...
As it is, we know this is true but only for triangular contours
...
Then one shows in
three steps, each involving a use of Goursat’s theorem (see Figure 4 on page
38 of [11]), that F (z0 + h) − F (z0 ) is precisely the contour integral over the
line segment connecting z0 and z0 + h
...
Corollary
6 (Cauchy’s theorem for a disc
...
Proof
...
Theorem 9 (Cauchy’s
theorem for a region enclosed by a “toy contour”)
...
2
Proof
...
The difficulty as the toy contour gets more complicated
is to make sure that the geometry works out when proving the existence of
the primitive — see for example the (incomplete) discussion of the case of
“keyhole contours” on pages 40–41 of [11]
...
26
7
7
CONSEQUENCES OF CAUCHY’S THEOREM
Consequences of Cauchy’s theorem
Theorem 10 (Cauchy’s integral formula)
...
Proof
...
It remains to deal with the case z ∈ D
...
The idea is now to consider instead the integral
I
I
Fz (w) dw =
Γ,δ
Γ,δ
f (w)
dw,
w−z
where Γ,δ is a so-called keyhole contour, namely a contour comprised of
a large circular arc around z0 that is a subset of the circle C , and another
smaller circular arc of radius centered at z , with two straight line segments
connecting the two circular arcs to form a closed curve, such that the width
of the “neck” of the keyhole is δ (think of δ as being much smaller than );
see Fig
...
Note that the function Fz (w) is holomorphic inside the region
enclosed by Γ,δ , so Cauchy’s theorem for toy contours (assuming you can be
convinced that the keyhole contour is a toy contour) gives that
I
Fz (w) dw = 0
...
So we can conclude that
I
I
Fz (w) dw =
Fz (w) dw
...
Integrating each term separately, we have for the first term
I
f (w) − f (z)
≤ 2π · sup |f (w) − f (z)|
dw
w−z
|w−z|=
C
= 2π sup |f (w) − f (z)| −−→ 0,
→0
|w−z|=
by continuity of f ; and for the second term,
I
1
f (z) ·
dw = f (z)
w−z
|w−z|=
I
|w−z|=
1
dw = 2πif (z)
w−z
(by aHstandard calculation, see Example 2 above)
...
that C 2πi
Example 3
...
0
In other words, we have proved:
Theorem 11 (the mean value property for holomorphic functions)
...
28
7
CONSEQUENCES OF CAUCHY’S THEOREM
Considering what the mean value property means for the real and imaginary parts of f = u + iv , which are harmonic functions, we see that they in
turn also satisfy a similar mean value property:
1
u(x, y) =
2π
Z
2π
u(x + r cos t, y + r sin t) dt
...
Theorem 12 (Cauchy’s integral formula, extended version)
...
(w − z)n+1
The fact that holomorphic functions are differentiable infinitely many
times is referred to by [11] as the regularity theorem
...
The key observation is that the expression on the right-hand side of
Cauchy’s integral formula for f (z) (which is the case n = 0 of the “extended”
version) can be differentiated under the integral sign
...
(w − z)n
Then
f (n−1) (z + h) − f (n−1) (z)
h
I
1
1
(n − 1)!
1
=
f (w) ·
−
dw
...
(The same claim without the
uniformity is just the rule for differentiation of a power function; to get the
uniformity one needs to “go back to basics” and repeat the elementary algebraic calculation that was originally used to derive this power rule — an
29
7
CONSEQUENCES OF CAUCHY’S THEOREM
illustration of the idea that in mathematics it is important not just to understand results but also the techniques used to derive them
...
Proof of Morera’s theorem
...
The regularity property now implies that the derivative F 0 = f is also holomorphic, hence f is holomorphic, which was the claim of Morera’s theorem
...
Theorem 13 (Cauchy inequalities)
...
Theorem 14 (Analyticity of holomorphic functions)
...
Proof
...
Each such function has a power series expansion
since it is, more or less, a geometric series, so the sum also has a power
series expansion
...
w − z0 n=0 w − z0
n=0
This is a power series in z − z0 that, assuming w ∈ CR (z0 ), converges absolutely for all z such that |z − z0 | < R (that is, for all z ∈ DR (z0 ))
...
Since infinite summations
that are absolutely and uniformly convergent can be interchanged with integration operations, we then get, using the extended version of Cauchy’s
integral formula, that
I
f (w)
dw
CR (z0 ) w − z
I
∞
X
1
=
(w − z0 )−n−1 (z − z0 )n dw
f (w)
2πi CR (z0 )
n=0
I
∞
X
1
n−1
=
f (w)(w − z0 )
dw (z − z0 )n
2πi
CR (z0 )
n=0
1
f (z) =
2πi
=
∞
X
f (n) (z0 )
n=0
n!
(z − z0 )n ,
which is precisely the expansion we were after
...
In the above proof, if we only knew the simple (n = 0) case of
Cauchy’s integral formula (and in particular didn’t know the regularity theorem that follows from the extended case of this formula), we would still
conclude from the penultimate expression in the above chain of equalities
P
n
that f (z) hasR a power series expansion of the form
n an (z − z0 ) , with
an = (2πi)−1 CR (z0 ) f (w)(w − z)−n−1
...
Theorem 15 (Liouville’s theorem)
...
Proof
...
31
7
CONSEQUENCES OF CAUCHY’S THEOREM
Theorem 16
...
Proof
...
We know that in a neighborhood of z0 , f has a convergent power series
expansion
...
It follows that for all k ,
am (wk − z0 )m (1 + g(wk )) = f (wk ) = 0,
but for large enough k this is impossible, since wk − z0 6= 0 for all k and
g(wk ) → g(z0 ) = 0 as k → ∞
...
But now we claim that that also implies that f is identically zero on all of
Ω, because Ω is a region (open and connected)
...
It is
obvious that U is open, hence also relatively open in Ω since Ω itself is open;
U is also closed, by the argument above; and U is nonempty (it contains z0 ,
again by what we showed above)
...
An alternative way to finish the proof is the following
...
Thus the discs {Dr(z) (z) : z ∈ Ω} form an open covering of
Ω
...
The open covering of Ω by discs is also an open covering of the
compact set γ[a, b] (the range of γ )
...
, m} (where we take
w = zm+1
...
It follows that we can get all the way
to the last disc Dr(w) (w)
...
Remark 5
...
In other words, the set of zeros of f
contains only isolated points
...
The condition that the limit point z0 be in Ω is needed
...
For example, consider the function e1/z − 1 — it has zeros in every
neighborhood of z0 = 0
...
If f, g are holomorphic on a region Ω, and f (z) = g(z) for z in a
set with limit point in Ω (e
...
, an open disc, or even a sequence of points zn
converging to some z ∈ Ω), then f ≡ g everywhere in Ω
...
Apply the previous result to f − g
...
If f is holomorphic on a
region Ω, and f+ is holomorphic on a bigger region Ω+ ⊃ Ω and satisfies
f (z) = f+ (z) for all z ∈ Ω, then f+ is the unique such extension, in the sense
that if f˜+ is another function with the same properties then f+ (z) = f˜+ (z) for
all z ∈ Ω+
...
In real analysis, we learn that “formulas” such as
1
,
2
1 + 2 + 4 + 8 + 16 + 32 +
...
=
don’t have any meaning
...
= −
1 − 2 + 3 − 4 +
...
4
33
7
CONSEQUENCES OF CAUCHY’S THEOREM
Such supposedly “astounding” formulas have attracted a lot of attention
in recent years, being the subject of a popular Numberphile video, a New
York Times article, a discussion on the popular math blog by Terry Tao, a
Wikipedia article, a discussion on Mathematics StackExchange, and more
...
A point z0 ∈ Ω is called a removable singularity of a function f : Ω → C∪{undefined} if f is holomorphic in a punctured
neighborhood of Ω, is not holomorphic at z0 , but its value at z0 can be redefined so as to make it holomorphic at z0
...
If f is holomorphic in Ω except at a point z0 ∈ Ω (where it may be undefined, or be defined
but not known to be holomorphic or even continuous)
...
Then f can be extended to a holomorphic function f˜ on all of Ω by defining (or redefining) its
value at z0 appropriately
...
Fix some disc D = DR (z0 ) around z0 whose closure is contained in Ω
...
Once we show this, we will set f˜(z0 ) to be
defined by the same integral representation, and it will be easy to see that
that gives the desired extension
...
After applying a limiting argument similar to the one
used in the proof of Cauchy’s integral formula, we get that
1
2πi
I
C
f (w)
1
=
w−z
2πi
I
C (z)
f (w)
1
+
w − z 2πi
I
C (z0 )
f (w)
...
The
second term can be bounded in magnitude using the assumption that f is
bounded in a neighborhood of z0 ; more precisely, we have
I
C (z0 )
f (w)
1
≤ 2π sup |f (w)| ·
−−→ 0
...
Finally, once we have the integral representation f˜ (defined only in terms
of the values of f (w) for w ∈ CR (z0 )), the fact that this defines a holomorphic function on all of DR (z0 ) is easy to see, and is something we implicitly
were aware of already
...
Another
approach is to show that integrating f˜ over closed contours gives 0 (which
requires interchanging the order of two integration operations, which will
not be hard to justify) and then use Morera’s theorem
...
Definition 3 (Uniform convergence on compact subsets)
...
Theorem 19
...
Proof
...
More precisely, note that
for each closed disc D = Dr (z0 ) ⊂ Ω we have fn (z) → f uniformly on D
...
γ
R
By Cauchy’s theorem, the integrals in this sequence are all 0, so γ f (z) dz
is also zero
...
This was true for any disc in Ω, and holomorphicity is a local
property, so in other words f is holomorphic on all of Ω
...
for z ∈ D
we have by Cauchy’s integral formula that
fn0 (z)
I
I
1
fn (w)
1
f (w)
− f (z) =
dw
−
dw
2πi ∂D (w − z)2
2πi ∂D (w − z)2
I
1
fn (w) − f (w)
=
dw
...
Finally, let K ⊂ D be compact
...
The family of discs
{Dz = Dr(z)/2 (z) : z ∈ Ω} is an open covering of K , so by the Heine-Borel
property of compact sets it has a finite subcovering Dz1 ,
...
We showed
that fn0 (z) → f 0 (z) uniformly on every Dzj , so we also have uniform convergence on their union, which contains K , so we get that fn0 → f 0 uniformly on
K , as claimed
...
z0 is a zero of a holomorphic function f if f (z0 ) = 0
...
If f is a holomorphic function on a region Ω that is
not identically zero and z0 is a zero of f , then f can be represented in the
form
f (z) = (z − z0 )m g(z)
in some neighborhood of z0 , where m ≥ 1 and g is a holomorphic function in
that neighborhood such that g(z) 6= 0
...
e, z0 will be described as a zero of
order m
...
Remark 7
...
Proof of the definition-lemma
...
That is, write the power series expansion (known to converge in a neighborhood of z0 )
f (z) =
∞
X
n
an (z − z0 ) =
n=0
= (z − z0 )m
∞
X
an (z − z0 )n
n=m
∞
X
n=0
am+n (z − z0 )n =: (z − z0 )m g(z),
36
8
ZEROS, POLES, AND THE RESIDUE THEOREM
where m is the smallest index ≥ 0 such that am 6= 0
...
On the other hand, given a representation of this form, expanding g(z) as a power series shows that m has to be the smallest index of
a nonzero coefficient in the power series expansion of f (z), which proves the
uniqueness claim
...
If f is defined and holomorphic in a punctured neighborhood of a point z0 , we say that it has a pole of order m at z0 if the function
h(z) = 1/f (z) (defined to be 0 at z0 ) has a zero of order m at z0
...
Remark 8
...
If f (z) is actually
holomorphic and nonzero at z0 (or has a removable singularity at z0 and can
be made holomorphic and nonzero by defining its value at z0 appropriately),
we define the order of the pole as 0 and consider f to have a pole of order 0
at z0
...
f has a pole of order m at z0 if and only if it can be represented
in the form
f (z) = (z − z0 )−m g(z)
in a punctured neighborhood of z0 , where g is holomorphic in a neighborhood
of z0 and satisfies g(z0 ) 6= 0
...
Apply the previous lemma to 1/f (z)
...
If f has a pole of order m at z0 , then it can be represented
uniquely as
f (z) =
a−m+1
a−1
a−m
+
+
...
...
Proof
...
a−m+1
+ (z−z
m−1 +
...
The coefficient a−1 is called the residue of f at z0 and denoted Resz0 (f )
...
The expansion
a−m
(z−z0 )m
37
8
ZEROS, POLES, AND THE RESIDUE THEOREM
Exercise 7
...
Denote the order of a zero of f at z0 by ordz0 (f )
...
Theorem 21 (The residue theorem (simple version))
...
Then
I
f (z) dz = 2πi Resz0 (f )
...
By the standard argument involving a keyhole contour, we see that
the circle C = ∂D in the integral can be replaced with a circle C = C (z0 ) of
a small radius > 0 around z0 , that is , we have
I
I
f (z) dz
...
Integrating
termwise gives 0 for the integral of G(z), by Cauchy’s theorem; 0 for the
integral powers (z − z0 )k with −m ≤ k ≤ −2, by a standard computation;
and 2πia−1 = 2πi Resz0 (f ) for the integral of r(z − z0 )−1 , by the same standard
computation
...
Theorem 22 (The residue theorem (extended version))
...
, zN ∈ D
...
38
9
MEROMORPHIC FUNCTIONS AND THE RIEMANN SPHERE
Proof
...
(Note: The above argument seems slightly dishonest to me, since it relies on the assertion that a multiple keyhole contour with arbitrary many
keyholes is a “toy contour”; while this is intuitively plausible, it will be undoubtedly quite difficult to think of, and write, a detailed proof of this argument
...
Assume that f is holomorphic in a region containing a toy contour γ (oriented
in the positive direction) and the region Rγ enclosed by it, except for poles
at the points z1 ,
...
Then
I
f (z) dz = 2πi
γ
N
X
Reszk (f )
...
Again, construct a multiple keyhole version of the same contour γ
(assuming that one can believably argue that the resulting contour is still a
toy contour), and then use a limiting argument to conclude that
I
f (z) dz =
γ
N I
X
k=1
f (z) dz,
C (zk )
for a small enough
...
9
Meromorphic functions, holomorphicity at ∞
and the Riemann sphere
Definition 8 (meromorphic function)
...
39
9
MEROMORPHIC FUNCTIONS AND THE RIEMANN SPHERE
Definition 9 (holomorphicity at ∞)
...
A function f : U → C
is holomorphic at ∞ if g(z) = f (1/z) (defined on a neighborhood D1/R (0) of
0) has a removable singularity at 0
...
Definition 10 (order of a zero/pole at ∞)
...
We say that a
function f : U → C has a zero (resp
...
pole) at z = 0, after appropriately defining the value of g at
0
...
Let’s define what that
phic functions with range in the Riemann sphere C
means
...
The Riemann sphere (a
...
a
...
sets in C
...
The identification is via stereographic projection, given explicitly by the
formula
(
x + iy =
(X, Y, Z) ∈ S 2 −
7 →
∞
X
1−Z
Y
+ i 1−Z
if (X, Y, Z) 6= (0, 0, 1),
if (X, Y, Z) = (0, 0, 1)
...
One can check that
this geometric identification is a homeomorphism between S 2 (equipped
b (with the one-point
with the obvious topology inherited from R3 ) and C
compactification topology defined above)
...
The details can be found in many
textbooks and online resources, and we will not discuss them in this
course
...
b ∪
Definition 12 (classification of singularities)
...
We classify
singularities into three types, two of which we already defined:
• Removable singularities: when f can be made holomorphic at z0 by
defining or redefining its value at z0 ;
• poles;
• any singularity that is not removable or a pole is called an essential
singularity
...
Theorem 24 (Casorati-Weierstrass theorem on essential singularities)
...
Proof
...
Then g(z) = f (z)−w
is a function that’s holomorphic and bounded in Dr (z0 )\
{z0 }
...
It then follows that
f (z) = w +
1
g(z)
has either a pole or a removable singularity at z0 , depending on whether
g(z0 ) = 0 or not
...
The logarithmic derivative of a holomorphic function f (z)
is f 0 (z)/f (z)
...
The logarithmic derivative of a product is the sum of the logarithmic derivatives
...
f
k (z)
k=1 fk
k=1
Proof
...
Theorem 25 (the argument principle)
...
Denote its zeros and poles inside D by z1 ,
...
Then
1
2πi
I
∂D
n
X
f 0 (z)
dz =
mk
f (z)
k=1
= [total number of zeros of f inside D]
− [total number of poles of f inside D]
...
Define
g(z) =
n
Y
(1)
(z − zk )−mk f (z)
...
, zn (so after
redefining its values at these points we can assume it is holomorphic on D )
...
k=1
Taking the logarithmic derivative of this equation gives that
n
f 0 (z) X mk
g 0 (z)
=
+
...
42
10
THE ARGUMENT PRINCIPLE
Remark 9
...
The proof above hides, as some slick mathematical proofs have a way of
doing, the fact that the formula (1) has a fairly simple intuitive explanation
...
Now note that the differential form dw/w has a special geometric meaning
in complex analysis, namely we have
dw
= “d (log w) ” = “d (log |w| + i arg w) ”
...
However, at least log |w| is well-defined for a curve that does not cross
0, so when integrating over the closed curve f ◦ γ , the real part is zero by
the fundamental theorem of calculus
...
e
...
Since the curve is closed, the total change in
the argument must be an integer multiple of 2π , so the division by 2πi turns
it into an integer
...
Connection to winding numbers
...
This number is more properly called the
winding number of f around w = 0 (also sometimes referred to as the
index of the curve around 0), and denoted
1
Ind0 (f ) =
2πi
I
γ
f (z)
dz
...
Note that winding number is a topological concept of planar geometry
that can be considered and studied without any reference to complex analysis; indeed, in my opinion that is the correct approach
...
Try to think what such a definition might look
like
...
Assume that f, g are holomorphic on a
region Ω containing a circle γ = C and its interior (or, more generally, a toy
contour γ and the region U enclosed by it)
...
Proof
...
Denote
1
nt =
2πi
I
γ
ft0 (z)
dz,
ft (z)
which by the argument principle is the number of “generalized zeros” (zeros
or poles, counting multiplicities) of ft in U
...
If we also knew that it was continuous, then it would have
to be constant (by the easy exercise: any integer-valued continuous function
on an interval [a, b] is constant), so in particular we would get the desired
conclusion that n1 = n0
...
For s, t ∈ [0, 1] satisfying |t − s| < δ , we can write
I
1
|nt − ns | ≤
|g(t, z) − g(s, z)| · |dz|
2πi γ
1
≤
len(γ) sup{|g(u, z) − g(v, z)| : z ∈ γ, u, v ∈ [0, 1], |u − v| < δ}
...
This is precisely what is needed to
show that t 7→ nt is continuous
...
The slogan to remember is “walking the dog”
...
You start at some point X
and go for a walk along some curve, ending back at the same starting point
X
...
Now imagine that you also have a dog that is walking alongside you in
some erratic path that is sometimes close to you, sometimes less close
...
Let M denote the dog’s winding
number around the pole at the origin
...
To see this, imagine that you had the dog
on a leash of variable length; if the distance condition was not satisfied, it
would be possible for the dog to reach the pole and go in a short tour around
it while you were still far away and not turning around the pole, causing an
entanglement of the leash with the pole
...
Exercise 8
...
You will probably forget the technical
details of the proof of Rouché’s theorem in a few weeks or months, but I
hope you will remember this intuitive explanation for a long time
...
Can you see
how?
11
Applications of Rouché’s theorem
Rouché’s theorem is an important tool in estimating the numbers of roots of
polynomials and other functions in regions of interests (see Exercises 28–29
on page 109)
...
Theorem 27 (the open mapping theorem)
...
Proof
...
What we need to show is that the image of any neighborhood D (z0 ) for > 0
contains a neighborhood Dδ (z0 ) of w0 for some δ > 0
...
The idea is now to apply Rouché’s theorem to F (z) and G(z)
...
Defining
δ = inf{|f (z) − w0 | : z ∈ D (z0 )},
we therefore have that δ > 0 and |f (z) − w0 | ≥ δ for z on the circle |z − z0 | =
...
e
...
The conclusion is that the equation h(z) = 0
46
12
SIMPLY-CONNECTED REGIONS AND CAUCHY’S THEOREM
(or equivalently f (z) = w) has the same number in solutions (in particular,
at least one solution) as the equation f (z) = w0 in the disc D (z0 )
...
Corollary 8 (the maximum modulus principle)
...
Proof
...
At the beginning of the course we discussed several proofs of the fundamental theorem of algebra
...
As a final demonstration of the power of Rouché’s theorem, problem 30 on page 109 asks you to use the theorem to make precise the idea of
the topological proof
...
Given a region Ω ⊂ C, two parametrized
curves γ1 , γ2 : [0, 1] → Ω (assumed for simplicity of notation to be defined on
[0, 1]) are said to be homotopic (with fixed endpoints) if γ1 (0) = γ2 (0),
γ1 (1) = γ2 (1), and there exists a function F : [0, 1] × [0, 1] → Ω such that
i) F is continuous
...
iii) F (1, t) = γ2 (t) for all t ∈ [0, 1]
...
v) F (s, 1) = γ1 (1) for all s ∈ [0, 1]
...
Intuitively, for each
s ∈ [0, 1] the function t 7→ F (s, t) defines a curve connecting the two endpoints
γ1 (0), γ1 (1)
...
Exercise 9
...
47
12
SIMPLY-CONNECTED REGIONS AND CAUCHY’S THEOREM
Definition 15 (simply-connected region)
...
Remark 10
...
The definition
of a simply-connected region then becomes a region in which any two closed
curves are homotopic
...
Theorem 28
...
γ0
γ1
Proof
...
(See
also pages 93–95 in [11] for a variant of the proof presented below
...
The strategy
of the proof is to show that there are values 0 = s0 < s1 < s2 <
...
f (z) dz =
f (z) dz =
...
To this end, for two fixed values 0 ≤ s <
s0 ≤ 1 that are close to each other (in a sense we will make precise shortly),
we break up the t-interval [0, 1] over which the curves s, s0 are defined into
very small segments by fixing points 0 = t0 < t1 <
...
γs0 ([tj−1 ,tj ])
We will now show that these two integrals are equal by exploiting our knowledge of local properties of f that follow from its analyticity
...
For each 0 ≤ j ≤ m, let ηj denote a straight line segment
(considered as a parametrized curve) from γs (tj ) to γs0 (tj ), and for each 1 ≤
j ≤ m let Γj denote the closed curve γs ([tj−1 , tj ]) + ηj − γs0 ([tj−1 , tj ]) − ηj−1 (the
concatenation of the four curves γs ([tj−1 , tj ]), “the reverse of ηj ”, γs0 ([tj−1 , tj ]),
and “the reverse of ηj−1 ”)
...
ηj
Summing this relation over j , we get that
Z
Z
f (z) dz −
γs
f (z) dz =
γs0
Z
m
X
j=1
=
j=1
f (z) dz −
f (z) dz −
f (z) dz
ηj
ηj−1
Z
f (z) dz −
η0
!
Z
Z
=
f (z) dz
γs0 ([tj−1 ,tj ])
γs ([tj−1 ,tj ])
Z
m
X
!
Z
f (z) dz = 0,
ηm
since in the next-to-last step the sum is telescoping, and in the last step
we note that η0 and ηm are both degenerate curves each of which simply
stays at a single point (γs (0) = F (s, 0) = γs0 (0), and γs (1) = F (s, 1) = γs0 (1),
respectively)
...
We still need to justify the assumption about the discs Dj
...
< tm = 1 are sufficiently densely spaced, using an argument
involving continuity and compactness
...
At each 0 ≤ t ≤ 1, by continuity of the homotopy function
F there exists a number δ > 0 and a disc Ds,t centered at γs (t) = F (s, t) such
that for any s0 , t0 ∈ [0, 1] with |s0 − s| < , |t0 − t| < δ , we have γs0 (t0 ) ∈ Ds,t
...
, Ds,tm
...
Finally, for each s denote by δ(s) the value of δ chosen above as a function
of s
...
This enables us to find a sequence 0 =
s0 < s1 <
...
, n (with sj−1 playing the role of s and sj playing the
role of s0 in the discussion above)
...
If f is holomorphic on a
simply-connected region Ω, then for any closed curve in Ω we have
I
f (z) dz = 0
...
Assume for simplicity that γ is parametrized as a curve on [0, 1]
...
Note that γ1 and γ2
have the same endpoints
...
γ2
Corollary 9
...
Exercise 10
...
Can you identify those inconsistencies? What additional work might
be needed to fix these problems? And why do you think the author of these
notes, and the authors of the textbook [11], chose to present things in this
way rather than treat the subject in a completely rigorous manner devoid of
any inaccuracies? (The last question is a very general one about mathematical pedagogy; coming up with a good answer might help to demystify for you
a lot of similar decisions that textbook authors and course instructors make
in the teaching of advanced material, and make the study of such topics a bit
less confusing in the future
...
It is easy to see that this
formula gives an inverse to the exponential function
...
This is called the principal branch
of the logarithm — basically a kind of standard version of the log function
that complex analysts have agreed to use whenever this is convenient (or
not too inconvenient)
...
When can this
be made to work? The answer is: precisely when Ω is simply-connected
...
Assume that Ω is a simply-connected region with 0 ∈
/ Ω, 1 ∈ Ω
...
ii) eF (z) = z for all z ∈ Ω
...
Proof
...
By the
general version of Cauchy’s theorem for simply-connected regions, this integral is independent of the choice of curve
...
It follows that
d
ze−F (z) = e−F (z) − zF 0 (z)e−F (z) = e−F (z) (1 − z/z) = 0,
dz
51
13
THE LOGARITHM FUNCTION
so ze−F (z) is a constant function
...
Finally, for real r close to 1 we have that F (z) = 1 w =
log r, which can be seen by taking the integral to be along the straight line
segment connecting 1 and r
...
Prove that the principal branch of the logarithm has the Taylor
series expansion
Log z =
∞
X
(−1)n−1
n=1
n
(z − 1)n
(|z| < 1)
...
Modify the proof above to prove the existence of a branch of
the logarithm function in any simply-connected region Ω not containing 0,
without the assumption that 1 ∈ Ω
...
Explain in what sense the logarithm functions F (z) = logΩ (z)
satisfying the properties proved in the theorem above (and its generalization
described in the previous exercise) are unique
...
Prove the following generalization of the logarithm construction above: if f is a holomorphic function on a simply-connected region Ω,
and f 6= 0 on Ω, then there exists a holomorphic function g on Ω, referred to
as a branch of the logarithm of f , satisfying
eg(z) = f (z)
...
On a simply-connected region Ω we can now define the power function z 7→ z α for an arbitrary α ∈ C
by setting
z α = eα logΩ z
...
) = e
1/n n
Note that if f (z) = z 1/n is one choice of an nth root function, then for any 0 ≤
k ≤ n − 1, the function g(z) = e2πik/n f (z) will be another function satisfying
g(z)n = z
...
52
14
14
THE EULER GAMMA FUNCTION
The Euler gamma function
The Euler gamma function (often referred to simply as the gamma function)
is one of the most important special functions in mathematics
...
It is a natural meromorphic function of a
complex variable that extends the factorial function to non-integer values
...
Most textbooks define the gamma function in one way and proceed to
prove several other equivalent representations of it
...
So, it seems more logical to start by simply
listing the various formulas and properties associated with it, and then proving that the different representations are equivalent and that the claimed
properties hold
...
There exists a unique function
Γ(s) of a complex variable s that has the following properties:
1
...
2
...
3
...
4
...
0
5
...
53
14
THE EULER GAMMA FUNCTION
6
...
1
2
+ 13 +
...
− log n = 0
...
Limit of finite products representation:
n! ns
n→∞ s(s + 1) · · · (s + n)
(s ∈ C)
...
Zeros: the gamma function has no zeros (so Γ(s)−1 is an entire function)
...
Poles: the gamma function has poles precisely at the non-positive integers s = 0, −1, −2,
...
The pole at
s = −n is a simple pole with residue
Ress=−n (Γ) =
(−1)n
n!
(n = 0, 1, 2,
...
The reflection formula (a surprising connection to trigonometry):
Γ(s)Γ(1 − s) =
π
sin(πs)
(s ∈ C)
...
This improper integral is easily seen to
converge absolutely for Re(s) > 0, since
Z
0
∞
−x s−1
e x
Z
dx ≤
0
∞
−x
e |x
s−1
Z
| dx =
0
∞
e−x xRe(s)−1 dx
...
Next, perform an integration by parts, to get that, again for Re(s) > 0, we
have
∞
Z
−x s
e x dx =
Γ(s + 1) =
0
Z
x=∞
−e−x xs x=0
∞
+
e−x sxs−1 dx = s Γ(s),
0
which is the functional equation
...
√
The special value Γ(1/2) = π follows immediately by a change of varix = u2 in√the integral and an appeal to the standard Gaussian integral
Rable
2
∞
e−u du = π :
−∞
Z
∞
Γ(1/2) =
−x −1/2
e x
Z
dx =
0
∞
−u2
e
0
Z
∞
2 du =
2
e−u du =
√
π
...
By the principle of analytic continuation this provides
a unique extension of Γ(s) to the region Re(s) > −1
...
Next, for Re(s) > −2 we define
Γ1 (s) =
Γ2 (s) =
Γ(s + 2)
Γ1 (s + 1)
=
,
s
s(s + 1)
a function that is holomorphic on Re(s) > −2, s 6= 0, −1, and coincides with
Γ1 (s) for Re(s) > −1, s 6= 0
...
The factors 1/s(s + 1) show that Γ2 (s) has a simple pole
at s = −1 with residue −1
...
, −n + 2, we now define
Γn (s) =
Γn−1 (s + 1)
Γ(s + n)
=
...
s
s(s + 1) · · · (s + n − 1)
55
14
THE EULER GAMMA FUNCTION
By inspection we see that this gives a meromorphic function in Re(s) > −n
whose poles are precisely at s = −n + 1,
...
An alternative way to perform the analytic continuation is to separate the
integral defining Γ(s) into
Z
1
−x s−1
e x
Γ(s) =
∞
Z
e−x xs−1 dx
dx +
1
0
and to note that the integral over [1, ∞) converges (and defines a holomorphic function of s) for all s ∈ C, and the integral over [0, 1] can be computed
by expanding e−x as a power series in x and integrating term by term
...
Thus, we have obtained not just an alternative proof for the meromorphic continuation of Γ(s), but a proof of the hybrid
series-integral representation of Γ(s), which also clearly shows where the
poles of Γ(s) are and that they are simple poles with the correct residues
...
For Re(s) > 0 we have
Z
Γ(s) = lim
n→∞
0
n
x n s−1
x dx
...
As n → ∞, theintegrand converges to e−x xs−1 pointwise
...
The claim
therefore follows from the dominated convergence theorem
...
For Re(s) > 0 we have
Z
0
n
x n s−1
n! ns
1−
x dx =
...
For n = 1, the claim is that
Z
0
1
(1 − x) xs−1 dx =
1
,
s(s + 1)
56
14
THE EULER GAMMA FUNCTION
which is easy to verify directly
...
Corollary 10
...
n→∞ s(s + 1) · · · (s + n)
Γ(s) = lim
Proof of the infinite product representation for Γ(s)
...
n
n=1
Γ(s)−1 = lim
We now check that the infinite product actually converges absolutely and
uniformly on compact subsets in all of C, so defines an entire function
...
Q∞Recall that for a sequence of complex numbers cn , the infinite
Qnproduct
n=1 cn is defined as the limit of finite (partial) products limn→∞
k=1 ck , if
the limit exists
...
One
point of terminology that differs slightly from the corresponding terminology
Q∞
for infinite series, is that we say the product n=1 cn converges if the limit of
the finite products exists and is non-zero
...
For a sequence of complex numbers (an )∞
n=1 , if an 6= −1 for all n
P∞
Q∞
and n=1 |an | < ∞ then the infinite product n=1 (1 + an ) converges
...
Under the assumption, there exists some large enough N0 ≥ 1 such
that |an | < 1/2 for all n ≥ N0 , which in particular implies that 1 + an =
exp(Log(1 + an )), where Log(z) is the principal branch of the logarithm function
...
k=N0
P∞
Pn
If we knew that the infinite series n=N0 Log(1 + an ) = limn→∞ k=N0 Log(1 +
ak ) converged, we could continue the above chain of equalities as
= exp
lim
n→∞
n
X
!
Log(1 + ak )
= exp
∞
X
!
Log(1 + an ) ,
n=N0
k=N0
Q∞
and since ez is never 0, that would mean that the product n=N0 (1+an ) exists
and is non-zero (that is, it converges), hence as was mentioned above the
Q∞
infinite product n=1 (1 + an ) also converges and the claim would be proved
...
In particular, there is some constant C > 0 such that
| Log(1 + w)| ≤ |w| + C|w|2
if |w| < 1/2
...
Lemma 9
...
If the
P∞
series
n=1 |fn | converges uniformly on compacts in Ω, the infinite product
Q∞
n=1 (1 + fn ) also converges uniformly on compact subsets in Ω to a nonzero
holomorphic function
...
Lemma 8 above implies that the infinite product n=1 (1 + fn (z)) converges to a nonzero limit for any z ∈ Ω
...
Proof that
Q∞
n=1
1+
z
n
e−z/n is an entire function
...
In particular, the convergence is uniform
on compacts on C
...
Then separately the factors (1 + z/n), n = 1,
...
, −N
...
Γ(s)Γ(1 − s) =
π
...
1
= Γ(s)−1 (−s)−1 Γ(−s)−1
Γ(s)Γ(1 − s)
∞
∞
Y
Y
−1
s −s/n
s s/n
γs
−γs
=
· se
1+
e
· (−s)e
1−
e
s
n
n
n=1
n=1
∞
2
Y
s
sin(πs)
sin(πs)
1− 2 =s
=
,
=s
n
πs
π
n=1
where we used the product representation sin(πz) = πz
the sine function derived in a homework problem
...
By analytic
continuation, it is enough to prove the formula for real s in (0, 1)
...
x
1
+
v
1
+
e
0
−∞
Γ(s)Γ(1 − s) =
So it is enough to prove that for 0 < s < 1 we have
Z
∞
−∞
π
esx
dx =
...
1, Chapter 3, pages 79–81 of [11] for the details
...
15
15
...
However, the Riemann zeta function is a lot more mysterious than the gamma
function, and remains the subject of many famous open problems, including
the most famous of them all: the Riemann hypothesis, considered by many
(including myself) as the most important open problem in mathematics
...
Its
study, and in particular the attempts to prove the Riemann hypothesis, have
also stimulated an unusually large number of important developments in
many areas of mathematics
...
Again, I will formulate this as a theorem asserting the
existence of the zeta function and its various properties
...
There exists a unique function, denoted ζ(s), of a complex variable s, having the following properties:
1
...
2
...
ζ(s) =
s
n
2
3
n=1
3
...
4
...
5
...
6
...
7
...
It is a simple pole with residue
1
...
The “Basel problem” and its generalizations: the values of ζ(s) at
even positive integers are given by Euler’s formula
ζ(2n) =
(−1)n−1 (2π)2n
B2n
2(2n)!
(n = 1, 2,
...
z
e − 1 m=0 m!
Many of the properties of these amazing numbers were discussed in
our homework problem sets
...
Values at negative integers: we have
ζ(−n) = −
Bn+1
n+1
(n = 1, 2, 3,
...
, since it is an
easy fact that the Bernoulli numbers satisfy B2k+1 = 0 for integer k ≥ 1
...
)
10
...
11
...
n=1
n=−∞
12
...
62
15
THE RIEMANN ZETA FUNCTION
13
...
Then
for non-integer x > 1,
ψ(x) = x −
X xρ
ρ
ρ
− log(2π),
where the sum ranges over all zeros ρ of the Riemann zeta function
counted with their respective multiplicities
...
Also the sum is only conditionally
convergent; refer to a book on analytic number theory for the proper
way to interpret it to get a convergent sum
...
In particular, proving that Re(s) has no zeros in Re(s) ≥ 1 will enable us to
prove one of the most famous theorems in mathematics
...
Let π(x) denote the number of prime
numbers less than or equal to x
...
x→∞ x/ log x
lim
Conjecture 1 (The Riemann hypothesis)
...
To begin the proof of Theorem 32, again, let’s take as the definition of
ζ(s) the standard representation
∞
X
1
...
In particular, it is uniform on
compact subsets, so ζ(s) is holomorphic in this region
...
It follows that Z(s) is well-defined, holomorphic and nonzero for
Re(s) > 1
...
We now prove that Z(s) = ζ(s)
...
)
1 − p−s p≤N
X
1
,
ns
j
j
n=p11 ···pkk
p1 ,
...
So, we have represented ζN (s) as a series of a similar form as the series defining ζ(s), but involving terms of the form n−s only
for those positive integers n whose prime factorization contains only primes
≤ N
...
s
n
n>N
Taking the limit as N → ∞ shows that Z(s) = limN →∞ ζN (s) = ζ(s)
...
Corollary 12
...
Proof
...
To prove the functional equation, we need a somewhat powerful tool from
harmonic analysis, the Poisson summation formula
...
For a sufficiently well-behaved
function f : R → C, we have
∞
X
n=−∞
f (n) =
∞
X
k=−∞
fˆ(k),
64
15
where
fˆ(k) =
THE RIEMANN ZETA FUNCTION
∞
Z
f (x)e−2πikx dx
...
Proof
...
Assume that f (x) is sufficiently well-behaved
(i
...
, decays fast enough as x → ±∞ so that g(x) is in turn well-behaved, and
has reasonable smoothness properties)
...
gˆ(k) =
0
In particular, setting x = 0 in the formula for g(x) gives the basic fact that
g(0) =
∞
X
gˆ(k)
...
On the other hand, the Fourier coefficient
gˆ(k) can be expressed in terms of the Fourier coefficients of the original
function f (x):
Z
gˆ(k) =
1
g(x)e
−2πikx
Z
dx =
0
=
1
∞
X
f (x + n)e−2πikx dx
0 n=−∞
∞
X
Z
1
−2πikx
f (x + n)e
n=−∞
Z ∞
dx =
0
∞ Z
X
n=−∞
n+1
f (u)e−2πiku du
n
f (u)e−2πiku du = fˆ(k)
...
65
15
THE RIEMANN ZETA FUNCTION
Theorem 35
...
Remark 11
...
Proof
...
With
the above substitution for f (x) and fˆ(k), the Poisson summation formula becomes precisely the functional equation for ϑ(t)
...
(a) Use the residue theorem to evaluate the contour integral
I
γN
2
e−πz t
dz,
e2πiz − 1
where γN is the rectangle with vertices ±(N + 1/2) ± i (with N a positive
integer), then take the limit as N → ∞ to derive the integral representation
Z
∞−i
ϑ(t) =
−∞−i
2
e−πz t
dz −
e2πiz − 1
Z
∞+i
−∞+i
2
e−πz t
dz
e2πiz − 1
for the Jacobi theta function
...
Evaluate the resulting infinite series, rigorously justifying
all steps, to obtain an alternative proof of the functional equation for ϑ(t)
...
The asymptotic behavior of ϑ(t) near t = 0 and t = +∞ is given
by
1
ϑ(t) = O √
t
ϑ(t) = 1 + O(e−πt )
(t → 0+),
(t → ∞)
...
The asymptotics as t → ∞ is immediate from
ϑ(t) − 1 = 2
∞
X
−πn2 t
≤2
e
n=1
∞
X
2e−πt
,
1 − e−πt
e−πnt =
n=1
−πt
which is bounded by Ce
if t > 10
...
Proof of the analytic continuation of ζ(s)
...
A linear change of variable x = πn2 t brings this to the
form
Z ∞
s
π −s/2 Γ
2
2
e−πn t ts/2−1 dt
...
gives π −s/2 Γ 2s ζ(s) — the function we denoted ζ ∗ (s) — adding the stronger assumption that Re(s) > 1
...
=
Z
∞
0
0
Z
∞
X
!
2t
e−πn
ts/2−1 dt
n=1
∞
=
0
ϑ(t) − 1 s/2−1
t
dt,
2
where the estimates in the lemma are needed to justify interchanging the order of the summation and integration, and show that the integral converges
for Re(s) > 1
...
Next, the idea is to use the functional
equation for ϑ(t) to bring this to a new form that can be seen to be welldefined for all s ∈ C except s = 1
...
67
15
THE RIEMANN ZETA FUNCTION
We can therefore write
Z
∗
1
ζ (s) =
ϕ(t)t
Z0 1
s/2−1
Z
∞
dt +
ϕ(t)ts/2−1 dt
1
−1/2
1 −1/2
t
2
t
s/2−1
Z
∞
ϕ(1/t) +
dt +
ϕ(t)ts/2−1 dt
− t
0
1
Z ∞
1
1
=−
− +
t−s/2−1/2 + ts/2−1 ϕ(t) dt
...
We have
therefore proved that ζ(s) can be analytically continued to a meromorphic
function on C
...
The zeta function satisfies the functional equation
ζ ∗ (1 − s) = ζ ∗ (s)
...
2
Proof
...
Corollary 14
...
1
Proof
...
Thus the poles of ζ (s) are simple poles at s = 0, 1 with
residues −1 and 1, respectively
...
(That is,
the pole of ζ ∗ (s) at s = 0 is cancelled out by the zero of Γ(s/2)
...
ζ(−n) = −Bn+1 /(n + 1) for n = 1, 2, 3,
...
Using the functional equation, we have that
ζ(−n) = 2−n π −n−1 sin(−πn/2)Γ(n + 1)ζ(n + 1)
= 2−n π −n−1 sin(−πn/2)n!ζ(n + 1)
...
We also know that
B2k+1 = 0 for k = 1, 2, 3,
...
If on the other hand n = 2k − 1 is odd, then sin(−π(2k − 1)/2) = (−1)k ,
and therefore we get, using the formula expressing ζ(2k) in terms of the
Bernoulli numbers (derived in the homework and in the textbook), that
ζ(−n) = (−1)k 2−2k+1 π −2k (2k − 1)!ζ(2k)
= (−1)k 2−2k+1 π −2k (2k − 1)!
=−
(−1)k−1 (2π)2k
B2k
2(2k)!
Bn+1
B2k
=−
,
2k
n+1
so again the formula is satisfied
...
The zeros of ζ(s) in the region Re(s) < 0 are precisely the
trivial zeros s = −2, −4, −6,
...
We already established the existence of the trivial zeros
...
Remark 12 (alternative approaches to the analytic continuation of ζ(s))
...
The technical name for this procedure, when it is done
in a more systematic way, is Euler-Maclaurin summation
...
s−1
1
69
15
THE RIEMANN ZETA FUNCTION
This representation is certainly valid for Re(s) > 1
...
Thus, the integral is actually an absolutely convergent integral in the larger region Re(s) > 0, and the representation we
derived gives an analytic continuation of ζ(s) to a meromorphic function on
Re(s) > 0, which has a single pole at s = 1 (a simple pole with residue 1) and
is holomorphic everywhere else
...
Another approach is to use the analytic continuation for Re(s) > 0
shown above, then prove that the functional equation ζ ( 1 − s) = ζ ∗ (s) holds in
the region 0 < Re(s) < 1, and then use the functional equation to analytically
continue ζ(s) to Re(s) ≤ 0 (which is the reflection of the region Re(s) ≥ 1
under the transformation s 7→ 1 − s)
...
2
A theorem on the zeros of the Riemann zeta function
Next, we prove a nontrivial and very important fact about the zeta function
that will play a critical role in our proof of the prime number theorem
...
ζ(s) has no zeros on the line Re(s) = 1
...
If you ever want to try solving this famous open problem, getting
a good understanding of its toy version seems like a good idea
...
For this proof, denote s = σ + it, where we assume σ > 1 and t is
real and nonzero
...
Consider the
following somewhat mysterious quantity
X = log |ζ(σ)3 ζ(σ + it)4 ζ(σ + 2it)|
...
Now
note that for z = a + ib with a > 1 and p prime we have |p−z | = p−a < 1, so
∞
X
p−mz
,
− Log(1 − p ) =
m
m=1
−z
and
∞
h
i
X
p−ma
− Re Log(1 − p ) =
Re cos(mb log p) + i sin(mb log p)
m
m=1
h
−z
i
∞
X
p−ma
=
cos(mb log p)
...
We can now use
a simple trigonometric identity
3 + 4 cos θ + cos(2θ) = 2(1 + cos θ)2 ,
to rewrite X yet again as
X=2
∞
X
n=1
cn n−σ (1 + cos θn )2
...
We now claim that this innocent-looking inequality is incompatible with the
existence of a zero of ζ(s) on the line Re(s) = 1
...
Then the three quantities ζ(σ),
ζ(σ + it) and ζ(σ + 2it) have the following asymptotic behavior as σ & 1:
1
+ O(1) (since ζ(s) has a pole at s = 1),
σ−1
|ζ(σ + it)| = O(σ − 1)
(since ζ(s) has a zero at s = 1 + it),
|ζ(σ + 2it)| = O(1)
(since ζ(s) is holomorphic at s = 1 + 2it)
...
In particular, eX → 0 as σ & 1, in contradiction to the result we proved above
that eX ≥ 1
...
Exercise 16
...
However, one can imagine a more direct approach that
starts as follows: denote x = p−σ and z = p−it = e−it log p
...
Since this is an elementary inequality,
it seems like it ought to have an elementary proof (i
...
, a proof that does not
involve logarithms and power series expansions)
...
(This
was the only number theory paper Riemann wrote in his career!) The history (including all the technical details) of these developments is described
extremely well in the classic textbook [4], which I highly recommend
...
Throughout the 20th century, mathematicians worked hard to find simpler ways to derive the prime
number theorem
...
Despite all the efforts
and the discovery of several paths to a proof that were simpler than the original approach, all proofs remained quite difficult
...
It is Newman’s proof (as presented in the short paper [13] by D
...
Define the weighted prime counting functions
π(x) = #{p prime : p ≤ x} =
X
1,
p≤x
X
ψ(x) =
pk ≤x
log x
,
log p =
log p
log
p
p≤x
X
with the convention that the symbol p in a summation denotes a prime number, and pk denotes a prime power, so that summation over p ≤ x denotes
summation over all primes ≤ x, and the summation over pk denotes summation over all prime powers ≤ x
...
Lemma 11
...
x
log x
is equivalent to the
73
16
THE PRIME NUMBER THEOREM
Proof
...
log p
log
p
log
p
p≤x
p≤x
p≤x
X
In the opposite direction, we have a similar (but slightly less elegant) inequality, namely that for any 0 < < 1 and x ≥ 2,
ψ(x) ≥
X
X
log p ≥
p≤x
log p ≥
x1−
X
log x1−
x1−
= (1 − ) log x π(x) − π(x1− ) ≥ (1 − ) log x π(x) − x1−
...
Then the first of the two bounds above
implies that
ψ(x)
,
log x
x
lim inf π(x)/
≥ 1
...
lim sup π(x)/
log x
x→∞
π(x) ≤
log x
x
=
1
...
Now assume that π(x) ∼ logx x , and apply the inequalities we derived above
in the opposite direction from before
...
x→∞
On the other hand,
ψ(x) ≥ (1 − ) log x(π(x) − x1− )
74
16
THE PRIME NUMBER THEOREM
implies that
log x
= 1 −
...
Comψ(x)
bining the two results about the lim inf and lim sup proves that limx→∞ x =
1, as claimed
...
For Re(s) > 1 we have
∞
ζ 0 (s) X
−
=
Λ(n)n−s
...
Using the Euler product formula and taking the logarithmic derivative (which is an operation that works as it should when applied to infinite
products of holomorphic functions that are uniformly convergent on compact
subsets), we have
ζ 0 (s) X
=
−
ζ(s)
p
=
=
X
p
∞
X
− p−s ) X log p · p−s
=
1 − p−s
1 − p−s
p
d
(1
ds
log p (p
−s
+p
−2s
+p
−3s
+
...
n=1
Lemma 13
...
Proof
...
Specifically, we have
that
22n
2n
X
2n
Y
2n
2n
= (1 + 1) =
>
≥
p = exp
k
n
n
X
= exp ψ(2n) − ψ(n) −
log p
...
!
X
n
log p
75
16
THE PRIME NUMBER THEOREM
√
(The estimate O( n log2 n) for the sum of log p for prime powers higher than
1 is easy and is left as an exercise
...
It follows that
ψ(2m ) = (ψ(2m ) − ψ(2m−1 ))
+ (ψ(2m−1 ) − ψ(2m−2 )) +
...
+ 20 ) ≤ C2 2m ,
so the inequality ψ(x) ≤ C2 x is satisfied for x = 2m
...
Theorem 37 (Newman’s tauberian theorem)
...
Define a function g(z) of a
complex variable z by
Z
∞
f (t)e−zt dt
g(z) =
0
(g is known as the Laplace transform of f )
...
Assume that g(z) has an analytic
continuation
to an open region Ω containing the closed half-plane Re(z) ≥ 0
...
Proof
...
Our goal is to show
that limT →∞ gT (0) = g(0)
...
Fix some large R > 0 and a small δ > 0 (which
depends on R in a way that will be explained shortly), and consider the contour C consisting of the part of the circle |z| = R that lies in the half-plane
Re(z) ≥ −δ , together with the straight line segment along the line Re(z) = −δ
connecting the top and bottom intersection points of this circle with the line
(see Fig
...
Assume that δ is small enough so that g(z) (which extends analytically at least slightly to the right of Re(z) = 0) is holomorphic in an open
76
16
C
THE PRIME NUMBER THEOREM
C+
C- '
C+
C-
R
R
δ
δ
(a)
(b)
R
(c)
0
Figure 6: The contours C , C+ , C− and C−
...
Then by Cauchy’s integral
formula we have
Z
1
z 2 dz
Tz
(g(z) − gT (z))e
g(0) − gT (0) =
1+ 2
2πi C
R
z
Z
Z
1
z 2 dz
Tz
=
+
,
(g(z) − gT (z))e
1+ 2
2πi
R
z
C+
C−
where we separate the contour into two parts, a semicircular arc C+ that
lies in the half-plane Re(z) > 0, and the remaining part C− in the half-plane
Re(z) < 0 (Fig
...
We now bound the integral separately on C+ and on
C−
...
So
in combination we have
Z
1
2πi
C+
Tz
(g(z) − gT (z))e
z 2 dz
2B
B
=
...
In the case of gT (z), the function is entire, so we
77
16
THE PRIME NUMBER THEOREM
0
= {|z| =
can deform the contour, replacing it with the semicircular arc C−
R, Re(z) < 0} (Fig
...
On this contour we have the estimate
T
Z
|gT (z)| =
f (t)e
−zt
Z
dt ≤ B
T
|e−zt | dt =
−∞
0
Be− Re(z)T
,
| Re(z)|
which leads using a similar calculation as before to the estimate
2
|dz|
B
z
T
z
gT (z)e
≤
...
Combining the above estimates, we have shown that
lim sup |g(0) − gT (0)| ≤
T →∞
2B
...
Consider now a very specific application of Newman’s theorem: take
f (t) = ψ(et )e−t − 1
(t ≥ 0),
which is bounded by the lemma we proved above, as our function f (t)
...
78
16
THE PRIME NUMBER THEOREM
Recall that −ζ 0 (s)/ζ(s) has a simple pole at s = 1 with residue 1 (because
ζ(s) has a simple pole at s = 1; it is useful to remember the more general
fact that if a holomorphic function h(z) has a zero of order k at z = z0 then
the logarithmic derivative h0 (z)/h(z) has a simple pole at z = z0 with residue
0 (z+1)
1
· ζζ(z+1)
has a simple pole with residue 1 at z = 0, and therefore
k )
...
Thus, the identity
ζ(z+1)
0 (z+1)
1
− z+1
· ζζ(z+1)
− z1 shows that g(z) extends analytically to a holomorphic
function in the set
{z ∈ C : ζ(z + 1) 6= 0}
...
Thus,
f (t) satisfies the assumption of Newman’s theorem
...
Proof of the prime number theorem
...
Assume by conψ(x)
ψ(x)
tradiction that lim supx→∞ x > 1 or lim inf x→∞ x < 1
...
For such values of x it then follows that
λx
ψ(t) − t
dt ≥
t2
λx
Z
λx − t
dt =
t2
λ
λ−t
dt =: A > 0,
t2
x
x
1
R∞
but this is inconsistent with the fact that the integral 1 (ψ(x) − x)x−2 dx
Z
Z
converges
...
79
17
17
INTRODUCTION TO ASYMPTOTIC ANALYSIS
Introduction to asymptotic analysis
In this section we’ll learn how to use complex analysis to prove asymptotic
formulas such as
n! ∼
√
2πn
n n
(Stirling’s formula),
(2)
(the Hardy-Ramanujan formula),
(3)
(asymptotics for the Airy function),
(4)
e
√
1
p(n) ∼ √ eπ 2n/3
4 3n
1 −1/4
2 3/2
Ai(x) ∼ √ x
exp − x
3
2 π
and more
...
Some related techniques (that are all minor
variations on the same theme) are Laplace’s method, the steepest descent
method and the stationary phase method
...
1
First example: Stirling’s formula
Our goal in this subsection is to prove a version of Stirling’s approximation
(2) for the factorial function n!
...
As for a lower bound, one can say similarly trivial things such as
n! ≥ (n/2)n/2 , but that is quite far from the upper bound
...
It now makes sense to try to get the best lower bound possible by looking for
the x where the lower-bounding function is maximal
...
It’s a great way to develop your intuition, and sometimes you discover that the easy
techniques solve the problem outright
...
e
...
Plugging this value into the inequality gives the bound
n! ≥ (n/e)n
(n ≥ 1)
...
The
point of this trivial calculation is that, as we shall see below, there is something special about the value x = n that resulted from this maximization operation; when interpreted in the context of complex analysis, it corresponds
to a so-called “saddle point,” since it is a local minimum of ex /xn as one
moves along the real axis, but it will be a local maximum when one moves
in the orthogonal direction parallel to the imaginary axis
...
Now let’s move on to the more powerful approach
...
Start with the power series expansion
z
e =
∞
X
zn
n=0
n!
...
It turns out that some values of r are better than
others when one is trying to do asymptotics
...
For convenience, rewrite this as
1
nn
=
n
e n!
2π
Z
π
exp n(eit − 1 − it) dt
...
n e
−1− √
2
n
n
√
Performing the change of variable and moving a factor of
side, the integral then becomes
√
nnn
1
=
n
e n!
2π
n to the left-hand
√
π n
√
iu
iu/ n
exp n e
−1− √
du
...
However, note that the O(u3 / n) esti√
mate holds whenever t = u/ n is in a neighborhood of 0, and since u actually
√
√
ranges in [−π n, π n], we need to be more careful to get a precise asymptotic result
...
Denote M = n1/10 , and let
√
π n
√
iu
iu/ n
I=
exp n e
−1− √
du = I1 + I2 ,
√
n
−π n
Z M
√
iu
iu/ n
du,
I1 =
exp n e
−1− √
n
−M
Z
√
iu
iu/ n
I2 =
exp n e
−1− √
du
...
For I1 , we have
M
2
3
u
u
du
exp − + O √
I1 =
2
n
−M
3
Z M
u
−u2 /2
e
exp O √
=
du
n
−M
3
Z M
Z
M −u2 /2
u
−u2 /2
−1/5
1+O √
=
e
du
e
du = 1 + O(n
)
n
−M
−M
Z ∞
Z ∞
2
−1/5
−2
e−u /2 du
= 1 + O(n
)
−∞
M
√
−1/5
−1/5
= 1 + O(n
)
2π − O exp −n
√
= 1 + O(n−1/5 ) 2π
...
7)
to infer further that
Z
√
π n
|I2 | ≤ 2
√
exp −u2 /8 du ≤ 2π n exp −n1/5 = O(n−1/5 )
...
The asymptotic relation
n n
√
n! = 1 + O(n−1/5 ) 2πn
e
holds as n → ∞
...
0
0
...
5
-1
...
17
...
A standard way to find the asymptotic behavior for an
n
as n → ∞ is to use Stirling’s formula
...
n
πn
(Note that this is not too far from the trivial upper bound 2n
≤ (1 + 1)2n =
n
2n
2
...
n
2πi |z|=r z n+1
By the same trivial method for deriving upper bounds that we used in the
case of the Taylor coefficients 1/n! of the function ez , we have that for each
x > 0,
2n
n
≤ (1 + x)2n /xn = exp (log(1 + x) − n log x)
...
This gives x = 1, the
location of the
point
...
84
17
INTRODUCTION TO ASYMPTOTIC ANALYSIS
Next, equipped with the knowledge of the saddle point, we set r = 1 in
the contour integral formula, to get
I
Z π
2n
1
(1 + z)2n
1
(1 + eit )2n e−int dt
=
dz =
2πi |z|=r z n+1
2π −π
n
Z π
1
it
=
exp n 2 log(1 + e ) − t dt
...
4
√
Again, we see that a change of variables u = t/ n will bring the integrand
to an asymptotically scale-free form
...
= √
exp − u + O
4
n
2π n −π
It is now reasonable to guess that in the limit as n → ∞, the pointwise limit
of the integrands translates to a limit of the integrals, so that we get the
approximation
Z ∞
4n √
2n
4n
4n
2
e−u /4 du = √ 2 π = √ ,
≈ √
n
2π n −∞
2π n
πn
as required
...
Exercise 17
...
n
Exercise 18
...
z n+1
85
17
INTRODUCTION TO ASYMPTOTIC ANALYSIS
2
5
1
4
10
20
30
40
3
-1
2
-2
1
-3
10
-4
-1
-5
-2
(a)
20
30
40
(b)
Figure 8: An illustration (with n = 40) of the random walks enumerated by
(a) the central binomial coefficients and (b) the central trinomial coefficients
...
Specifically,
an and bn correspond to the numbers of random walks on Z that start and end
at 0 and have n steps, where in the case of the central binomial coefficients
the allowed steps of the walk are −1 or +1, and in the case of the central
trinomial coefficients the allowed steps are −1, 0 or 1; see Fig
...
Using a saddle point analysis, show that the asymptotic behavior of bn as
n → ∞ is given by
√
3 · 3n
...
3
A conceptual explanation
In both the examples of Stirling’s formula and the central binomial coefficient we analyzed above, we made what looked like ad hoc choices regarding how to “massage” the integrals, what value r to use for the radius of the
contour of integration, what change of variables to make in the integral, etc
...
Note that the quantities we were trying to estimate took a particular form,
where for some function g(z) our sequence of numbers could be represented
86
17
INTRODUCTION TO ASYMPTOTIC ANALYSIS
in the form
I
I
dz
1
1
e−ng(z) dz
=
a(n) =
exp
−
n(g(z)
+
log
z)
2πi |z|=r z n
z
2πi |z|=r
z
Z π
1
it
=
e−ng(re ) r−n e−int dt
2π −π
Z π
1
it
=
exp − n(g(re ) + it − log r) dt
...
) The idea that is key to making the method work turns out to be
to choose the contour radius r as the solution to the equation
1
d
(g(z) + log z) = g 0 (z) + = 0
...
One is then left with a constant term, that can be pulled
outside of the integral; a second order term, which (in favorable circumstances where this technique actually works) causes the integrand to be
2
well-approximated by a Gaussian density function e−u /2 near z = r ; and
lower-order terms which can be shown to be asymptotically negligible
...
” This phenomenon is illustrated with many beautiful
examples and graphical figures in the lecture slides [6] prepared by Flajolet
and Sedgewick as an online resource to accompany their excellent textbook
Analytic Combinatorics [5]
...
It is instructive to see an example where the saddle point analysis fails if applied mindlessly without checking that the part of the integral
that is usually assumed to make a negligible contribution actually behaves
that way
...
87
17
INTRODUCTION TO ASYMPTOTIC ANALYSIS
Clearly any analysis, asymptotic or not, needs to address and take into account the fact that bn behaves differently according to whether n is even or
odd
...
The method fails, but the failure can easily be turned into a success
by noting that there are actually two saddle points, each of which makes a
contribution to the integral, in such a way that for odd n the contributions
cancel and for even n they reinforce each other
...
Exercise 20
...
Can you succeed in deriving an asymptotic formula
for the constant function 1?
17
...
Our next goal is to prove a stronger version of Stirling’s formula that gives
an asymptotic formula for Γ(t), the extension of the factorial function to noninteger arguments
...
Theorem 39 (Stirling’s approximation for Γ(t))
...
Proof
...
Start with the integral formula
Z
∞
Γ(t) =
e−x xt−1 dx
0
Performing the change of variables x = tu in the integral gives that
Z
∞
Γ(t) = t
e
−tu t−1
u
t −t
Z
∞
e−tu+t ut−1 du
0
0
t
Z ∞
Z ∞
t
−tu+t t−1
t −t
−tΦ(u) du
t −t
=te
e
u du = t e
e
=
I(t),
u
e
0
0
t
du = t e
88
17
INTRODUCTION TO ASYMPTOTIC ANALYSIS
where we define
Φ(u) = u − 1 − log u,
Z ∞
du
e−tΦ(u)
...
) Our goal is to prove that
r
I(t) =
2π
+ O(t−7/10 )
t
as t → ∞
...
The idea is that for large t, the bulk of the contribution to
the integral comes from a region very near the point where Φ(u) takes its
minimum
...
See Fig
...
Denote
Z
1/2
du
,
u
0
Z 2
du
I2 =
e−tΦ(u) ,
u
1/2
Z ∞
du
e−tΦ(u) ,
I3 =
u
2
I1 =
e−tΦ(u)
so that I(t) = I1 + I2 + I3
...
Expanding Φ(u) in a Taylor series around u = 1, we have
(u − 1)2
+ O((u − 1)3 )
2
(u−1)2
for u ∈ [1/2, 2] (in fact the explicit bound Φ(u) − 2 ≤ (u − 1)3 on this
Φ(u) =
interval can be easily checked)
...
0
2
...
0
1
...
0
0
...
√
we see that it is natural to apply a linear change of variables v = t(u − 1) to
bring the integrand to a scale-free, centered form
...
√
2
t − 12 t
t
t
1
I2 = √
t
Z
As before, we actually need to split
√ up this integral into two parts to take into
3
account the fact that the O(v / t) term can blow up when v is large enough
...
For J1 we have
Z M
1
v
1
√ dv
exp −tΦ 1 + √
J1 = √
t −M
t
1 + v/ t
3
Z M
1
v
v
−v 2 /2
=√
e
1+O √
1+O √
dv
t −M
t
t
r
Z M
1
2π
2
= √ 1 + O(t−1/5 )
e−v /2 dv =
1 + O(t−1/5 ) ,
t
t
−M
in the last step using a similar estimate as the one we used in our proof of
Stirling’s approximation for n!
...
t
as in our earlier proof
...
t
Next, we bound I1
...
Considering first a truncated integral over [ε, 1/2] and performing an
integration by parts, we have
Z
1/2
−tΦ(u) du
e
ε
1
=−
u
t
Z
1/2
d −tΦ(u) 1
e
du
du
Φ0 (u)u
ε
u=1/2
Z
1 1/2 −tΦ(u) du
1 e−tΦ(u)
=−
−
e
...
Finally, I leave it as an exercise to obtain a similar estimate I3 = O(1/t)
for the remaining integral on [2, ∞)
...
t
The proof above is a simplified version of the analysis in Appendix A
of [11]
...
Specifically, they prove
that for complex s in the “Pac-Man shaped” region
Sδ = {z ∈ C : | arg z| ≥ π − δ}
91
17
INTRODUCTION TO ASYMPTOTIC ANALYSIS
(for each fixed 0 < δ < π ) the gamma function satisfies
√
Γ(s) = 1 + O(|s|−1/2 ) 2πss−1/2 e−s
as |s| → ∞, s ∈ Sδ
...
This sort of approximation is
important in certain applications of complex analysis, for example to analytic
number theory
...
Below is a list of basic formulas in complex analysis (think of them as
“formulas you need to know like the back of your hand”)
...
In the formulas below, a, b, c, d, t, x, y denote arbitrary real numbers; w, z
denote arbitrary complex numbers
...
(a + bi)(c + di)
= (ac − bd) + (ad + bc)i
b
...
z = Re(z) + i Im(z)
c
...
z = Re(z) − i Im(z)
f
...
h
...
j
...
z+z
Re(z) =
2
z−z
Im(z) =
2i
2
|z| = zz
1
z
= 2
z
|z|
1
x − iy
= 2
x + iy
x + y2
w·z =w·z
l
...
|w| − |z| ≤ |w + z| ≤ |w| + |z|
n
...
|ez | = eRe(z)
p
...
eit = cos(t) + i sin(t)
r
...
sin(t) =
2i
πi
u
...
cos(t) =
v
...
e2πi = 1
2
...
2, 1
...
1, 2
...
Try to spend some time thinking
about the answers yourself before looking them up
...
real part
f
...
imaginary part
g
...
complex conjugate
h
...
modulus
i
...
argument
j
...
region
l
...
Cauchy sequence
complex variable)
r
...
analytic function
n
...
entire function
o
...
meromorphic function
p
...
harmonic function (of two
variables)
q
...
For each of the following functions, determine for which z it is analytic
a
...
f (z) = |z|2
b
...
f (z) = z
c
...
f (z) = 1/z
4
...
(Hint4 )
a
...
u(x, y) = x4 −6x2 y 2 +3x+y 4 −2
b
...
u(x, y) = cos x cosh y
5
...
94
PROBLEMS
3
2
1
-3
-2
1
-1
2
3
-1
-2
-3
under each of the following maps w = f (z):
a
...
w = (2 + i)z − 3
b
...
w = 1/z
c
...
w = z 2 − 1
6
...
7
...
8
...
+ a0 ,
(where a0 ,
...
, zn ∈ C (these are the roots of p(z) counted with multiplicities)
...
, an are real has a factorization
p(z) = an Q1 (z)Q2 (z)
...
e
...
9
...
+ a0 be a complex polynomial of degree n
(that is, a0 ,
...
, zn are the roots of p(z) counted with multiplicities
...
, wn−1 denote the roots of p0 (z)
...
, wn−1 all
lie in the convex hull of z1 ,
...
That
is, each wk can be expressed as a convex combination
wk = α1 z1 + α2 z2 +
...
, αn are nonnegative real numbers and
clear, there are different coefficients for each k
...
(To be
Hint
...
(Note: the expression p0 /p is known as
the logarithmic derivative of p
...
10
...
Let p(z) = az 3 +
bz 2 + cz + d, with a, b, c, d ∈ C, a 6= 0
...
e
...
(a) Show that the substitution w = z −
simpler form
b
3a
brings the equation to the
w3 + pw + q = 0
(5)
for some values of p, q (find them!) given as functions of a, b, c, d
...
and w1 = 0
...
586i,
tive
...
...
336 − 0
...
414 + 1
...
(b) Show that assuming a solution to (5) of the form w = u + v , the
equation (5) for w can be solved by finding a pair u, v of complex
numbers such that the equations
p = −3uv,
q = −(u3 + v 3 )
(6)
(7)
are satisfied
...
More precisely, any solution of (6)–(7) can be obtained from
some (easily determined) solution of (8)–(9)
...
(10)
97
PROBLEMS
(e) Using the above reductions, show that the three solutions of the
simplified cubic (5) can be expressed as
w1 = u + v,
w2 = ζ u + ζ v,
w3 = ζ u + ζ v,
√
where ζ = e2πi/3 = 21 (−1 + i 3) (a cube root of unity) and u, v are
properly chosen cube roots of R, S obtained as solutions to (10)
...
Bring the formulas to a form that makes it clear that the roots are
real numbers
...
Let A =
a b
be a 2 × 2 real matrix
...
(Here hw1 , w2 i denotes the standard inner product in R2 , and |w| = hw, wi1/2 is the usual two-dimensional norm of
a vector in R2
...
−b a
cos θ − sin θ
(c) A takes the form A = r
for some r > 0 and θ ∈ R
...
)
12
...
However, if we
think of the equations
z = x + iy,
z = x − iy
98
PROBLEMS
as representing a formal change of variables from the “real coordinates” (x, y) to the “complex conjugate coordinates” (z, z), then it may
make sense to think of f as a function of the two variables z and z
(pretending that those are two independent variables)
...
Show that, from this somewhat strange point of view, the CauchyRiemann equations
∂v
∂u
=
,
∂x
∂y
∂u
∂v
=−
∂y
∂x
can be rewritten in the more concise equivalent form
∂f
= 0,
∂z
assuming that it is okay to apply the chain rule from multivariable calculus; and moreover, that in this notation we also have the identity
f 0 (z) =
∂f
...
Let f : Ω → C be a function defined on a region Ω such that both
the functions f (z) and zf (z) have real and imaginary parts that are
2
2
harmonic functions (i
...
, satisfy the Laplace equation ∂∂xu2 + ∂∂yu2 = 0)
...
14
...
Given a sequence of
functions fn : Ω → C, n ≥ 1, defined on a region Ω ⊂ C, we say that fn
converges uniformly on compact subsets to a limiting function f : Ω →
C if for any compact subset K ⊂ Ω, fn (z) → f (z) as n → ∞, uniformly
on z ∈ K
...
(b) Prove that if fn are holomorphic functions, fn → f uniformly on
compact subsets, fn0 → g uniformly on compact subsets, and g is
continuous, then f is holomorphic and f 0 = g
...
Fix some z0 ∈ C
...
P∞
n
(c) Prove that a power series
n=0 an z converges uniformly on compact subsets in its disk of convergence, and that the function it
defines is continuous
...
It is enough (explain why) to prove uniform convergence on
any closed disk of the form Dr (0) where 0 < r < R and R is the
radius of convergence of the series
...
Remark
...
This
is a surprising and nontrivial fact, as illustrated for example by the observation that the analogous statement in real analysis is false (e
...
, by
the Weierstrass approximation theorem, any continuous function on a
closed interval is the uniform limit of a sequence of polynomials)
...
Cauchy’s theorem and irrotational vector fields
...
By the funda-
mental theorem of calculus for line integrals, for such a vector field we
have
I
F · ds = 0
γ
for any closed curve γ
...
The converse also holds under suitable conditions:
curl F = ∂Q
∂x
∂y
if the region Ω is simply-connected (a concept we will discuss later in
100
PROBLEMS
the course), then a theorem in vector calculus says that an irrotational
vector field is also conservative
...
(This is, of course, Cauchy’s theorem
...
The Bernoulli numbers
...
(a) Convince yourself that f (z) is analytic in a neighborhood of 0
...
(b) One of the basic complex analysis theorems we will discuss is that
analytic functions have a power series expansion
...
For example, the first three Bernoulli numbers are B0 = 1, B1 =
−1/2, B2 = 1/6
...
B2k+1 = 0 for k = 1, 2,
...
P∞
n
Hint
...
= 0 if and only if g(z) = g(−z), i
...
, g(z) is an even function
...
(n + 1)Bn = −
Pn−1
k=0
iii
...
2n
B2k B2n−2k , (n ≥ 2)
...
Show that the function g(z) = f (z) + z/2 satisfies the
equation
g(z) − zg 0 (z) = g(z)2 − z 2 /4
...
z X B
z
2n 2n
coth
=
z
...
Assuming
this, deduce that
Bn 1/n
lim sup = 1/R,
n!
n→∞
where R is the number you found in part (a)
...
)
17
...
The Bessel functions are a family of functions (Jn )∞
n=−∞
of a complex variable,defined by
Jn (z) =
∞
X
k=0
(−1)k z 2k+n
...
) Find the radius of convergence of the series defining Jn (z),
and prove that the Bessel functions satisfy the following properties:
i
...
2n
Jn (z) − Jn−1 (z)
...
Bessel’s differential equation: z 2 Jn00 (z) + zJn0 (z) + (z 2 − n2 )Jn (z) = 0
...
Summation identity:
Jn (z) = 1
...
Recurrence relation:
Jn+1 (z) =
v
...
0
Hint for the last equation: cos(a − b) = cos(a) cos(b) + sin(a) sin(b)
...
The Bessel functions are very important functions in
mathematical physics, and appear naturally in connection with
various problems in diffusion, heat conduction, electrodynamics,
quantum mechanics, Brownian motion, probability, and more
...
Their properties as analytic functions of a complex variable are
also a classical, though no longer very fashionable, topic of study
...
Show that Liouville’s theorem (“a bounded entire function is constant”)
can be proved directly using the “simple” (n = 0) case of Cauchy’s integral formula, instead of using the case n = 1 of the extended formula
as we did in the lecture
...
For an arbitrary pair of complex numbers z1 , z2 ∈ C, show that
|f (z1 ) − f (z2 )| = 0
...
Show that Liouville’s theorem can in fact be deduced even just from
the mean value property of holomorphic functions, which is the special
case of Cauchy’s integral formula in which z is taken as the center of
the circle around which the integration is performed
...
Here it makes sense to consider a modified version of the mean
value property (that follows easily from the original version) that says
that f (z) is the average value of f (w) over a disc DR (z) (instead of a
circle CR (z))
...
Explain why this formula holds, then use it to again bound |f (z1 )−f (z2 )|
from above by a quantity that goes to 0 as R → ∞
...
Prove the following generalization of Liouville’s theorem: let f be an
entire function that satisfies for all z ∈ C the inequality
|f (z)| ≤ A + B|z|n
for some constants A, B > 0 and integer n ≥ 0
...
21
...
+ a0 is a polynomial of degree n such
that
|an | >
n−1
X
|aj |,
j=0
prove that p(z) has exactly n zeros (counting multiplicities) in the unit
disc |z| < 1
...
Use the fundamental theorem of algebra
...
This is a special case of a less elementary fact that can be proved
using Rouché’s theorem; see problem 28
22
...
Solve exercises 11–12 (pages
66–67) in Chapter 2 of [Stein-Shakarchi], which explore this connection, and more generally the connection between holomorphic and harmonic functions
...
Spend at least 5–10 minutes thinking about the concept of a toy contour
...
Even better, sketch a proof of
the key result that a function holomorphic in such a region (and therefore having the property that its contour integral along triangles and
rectangles vanish) has a primitive
...
Characterizing some important families of holomorphic funcb
...
M = the set of Möbius transformations z 7→
cz + d
Note the containment relations K ⊂ L ⊂ P ⊂ R ⊃ M ⊃ L ⊃ K
...
(b) Prove that the set of entire functions f : C → C that have a
nonessential singularity at ∞ is P , the polynomials
...
(d) Prove that the set of meromorphic, one-to-one and onto functions
b→C
b is M\K, the set of nonconstant Möbius transformations
...
Use the characterization in problem 24c above
...
For example (I’m not sure if
this is the simplest argument): argue that if z0 is a complex number
such that q(z0 ) 6= 0 and such that p0 (z0 )q(z0 ) − p(z0 )q 0 (z0 ) 6= 0, and
w0 = f (z0 ), then the equation f (z) = w0 must have more than one
solution in z , unless p(z), q(z) are linear functions
...
105
PROBLEMS
Hint
...
See the
guidance for exercise 14 on page 105 of [Stein-Shakarchi]
...
Given a region Ω ⊂ C, or more generally a Riemann surface
Σ, complex analysts are interested in understanding the structure of
its set of holomorphic functions (C-valued holomorphic functions on Σ);
b -valued holomorphic functions on
its set of meromorphic functions (C
Σ); and its set of holomorphic automorphisms (holomorphic, one-to-one
and onto mappings from Σ to itself)
...
(i) The constant functions are the only holomorphic functions on C
b
...
(iv) The nonconstant Möbius transformations are the holomorphic aub
...
(Try to prove that any such map is indeed an automorphism of H; the reverse implication that all automorphisms
of H are of this form is a bit more difficult and requires a result
known as the Schwarz lemma
...
k
...
entire functions) and the set of meromorphic functions on C are much larger
families of functions that do not have such a simple description
as the functions in the relatively small families L, P, R, M
...
25
...
In this
multipart question you are asked to prove a well-known infinite summation identity (equation (12) below), known as the partial fraction expansion of the cotangent function
...
106
PROBLEMS
(a) Let z ∈ C \ Z
...
Take the limit as N → ∞ to
deduce the identity
∞
X
1
π2
=
(sin πz)2 n=−∞ (z + n)2
(z ∈ C \ Z)
...
This is not a trivial exercise, but is not very difficult when
broken down into the following elementary substeps:
i
...
This provides
some good practice with residue computations
...
Use the residue theorem to obtain an expression for the contour integral IN defined above
...
Separately, obtain estimates for IN that can be used to show
that IN → 0 as N → ∞
...
iv
...
(b) Integrate the identity (11) to deduce (using some additional fairly
easy reasoning) the formulas
π cot(πz) = lim
N →∞
N
X
n=−N
∞
1
1 X 2z
= +
z+n
z n=1 z 2 − n2
(z ∈ C \ Z)
...
Consequences of the partial fraction expansion of the cotangent
function
...
(13)
Note that the function on the right-hand side is (or can be easily
checked to be) an entire function of z with a simple zero at any
integer z = n ∈ Z, and whose Taylor expansion around z = 0 starts
with πz + O(z 3 ); thus it is a natural guess for an infinite product
expansion of sin(πz), although the fact that this guess is correct is
far from obvious; for example one can multiply the right-hand side
by an arbitrary function of the form eg(z) and still have an entire
function with the same set of zeros
...
Compute the logarithmic derivatives of both sides of (13)
...
(Spoiler alert: pages 142–144 contain a solution to this subexercise, starting with an independent proof of (12) and proceeding
with a derivation of (13) along the same lines as I described above
...
2
1 3 3 5 5 7 7 9
(c) By comparing the first terms in the Taylor expansion around z = 0
of both sides of (13), derive the well-known identities
∞
X
1
π2
=
,
2
n
6
n=1
∞
X
1
π4
=
...
) = 1 +
1
1
1
+
+
+
...
To see this, first, rewrite
(12) as
∞
X
1
1
1
−
π cot(πz) = +
z n=−∞ z + n n
(z ∈ C \ Z)
...
Compare coefficients and simplify to get the
formula
ζ(2k) =
(−1)k+1 (2π)2k
B2k
...
(e) Show that ζ(2k) = 1 + O(2−2k ) as k → ∞, and deduce that the
asymptotic behavior of the Bernoulli numbers is given by
B2k = (1 + O(2−2k ))(−1)k+1
2(2k)!
,
(2π)2k
k → ∞
...
2π
109
PROBLEMS
27
...
Prove that the sum
of the residues of f (z) over all its poles is equal to 0
...
(A generalization of the result from problem 21) If p(z) = an z n +an−1 z n−1 +
...
29
...
stackexchange
...
30
...
This proof is one way to make
precise the intuitively compelling “topological” proof idea we discussed
at the beginning of the course
...
(a) Draw a simply-connected region Ω ⊂ C such that 0 ∈
/ Ω, 1, 2 ∈ Ω,
and such that there exists a branch F (z) of the logarithm function on Ω
satisfying
F (1) = 0,
F (2) = log 2 + 2πi
(where log 2 = 0
...
is the ordinary logarithm of 2 in the usual
sense of real analysis)
...
If we were to replace the above condition
F (2) = log 2+2πi with the more general condition F (2) = log 2+2πik but
keep all the other conditions, would an appropriate simply-connected
region Ω = Ω(k) exist to make that possible? If so, what would this
region look like, roughly, as a function of k ?
32
...
Values at half-integers:
Γ n+
1
2
=
(2n)! √
π
4n n!
(n = 0, 1, 2,
...
The duplication formula:
√
Γ(s)Γ(s + 1/2) = 21−2s πΓ(2s)
...
* The multiplication theorem:
Γ s Γ s + k1 Γ s + k2 · · · Γ s +
k−1
k
= (2π)(k−1)/2 k 1/2−ks Γ(ks)
...
For n ≥ 1, let Vn denote the volume of the unit ball in Rn
...
Rn
exp − 12
n
X
!
x2j
dx1 dx2
...
Γ n2 + 1
Note
...
The solution can be found on this Wikipedia page
...
The beta function is a function B(s, t) of two complex variables, defined
for Re(s), Re(t) > 0 by
Z
B(s, t) =
1
xs−1 (1 − x)t−1 dx
...
(b) Show that B(s, t) can be expressed in terms of the gamma function
as
B(s, t) =
Γ(s)Γ(t)
...
Start by writing Γ(s)Γ(t) as a double integral on the positive
quadrant [0, ∞)2 of R2 (with integration variables, say, x and y );
then make the change of variables u = x + y , v = x/(x + y) and
use the change of variables formula for two-dimensional integrals
to show that the integral evaluates as Γ(s + t)B(s, t)
...
Note the similarity of theidentity relating the gamma
n
and beta functions to the formula nk = k!(n−k)!
; indeed, using the
111
PROBLEMS
relation Γ(m+1) = m! and the functional equation Γ(s+1) = s Γ(s),
we see using the above relation that for nonnegative, integervalued arguments we have
B(n, m)
−1
Γ(n + m + 1)
nm n + m
nm
·
=
...
35
...
(a) Show that ψ(s) has the convergent series expansions
∞
s
1 X
+
s n=1 n(n + s)
∞
X
1
1
= −γ +
−
n+1 n+s
n=0
ψ(s) = −γ −
(s 6= 0, −1, −2,
...
(b) Equivalently, show that ψ(s) can be expressed as
ψ(s) = − lim
n
X
n→∞
k=1
!
1
− log n
...
s
(d) Show that
ψ(n + 1) = −γ +
n
X
1
k=1
k
(n = 0, 1, 2,
...
112
PROBLEMS
(e) Show that ψ(s) satisfies the reflection formula
ψ(1 − s) − ψ(s) = π cot(πs)
...
Consider
the sequence of polynomials
Pn (x) = x(x − 1)
...
)
and their derivatives
Qn (x) = Pn0 (x)
...
Denote this root by k + αn,k ,
so that the numbers αn,k (the fractional parts of the roots of Qn (x))
are in (0, 1)
...
, n − 1 numerically, say for n = 50 (Figure 11(a))
...
This is correct, and in fact the following precise statement
can be proved
...
Let t ∈ (0, 1)
...
Then we have
lim αn,k(n)
n→∞
1
= R(t) := arccot
π
1
log
π
1−t
t
...
The limiting
function R(t) is shown in Figure 11(b)
...
Guidance
...
This will give an equation with a sum of terms
...
Take the limit as n → ∞, then simplify using the
reflection formula (part (c))
...
0
1
...
8
0
...
6
0
...
4
0
...
2
0
...
0
0
...
4
(a)
0
...
8
1
...
0
0
...
6
-200
0
...
2
-600
0
...
2
0
...
6
0
...
0
(c)
(d)
Figure 11: (a) A plot of the fractional parts of the roots of Qn (x) for n = 50
...
(c) The two previous plots combined
...
Note that the roots of Q7 (x) correspond to the local minima
and maxima of P7 (x), which are highlighted
...
Given two integrable functions f, g : R → C (of a real variable), their
convolution is the new function h = f ∗ g defined by the formula
Z
∞
f (t)g(x − t) dt
h(x) = (f ∗ g)(x) =
(x ∈ R)
...
For α > 0 define the gamma density with parameter α, denoted γα :
R → R, to be the function
γα (x) =
1 −x α−1
e x 1[0,∞) (x)
Γ(α)
(x ∈ R)
(where 1A (x) denotes the characteristic function of a set A ⊂ R, equal
to 1 on the set and 0 outside it)
...
0
0
...
6
0
...
2
2
α=4
α=5
4
6
8
10
12
Figure 12: The gamma densities γα (x) for α = 1, 2, 3, 4, 5
...
See Figure 12 for an
illustration
...
That is, the family of density functions (γα )α>0 is closed under the convolution operation
...
37
...
If you’re feeling especially
energetic, derive the more detailed expansion
1
Γ(s) = − γ +
s
γ 2 π2
+
2
12
s + O(s2 )
and proceed to derive (by hand, or if you prefer using a symbolic math
software application such as SageMath or Mathematica) as many additional terms in the expansion as you have the patience to do
...
s−1
115
PROBLEMS
38
...
39
...
n=1
40
...
2s 3s 4s
(a) Prove that the series defining D(s) converges uniformly on any
half-plane of the form Re(s) ≥ α where α > 0, and conclude that
D(s) is defined and holomorphic in the half-plane Re(s) > 0
...
(c) Using this relation, deduce a new proof that the zeta function can
be analytically continued to a meromorphic function on Re(s) > 0
that has a simple pole at s = 1 with residue 1 and is holomorphic
everywhere else in the region
...
Let ψ(x) =
pk ≤x log p denote von Mangoldt’s weighted prime counting function
...
, n), where for integers
a1 ,
...
, ak ) denotes the least common multiple of a1 ,
...
Note that this implies that an equivalent formulation of the prime number theorem is the interesting statement that
lcm(1,
...
116
PROBLEMS
42
...
(b) Pass to the logarithm and deduce that for some constant K > 0 we
have the bound
X1
p≤x
p
≥ log log x − K
(x ≥ 1)
...
P
43
...
Recall that in Subsection 15
...
t
(15)
(a) Use the residue theorem to evaluate the contour integral
I
γN
2
e−πz t
dz,
e2πiz − 1
where γN is the rectangle with vertices ±(N +1/2)±i (with N a positive
integer), then take the limit as N → ∞ to derive the integral representation
Z
Z
2
2
∞−i
ϑ(t) =
−∞−i
e−πz t
dz −
e2πiz − 1
∞+i
−∞+i
e−πz t
dz
e2πiz − 1
for the function ϑ(t)
...
Evaluate the resulting infinite series, rigorously justifying all steps, to obtain an alternative proof of the functional
equation (15)
...
(a) Reprove Theorem 36 (the “toy Riemann hypothesis” — the result
that the Riemann zeta function has no zeros on the line Re(s) = 1) by
considering the behavior of
ζ 0 (σ + it) ζ 0 (σ + 2it)
ζ 0 (σ)
−4
−
Y = Re −3
ζ(σ)
ζ(σ + it)
ζ(σ + 2it)
for t ∈ R \ {0} fixed and σ & 1, instead of the quantity
X = log |ζ(σ)3 ζ(σ + it)4 ζ(σ + 2it)|
...
(b) Try to reprove the same theorem in yet a third way by considering
Z = log |ζ(σ)10 ζ(σ + it)15 ζ(σ + 2it)6 ζ(σ + 3it)|,
and attempting to repeat the argument involving expanding the logarithm in a power series and deducing that Z ≥ 0
...
(a + b)6 = a6 + 6a5 b + 10a4 b2 + 15a3 b3 + 10a2 b4 + 6ab5 + b6
...
Define arithmetic functions taking an integer argument n, as follows:
(
(−1)k if n = p1 p2 · · · pk is a product of k distinct primes,
µ(n) =
0
otherwise,
(the Möbius µ-function),
d(n) =
X
σ(n) =
X
1,
(the number of divisors function),
d,
(the sum of divisors function),
d|n
d|n
φ(n) = #{1 ≤ k ≤ n − 1 : gcd(k, n) = 1}, (the Euler totient function),
(
log p if n = pk , p prime,
Λ(n) =
(the von Mangoldt Λ-function)
...
n=1
Use the Euler product formula for the zeta function or other elementary
manipulations to prove the following identities (valid for Re(s) > 1):
0
ζ (s) = −
1
=
ζ(s)
ζ(s)
=
ζ(2s)
∞
X
log n · n−s ,
n=1
∞
X
µ(n)n−s ,
n=1
∞
X
|µ(n)|n−s
...
n=1
46
...
(z − zn ) is a complex polynomial whose roots zj , j = 1,
...
(a) Prove the conjecture for the case n = 2 of quadratic polynomials
...
(c) Prove the conjecture for the case n = 3 of cubic polynomials
...
119
PROBLEMS
Suggested topics for course projects
1
...
Let an denote
the number of self-avoiding walks of length n in the square lattice Z2
(that is, lattice paths with n steps) starting from (0, 0)
...
638
...
However, for
the hexagonal lattice, a precise analogous result is known: if bn denotes
the number
p of self-avoiding walks of length n starting at the origin, then
bn −−−→
n→∞
2+
√
2 ≈ 1
...
This is proved in the paper [3]
...
See
this Wikipedia article for more details
...
Complex analysis and the art of M
...
Escher
...
1 on page 3
and the references [8], [10]
...
Elliptic functions and the characterization of meromorphic maps
on the complex torus
...
The case when N is the Riesurfaces (such as M = C, N = C
ˆ and M is a complex torus C/Λ where Λ = Z + τ Z for
mann sphere C
some complex number τ ∈ C \ R is an especially interesting example
of such a question: in that case the relevant family turns out to be the
family of elliptic functions (also known as doubly periodic functions)
...
9] and [1,
Ch
...
4
...
This conjecture is described in Problem 46, and
can be a suitable topic for a project
...
Advanced topic in analytic combinatorics
...
The book [5] is an excellent reference for this
subject (see also [6]), and provides many topics suitable for a project
...
The Jordan curve theorem
...
The theorem plays
120
PROBLEMS
an important role in complex analysis, discussed in Appendix B of [11]
...
A project could present one of those proofs,
and discuss additional related topics such as the significance of the
theorem for complex analysis, various generalizations of it (for example its higher-dimensional analogues), and related theorems and open
problems in topology
...
121
PROBLEMS
References
[1] T
...
Modular Functions and Dirichlet Series in Number Theory
...
[2] G
...
Cohen, G
...
Smith
...
Amer
...
Monthly 95 (1988), 734–737
...
Duminil-Copin,
p S
...
The connective constant of the honeycomb
lattice equals
2+
√
2
...
Math
...
[4] H
...
Edwards
...
Dover Publications, 2001
...
Flajolet, R
...
Analytic Combinatorics
...
[6] P
...
Sedgewick
...
Online resource:
https://ac
...
princeton
...
Accessed March 20, 2020
...
Ivi´
c
...
Dover Publications, 2003
...
de Smit, H
...
Lenstra Jr
...
C
...
Notices Amer
...
Soc
...
[9] R
...
The Jordan curve theorem via the Brouwer fixed point theorem
...
Math
...
[10] D
...
The mathematical side of M
...
Escher
...
Math
...
57 (2010), 706–718
...
M
...
Shakarchi, Complex Analysis
...
[12] H
...
A proof of the Jordan curve theorem
...
London
Math
...
12 (1980), 34–38
...
Zagier
...
Amer
...
Monthly 104 (1997), 705–708
...
C
...
,
52
equivalent definitions, 52
interpolating factorials, 52
reflection formula, 58
Stirling’s formula, 87
geometric series, 29
Goursat’s theorem, 22
rectangles, 24
Hardy-Ramanjuan formula, 2
Hardy-Ramanjuan’s formula, 79
harmonic function, 12
Heine-Borel-property, 31
holomorphic
at ∞, 39
holomorphic function, 7
mean value property, 27
holomorphic implies analytic, 29
homotopic curves, 46
homotopy, 46
homotopy invariance of integrals, 47
index of a curve, 42
inequality, 15
infinite products
what you need to know, 56
124
infinte products
convergence criteria, 56
inner product, 8
integral
homotopy-invariant, 47
isolated point, 32
isolated zero, 32
Jacobi ϑ-function, 65
Jacobian matrix, 13
keyhole contour, 26
Λ(n), 72
Laplace’s equation, 12
Laplace’s method, 79
limit points
common, 32
lim sup, 14
line integral
two types, 17
logarithm function, 50
logarithmic derivative, 41
of a product, 41
Log z , 50
log0 z , 50P
(s)
−s
= ∞
− ζζ(s)
n=1 Λ(n)n , 74
M
...
Escher, 2
Mandelbrot set, 4
Mangoldt function Λ(n), 72
maximum modulus principle, 46
mean value property, 27
Mellin transform, 52
meromorphic function, 38
Morera’s theorem, 21, 22
proof, 29
Newman’s tauberian theorem, 75
n!
asymptotic relation, 82
n-th root, 51
INDEX
open mapping theorem, 45
order
of pole, zero, at ∞, 39
orthogonal level curves, 11
partial differential equation, 9
physics problems, 2
π(x) ∼ logx x , 72
Poisson summation formula, 63
polar decomposition, 7
pole
of order m, 36
representation Lemma, 36
simple, 36
polynomial equation, 1
polynomial root
existence, 3
Fundamental Theorem of Algebra,
3
power function, 51
power series, 13
divergent, 14
prime counting functions
weighted, 72
prime number theorem, 1, 62
equivalences, 72
history, seealso [4], 71
Newman’s proof, 72
prime number theory
proof, 78
primitive
existence criteria, 20
existence on a disc, 25
of holomorphic function, 49
synonym of anti-derivative, 49
with log-properties, 50
primitive, anti-derivative, 19
principal branch
of logarithm, 50
principal part, 36
ψ(x) ∼ x, 72
125
punctured neighborhood, 32
radius of convergence, 14
region
open connected set, 19
simply-connected open set, 22
regularity theorem, 28
removable singularity, 33
theorem, 33
residue of
...
, 35
simple, 35
zeta function
as contour-integral, 61
axiomatic definition, properties, 60
Basel problem, 60
Bernoulli numbers, 60
126
Euler’s product formula, 60
Euler-Maclaurin summation, 68
functional equation, 61, 67
in prime counting, 62
location of zeroes, 60
Mellin transformed, 61
no zeroes with Re(s) = 1, 69
pole, 67
INDEX