Complex Variables
Patrick Gallagher
Transcribed by Ron Wu
This is an undergraduate course. O↵ered in Spring 2014 at
Columbia University. Reference: Ahlfors, Complex Analysis.
Office hours: TuWe 5:30-6:30.
Contents
1 Complex Numbers and Complex Functions
1.1 Complex Numbers . . . . . . . . . . . . . .
1.2 Polynomials . . . . . . . . . . . . . . . . . .
1.3 Point Set Topology: Derivatives . . . . . . .
1.4 Analytic Functions . . . . . . . . . . . . . .
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5
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8
12
16
2 Complex Integration
2.1 Integral and Path Integral . .
2.2 Cauchy Integral Theorem . .
2.3 Cauchy’s Integral Formula . .
2.4 Cauchy’s Derivative Formulas
2.5 Cauchy-Taylor Formulas . . .
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20
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28
31
34
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3 Little Analysis
38
3.1 Uniform Convergence . . . . . . . . . . . . . . . . . . . . 38
3.2 Cauchy Criterion . . . . . . . . . . . . . . . . . . . . . . . 42
3.3 Exponential, Hyperbolic, Trigonometric Functions . . . . 46
4 Residue Theorem
49
4.1 Zeros . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
4.2 Poles & Singularities . . . . . . . . . . . . . . . . . . . . . 54
4.3 Cauchy’s Residue Theorem . . . . . . . . . . . . . . . . . 58
1
5 Applications of Residue Theorem
5.1 Fresnel’s Integrals . . . . . . . .
5.2 Fourier Transforms . . . . . . . .
5.3 Cauchy Principal Values . . . . .
5.4 Rouche’s Theorem . . . . . . . .
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63
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6 More Analysis
70
6.1 Jensen’s Theorem . . . . . . . . . . . . . . . . . . . . . . . 70
6.2 Euler’s Formulas . . . . . . . . . . . . . . . . . . . . . . . 70
6.3 Stirling Approximation . . . . . . . . . . . . . . . . . . . . 70
7 Intro to Analytic Number
7.1 Elliptic Functions . . . .
7.2 Riemann Zeta Functions
7.3 Gamma Functions . . .
Theory
70
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2
From a truly dedicated teacher to his immature students:
A Warm Note to the Mermaid
by Patrick Gallagher
There’s always danger in excess
Too much of anything can make a mess
Too many notes, with thoughts too subtle
Tossed out to sea, each in a bottle
Too late I wish I’d written less
A rain of asteroids would wreck no worse
Oh, what a mess, poor planet earth
All your blue seas boil out to space. Then
Higher forms come from some place
With waddling walks, and eyes on stalks
From every salty, bubbling puddle
They’ll fish out a little bottle
They’ll read the notes, and know the thoughts
That only you were meant to see
They’ll feel so blue, not knowing you
They’ll know how much you meant to me.
3
Course Overview
Lecture 1
(1/22/14)
For those interested in the historical development of the subject may
look up brief biographies of
Gauss, Cauchy,
Abel, Jacobi, Liouville, Riemann, Weierstrass, Picard
Hadamard, Nevanlinna, Ahlfors
We will cover the following topics, tentative
1. complex numbers,
2. polynomials and their toots. fundamental theorem of algebra
3. open sets, paths, regions, convex set, analytic functions
4. derivatives of polynomials & rational functions, Gauss-Lucas theorem
5. path length and path integral
6. Cauchy integral theorem
7. Cauchy integral formula and Cauchy derivative formula
8. Morera theorem, Liouville theorem, Cauchy-Taylor formula
9. uniform convergence
10. Weierstrass test and Abel theorem
11. exponential, hyperbolic and trigonometric functions
12. zeros of analytic functions, identity theorem, reflection principle
13. isolated singularities of analytic functions
14. Cauchy residue theorem
15. Fourier transform and Fresnel integral
16. example of integrals using residue theorem
17. counting zeros, Rouche-Glicksberg theorem, local mapping
18. analytic functions without zeros, Jenson’s formula
19. Euler formula, Weierstrass product, Stirling formula
20. elliptic functions, Liouville theorem and Weierstrass P-function
4
21. ODE for the P-function and Jacobi identities
option 1: advanced complex analysis
22. Riemann sphere and Nevanlinna first affinity theorem
23. Ramification and spherical derivative, Nevanlinna second affinity
theorem
24. Nevanlinna main theorem, multiplicity theorem, and Picard theorem
25. Ahlfor’s proof of Nevanlinna’s main theorem
option 2: number theory
26. Elementary number theory
27. Theorem of Tcebychew, Mertens on prime numbers
28. Riemann zeta function
29. Modern proof of prime number theorem
1 Complex Numbers and Complex Functions
1.1 Complex Numbers
P is the set of all prime numbers, and H is the set of all quaternions,
invented by Hamilton, but it doesn’t have much usage.
P⇢N⇢Z⇢Q⇢R⇢C⇢H
Around 1800, Gauss conjectured (when he was 15-16 years old) prime
number theorem: an approximate formula that counts the number of
prime numbers up to n. But he couldn’t prove it. He was also the first
person who used complex numbers professionally.
1800-1900 was the most active period in the development of mathematics. The two unrelated fields P and C finally married.
5
1800
1825
1850
1875
1900
1950
P
Gauss states PNT
C
Gauss used C
Cauchy studied curves, integral in space
replaced line in R by curves in R2
He invented Riemann zeta function
Riemann proved weaker
first time used complex varibles
verison PNT
in number thoery
Wiesstrass maked complex
analysis rigiuous
One hundred later Hadamard proved PNT using complex analysis
Selberg, Erdos found simpler proof. So far all proofs of PNT
used complex analysis
Definition 1. A complex number z is a point (x, y) in the xy plane,
called complex plane and denoted by C. The real coordinates x, y are
called the real part and imaginary part of z,
x = Rez
y = Imz
Among the complex numbers are those with Imz = 0, i.e. z = (x, 0).
We regard these points as real numbers and write simply (x, 0) = x, e.g.
(0, 0) = 0. called 0 complex number. Thus the set R of all real numbers
is just the x axis in C, called the real axis.
Each point z = (x, y) also has polar coordinates r > 0 and ✓ 2 R, except
for z = (0, 0) . r is called absolute value of z and ✓ is argument of z,
written 0
r = |z| ✓ = arg z
and ✓ is only determined up to an integer multiple of 2⇡.
Definition 2. The sum z1 + z2 of complex numbers z1 , z2 is defined
using Cartesian coordinates
(x1 , y1 ) + (x2 , y2 ) = (x1 + x2 , y1 + y2 )
Geometrically addition of complex numbers is vector addition z1 + z2
is the point in C for which 0 z1 , z1 + z2 , z2 are the four vertices of a
parallelogram.
Proposition 3. For z1 ,z2 ,z3 ,z 2 C
1) z1 + z2 = z2 + z1
6
2) z1 + (z2 + z3 ) = (z1 + z2 ) + z3
3) z + 0 = z
4) z + ( z) = 0 with
z = ( x, y) for z = (x, y)
Proof using Cartesian coordinates.
Definition 4. The product z1 z2 of nonzero complex numbers is defined
using polar coordinates,
(r1 , ✓1 )(r2 , ✓2 ) = (r1 r2 , ✓1 + ✓2 )
If one of z1 , z2 is 0, then the product is 0.
Proposition 5. For z1 ,z2 ,z3 ,z 2 C
1) z1 z2 = z2 z1
2) z1 (z2 z3 ) = (z1 z2 )z3
3) z1 = z
4) zz
1
= 1 if z 6= 0 with z
1
= (r
1,
✓) for z = (r, ✓)
Proof using polar coordinates
Proposition 6. For z1 ,z2 ,z 2 C
z(z1 + z2 ) = zz1 + zz2
(z1 + z2 )z = z1 z + z2 z
Definition 7. The complex number i is the point (0, 1). Equivalently
in polar
|i| = 1
arg i = ⇡/2
Proposition 8.
i2 = 1
Proof.
⇡
⇡
i2 = i · i = (1, )(1, ) = (1, ⇡) =
2
2
Proposition 9. Each z in C satisfies
z = x + iy with x = Rez, y = Imz
7
1
Proof. z = (x, y) = (x, 0) + (0, y) = x + iy.
Definition 10. The complex conjugate z̄ of a complex number z is
defined by
z̄ = x iy for z = x + iy
Equivalent in polar
|z̄| = |z| and arg z̄ =
arg z for z 6= 0, and z = 0 then z̄ = 0
Proposition 11. For z1 ,z2 ,z 2 C
1) Rez = 12 (z + z̄) Imz =
1
2i (z
z̄)
2) z is real i↵ z̄ = z
3) z̄¯ = z
4)
5) z
z̄ =
1
z̄
= z̄
1
for z 6= 0
6) z1 + z2 = z1 + z2
7) z1 z2 = z1 · z2
8) |z|2 = z z̄
9) |z1 ± z2 |2 = |z1 |2 ± 2Re(z1 z2 ) + |z2 |2
10) For each parallelogram the sum of the square of the lengths of the
two diagonals equals the sum of the square of the length of the four sides.
11) Re(z1 z2 ) = 14 (|z1 + z2 |2
|z1
z2 | 2 )
12) z1 z2 = 14 (|z1 + z2 |2 + i |z1 + iz2 |2
13) |z1 + z2 |  |z1 | + |z2 |
|z1
z2 | 2
i |z1
iz2 |2 )
1.2 Polynomials
Lecture 2
(1/27/14)
Definition 12. A polynomial is a complex valued function f of a complex variable z of the form
f (z) = c0 z n + c1 z n
1
+ ... + cn
(1.1)
with complex coefficients c0 , ..., cn . The terms are c0 z n , c1 z n 1 ,... The
constant term is cn . If c0 6= 0, then f has degree n and c0 is the
leading coefficient. The polynomials of degree 0 are the nonzero constant
function with c0 6= 0. The f = 0 function has no degree and no leading
coefficient.
8
Definition 13. For f a polynomial and z0 2 C, if
f (z0 ) = 0 and f 6= 0
then z0 is a root of f . The f = 0 function has no root.
Proposition 14. If f is a polynomial of degree n and z0 is a root of f ,
then n 1 and
f (z) = (z z0 )f1 (z)
for all z in C, with f1 is a polynomial of degree n
leading coefficient as f .
1 with the same
Proof. Suppose f is a polynomial given by (1.1). Since z0 is a root of f ,
it follows
c0 z n + c1 z n 1 + ... + cn = 0
Subtracting from (1.1)
c0 (z n
z0n ) + c1 (z n
z0n
1
1
) + ... + cn
1 (z
z0 ) = f (z)
(1.2)
Use the identity
ak
bk = (a
b)(ak
1
+ ak
2
b + ... + bk
valid for all complex numbers a, b and all integer k
z z0 from (1.2)
f1 (z) = c0 (z n
= c0 z
n 1
1
+ zn
2
z0 + ... + z0n
1
1
)
1. Factor out
) + terms of degee < n
+ terms of degee < n
1
1
showing f1 is a polynomial of degree n 1 with leading coefficient c0 .
Theorem 15. (Fundamental Theorem of Algebra, Gauss 1800) Each
polynomial f of degree 1 with all coefficients in C has a root in C.
Proof. Let f be a polynomial of degree n
1. Assume that |f | has a
minimum (we will prove that later) for some z0 2 C, i.e.
|f (z)|
|f (z0 )| 8 z 2 C
Assume f (z0 ) 6= 0 and put
g(z) =
f (z + z0 )
f (z0 )
9
then g is also a polynomial of degree n satisfying
g(0) = 1 8 z 2 C
|g(z)|
(1.3)
So the constant term of g is 1, and there is at least one other nonzero
coefficient. Let cz k be the term in g of least positive degree with nonzero
coefficient, then
g(z) = 1 + cz k + h(z)
so that h(z) is a polynomial with no term of degree  k, h may be 0.
We can choose z 6= 0 with |z| small s.t.
cz k =
cz k
and
cz k < 1 and |h(z)| < cz k
They imply
|g(z)|  1 + cz k + |h(z)| < 1
| {z }
=1 |cz k |
Contradicting to (1.3). Hence f (z0 ) = 0.
Corollary 16. Each polynomial f of degree n
f (z) = c0 (z
z1 )(z
z2 )...(z
1 factors as follows
zn )
(1.4)
for some z1 , z2 ... 2 C.
Proof by induction and use proposition 14 and theorem 15.
Definition 17. In corollary 16, the zk may not be distinct. Suppose
z1 , ..., zd are distinct and the remaining zk are repetitions of these. For
each j = 1, ..., d, denoted by mj the number of k = 1, ..., n with zk = zj
we call mj the multiplicity of the root zj . So (1.4) becomes
f (z) = c0 (z
z1 )m1 ...(z
zd )md , with n = m1 + ... + md
(1.5)
We still say that f has n roots, counted according to multiplicity.
Definition 18. For each positive integer n, the nth roots of unity are
the solutions in C of the equation
zn = 1
10
Proposition 19. For each n in N, there are exactly n distinct nth roots
of unity.
For n 3. They are vertices of the polygon with n equal sides inscribed
in the circle of radius 1 centered at 0 with one vertex at 1.
Proof. For ⇠ in C, ⇠ n = 1 means
|⇠ n | = 1 and arg ⇠ n = 0
so
|⇠|n = 1 and n arg ⇠ = 0
so
|⇠| = 1 and arg ⇠ =
2⇡k
n
for some integer k
Exercise 20. Prove for each integer n > 1
1) the sum of all the nth roots of unity is 0
2) the product of the distance from 1 to the other nth roots of unity is
n.
Proposition 21. For each n in N and each nonzero d in C, there are
exactly n distinct roots of z n d. If is any one of them, then they are
the n complex numbers of the form ⇢ = ⇠ , where ⇠ is an nth root of
unity.
Example 22. Solve polynomials of degree 3.
A special case (to which the general case easily reduces) is
z3
with z = w + w
1,
3z
2b = 0 for b 2 C
(1.6)
thus w is root of
w2
wz + 1 = 0
the binomial theorem gives
z 3 = w3 + 3w + 3w
1
+w
or
z3
3z = w3 + w
11
3
3
so (1.6) becomes
w6
2bw3 + 1 = 0
so
w3 = b ±
Since
(b +
p
b2
p
1)(b
b2
p
1
b2
(1.7)
1) = 1
So the six solutions of w to (1.7) are really three and their reciprocals,
giving at most three distinct values for z. If b = ±1, there are just two
distinct roots, one of multiplicity 2.
1.3 Point Set Topology: Derivatives
Lecture 3
(1/29/14)
In complex variable one studies complex functions of a complex variable,
defined in regions of C and taking values in C, their derivatives, and their
integrals over paths.
Definition 23. Let D be a subset C. We say D is open if for each point
p in D, there is an r > 0 so that the open disc
|z
p| < r
of radius r centered at p is entirely contained in D.
Example 24. (1) C itself is open, since every open disc is contained in
C; (2) For every choice of z1 , z2 , ..., zn in C, the set C {z1 , z2 , ..., zn }
is open. (3) Each open disc is open (4) the slit plane C {( 1, 0]} is
open.
Recall in real function of real variable, the concepts of limit, continuous,
derivative are defined on a closed interval [a, b] of R, or on a puncture
interval [a, b] {t0 }. Here we will do the same:
Definition 25. A complex valued function = (t) with real variable,
where t is defined on a closed interval [a, b] of R, or on a puncture interval
[a, b] {t0 }. For such defined on [a, b] {t0 }, and for L in C, we write
the limit L as
(t) ! L for t ! t0
if for each real ✏ > 0, 9 real
| (t)
> 0 so that
L| < ✏ for all t in [a, b] with 0 < |t
12
t0 | <
For
defined on [a, b] we say
is continuous if for each t0 in [a, b],
(t) ! (t0 ) for t ! t0
and
has derivative
(t)
t
0
if for each t0 in [a, b]
(t0 )
!
t0
0
(t0 ) for t ! t0 , t 6= t0
Definition 26. Let p, q be points in C. A smooth path from p to q is
any complex valued function = (t) defined on [a, b] with a continuous
derivative 0 and satisfying (a) = p and (b) = q..
Notice that a smooth path is the function , not the set of points (t),
though we sometimes use the same notation for both.
Example 27. (1) For any point p and q in C consider the line segment
[p, q] in C. To get a smooth path from p to q we usually take
t)p + tq for t 2 [0, 1]
(t) = (1
and denote this path by [p, q]. Here
uous.
0
=q
p is constant, so is contin-
(2) Let C be the circle centered at z0 with radius r. The usual smooth
path is given by
(✓) = z0 + r cos ✓ + ir sin ✓ for ✓ 2 [0, 2⇡]
with
0
(✓) =
r sin ✓ + ir cos ✓ = i(
z0 )
here p = q = z0 + r.
(3) For each smooth path from p to q, defined on [a, b] the opposite
smooth path
from q to p is defined on [ b, a] by
(t) = ( t).
Definition 28. A piecewise smooth path from p to q is a finite sequence of n 1 smooth paths 1 , ..., n from z0 to z1 , ..., from zn 1 to
zn with z0 = p and zn = q. We write
=
1
+ ... +
n
From now on path means piecewise smooth path.
Definition 29. A region is an open subset D of C which is connected,
i.e. for each p and q in D there is a path in D from p to q.
13
Definition 30. A subset D of C is convex if for each p and q in D, the
line segment [p, q] is in D.
Exercise 31. Prove that each convex open set is a region.
Exercise 32. Prove C
{0} is a region but not convex.
Exercise 33. Prove C
R is not a region.
Definition 34. A complex valued function f = f (z) with complex
variable, defined on an open set D, or a puncture open set D {z0 }
with z0 2 D. For such f defined on such D {z0 } and L in C, we say
L is the limit
f (z) ! L as z ! z0
if for each ✏ > 0 there is
|f (z)
> 0 so that
L| < ✏ for all z in D with 0 < |z
z0 | <
For f defined on such D, we say f is continuous if for each z0 in D,
f (z) ! f (z0 ) for z ! z0
and f has derivative f 0 in D if for each z0 2 D
f (z)
z
f (z0 )
! f 0 (z0 ) for z ! z0 , z 6= z0
z0
Proposition 35. Let D be an open set in C, let z0 be a point in D,
and let f , g be complex-valued functions defined on the punctured set
D {z0 }. Assume that for z ! z0
f (z) ! L and g(z) ! M
(1.8)
the for z! z0
f (z) + g(z) ! L + M
f (z)g(z) ! LM
If in addition g(z) 6= z for z in D
(1.9)
{z0 } and M 6= 0 then
f (z)
L
!
g(z)
M
14
(1.10)
Proof. let’s prove (1.9). Using an identity
f (z)g(z)
LM = (f (z)
L)M + L(g(z)
M ) + (f (z)
L)(g(z)
M)
and using triangle inequality
|z1 + z2 + z3 |  |z1 | + |z2 | + |z3 |
we have
|f (z)g(z)
LM |  |f (z)
L| |M |+|L| |g(z)
M |+|f (z)
L| |g(z)
M|
To make the right side of this < ✏ we make each of the three terms
< ✏/3, i.e. we choose
|f (z)
and
|f (z)
L| <
✏/3
, |g(z)
|M | + 1
M| <
p
✏/3, |g(z)
M| <
L| <
they can be met because of (1.8).
✏/3
|L| + 1
p
✏/3
Now prove (1.10). It suffices to show
1
1
!
g(z)
M
start with
1
g(z)
1
M g(z)
=
for z 2 D
M
g(z)M
z0
for small |z z0 | 6= 0 we have |g(z)|
|M | /2 since g(z) ! M and
M 6= 0. It follows that for small |z z0 | =
6 0
1
g(z)
1
2 |g(z) M |

M
M2
The right side will be < ✏ provided |g(z)
met because g(z) ! M for z ! z0 .
M | < |M |2 ✏/2 which can be
Theorem 36. If f,g are complex-valued functions with derivatives f 0 , g 0 on
an open set D then f + g and f g have derivatives on D given by
(f + g)0 = f 0 + g 0
(f g)
0
0
= f g + fg
(1.11)
0
If g(z) 6= 0 for z 2 D then f /g has a derivative
✓ ◆0
f
gf 0 f g 0
=
g
g2
15
(1.12)
(1.13)
Proof. Prove (1.12), use (1.9). For z0 2 D and z 2 D
{z0 }
(f g)(z) (f g)(z0 ) = (f (z) f (z0 ))g(z0 )+f (z0 )(g(z) g(z0 ))+(f (z) f (z0 ))(g(z) g(z0 ))
or
(f g)(z)
z
(f g)(z0 )
f (z)
=
z0
z
f (z0 )
g(z)
g(z0 )+f (z0 )
z0
z
g(z0 ) f (z)
+
z0
z
f (z0 ) g(z)
z0
z
then z ! z0 , we have (1.12).
Similarly to show (1.13), suffice
✓ ◆0
1
=
g
and use
1
g(z)
z
1
g(z0 )
=
z0
g(z)
z
g0
g2
g(z0 ) 1
1
z0
g(z) g(z0 )
Exercise 37. Let
(t) = t2 sin
1
for 0¡ |t|  1 and (0) = 0
t
Graph show that has derivative on [ 1, 1] with 0 (0) = 0 but
is not continuous on [ 1, 1] : i.e. 0 (t) has no limit for t ! 0.
0 (0)
1.4 Analytic Functions
Lecture 4
(2/3/14)
Definition 38. A complex valued function of a complex variable defined
on a region D and having a derivative there is said to be analytic in D.
Proposition 39. (1) Each polynomial (1.1) is analytic in C, its derivative is
f 0 (z) = ncn z n 1 + ... + c1
(2) Each rational function f = g/h, g, h are polynomial, h 6= 0 is analytic in the region D = C {z1 , ..., zn } where z1 , ..zn are the roots of h.
The derivative of f is (1.13) is also an analytic rational function in the
same D.
16
g(z0 )
(z z0 )
z0
Proof. Let’s prove part (1). It suffices in view of (1.11) to observe that
c0 = 0 for each constant c and to show that (cz k )0 = kcz k 1 for constant
c and k 2 N. In fact for z0 and z 2 C
cz k
cz0k = c(z
z0 )(z k
1
+ z0 z k
2
+ ... + z0k
1
)
for z ! z0 with z 6= z0 ,
cz k
z
cz0k
! c(z k
z0
1
+ z0 z k
2
+ ... + z0k
1
) = kcz0k
1
Proposition 40. Let k > 1. If f1 , ..., fk are analytic in D, then so is
f1 · ... · fk and
(f1 · ... · fk )0 = f10 f2 ...fk + ... + f1 f2 ...fk0
(1.14)
in particular if f is analytic in D, so is f k and
(f k )0 = kf k
1 0
f
Proof. The case k = 2 is the product rule, then use induction.
Proposition 41. Let z0 be a root of multiplicity m of a polynomial f ,
(a) if m = 1, then f 0 (z0 ) 6= 0. (b) if m > 1, then z0 is a root of f 0 of
multiplicity m 1.
Proof. By (1.5), f (z) = (z
g(z0 ) 6= 0, i follows that
f 0 (z) = m(z
z0 ) m
1
z0 )m g(z), where g is a polynomial with
g(z) + (z
z0 )m g 0 (z) = (z
z0 ) m
1
h(z)
with h(z) = mg(z)+(z z0 )g 0 (z). Thus h is a polynomial with h(z0 ) 6= 0.
Therefore if m = 1 then f 0 (z0 ) 6= 0, while if m > 1 then z0 is a root of
f 0 of multiplicity m 1.
Proposition 42. If f is a polynomial of degree n, then there are fewer
than n distinct complex numbers w for which f w has a multiple root
i.e. a root of multiplicity > 1.
Proof. If w1 , ..., wn are distinct, and f (z1 ) w1 = 0, ..., f (zn ) wn = 0,
then z1 , ..., zn are distinct. If in addition zk is a multiple root of f wk
for each k, then each zk is a root of f 0 , too many roots for a polynomial
of degree n 1.
17
1 which is real on R
Exercise 43. Let f be a polynomial of degree n
Prove that
(1) f has at most one more real root than f 0
(2) f 0 has no more non-real roots than f
(3) if all the roots of f are real, then all the roots of f 0 are real.
Lemma 44. For each polynomial f of degree n
1, with z1 , ..., zn
n
X 1
f0
(z) =
for z 6= z1 , ..., zn
f
z zk
(1.15)
k=1
Proof. From f (z) = c0 (z
f 0 (z) = c0
z1 )...(z
n
X
zn ) and (1.14)
z1 )...(z\
zk )...(z
(z
zn )
k=1
ŵ means that the factor w is to be omitted. Dividing the formula for
f 0 (z) by the formula for f (z) gives, after cancellation, the formula for
(f 0 /f )(z).
Definition 45. The convex hull H of complex numbers z1 , ..., zn (not
necessarily distinct) is the set of all points z of the form
z=
1 zz
+ ... +
n zn
with all
0 and
k
1
+ ... +
n
=1
(1.16)
Exercise 46. Prove that the convex hull of z1 , ..., zn is contained in
every convex set which contains z1 , ..., zn .
Theorem 47. (Gauss-Lucas) Let f be a polynomial of degree n
1.
0
Then each root of f is contained in the convex hull H of the roots of f .
Proof. By lemma 44 let z1 , ..., zn be the roots of f . Let z ⇤ be the roots
of f 0 . If z ⇤ is one of z1 , ..., zn , then z ⇤ is in H, set one k = 0 the rest 0.
If z ⇤ is not one of z1 , ..., zn , we put z = z ⇤ in (1.15) gives
n
X
k=1
1
z⇤
zk
=0
conjugating this, and using 1/w̄ = w/ |w|2 for w 6= 0 gives
n
X
z⇤
|z ⇤
k=1
zk
=0
zk | 2
18
writing this sum as a di↵erence of two sums, and factoring z ⇤ from the
numerators of the first sum gives
n
X
⇤
z =
k zk
k=1
with
k
for k = 1, ..., n. Clearly
=
1
|z ⇤ zk |2
Pn
1
k=1 |z ⇤ zk |2
satisfies (1.16).
k
Exercise 48. Prove if f is a polynomial of degree n with n distinct
real roots x1 < ... < xn then 9 a root x⇤j of f 0 with xj 1 < x⇤j < xj for
j = 2, ..., n.
Theorem 49. (Marcel Riesz 1926) Let f be a polynomial of degree
3 with n distinct real roots separated as in above exercise by the n 1
distinct real roots of f 0 . Then the smallest di↵erence between consecutive
roots of f 0 is greater than the smallest di↵erence between consecutive
roots of f .
Proof. By (1.16) with the zk = xk and with z = x⇤j or x⇤j
n
X
1
x⇤
k=1 j
xk
n
X
1
x⇤
k=1 j 1
xk
1,
=0
for all j = 3, ..., n. Combining the k = 2, ..., n terms in the first sum
with the k = 1, ..., n 1 terms in the second sum gives for all j = 3, ..., n
!
n
X
1
1
1
1
+
+
=0
(1.17)
x⇤j x1
x⇤j xk
x⇤j 1 xk 1
xn x⇤j 1
k=1
Note that the two terms at the ends are positive. If the assertion of the
theorem is false, then
x⇤j
x⇤j
1
 xk
xk
1
(1.18)
for some j = 3, ..., n and all k = 2, ..., n, from this we will show that for
some j = 3, ..., n all the terms in the middle sum in (1.17) are 0 giving
a contradiction.
In fact (1.18) is
x⇤j
xk  x⇤j
19
1
xk
1
for some j = 3, ..., n and all k = 2, ..., n. For k < j we have x⇤j xk > 0
and for k j we have k 1 j 1 so x⇤j xk < 0. Thus in both cases
the two di↵erences x⇤j xk and x⇤j 1 xk 1 have the same sign. Now
we use the fact that if u, v have same sign, and u  v then 1/u 1/v,
so
1
1
0
⇤
⇤
xj xk
xj 1 xk 1
for some j = 3, ..., n and all k = 2, ..., n.
2 Complex Integration
2.1 Integral and Path Integral
Lecture 5
(2/5/14)
We define the integral
ˆ
b
=
a
b
ˆ
(t)dt
a
of a complex-valued continuous function
on a closed interval [a, b] ⇢ R.
Definition 50. Given a closed interval [a, b] ⇢ R, a partition of [a, b] is
any finite increasing sequence in R from a to b
P : a = t0  t1  ...  tn = b
the norm kPk is defined by
kPk := max{t1
t0 , ..., tn
tn
1}
Definition 51. Given a partition P of a closed interval [a, b] of R, we
call intermediate points any finite sequence
P ⇤ := t⇤1 , ..., t⇤n with tk
1
 t⇤k  tk for k = 1, ..., n
Definition 52. Given a complex-valued function defined on a closed
interval [a, b], a partition P of [a, b] and a sequence P ⇤ of intermediate
points, the corresponding Riemann sum S( ) is defined by
S( ) =
n
X
(t⇤k )(tk
tk
1)
k=1
observe that S( ) depends on all three of
20
, P, P ⇤ .
Proposition 53. (also served as a definition for integral) For each continuous on [a, b] as above there is a unique complex number called the
integral of and denoted as above which has the following property: For
each ✏ > 0 9 > 0 so that
ˆ b
S( )
< ✏ for all those P, P ⇤ as above for which kPk <
a
In this sense, the integral is the limit of Riemann sums as kPk ! 0.
Proof see introduction to modern analysis I notes.
Proposition 54. For continuous
ˆ
a
b
and
on [a, b] and constant c
ˆ b
ˆ b
( + ) =
+
a
a
ˆ b
ˆ b
c
= c
a
a
ˆ b
ˆ c
ˆ b
=
+
a
a
c
ˆ b
ˆ b

| |
a
for a  c  b
(2.1)
a
ˆ
b
1 = b
a
a
One can prove those using Riemann sums.
Corollary 55. For value-valued continuous
ˆ
b
ˆ
b
a
a
0 if
ˆ b
and
0on [a, b]
if
on [a, b]
on [a, b]
(2.2)
(2.3)
a
0 implies | | = . (2.3) follows
Proof. (2.2) follows from (2.1), since
from (2.2) on replacing by
.
Definition 56. For each smooth path , (here smooth doesn’t mean
piecewise smooth) the length L = L( ) is defined by
L :=
ˆ
b
a
21
0
(t) dt
Example 57. Prove the opposite path
has the same length as .
Recall if
is defined on [a, b], then
is defined on [ b, a], and
( )(t) = ( t), thus the length of
is
ˆ a
ˆ a
ˆ b
0
0
0
( t) dt =
(q) dq =
(q) dq
b
b
a
Definition 58. For each piecewise smooth path
with length L1 , ..., Ln , the length of is
=
1
+
2
+ ... +
n
L1 + L2 + ... + Ln
Definition 59. For each region D in C each complex-valued f defined
and continuous on D and each piecewise smooth path in D the path
integral
ˆ
ˆ
f=
of f over
ˆ
and if
f (z)dz
is defined by
ˆ
f = f ( (t)) 0 (t)dt for smooth
is piecewise smooth
ˆ
ˆ
ˆ
f :=
f + ... +
1
f, for
defined on [a, b]
1 , ..., n
each smooth
n
Proposition 60. For all D, f, g, ↵, , , L as above and c in C
ˆ
ˆ
ˆ
(f + g) =
f+ g
ˆ
ˆ
cf = c f
ˆ
ˆ
ˆ
f =
f + f if =↵+
↵
ˆ
f
 M L where M is the maximum of |f (z)| for z on(2.4)
Proof. We prove (2.4). If is a smooth path, then
ˆ
ˆ b
f =
f ( (t)) 0 (t)dt

ˆ
a
b
a
|f ( (t))|
0
(t) dt 
22
ˆ
a
b
M
0
(t) dt = M L
If
ˆ
is piecewise smooth, then
ˆ
ˆ
f =
f + ... +
f
n
1

ˆ
f + ... +
1
ˆ
n
f  M1 L1 + ... + Mn Ln  M (L1 + ... + Ln ) = M L
Proposition 61. For each counterclockwise circle C centered at z0
ˆ
dz
= 2⇡i
z0
C z
Proof. By example 27(2) 0 = i(
z0 ) for C, so
ˆ
ˆ 2⇡ 0
ˆ 2⇡
dz
(✓)d✓
=
=
id✓ = 2⇡i
z0
(✓) z0
C z
0
0
Below is our first version of fundamental theorem of calculus for path
integrals:
Proposition 62. Let D be a region in C, f a complex-valued continuous
function on D, and a path in D from p to q. Assume that 9 a function
F defined on D with F 0 = f i.e. f has an anti derivative F , then
ˆ
f = F (q) F (p)
If besides f having an anti derivative, is a closed path i.e. with p = q
then
ˆ
f =0
Proof. For smooth
ˆ
ˆ b
f =
F 0 ( (t)) 0 (t)dt
a
=
ˆ
b
(F ( (t)))0 dt (by calculus chain rule)
a
= F ( (b))
= F (q)
F ( (a)) (by calculus fundamental thm)
F (p)
Then complete the proof for piecewise smooth .
23
Exercise 63. Prove that there is no F define on D = C {z0 } satisfying
F 0 (z) =
Exercise 64. Prove that
path .
´
1
z
z0
on D
(2.5)
f = 0 for each polynomial f and each closed
2.2 Cauchy Integral Theorem
Lecture 6
(2/10/14)
This section is the most important section, from which all the rest of
the course follows. The Cauchy’s integral theorem (1820) is related to
Green’s theorem (also found in 1820) in vector integral in R2 . Here we
derive it by the method of Goursat, which requires weaker hypotheses
than those for Green’s theorem stated in Calculus.
Theorem 65. (Cauchy’s integral theorem) Let D be a convex region in
C, and f is analytic in D, and is a closed path in D then
ˆ
f (z)dz = 0
The proof depends on two lemmas.
Lemma 66. (Goursat, 1900) If D is a convex region in C and f is
analytic in D and
is a triangle in D then
ˆ
f (z)dz = 0
Proof. Starting with we get four triangles (1) , (2) , (3) , (4) each
quarter the size of by connecting the midpoints of the sides of . After
adding canceling integrals in both directions over the new line segments,
we find that the path integral is equal to
ˆ
ˆ
ˆ
ˆ
ˆ
f=
f+
f+
f+
f
(1)
Let
´
(1)
(2)
(3)
f be the largest of the four i.e.
ˆ
f
(1)
ˆ
24
f
(2,3,4)
(4)
then by triangle inequality
ˆ
f 4
ˆ
f
(1)
then repeat this process for (1) ,´ cut it into four triangles (with relabeling) (1) , (2) , (3) , (4) with
be the largest of the four, etc,
(2) f
thus
ˆ
ˆ
f  4n
f n2 N
(2.6)
(n)
we have the length of
(n)
Ln =
L
2n
(2.7)
Now we use a fact to be proven later: As n ! 1,9 z ⇤ is contained in
all n (possible on some edges n ). Since z ⇤ 2 n ⇢ ⇢ D, and f is
analytic in D
f (z) f (z ⇤ )
! f 0 (z ⇤ ) for z ! z ⇤
z z⇤
define
(
0
z = z⇤
r(z) = f (z) f (z ⇤ )
f (z ⇤ ) z 6= z ⇤
z z⇤
then
f (z) = f (z ⇤ ) + f 0 (z ⇤ )(z
z ⇤ ) + r(z)(z
z⇤)
and
r(z) ! 0 as z ! z ⇤
(2.8)
Thus
ˆ
f (z) =
ˆ
⇤
0
⇤
f (z ) + f (z )(z
⇤
z ) dz +
ˆ
r(z)(z
z ⇤ )dz
the first integral on the right is 0, for the integrand is a polynomial of
degree 1 in z. We estimate the second integral on the right
ˆ
f (z)dz  Mn L2n
(n)
where Mn is the maximum of |r(z)| for z on n . Here we have used that
|z z ⇤ |  Ln for z on n . By (2.8) and Ln ! 0 for n ! 1, we have
Mn ! 0 as n ! 1
25
(2.9)
By (2.6), (2.7)
ˆ
f  Mn L 2
ˆ
f (z)dz = 0
then by (2.9)
Lemma 67. Let D be a convex region in C, let f be continuous on D,
and assume that
ˆ
f =0
for every triangle
by
in D. Choose any point p in D, and define F in D
F (z) :=
ˆ
f
[p,z]
with [p, z] the line segment from p to z. Then F ” = f in D. In particular
f has an analytic anti derivative in D.
Proof. Because D is convex the line segment [p, z] is contained in D
for every z 2 D. Therefore F is well-defined. It remains to show that
F 0 = f . For z0 and z in D consider the triangular path
= [p, z] + [z, z0 ] + [z0 , p]
Since D is convex,
´
is contained in D, so by assumption
ˆ
ˆ
ˆ
f+
f+
f =0
[p,z]
[z,z0 ]
f = 0 i.e.
[z0 ,p]
which we may write as
F (z)
F (z0 ) =
ˆ
f=
[z,z0 ]
ˆ
f (z0 ) +
[z,z0 ]
ˆ
(f
f (z0 ))
[z,z0 ]
From which
|F (z)
i.e.
F (z)
z
F (z0 )
F (z0 )
z0
f (z0 )(z
z0 )|  M (z, z0 ) |z
z0 |
f (z0 )  M (z0 , z) for z 6= z0
26
where M (z0 , z) is the maximum of |f f (z0 )| on the line segment [z0 , z].
Since f is continuous in D, we have M (z0 , z) ! 0 for z ! z0 , giving
F 0 (z0 ) = f (z0 ) for all z0 in D
Now we prove Cauchy’s integral theorem.
Proof. By above two lemmas, each function f analytic in convex regions
D has an anti derivative F in D. It follows from proposition 62.
Use anti derivative lemma one can define log. Recall natural logarithm
of calculus is defined by
ˆ x
d⇠
log x =
for real x > 0
1 ⇠
The following 4 exercise introduce complex log, which is defined and
analytic in the slit plane S = C {( 1, 0]} and agrees with the log of
calculus on (0, 1).
Exercise 68. Define log by
log(z) =
ˆ
[1,z]
d⇣
for z in S
⇣
[1, z] is a straight path on the complex plant from (1, 0) to complex z.
Since the integrand 1/⇣ is continuous in C {0} and therefore in S, and
since [1, z] is in S the integral is defined. Prove that
(log z)0 =
1
for z in S
z
Hint: although S is not convex, we can take a smaller convex region.
Exercise 69. With log defined above, prove that
ˆ
d⇣
log z =
for z in S
⇣
where
is the piecewise smooth path consisting of the line segment
[1, |z|] following by an arc from |z| to z of the circle of radius |z| centered
at 0.
27
Exercise 70. Use exercise 69 to prove that
log z = log |z| + i arg z for z in S
where log in log |z| is the log of calculus and we take
⇡ < arg z < ⇡.
Exercise 71. Use exercises 70 to show that
log i = i⇡/2 and log i =
and
log z !
(
i⇡
i⇡
i⇡/2
Imz > 0
for z !
Imz < 0
1
this shows log cannot be extended continuously from S to C
{0}.
2.3 Cauchy’s Integral Formula
Theorem 72. (Cauchy’s Integral Formula) Let D be a convex region,
let f be a function analytic in D, let C be a counterclockwise circle in
D and let z0 be any point inside C then
ˆ
1
f (z)
f (z0 ) =
dz
2⇡i C z z0
Corollary 73. Each such f (z0 ) is determined by the values of f for z
on C.
Proof. Of the theorem. We first show that all sufficiently small r > 0
ˆ
ˆ
f (z)
f (z)
dz =
dz
(2.10)
z0
z0
C z
Cr z
where Cr is the counterclockwise circle of radius centered at z0 and
inside C. In fact we can connect C and Cr with some paths so the
annulus looks like consists of some fan-shaped regions. The internal
path integrals cancel.
ˆ
ˆ
ˆ
ˆ
f (z)
f (z)
f (z)
f (z)
dz =
dz +
dz +
dz + ...
z0
z0
z0
z0
C z
Cr z
F an1 z
F an2 z
One then applies Cauchy’s integral theorem to each fan. because f (z)/(z
z0 ) is analytic in a convex region containing the fan. The convex region
could be the
of D with the half plane whose edge across z0 .
´ intersection
f (z)
Thus each F an z z0 dz = 0.
28
Next show
ˆ
Cr
f (z)
dz ! 2⇡if (z0 ) for r ! 0
z z0
In fact, using (2.5) for Cr gives
ˆ
ˆ
f (z)
f (z)
dz = 2⇡if (z0 ) +
z0
z
Cr
Cr z
We have
ˆ
Cr
f (z)
z
f (z0 )
dz
z0
(2.11)
(2.12)
f (z0 )
dz  Mr Lr
z0
with Lr = 2⇡r and
Mr = max
f (z)
z
f (z0 )
for z on Cr
z0
Since f is analytic in D and z0 is in D, we have Mr ! |f 0 (z0 )| for r ! 0,
with Lr ! 0 for r ! 0, we conclude that the integral on the right in
(2.12) has limit 0. Thus combining (2.10), (2.11), prove the theorem.
Exercise 74. In the proof we assumed (1), so prove the following
(1) intersection of two convex sets is a convex set;
(2) intersection of two open sets is an open set;
(3) intersection of two convex regions is a convex region.
Exercise 75. Use Cauchy’s Integral theorem and Cauchy’s Integral Formula to show for each counterclockwise circle C and each z0 not on C
(
ˆ
2⇡i z0 inside C
dz
=
z0
0
z0 outside C
C z
Lecture 7
(2/12/14)
Definition 76. An entire function is a function analytic in all of C.
Thus each polynomial is entire, but log is not.
Definition 77. A function f on D is bounded if for some positive B
|f (z)|  B
for all z 2 D.
Lemma 78. No non constant polynomial f is bounded. In fact for each
such f
|f (z)| ! 1
(2.13)
for |z| ! 1.
29
Proof. For each such f = c0 z n + ... + cn ,
f (z)
c1
cn
= c0 +
+ ... + n
n
z
z
z
which ! c0 for |z| ! 1, from which (2.13) follows.
Theorem 79. (Liouville 1840) Each bounded entire function is constant.
Proof. Let f be entire and suppose |f (z)|  B for all z in C. Let p, q be
any two points in C. For each r > max{|p| , |q|}, Cauchy integral formula
gives
◆
ˆ ✓
1
1
1
f (q) f (p) =
f (z)dz
2⇡i Cr z q z p
where Cr is the counterclockwise circle of radius r centered at 0.
For each z on Cr we have |z| = r so |z p| r |p| and |z q| r |q| ,
so
1
1
q p
|q p|
=

z q z p
(z q)(z p)
(r |q|)(r |p|)
It follows by the ML inequality, cf (2.4),
|f (q)
Thus f (q)
f (p)| 
B |q p|
2⇡r ! 0 for r ! 1
(r |q|)(r |p|)
f (p) doesn’t depend on r, so f is constant.
One of the application of Liouville’s theorem is a short proof of the
fundamental theorem of algebra (already proved before theorem 15.
Proof. (new proof) suppose f is a non constant polynomial with no
root. Put g = 1/f . then g is an entire function. By the lemma above,
it follows that
g(z) ! 0
(2.14)
for |z| ! 1. Therefore g is bounded. (In more detail, (2.14) implies
|g(z)|  1 for |z| > R for some large constant R. However |g(z)| being
continuous, by a general theorem in analysis, is also bounded for |z|  R)
By Liouville’s theorem g as a bonded entire function must be constant.
By (2.14) the constant must be 0, an impossible value for a reciprocal.
Contradiction.
30
Finally we show another proof of fundamental theorem of algebra, due
to N. Ankeny (1950’s). His proof uses the Cauchy integral theorem, but
not the Cauchy integral formula:
Proof. For each non constant polynomial f (z) = c0 z n + ... + cn put
f ⇤ (z) = c̄0 z n + ... + c̄n
then
f (z) = f ⇤ (z̄)
(2.15)
f⇤
so the roots of
are the conjugates of the roots of f. Thus if f has no
roots, neither does f ⇤ so if we put h = f f ⇤ then h is a polynomially with
no roots, and again by (2.15)
h(x) = f (x)f ⇤ (x) = f (x)f (x) = |f (x)|2 > 0, for all real x
also since h has degree
|h(z)|
(2.16)
2, 9 constant c > 0 s.t.
c |z|2 for all sufficiently large |z|
(2.17)
Since h is entire and has no roots, 1/h is also entire. By Cauchy’s
integral theorem, for each r > 0
ˆ
ˆ
1
1
dx =
dz
h(z)
[ r,r] h(x)
r
where r is the clockwise semicircle connecting r, r. By (2.16) the
integral on the left is a positive increasing function of r for r > 0. By
(2.17) and the ML inequality the integral on the right for sufficiently
large r has absolute value
ˆ
1
⇡r
dz  2 ! 0 for r ! 1
h(z)
cr
r
not possible for a positive increasing function. The contradiction shows
that a non constant polynomial with no roots is impossible.
2.4 Cauchy’s Derivative Formulas
The Cauchy’s integral formula can be used to reveal a bunch of pleasant
properties of analytic functions. For example we can take derivative of
it (for now not worrying about strict rigor)
ˆ
ˆ
ˆ
d 1
f (z)
1
d f (z)
1
f (z)
0
f (z0 ) =
dz =
dz =
dz
dz0 2⇡i C z z0
2⇡i C dz0 z z0
2⇡i C (z z0 )2
(2.18)
31
The pass of derivative and integral will be make rigor later.
Let’s look at this in a bit more detail: for z0 ,z1 both inside C,
◆
ˆ ✓
ˆ
1
1
1
1
z1 z0
f (z1 ) f (z0 ) =
f (z)dz =
f (z)dz
2⇡i C z z1 z z0
2⇡i C (z z1 )(z z0 )
so for z0 ,z1 both inside C, and z0 6= z1
ˆ
f (z1 ) f (z0 )
1
=
z 1 z0
2⇡i C (z
f (z)
z1 )(z
z0 )
dz
All the remains is to let z1 ! z0 and say that
ˆ
ˆ
f (z)
f (z)
dz !
dz
z1 )(z z0 )
z0 ) 2
C (z
C (z
this requires uniform convergence, which we will study later.
Theorem 80. (Cauchy’s Derivative Formulas) Let f be analytic in any
region D. Then for n = 1, 2, ...,
1. f (n) is analytic in D. In particular f 0 is analytic in D.
2. For D is convex, C a counterclockwise circle in D and z0 inside C
ˆ
n!
f (z)
(n)
dz
f (z0 ) =
2⇡i C (z z0 )n+1
Proof. With the usual convention f (0) = f, prove by induction n = 0 is
Cauchy’s integral formula. To go from n 1 to n, we use similar thing
in (2.18)
ˆ
ˆ
d (n 1) (n 1)!
d
f (z)
n!
f (z)
f (n) (z0 ) =
f
=
dz
=
dz
n
dz0
2⇡i
z0 )
2⇡i C (z z0 )n+1
C dz0 (z
the passing of derivative will be justified later. Because we have been
able to calculate f (n) at each point inside C, it follows that f (n 1) is
analytic in this open disc. However each point in D is inside some circle
C in D (because D is open), so f (n 1) is analytic in D.
There are three well-known consequences of Cauchy’s Derivative Formulas: theorem 81, theorem 84 & Taylor’s formula theorem 85.
Theorem 81. (Morera) Let f be continuous in a convex region D. If
ˆ
f =0
for each triangle
in D, then f is analytic in D.
32
Proof. The hypothesis implies that f = F 0 for some F, by anti derivative
lemma 67. F is analytic, so f is analytic by theorem 80.
Corollary 82. For f continuous in a convex region D. f is analytic in
D i↵
ˆ
f =0
for each triangle
in D.
Proof. In one direction this follows from Cauchy’s integral theorem, and
in the other direction it is Morera’s theorem.
Corollary 83. For f continuous in a convex region D. f is analytic in
D i↵
ˆ
f =0
for each closed path
in D.
Theorem 84. (Cauchy’s Derivative Estimates) Let f be analytic in a
convex region D, let C be a circle in D of radius r centered at z0 . Then
f (n) (z0 )
B
 n
n!
r
for n = 0, 1, 2, ... where B is the maximum of |f (z)| for z on C.
Proof. By the Cauchy derivative formula
f (n) (z0 )
1
=
n!
2⇡i
Since |z
ˆ
C
(z
f (z)
dz
z0 )n+1
z0 | = r for z on C, the ML inequality gives
f (n) (z0 )
1 B
B

2⇡r = n
n+1
n!
2⇡i r
r
33
2.5 Cauchy-Taylor Formulas
Lecture 8
(2/17/14)
Theorem 85. (Cauchy-Taylor Formulas) Let f be analytic in a convex
region D and let C be a circle in D centered at z0 . Then for all z inside
C
1
X
f (n) (z0 )
f (z) =
(z z0 )n
n!
n=0
It follows that the Taylor series for f about z0 converges to f in each
open disc in D centered at z0 .
Proof. We deal with the case z0 = 0. Let C be a counterclockwise circle
in D centered at 0. For z in C the Cauchy integral formula (with z0 and
z replaced by z and ⇣) reads
ˆ
1
f (⇣)
f (z) =
d⇣
(2.19)
2⇡i C ⇣ z
wN = (1
Using the identity 1
1
1
=
w
N
X1
w)(1 + w + ... + wN
wn +
n=0
1)
in the form
wN
for w 6= 1
1 w
with w = z/⇣ and dividing by ⇣ gives
N
1
⇣
z
=
X zn
1
1
z
1
=
+ ( )N
⇣ 1 z/⇣
⇣ n+1
⇣ ⇣ z
(2.20)
n=0
for ⇣ on C and z inside C.
Substituting (2.20) into (2.19) gives
f (z) =
N
X1
n=0
1
2⇡i
ˆ
C
N
X1 f (n) (0)
f (⇣)
n
d⇣z + rN =
z n + rN
⇣ n+1
n!
n=0
by the Cauchy derivative formulas for n = 0, 1, ..., N
ˆ
1
z
f (⇣)
rN =
( )N
d⇣
2⇡i C ⇣ ⇣ z
1 and with
Let B be the maximum of |f (⇣)| for ⇣ on C. Using also |⇣| = r (the
radius of C) and |z| < r for z inside C, the ML inequality gives
|rN | 
r
Br |z| N
( ) ! 0 for N ! 1
|z| r
34
thus
f (z) =
1
X
f (n) (0)
n=0
n!
z n for z inside C
For z in any open disc in D centered at z0 there is a circle C in D
centered at z0 with z inside C to which the series formula above applies.
Finally the general case, with z0 not necessarily equal to 0: Put g(z) =
f (z + z0 ), so that g is analytic in the convex region D z0 gotten by
translating D by z0 , we have g (n) (0) = f (n) (z0 ), so for z inside any
open disc in D centered at z0 , i.e. z z0 inside the translated disc in
D z0 centered at 0,
f (z) = g(z
z0 ) =
1
X
g (n) (0)
n=0
n!
z)n0
(z
=
1
X
f (n) (z0 )
n=0
n!
(z
z0 ) n
Considering the Cauchy-Taylor Theorem for R, we may wonder if
h(x) =
1
X
h(n) (0)
n=0
n!
xn for some x 6= 0
could be proved for each real valued function h of a real variable, assumed
to be defined and with derivatives of all orders on all of R.The answer
is surprisingly NO. This is the goal of the next four exercises.
Exercise 86. Using the fact that
ex =
1
X
xn
=0
n!
prove that
xn  n!ex
for x
0 and each non negative integer n.
Exercise 87. Replacing n by n + 1 in the inequality above, prove that
xn e
x
!0
for x ! 1, for each non negative integer n.
35
Exercise 88. Replacing x by 1/x2 in the limit above to prove that
1/x2
e
!0
x2n
for x ! 0, x 6= 0 for each non negative integer n.
Exercise 89. Put
h(x) =
(
1/x2
e
0
x 6= 0
x=0
(1) prove that h is continuous on R, using the case n = 0 in exercise 88
to deal with x = 0.
(2) prove by induction that
1
h(n) (x) = pn ( )e
x
1/x2
for x 6= 0 for each non negative integer n, for some polynomials pn and
calculate p1 (u), p2 (u), p3 (u), p4 (u).
(3) Conclude from (2) and exercise 88 that h has derivative of all orders
at each point of R and
h(n) (0) = 0
for each non negative integer n, so
1
X
h(n) (0)
n=0
n!
xn = 0
hence it is not equal to h(x).
Next a generalization of Liouville’s theorem, proved by combining the
Cauchy-Taylor formula with the Cauchy derivative estimates
Theorem 90. (generalized Liouville) Let f be an entire function, and
let N 0 be an integer, then
f is a polynomilal of degree  N
(2.21)
i↵ there is a constant A > 0 s.t.
|f (z)|  A |z|N for |z|
36
1
(2.22)
Proof. Suppose first that f is entire and satisfies (2.21). Then
f (z) = c0 z N + ... + cN
By the triangle inequality
|f (z)|  |c0 | |z|N + ... + |cN |
For |z|
1 we have |z|n  |z|N for n = 0, ..., N so f satisfies (2.22) with
A = |c0 | + ... + |cN |
Suppose f is entire and satisfies (2.22). Applying the Cauchy derivative
estimates for r 1 with z0 = 0 gives for each integer n 0
f (n) (0)
ArN
 n
n!
r
for each n > N , the upper bound ! 0 for r ! 1 so
f (n) (0) = 0 for n > N
Thus the Cauchy-Taylor formula with z0 = 0 reduces to
f (z) =
N
X
f (n) (0)
n=0
n!
zn
a polynomial of degree  N .
Exercise 91. Apply Cauchy-Taylor to prove that for |z| < 1,
log(1 + z) = z
z2 z3
+
2
3
z4
+ ...
4
(2.23)
Actually above is true for |z|  1 except for z = 1. To show that, we
start with
ˆ
ˆ
dw
d⇣
log(1 + z) =
=
(2.24)
[1,1+z] w
[0,z] 1 + ⇣
for 1 + z in the slit plane. For ⇣ 6= 1 and each integer N
1
=1
1+⇣
⇣ + ⇣2
⇣ 3 + ... ⌥ ⇣ N
1
1,
⇣N
1+⇣
(2.25)
⇣N
d⇣
1+⇣
(2.26)
±
substituting (2.25) into (2.24) gives
log(1 + z) = z
z2 z3
+
2
3
zN
... ⌥
±
N
37
ˆ
[0,z]
For |z|  1 with z 6= 1 the ML inequality does not show that the integral
in (2.26) ! 0 since for |z| = 1 the bound M doesn’t ! 0. However the
integral does ! 0 because putting ⇣ = tz with t 2 [0, 1] we get
ˆ
[0,z]
⇣N
d⇣ =
1+⇣
ˆ
0
1
tN z N
1
zdt 
1 + tz
m
ˆ
1
tN dt
(2.27)
0
with m the minimum of |1 + tz| for t in [0, 1] and the integral on the
right in (2.27) is 1/(N + 1) ! 0 proving (2.23).
One application of above is that for
✓
log(2 cos ) = cos ✓
2
⇡<✓<⇡
cos 2✓ cos 3✓
+
2
3
cos 4✓
+ ...
4
(2.28)
and
✓
sin 2✓ sin 3✓ sin 4✓
= sin ✓
+
+ ...
2
2
3
4
To see these, for |z| = 1 with z 6= 1, put arg z = ✓ with
For the isosceles triangle of vertics 0, 1 and 1 + z
arg(1 + z) =
=
so
⇡ < ✓ < ⇡.
✓
2
and
|1 + z| = 2 cos
(2.29)
= 2 cos
✓
2
✓
✓
log(1 + z) = log(2 cos ) + i
2
2
Since |z n | = 1 and arg(z n ) = n✓, equation (2.28), (2.29) express the
equality of the real, imaginary parts of the two sides of (2.23).
Observe that for ✓ = ±⇡ the series in (2.28) diverges to 1 so (2.28)
still holds provided we interpret log 0 as 1. For ✓ = ±⇡ the terms
of series (2.29) are all 0, the series converges to 0 so (2.29) is not true
under any interpretation. However there is a rationale for what happens
with (2.29) at ±⇡ about which later we study Fourier series.
3 Little Analysis
3.1 Uniform Convergence
Lecture 9
(2/24/14)
First we adapt the definition of convergence to complex sequences.
38
Definition 92. Let sn and s be complex numbers, n 2 N then sn
converges to s written sn ! s if for each real ✏ > 0 there is a positive
integer N s.t. |sn s| < ✏ for all n > N . s is the limit of the sequence.
Definition 93. Let fn and f be complex functions defined on a set S,
n2N
(1) fn ! f pointwise on S means fn (z) ! f (z) for each z 2 S i.e.
8 z 2 S, 8 ✏ > 0, 9 N✏,z 2 N : |fn (z)
f (z)| < ✏, 8 n > N
(2) fn ! f uniformly on S means
8 ✏ > 0, 9 N✏ 2 N : |fn (z)
f (z)| < ✏, 8 n > N, 8 z 2 S
Example 94. Let fn = xn for 0  x  1. Then fn ! f pointwise on
[0, 1] with
(
0 x 2 [0, 1)
f (x) =
1 x=1
Recall the definition of continuity
Definition 95. Let S be a subset of C, f a complex valued function
defined on S. Let z0 be a point in S. We say f is continuous at z0 if
8 ✏ > 0, 9 > 0 : |f (z)
f (z0 )| < ✏, 8 z 2 S with |z
z0 | <
we say f is continuous on S if f is continuous at each point of S.
Example 96. Continue from example 94, each of the fn is continuous
on [0, 1] but the pointwise limit f is not. Thus
if fn ! f pointwise on Sand each fn is continuous on S
then f is continuous on S
is in general false. Uniform convergence is better than pointwise convergence.
Proposition 97. If fn and f are defined on S ⇢ C if fn ! f uniformly
on S, and if each fn is continuous on S, then f is continuous on S. That
is
A uniform limit of continuous functions is continuous.
39
Proof. Let z0 2 S. To show f is continuous at z0 . For each n 2 N and
z2S
f (z)
f (z0 ) = f (z)
fn (z) + fn (z)
fn (z0 ) + fn (z0 )
fn (z)| + |fn (z)
fn (z0 )| + |fn (z0 )
f (z0 )
by triangle inequality
|f (z)
f (z0 )|  |f (z)
f (z0 )|
Since fn ! f uniformly on S, we have the first and third absolute
di↵erences < ✏ and fn is continuous so the second absolute di↵erences
is too < ✏.
Proposition 98. If fn and f are complex functions defined on [a, b]
in R with each fn continuous and fn ! f uniformly on [a, b] then f is
continuous on [a, b] and
ˆ b
ˆ b
fn !
f
a
a
That is
The integral of a uniform limit is the limit of the integrals
Proof. The continuity of f is a special cases S = [a, b] of the previous
proposition. For each n 2 N
ˆ b
ˆ b
ˆ b
fn
f =
(fn f )  Mn (b a)
a
a
a
with Mn equal to the maximum of |fn f | on [a, b]. Since fn ! f
uniformly on [a, b] it follows that Mn ! 0.
Corollary 99. If fn are complex functions defined on a region D in C
with each fn continuous and fn ! f uniformly on D then f is continuous on D and
ˆ
ˆ
fn (z)dz ! f (z)dz for each path in D
Proof. The continuity of f on D follows from previous proposition. If
is smooth, i.e.
complex function on [a, b] with values in D and
continuous derivative, then
ˆ
ˆ
ˆ b
fn
f =
(fn ( (t)) f ( (t))) 0 (t)dt  Mn L
a
40
where L is the length of and Mn is the maximum of |fn f | on ,
since fn ! f uniformly on D, it follows that Mn ! 0 giving that we
want.
Then the case that
is piecewise smooth is easily followed
Proposition 100. Let fn and f be complex functions on a region D
in C with each fn analytic in D and fn ! f uniformly on E for each
closed disc E in D. Then f is also analytic in D. That is
A uniform limit of analytic funtions is analytic
Proof. Since each z in D is inside some closed disc E in D, it suffices to
show that f is analytic inside each closed disc E in D. For each triangle
inside E Cauchy integral theorem gives
ˆ
fn = 0 for each n 2 N
(3.1)
Since fn ! f uniformly on E previous proposition shows that f is
continuous on E and
ˆ
ˆ
fn !
f
(3.2)
Combining (3.1) & (3.2) gives
ˆ
f = 0 for each triangle
inside E
(3.3)
By Morera’s theorem it follows (3.3) that f is analytic inside E.
Proposition 101. Same hypotheses as in proposition 100.
fn0 ! f 0 uniformly on each closed disc E in D
That is
The derivative of a uniform limit is the uniform limit of the derivatives.
Proof. For each closed disc E in D there is (by analysis) a slightly larger
concentric closed disc F in D, let C be the counterclockwise boundary
circle of F .
By Cauchy’s derivative formula for fn0 and f 0 ,
ˆ
1
fn (⇣) f (⇣)
fn0 (z) f 0 (z) =
d⇣
2⇡i C (⇣ z)2
41
for z inside C. Let r and s be the radii of E and F . Then |⇣
for ⇣ on C and z on E, so
fn0 (z)
f 0 (z) 
with Mn the maximum of |fn
uniformly on E.
z|
s
r
Mn s
for z on E
(s r)2
f | on C. Since Mn ! 0 this gives fn0 ! f 0
3.2 Cauchy Criterion
Lecture 10
(2/26/14)
In this section we discuss a subtle “completeness” property of R and
C, which was recognized explicitly only in the late 19th century. It is
expressed by the Axiom below. In treatments where R is “constructed”
(see Intro to Modern Analysis I notes), this Axiom is a theorem.
Definition 102. Let S be a subset of the set of real numbers, i.e. S ⇢ R.
A real number B is an upper bound (u.b.) for S if x  B for each x 2 S.
Lower bound (l.b.) is defined similarly.
Remark 103. (1) It is not required of an u.b. for S that it belongs to S.
(2) some subsets S of R have no u.b. e.g. N and R itself have none.
(3) The empty set ? has o elements has every B 2 R as an u.b. (Does
every cat which plays the violin bark rather than meow?)
Definition 104. Let S ⇢ R, a least upper bound (l.u.b.) for S is an
u.b. B ⇤ satisfying B ⇤  B for all u.b. B. Greatest lower bound (g.l.b.)
is defined similarly.
Exercise 105. Prove if S ⇢ R has a least upper bound, then it is
unique.
Axiom 106. Each nonempty S ⇢ R which has an u.b. has a unique
l.u.b.
Proposition 107. If sn and B are real numbers with
s1  s2  ...  B
then sn ! s for some real s (with s  B). That is
Each bounded increasing sequence of real numbers converges.
42
Proof. The set S of all sn has B as u.b. let s be the l.u.b for S, guaranteed by the Axiom. Since s is an u.b for S, we have sn  s for all
n. For each ✏ > 0, the number s ✏ which is < s, is not an u.b. for
S, because s was least so there is an N in N s.t. sN > s ✏. Since the
sequence is increasing, it follows that sn > s ✏ for all n > N . Thus for
each ✏ > 0 there is an N in N so that |sn s| < ✏ for all n > N , showing
sn ! s.
Definition 108. Let s1 , s2 , ... be any sequence. For each strictly increasing sequence of positive integers n1 < n2 < ..., we get a subsequence
sn1 , sn2 , ... of the sequence s1 , s2 , ....
Proposition 109. Let sn 2 [A, B] then for some n1 < n2 < n3 < ...
and some s 2 [A, B] we have snk ! s. That is
Each bounded sequence of real numbers has a convergent subsequence.
Proof. [A1 , B1 ] = [A, B] and choose n1 2 N arbitrarily. Then sn1 2
[A1 , B1 ]. Let M1 be the midpoint of segment [A1 , B1 ]. Either the left half
[A1 , M1 ] or the right half [M1 , B1 ] of [A1 , B1 ] contains sn for infinitely
many n or perhaps both halves do. Choose such a half call it [A2 , B2 ]
and choose any n2 > n1 with sn2 2 [A2 , B2 ].
Repeat this process again and again, getting a sequence of intervals
[A1 , B1 ], [A2 , B2 ], ...
each after the first either the left half or the right half of the proceeding
one, and a subsequence
sn1 , sn2 , ... of s1 , s2 , s3 , ... with skk 2 [Ak , Bk ] for all k 2 N
Since A1  A2  ...  B, 85 gives Ak ! s for some s 2 [A, B]. Since
Bk Ak = (B A)/2k 1 we have Bk Ak ! 0 so Bk ! s also. Since
snk 2 [Ak , Bk ] it follows that snk ! s too.
Proposition 110. Let sn 2 C satisfy |sn |  B for some B
0, then
some subsequence snk ! s for some s 2 C with |s| 2 B. That is
Each bounded sequence of complex numbers has a convergent subsequence.
Proof. Let sn = xn + iyn . Since each |xn |  B, some subsequence of the
xn converges to some x 2 R. After discarding the unchosen indices and
reindexing the remaining sn by 1, 2, 3, ... we may assume that xn ! x.
43
Working with this new sequence of sn , we may choose by a similar
argument a subsequence for which also ynk ! y for some y 2 R giving
snk ! s = x + iy. Clearly |s|  B.
Definition 111. A subset F of C is closed if F contains the limit of
each convergent sequence in F .
Exercise 112. Let F be a subset of C and let D be the complement
of F in C, i.e. D is the set of all points in C which are not in F. Prove
that F is closed i↵ D is open.
Exercise 113. Let f be a complex function defined on a subset D of
C and let p be any point in D. Prove that f is continuous at p i↵
f (zn ) ! f (p) for each sequence of points zn in D with zn ! p.
Proposition 114. Let F be a nonempty closed and bounded subset of C.
Then each real continuous function on F has a maximum, i.e. there
is a point p in F with (z)  (p) for all z in F .
Proof. Let S be the set of all values (z) of for all z in F . We show
first that S has an u.b. If not then for each n in N there is a point
zn in F with (zn ) > n. Since F is bounded, zn is bounded , so some
subsequence znk ! p for some p in C by the previous proposition. Since
F is closed, p is in F . Since is continuous at p. it follows by exercise
113 that (znk ) ! (p), contrary to (znk ) > nk and nk ! 1.
Because F is not empty, S is not empty, so since S has an u.b. S has a
l.u.b B ⇤ . Since B ⇤ is the l.u.b for S, it follows that for each n 2 N there
is a point zn in F with
1
(zn ) > B ⇤
n
Since F is bounded and closed, some subsequence znk !some p in F .
Since is continuous at p, exercise 113 gives (znk ) ! (p), so (p)
B ⇤ . Since B ⇤ is an u.b. for the values of , we conclude that (p) = B ⇤ ,
so (p) is the maximum value of .
Definition 115. A complex sequence s1 , s2 , ... satisfies Cauchy’s criterion if for each ✏ > 0 9 N 2 N s.t.
|sm
sn | < ✏, 8 m, n > N
Cauchy’s criterion is easier to verify than convergence since Cauchy’s
criterion doesn’t involve s. Nevertheless
44
Proposition 116. A complex sequence satisfies Cauchy’s criterion i↵
it converges to some s 2 C.
Proof. First assume sn ! s. To show Cauchy’s criterion is satisfied,
since
|sm sn |  |sm s| + |sn s|
and each term on the right is < ✏/2 for m, n > N✏/2 we get
|sm
sn | < ✏ for m, n > N✏/2
showing that Cauchy’s criterion is satisfied.
Second assume only that Cauchy’s criterion is satisfied. To show sn !some
s in C. We show first that Cauchy’s criterion implies the sequence is
bounded: Cauchy’s criterion with ✏ = 1 gives
|sm
sn | < 1 for all m, n > N = N1
it follows that |sn |  B for all n, with
B = max{|s1 | , ..., |sN | , |sN +1 | + 1}
By theorem 110 some subsequence snk ! s for some s 2 C. Using
Cauchy’s criterion again we show that sn ! s. In fact for n, k 2 N
|sn
s|  |sn
snk | + |snk
s|
For n, nk sufficiently large, the first term on the right is < ✏/2, and for
k sufficiently large the second term on the right is also < ✏/2. Since we
are free to choose k as large as we want, it follows that |sn s| < ✏ for
sufficiently large n. Thus sn ! s.
P
Definition 117. An infinite series
an (z) whose terms are complex
functions an = an (z), all defined on a subset E of C, converges uniformly
on E if the sequence of partial sums sn = a1 +...+an converges uniformly
on E.
Proposition
P 118. (WeierstrassPTest) If |an (z)|  bn for z in E and n
in N and
bn converges, then
an (z) converges uniformly on E.
Proof. Let sn = a1 + ... + an and tn = b1 + ... + bn . For m > n
sm
sn = an+1 + ... + am
45
so for all z in E
|sm (z)
sn (z)|  |an+1 (z)| + ... + |am (z)|  bn+1 + ... + bm = tm
tn
P
Since bn converges, proposition 116 implies that tm tn < ✏ for m, n >
N✏ , with the same bound for |sm (z) sn (z)|. Thus for each z in E the
sequence of sn (z) satisfies Cauchy’s criterion and therefore converges,
by proposition 116. From sm (z) ! s(z) we get |sn (z) s(z)|  ✏ for
n > N✏ and all z in E, giving uniform convergence.
3.3 Exponential, Hyperbolic, Trigonometric Functions
Lecture 11
(3/3/14)
Now we define by power series and get the main properties of five closely
related entire functions: exp, cosh, sinh, cos & sin.
Definition 119. For each z 2 C
exp z =
1
X
zn
n=0
cosh z =
1
1
X
X
z 2n
z 2n+1
& sinh z =
(2n)!
(2n + 1)!
n=0
cos z =
1
X
n!
( 1)n
n=0
n=0
1
X
z 2n
z 2n+1
& sin z =
( 1)n
(2n)!
(2n + 1)!
n=0
The five series reduce on the real line, i.e. for z 2 R, to the series derived
in Calculus for the real functions on R.
Proposition 120. Each of the above five series converges for all z 2 C,
uniformly in each closed disc |z|  r to an entire function.
Proof. Here we prove the exponential series. For each r > 0, the series
for er converges by the ratio test, since
r n+1
(n+1)!
rn
n!
since
=
r
! 0 for n ! 1
n+1
zn
rn

for |z|  r
n!
n!
46
Weierstrass test with bn = rn /n! shows tat the series for ez converges
uniformly for |z|  r.
By proposition 100 the function ez is analytic in each open disc |z| < r.
Therefore ez is an entire function.
Proposition 121. For all z 2 C
cosh z =
cos z =
Proof. Replace z by
ez + e
2
eiz + e
2
z
& sinh z =
iz
& sin z =
ez
e
z
2
eiz
e
2i
iz
z in the exponential series. Now both
exp z =
1
X
zn
n=0
n!
exp z =
1
X
( 1)n
n=0
zn
n!
adding or subtracting gives the series for 2 cosh z or 2 sinh z.
Similarly replacing z by iz in the exponential series and adding or subtracting gives 2 cos z or 2i sin z.
Corollary 122. (Euler) For all real ✓
ei✓ = cos ✓ + i sin ✓
i.e.
ei✓ = 1 and arg ei✓ = ✓
Corollary 123.
ei⇡ = 1
What are some applications of these things? Corollary 122 may be
applied to give an improved version of the story about Euler and Diderot
in the Court of Catherine the Great–The proof of existence of God.
Lemma 124. (Binomial) For all integers n
are nonnegative integers.
X a l bm
(a + b)n
=
n!
l! m!
l+m=n
Proposition 125. For all a, b 2 C,
ea+b = ea eb
47
0 and all a, b 2 C. l, m
Proof. We cannot say “rule of exponents”, since that rule was proved in
Calculus only works for real a, b. We use Binomial lemma instead
2N
X a l X bm
X a l bm X
=
=
l!
m!
l! m!
lN
mN
n=0
l,mN
X
l+m=n
l, m  N
a l bm
l! m!
Bounding the inner sums with n > N crudely and using the binomial
lemma for 0  n  2N , we get
X a l X bm
l!
m!
lN
mN
1
X (a + b)n
X
(|a| + |b|)n

n!
n!
nN
n=N +1
For N ! 1, the sums on the left ! ea , eb , ea+b while the right series
! 0.
Corollary 126. For all z = x + iy with x, y 2 R
ez = ex cos y + iex sin y
and
|ez | = ex and arg ez = y
Proof. These follow from ez = ex eiy and corollary 122 with ✓ = y.
Corollary 127. For all a, b 2 C
cosh(a + b) = cosh a cosh b + sinh a sinh b
sinh(a + b) = sinh a cosh b + cosh a sinh b
Corollary 128. For all z 2 C
cosh2 z
sinh2 z = 1 & cos2 z + sin2 z = 1
Corollary 129. For all z = x + iy, x, y 2 R
cosh z = cosh x cos y + i sinh x sin y
sinh z = sinh x cos y + i cosh x sin y
|cosh z|2 = sinh2 x + cos2 y
|sinh z|
2
2
2
= sinh x + sin y
48
(3.4)
Exercise 130. For each nonconstant analytic function f, the lines or
curve on which f (z) is real, i.e. Imf (z) = 0 separate the regions where
Imf (z) > 0 from those where Imf (z) < 0 and the lines or curves on
which |f (z)| = 1, separate the regions where |f (z)| > 1 from those
where |f (z)| < 1. Do that for
f (z) = z 2 , f (z) = ez , f (z) = cosh z
4 Residue Theorem
4.1 Zeros
Recall in exercise 89 we had a real function h defined on R, positive
except at 0 with derivatives of all orders at every pint of R all of which
are 0 at the point 0: h(n) (0) = 0 for n = 1, 2, .... Nothing like this can
happen with analytic functions:
Proposition 131. Let f be analytic in a region D in C. Suppose there
is a point p in D for which
f (n) (p) = 0 for n = 0, 1, 2, ...
(4.1)
Then f = 0 i.e. f (z) = 0 for all z 2 D.
The proof uses a lemma:
Lemma 132. (Disc-Chain) Let D be a region in C, and p, q pints in
D. Then there is a finite sequence of points z0 , ..., zk in D, with z0 = p,
zk = q and open discs D0 , ..., Dk 1 in D so that Dj is centered at zj and
contains zj+1 for j = 0, ..., k 1.
The proof of the lemma uses the following two lemmas:
Lemma 133. (Bowstring) For each path
q we have |q p|  L. That is
of length L going from p to
The bowstring is not longer than the bow.
Proof. If
is a smooth path from p to q, then
q
p = (b)
(a) =
ˆ
a
49
b
0
(t)dt
so
|q
p| =
ˆ
a
b
0
(t)dt 
ˆ
b
0
(t) dt = L
a
In the general case of piecewise smooth , we have = 1 + ... + s
with each r smooth of length Lr from zr 1 to zr for r = 1, ..., s 1 and
z0 = p, zs = q. it follows that
q
p = zr
z0 = (z1
z0 ) + ... + (zr
zr
1)
so
|q
Lecture 12
(3/5/14)
p|  |z1
z0 | + ... + |zr
zr
1|
 L1 + ... + Lr = L
Lemma 134. (Worm) For each path in a region D of C, there is
an r > 0 so that each open disc of radius r centered at a point on is
contained in D.
Proof. Since each smooth piece of is continuous, and the pieces attach
together at the ends, we may suppose that is a continuous function
on a closed interval [a, b] of R with all values in D. If there is no r > 0
so that each open disc of radius r centered on is contained in D, then
choosing r = 1, 1/2, 1/3, ... we can find points z1 , z2 , ... on and points
z10 , z20 , ... not in D s.t. |zn0 zn | < 1/n for n = 1, 2, 3, ...
We have zn = (tn ) for a sequence t1 , t2 , ... in [a, b]. Because it is a
bounded sequence, there is a subsequence tn1 , tn2 , ... which converges to
a point t⇤ 2 [a, b]. Since is continuous, znk ! z ⇤ := (t⇤ ). it follows
that zn0 k ! z ⇤ too.
Since z ⇤ is on , which is in D, and D is open, the open disc |z z ⇤ | < ✏
is contained in D for some positive ✏. Since zn0 k z ⇤ < ✏ for sufficiently
large k, it follows that zn0 k 2 D for these k. Contradiction.
Now we prove Disc-Chain Lemma
Proof. Let p, q be points in a region D. Since D is connected, there is
path in D from p to q. By Worm, there is an r > 0 s.t. each open
disc of radius r centered at a pint of is contained in D. Choose points
z0 , ..., zk on with z0 = p, zk = q, and s.t. that part of between zj and
zj+1 has length < r for j = 0, ..., k 1, the disc Dj of radius r centered
at zj contains zj+1 and is contained in D.
50
We now prove proposition 131.
Proof. Let D0 be any open disc in D centered at p, by the Cauchy-Taylor
theorem
1
X
f (n) (p)
f (z) =
(z p)n for all z 2 D0
(4.2)
n!
n=0
It follows from (4.1), (4.2) that f (z) = 0 for all z 2 D0 . If D0 = D, we
are done. If not, use Disc-chain lemma. (4.1) implies f = 0 in D0 , so
f (n) = 0 in D0 for all integers n
0. In particular, f (n) (z1 ) = 0 for
all n 0 so f = 0 in D1 by the previous argument. After k such steps
we reach f = 0 in Dk 1 . In particular f (q) = 0. Thus (4.1) implies
f (q) = 0 for all q 2 D.
Definition 135. Let f 6= 0 be analytic in a region D, and let z0 be a
point in D. If f (z0 ) = 0 then z0 is a zero of f .
Remark 136. Thus f = 0 has no zeros, and the zeros of a polynomial
are its roots.
Proposition 137. Let f be analytic in a region D and let z0 be a zero
of f in D. Then there is a positive integer m called the order of z0 so
that
f (z0 ) = 0, ..., f (m 1) (z0 ) = 0, but f (m) (z0 ) 6= 0
(4.3)
Proof. Since f has a zero, f 6= 0 so by proposition 131 f (n) (z0 ) 6= 0
for some positive integer n. Then (4.3) holds, with m the least such
integer.
Lemma 138. Assume the convergence, for some z1 6= z0 , of the power
series
1
X
g=
cn (z z0 )n
(4.4)
n=0
Let D1 be the open disc centered at z0 with radius r1 = |z1
z0 |, then
(1) the series (4.4) also converges for all z in D.
(2) it converges uniformly in each closed disc E centered at z0 with radius
r < r1
(3) its sum g is analytic in D1
(4) it is the Cauchy-Taylor series of g in D1 . That is
Each convergent power series is the Cauchy-Taylor series of its sum.
51
Proof. (1) follows from (2) since each z in D1 is in some E.
Prove (2). For z in E
z0 )n |  bn := |an | (r/r1 )n , with an = cn (z1 z0 )n
P
since the series an converges,
P its sequence of terms a0 , a1 , ... is bounded.
Since the geometric series (r/r1 )n converges, it follows by Weierstrass
test that (4.4) converges uniformly in E.
|cn (z
(3) follows from (2) by proposition 100.
Prove (4). By proposition 101 it follows that for each integer k
g (k) (z) =
1
X
n(n
1)...(n
k + 1)cn (z
z0 ) n
n=k
k
0
for all z 2 D1
For z = z0 this gives g (k) (z0 ) = k!ck i.e. (4) is the Cauchy-Taylor series
of g.
Proposition 139. Let f be analytic in a region D, and let z0 be a zero
of f , of order m, then for each open disc D0 in D centered at z0
f (z) = (z
z0 )m f0 (z), 8 z 2 D0
(4.5)
for some f0 analytic in D0 with f0 (z0 ) 6= 0.
Proof. By the Cauchy-Taylor formula, for z 2 D0
f (z) =
1
X
f (n) (z0 )
(z
n!
n=m
z0 )n = (z
z0 ) m
1
X
f (n) (z0 )
(z
n!
n=m
z0 ) n
m
giving (4.5) with f0 the power series on the right, convergent in D0 and
therefore by lemma 138 it’s analytic in D0 and satisfying f0 (z0 ) 6= 0.
Theorem 140. (isolation of zeros) Let f be analytic in a region D.
Then each zero z0 of f is isolated, i.e. there is a > 0 so that f has no
zeros in the punctured disc of radius centered at z0 .
Proof. The function f0 in proposition 139 is analytic in D0 . In particular
f0 is continuous at z0 . Since f0 (z0 ) 6= 0 there is a > 0 s.t.
|f0 (z)
f0 (z0 )| < |f0 (z0 )| for |z
In particular f0 has no zero in |z
|z z0 | < .
z0 | <
z0 | < , so f has no zero in 0 <
52
Theorem 141. (Identity theorem) Let f1 , f2 be analytic in a region D.
If f1 = f2 on some line segment [p, q] in D with p 6= q then f1 = f2 in
D.
Proof. Put f = f1 f2 . then f is analytic in D and f = 0 on [p, q]. In
particular f (p) = 0. If f 6= 0, then p is a zero of f. Since f = 0 on [p, q],
the zero at p is not isolated, contrary to theorem 140.
Theorem 142. (reflection principle) Let f be an entire function which
is real on R, i.e. f (x) is real for all real x, then
f (z̄) = f (z)
8z 2 C
Proof. Since f is entire, we have
f (z) =
1
X
f (n) (0)
n=0
n!
z n ,8 z 2 C
since f is real on R, so are f (n) for all n (see exercise below). In particular
all f (n) (0) are real, so
f (z) =
1
X
f (n) (0)
n=0
n!
z n = f (z̄)
Exercise 143. Show that if f is an entire function real on R, then f 0
is also real on R. Hint: show first that the derivative of an analytic
function is its partial derivative with respect to x.
Exercise 144. Show for each root of a polynomial
multiplicity=order
Exercise 145. For ez show it has no zeros, and it has period 2⇡i, i.e.
ez+2⇡i = ez , 8 z 2 C
Exercise 146. Find all zeros in C of cosh,cos, sinh, sin and show that
all these zeros have order 1. Also show that cos, sin are unbounded on
the imaginary axis.
Exercise 147. Use the series definitions to find the derivatives of exp,cosh,cos,
sinh, and sin.
53
4.2 Poles & Singularities
Lecture 13
(3/10/14)
Definition 148. If f is analytic in a punctured region D
z0 is an isolated singularity of f .
{z0 }, then
Isolated singularities come in three flavors:
Definition 149. Let f be defined and analytic in a punctured region
D {z0 }.
(i) If some choice of f (z0 ) makes f analytic in D, then z0 is a removable
singularity of f.
(ii) If |f (z)| ! 1 for z ! z0 z 6= z0 then z0 is a pole of f.
(iii) If z0 is neither removable nor a pole, then z0 is an essential singularity of f .
Example 150. The function f defined by
f (z) =
sin z
z
is defined and analytic in C {0}, for z 6= 0
✓
◆
1
z3 z5
f (z) =
z
+
+ ... = 1
z
3!
5!
z2 z4
+
+ ...
3!
5!
The power series on the right converges for all z 2 C to an entire function
with value 1 at 0. Therefore the isolated singularity z0 = 0 of f is
removed by choosing f (0) = 1.
Example 151. The function
f (z) =
is defined and analytic in C
1
z
{0} with a pole at z0 = 0.
Example 152. The function
f (z) = exp
1
1
1 1
=1+ +
+ ...
z
z 2! z 2
is defined and analytic in C {0}, since each term of the series is analytic
there and the series converges uniformly for |z| > s for each s > 0. In
this example, f is real on R {0}. To see z = 0 is not a pole
f (x) ! 1 for x ! 0+
54
f (x) ! 0 for x ! 0
One can also find the behavior of f (z) along another punctured line and
punctured circles.
From above examples we understand an essential singularity is a tiger,
a removable singularity is a lamb. To get some poetic feeling of these,
read William Blake “The Tyger”, “The Lamb”.
Exercise 153. Prove if f has a removable singularity at z0 , there is
only one way for f (z0 ) which removes the singularity.
Proposition 154. (on removable singularity) If f is analytic and bounded
in 0 < |z z0 | < r1 then z0 is a removable singularity of f .
Proof. Let be any counterclockwise circle centered at z0 with radius
< r1 . We will show first that for each z 6= z0 inside and for sufficiently
small counterclockwise circles C and C0 centered at z and z0 ,
ˆ
ˆ
ˆ
f (⇣)d⇣
f (⇣)d⇣
f (⇣)d⇣
=
+
⇣ z
z
z
C ⇣
C0 ⇣
this can be done by the same kind of deformation made in the proof of
corollary 73. By Cauchy’s integral formula
ˆ
f (⇣)d⇣
= 2⇡if (z)
z
C ⇣
and since f (⇣)/(⇣
thus
z) is bounded for ⇣ near to z0 , so
ˆ
f (⇣)d⇣
!0
z
C0 ⇣
1
f (z) =
2⇡i
ˆ
f (⇣)d⇣
for z inside
⇣ z
with z 6= z0
(4.6)
Of course (4.6) doesn’t hold at z = z0 since f is not defined at z0 .
However if we now choose f (z0 ) to be the right side of (4.6) for z =
z0 , then we have extended the definition of f to the unpunctured disc
inside . Now the same di↵erentiation argument which proves Cauchy’s
derivative formula shows that this extended f which satisfies (4.6) for
all z in this disc inside is analytic in this disc. Thus we have removed
the singularity of f at z0 .
55
Proposition 155. (on poles) If f is analytic in D {z0 } with a pole
at z0 then
f0 (z)
f (z) =
for z 2 D {z0 }
(4.7)
(z z0 )m
with f0 analytic in D with f0 (z0 ) 6= 0 and m 2 N the order of the pole.
Also
✓
◆
1
f (z) = P
+ g(z) for z 2 D {z0 }
(4.8)
z z0
with P a polynomial of degree m with constant term 0, and g analytic in
D.
The first term on the right is the principal part of f at the pole. The
order m and the three functions f0 , P, g are uniquely determined by f
and z0 .
Proof. Since |f (z)| ! 1 for z ! z0 , z 6= z0 there is an r > 0 so that the
punctured disc |z z0 | < r belongs to D {z0 } and contains no zeros
of f .
Put = 1/f in this punctured disc. Thus is also analytic and without
zeros in the punctured disc, and (z) ! 0 for z ! z0 , z 6= z0 . By
proposition 154
has removable singularity at z0 . The only choice
of (z0 ) is 0, so after removing the singularity, z0 is zero of . By
proposition 139
(z) = (z
with
0
z0 ) m
0 (z)
in the disc |z
z0 | < r
analytic and without zeros in this disc.
Put f0 = 1/ 0 in this disc. Then f0 is analytic and without zeros in the
disc, and (4.7) holds in the unpunctured disc. To get (4.7) throughout
D z0 supplement the definition of f0 by
z0 )m f (z) for z 2 D
f0 (z) = (z
{z0 }
In the disc, f0 is the sum of a Cauchy-Taylor series:
f0 (z) =
1
X
z0 )n with c0 6= 0
cn (z
n=0
Combined with this gives in the punctured disc
f (z) =
m
X1
n=0
(z
cn
z0 ) m
n
+
1
X
cn (z
n=m
56
z0 ) n
m
=P
✓
1
z
z0
◆
+ g(z)
with P a polynomial of degree m constant term 0 and g analytic in the
disc. To get (4.8) throughout D {z0 } supplement the definition of g
by
✓
◆
1
g(z) = f (z) P
for z 2 D {z0 }
z z0
The uniqueness part of the proof is left as an exercise: show (1) there
is only one pair m, f0 with m 2 N and f0 analytic in D {z0 } with
f (z0 ) 6= 0 which satisfies (4.7); (2) there is only one pair P, g with P a
polynomial with constant term 0, and g analytic in D which satisfies
(4.8).
The next proposition gives an “additive” representation of an arbitrary
rational function analogous to the “multiplicative” factorization of an
arbitrary nonzero polynomial. Analogous to the powers of linear factors
corresponding to the distinct roots and the leading coefficient for a polynomial are the principal parts at the distinct poles, and a polynomial
P1 for a rational function.
Proposition 156. (partial fractions) For each rational function f with
distinct poles z1 , ..., zr
f (z) =
r
X
j=1
Pj
1
z
zj
+ P1 (z) for z 2 C
{z1 , ..., zr }
where for j = 1, ..., r the jth term in the sum is the principal part of f
at the pole zj and P1 is a polynomial.
Proof. The principal part at zj is analytic in C {zj }. By proposition
155
r
X
1
P1 (z) = f (z)
Pj
z zj
j=1
has removable singularities at z1 , ..., zr . After removing them, P1 is an
entire rational function, and is therefore a polynomial.
Proposition 155 shows that the blowup of an analytic function at a
pole no worse than the behavior of a polynomial for large z. The next
theorem shows that the behavior of an analytic function at an essential
singularity is essentially more chaotic:
57
Theorem 157. (on essential singularities -Casorati & Weierstrass) If
f is analytic in D {z0 } with an essential singularity at z0 , then
8 ✏ > 0, 8 > 0, 8 w 2 C, 9 z 2 D
{z0 } : |z
z0 | <
& |f (z)
w| < ✏
that is
f (z) is arbitrarily close to each w 2 C at points arbitrarily close to z0 .
Proof. Suppose f doesn’t satisfy the conclusion,
9 ✏ > 0, 9 > 0, 9 w 2 C : 8 z 2 D {z0 }, if |z
we may choose
then
g=
z0 | < , then |f (z)
> 0 s.t. the open disc of radius
1
f
w
satisfies |g(z)| 
1
for 0 < |z
✏
w|
is contained in D,
z0 | <
By proposition 154 g has a removable singularity at z0 . After removing
the singularity,
(i) if g(z0 ) 6= 0, then f (z)
removable singularity of f
w is bounded for z near z0 , so z0 is a
(ii) if g(z0 ) = 0, then |f (z) w| ! 1 for z ! z0 so z0 is a pole of f .
Either way f doesn’t satisfy the hypothesis.
4.3 Cauchy’s Residue Theorem
Lecture 14
(3/12/14)
Cauchy’s Residue Theorem is an important generalization of Cauchy’s
integral theorem which ma be applied to evaluate certain definite integrals over R and to study the distribution of zeros and poles of meromorphic functions. This results was also used by Riemann to uncover
the connection between set of all prime numbers and the set of all zeros
of the Riemann zeta functions.
Definition 158. Let f be analytic in a punctured disc centered at z0 ,
with a pole of order m at z0 . Writing the principal part of f at the pole
z0 as
1
a1
am
P
=
+ ... +
(4.9)
z z0
z z0
(z z0 )m
the residue of f at z0 is the coefficient a1 :
Res(f, z0 ) = a1
58
✏
Definition 159. Let D be a region and z1 , ..., zn be distinct points in
D. A function f , analytic in D {z1 , ..., zn }, with a pole at each zk is
said to be meromorphic in D.
For example, each rational function is meromorphic in C.
Definition 160. A closed path
simple if
from p to q in C defined on [a, b] is
(t1 ) = (t2 ) with a  t1 < t2  b only for t1 = a, t2 = b
Remark 161. As with circles and rectangles, each simple closed path
is either clockwise or counterclockwise, and each such separates C
into two regions, made up of the points inside and the points outside
(Jordan Curve Theorem). Although precise definitions of these terms
and a proof of Jordan Curve Theorem are difficult. We will use only
circles, rectangles, halfmoons, and other easy , made up of a few line
segments and arcs of circles, for which the meanings of these terms and
the truth of Jordan Curve Theorem are clear.
Theorem 162. If f is meromorphic in a convex region D, and is a
counterclockwise simple closed curve in D not passing through any pole
of f , then
ˆ
X
f (z)dz = 2⇡i
Res(f, zj )
The sum is over those poles zj of f which are inside .
Proof. We will suppose is a rectangle, and leave the proof for a circle
or halfmoon as exercises. Let z1 , ..., zn be the distinct poles of f in D,
put
n
X
1
g(z) = f (z)
P
for z 2 D {z1 , ..., zn }
z zj
j=1
where the jth term in the sum is the principal part of f at zj . Then g
has a removable singularity at each zj . We denote by the same symbol
g the function gotten by removing all these removable singularities. By
Cauchy’s integral theorem, the integral of g around is 0, so
ˆ
f (z)dz =
n ˆ
X
j=1
59
P
1
z
zj
dz
To finish, it suffices to show that for each z0 in D
and each P as in
(4.9),
ˆ
1
P
dz = 0 for z0 outside , and 2⇡ia1 for z0 inside
z zj
If z0 is outside , then we draw a convex region containing but not
z0 . In this region the integrand is analytic, so the integral is 0 . If z0 is
inside , we draw a counterclockwise circle C inside and centered at
z0 , then
ˆ
ˆ
1
1
P
dz =
P
dz
z zj
zj
C z
It remains to show that for positive integers l,
(
ˆ
2⇡i l = 1
dz
=
l
z0 )
0
l>1
C (z
For l = 1, this is proposition 61. For l > 1, the integrand has an
antiderivative in C {z0 }
1
(z
z0 ) l
=
1
1
l
✓
1
z0 ) l
(z
1
◆0
so the integral is 0 by proposition 62.
Proposition 163. (i) If f has a simple pole at z0 , i.e. a pole of order
1, then
Res(f, z0 ) = lim(z z0 )f (z) for z ! z0 , z 6= z0
(ii) If f = g/h with g, h analytic in a region containing z0 , g(z0 ) 6= 0
and z0 is a simple zero, i.e. a zero of order 1, of h, then
Res(f, z0 ) =
g(z0 )
h0 (z0 )
(iii) If f has a pole of order m at z0 then
Res(f, z0 ) =
1
(m
1)!
✓
d
dz
◆m
1
{(z
z0 )m f (z)}
at z0 . Where {...} means with removable singularity at z0 removed.
60
Proof. (i) By hypothesis in some punctured disc centered at z0
f (z) =
z
a1
+ h(z)
z0
with h analytic in the unpunctured disc. Since h is bounded near z0 ,
(z
z0 )h(z) ! a1 for z ! z0 , z 6= z0 .
z0 )f (z) = a1 + (z
(ii) Apply (i) the hypothesis imply f has a simple pole at z0 , so for
z 6= z0
(z
z0 )f (z) =
g(z)
h(z) h(z0 )
z z0
!
g(z0 )
for z ! z0 , z 6= z0 .
h0 (z0 )
(iii) By (4.8) and (4.9)
f (z) =
am
a1
+ ... +
+ c0 + c1 (z
m
(z z0 )
z z0
z0 ) + ...
in some punctured disc centered at z0 , so
(z
z0 )m f (z) = am + ... + a1 (z
z0 ) m
1
+ c0 (z
z0 )m + ...
with the same formula for {(z z0 )m f (z)} in the unpunctured disc.
Thus a1 is the (m 1)st Cauchy-Taylor coefficient for this function at
z0 .
Example 164. Let us calculable some residue
f (z) =
(z 2
1
, m2N
+ 1)m
with poles of order m at ±i. Let us do residue at z0 = i
(z
i)m f (z) =
1
(z + i)m
for z 6= ±i. Using (iii) above
Res(f, i) =
=
1
( m)( m
1)...( m (m 2))
(m 1)!
(z + i)2m 1
✓
◆
( 1)m 1 m(m + 1)...(2m 2)
1 1
2m 2
=
m 1
(m 1)!
(2i)2m 1
2i 4m 1
61
Exercise 165. Prove if f has a zero of order m at z0 , then 1/f has a
pole of order m at z0 .
Example 166. Let us use residue theorem to compute some integral
ˆ 1
dx
=⇡
(4.10)
2
1 x +1
This was done in Calculus,
ˆ
R
R
dx
= arctan R
x2 + 1
arctan R = 2 arctan R ! 2
Now we use residue and upper halfmoon path
✓
R (✓) = Re , 0  ✓  ⇡ to
f (z) =
thus
ˆ
R
f (z)dz =
ˆ
f (z)dz +
[ R,R]
The ML inequality gives
ˆ
f (z)dz 
R
ˆ
z2
= [ R, R] +
R
1
R
with
1
+1
f (z)dz = 2⇡iRes(f, i) =
1
R2
R
⇡
=⇡
2
2⇡i
=⇡
2i
⇡R ! 0 as R ! 1
Remark 167. The pole at i “explains” the evaluation (4.10) without
using an antiderivative. The pole at i is also behind the fact that for
real x
1
1 x2 + x4 ... = 2
x +1
only works for 1 < x < 1 even though the function on the right has
derivatives of all orders for all real x.
Proposition. Let f be a rational function with no real poles, i.e. no
poles on R. If f = g/h with g, h polynomials of degree k and l with
l k + 2, then
ˆ 1
X
f (x)dx = 2⇡i
Res(f, zj )
1
the sum is over the poles of f in the upper half plane.
62
Proof. As in the proceeding example, we need to show
ˆ
f (z)dz ! 0
R
Let bz k and cz l be the highest degree terms in g, h. Then g(z) ⇠ bz m
l
|z| ! 1, meaning that g(z)
bz m ! 1 as |z| ! 1. Similarly h(z) ⇠ cz . It
d
follows that f (z) ⇠ az with a = b/c, d = k l. The maximum MR of
|f (z)| for z 2 R satisfies MR ⇠ |a| Rd for R ! 1. For sufficiently large
R
ˆ
f (z)dz  MR ⇡R
R
since d 
2, so
´
R
f (z)dz ! 0.
Example 168. Thus example 164 gives
✓
◆
ˆ 1
dx
⇡
2m 2
= m 1
2
m
m 1
4
1 (x + 1)
a generalization of (4.10).
5 Applications of Residue Theorem
5.1 Fresnel’s Integrals
Lecture 15
(3/24/14)
We begin by related evaluations of two integrals
ˆ 1
2
G =
e ⇡x dx
1
ˆ 1
2
F =
ei⇡x dx
1
G relates to the Gaussian distribution and the Laplace central limit
theorem in probability theory; F was introduced by Fresnel to study
di↵raction of light and used also by Riemann and Debye in the method
of stationary phase.
Proposition 169.
G=1
63
Proof. G 2 is
ˆ 1
e
⇡x2
dx
1
ˆ
1
e
⇡y 2
¨
dy =
1
R2
1 ˆ 2⇡
ˆ
=
(x2 +y 2 )
e
0
e
dxdy
⇡r2
d✓rdr = 1
0
The conversion between R2 Cartesian to polar can be more precise. For
R>0
✓ˆ R
◆2 ˆ
2
⇡x2
e
dx =
e ⇡r dA
R
Square
Square
= [ R, R] ⇥ [ R, R]. It is bounded be two discs of radii R and
p
2R thus
ˆ
ˆ
ˆ
2
⇡r2
⇡r2
e
dA <
e
dA <
e ⇡r dA
p
Square
| discR {z
}
| disc 2R{z
}
1 e
⇡R2
and both ! 1, so
´
1 e
Square e
⇡r2 dA
2⇡R2
! 1.
Since G > 0, G = 1.
Proposition 170.
F = ⇣8 := ei⇡/4
Proof. By Cauchy’s integral theorem,
ˆ
f (z)dz = 0
for every entire function f and every closed path
2
e ⇡z and the counterclockwise eighthmoon path
= [0, R] +
R
+ [⇣8 R, 0] with
R (✓)
in C, for f (z) =
= Rei✓ , 0  ✓  2⇡/8.
We get
ˆ
0
R
e
⇡x2
dx +
ˆ
e
⇡z 2
dz
ˆ
R
e
⇡(⇣8 r)2
⇣8 dr = 0
0
R
The first integral ! 1/2 by proposition 169.
Since (⇣8 )2 = i, the last integral ! ⇣8 F̄/2, assuming it has a limit, i.e.
assuming the integral defining F converges.
64
For z = Rei✓ , we have Rez 2 = R2 cos 2✓, so
ˆ
Put
⇡z 2
e
R
= ⇡/2
ˆ
e
dz 
ˆ
2⇡/8
⇡R2 cos 2✓
e
0
R
R
dz 
2
ˆ
ˆ
⇡/2
e
⇡R2 cos
d
0
2↵/⇡ for 0  ↵  ⇡/2
↵, use sin ↵
⇡z 2
R
Rd✓ =
2
⇡/2
e
⇡R2 sin ↵
0
showing that the integral over
R
R
d↵ 
2
ˆ
1
e
2R2 ↵
0
d↵ 
1
4R
goes to 0 for R ! 1.
It follows that the integral defining F does converge, and
G = ⇣8 F̄ =) F̄ = 1/⇣8 =) F = ⇣8
5.2 Fourier Transforms
Definition 171. For each complex valued F continuous on R,and satisfying
ˆ 1
|F (x)| dx < 1
1
the Fourier transform F̂ is defined on R by
ˆ 1
F̂ (t) =
F (x)e2⇡ixt dx for t 2 R
1
Proposition 172. If
F (x) =
1
,x2R
1 + x2
then
F̂ (t) = ⇡e
2⇡|t|
, t2R
Proof. For each t 2 R the residue theorem applied to e2⇡izt /(1 + z 2 ) and
the halfmoon path, with R > 1 gives
ˆ
R
R
e2⇡ixt
dx +
1 + x2
ˆ
R
e2⇡izt
e 2⇡t
dz
=
2⇡i
= ⇡e
1 + z2
2i
65
2⇡t
For t
R>1
0 we have e2⇡izt = e
0 and y
ˆ
R
2⇡yt
 1, so for t
0 and
e2⇡izt
⇡R
dz  2
!0
1 + z2
R
1
giving what we want for t 0, for t  0 either use the lower halfmoon
path or use the following general fact:
Lemma 173.
¯
F̂ ( t) = F̄ˆ (t) for all t 2 R
Theorem 174. If
F (x) = e
⇡x2
F̂ (t) = e
⇡t2
then
x2R
, t2R
Proof. To show:
ˆ
x2 + 2ixt =
ˆ 1
Since
that
1
⇡x2 2⇡ixt
e
e
dx = e
⇡t2
1
e
(5.1)
it)2
(x
⇡(x
1
, for t 2 R
t2 it suffices by proposition 169 to show
ˆ 1
2
it)2
dx =
e ⇡x dx, for t 2 R
(5.2)
1
to prove it, apply Cauchy’s integral theorem to the function exp ⇡z 2
and the rectangular path with corners R it, R it, R, R to get
ˆ 1⇣
ˆ t⇣
⌘
⌘
2
2
⇡(x it)2
⇡x2
e
e
dx = i
e ⇡(R iy)
e ⇡( R iy) dy
1
0
(5.3)
For y between 0 and t,
e
⇡(±R iy)2
=e
⇡(R2 y 2 )
 e⇡(t
2
R2 )
so for R ! 1, the integral on the right in 5.3! 0, giving 5.2.
Theorem 175. If
F (x) =
1
, x2R
cosh ⇡x
F̂ (t) =
1
, t2R
cosh ⇡t
then
66
Proof. For each t 2 R, the residue theorem applied to the function
e2⇡izt / cosh ⇡z and the counterclockwise rectangular path R with corners ±R, ±R + i gives
✓ 2⇡izt
◆
ˆ
e
i
e ⇡t
2⇡izt
F (z)e
dt = 2⇡iRes
,
= 2⇡i
= 2e ⇡t
cosh
⇡x
2
⇡
sinh(⇡i/2)
R
Since the simple poles of the integrand are at ±i/2, ±3i/2, ±5i/2,...
with i/2 inside the rectangle, and all the others outside.
The contribution of the bottom side of the rectangle is
ˆ
R
R
e2⇡ixt
dx
cosh ⇡x
the integral over the top side of the rectangle is
ˆ
R
R
e2⇡i(x+i)t
dx = e
cosh ⇡(x + i)
since cosh ⇡(x + i) =
2⇡t
ˆ
R
R
e2⇡ixt
dx
cosh ⇡x
cosh ⇡x.
The two vertical sides of the rectangle contribute
±i
ˆ
0
1
e2⇡i(±R+iy)t
dy
cosh ⇡(±R + iy)
The identity (3.4) implies
|cosh ⇡(±R + iy)|
sinh ⇡R
It follows that for each t 2 R, the absolute value of each vertical contribution is
e2⇡|t|

! 0 for R ! 1
sinh ⇡R
Combining all the contributions and letting R ! 1 gives
(1 + e
2⇡t
)F̂ (t) = 2e
⇡t
=) F̂ (t) =
1
cosh ⇡t
Let’s look at an example, the average of some periodic function, how
residue theroem can be used for computing Fourier coefficeints.
67
p
Example 176. For a > 1, put b = a2 1, then
ˆ 2⇡
1
d✓
1
=
2⇡ 0 a cos ✓
b
Proof. The left side is
ˆ 2⇡
ˆ 2⇡
ˆ
1
d✓
1
2iei✓ d✓
1
=
=
2⇡ 0 a ei✓ +e i✓
2⇡ 0 (ei✓ )2 2aei✓ + 1
2⇡i C z 2
2
2dz
2az + 1
where C is the counterclockwise circle of radius 1 centered at 0. The
poles of the integrand are a ± b. Since the poduc of the root is 1, we
have
0<a b<1<a+b
the smaller root inside C, the larger one outside C, it follows that
ˆ 2⇡
1
d✓
2
1
=
=
2⇡ 0 a cos ✓
2z 2a z=a b
b
Similarly we can do Fourier
1
2⇡
ˆ
0
2⇡
ein✓ d✓
1
=
a cos ✓
2⇡i
ˆ
C
z2
2z n dz
2z n
=
2az + 1
2z 2a
=
z=a b
By Fourier series,
1
a
X
b
=
(a
cos ✓
1
b)|n| ein✓
5.3 Cauchy Principal Values
Lecture 16
(3/26/14)
Recall in Calculus we studied improper
(
ˆ 1
1
dx
1 ↵
=
↵
1
0 x
(
ˆ 1
1
dx
↵ 1
=
x↵
1
1
68
integrals e.g.
0↵<1
↵>1
↵>1
0↵<1
(a
b)|n|
, 8n 2 Z
b
Definition 177. For f continuous on (a, b] but not necessarily continuous, or even defined at a, the improper integral of f over (a, b] converges
if the intgral of f over [a + , b] has a finite limit as ! 0, > 0 in whcih
case this limit is the value of the integral; the intgral diverges otherwise.
Integral over [a, b) are treated analogously.
SImilarly for f continuous on [a, 1), the improper integral of f over
[a, 1) converges if the intgral of f over [a, b] has a finite limit as b ! 1
in whcih case this limit is the value of the integral; the intgral diverges
otherwise. Integral over ( 1, b) are treated analogously.
Example 178. The simplest divergent impropter integrals are
ˆ 1
ˆ 1
dx
dx
&
both diverging to 1
x
x
0
1
ˆ 0
ˆ 1
dx
dx
&
both diverging to -1
1 x
1 x
Definition 179. Let f be continuous on R. The Cauchy principal value
of the integral of f over R is
P.V.
ˆ
1
f (x)dx = lim
R!1
1
ˆ
R
f (x)dx
R
More generally, for f continuous on R except at x1 < ... < xr ,
ˆ 1
ˆ
P.V.
f (x)dx =
lim
f (x)dx
1
R ! 1 IR,
!0
where IR, is [ R, R] with (xj
, xj + ) removed, for j = 1, ..., r In
each case, if there is no limit there is no Cauchy principal value.
Exercise 180. Using defintion to show
ˆ 1
dx
P.V.
=0
1 x
69
5.4 Rouche’s Theorem
6 More Analysis
6.1 Jensen’s Theorem
6.2 Euler’s Formulas
6.3 Stirling Approximation
7 Intro to Analytic Number Theory
7.1 Elliptic Functions
7.2 Riemann Zeta Functions
7.3 Gamma Functions
Lecture 17
(4/7/14)
Lecture 18
(4/9/14)
Lecture 19
(4/14/14)
Lecture 20
(4/16/14)
Lecture 21
(4/21/14)
Lecture 22
(4/23/14)
Lecture 23
(4/28/14)
Lecture 24
(4/30/14)
Lecture 25
-Last Lec(5/5/14)
70