mathematics and history of complex variables

1

Complex Variables

SOLO HERMELIN

Updated: 11.05.07

28.10.12

2

SOLO Complex Variables

Table of Contents Set of Numbers – Examples Fundamentals Operations with Complex Numbers z = x + i y Axiomatic Foundations of the Complex Number System R babaz ,,

History of Complex Numbers Derivatives

Cauchy-Riemann EquationsHarmonic FunctionsOrthogonal FamiliesSingular Points

Complex Line IntegralsSimply and Multiply Connected Regions Green’s Theorem in the PlaneConsequences of Green’s Theorem in the PlaneCauchy’s TheoremCauchy-Goursat TheoremConsequences of Cauchy-Goursat Theorem


Table of Contents (continue - 1)

Cauchy’s Integral Formulas and Related Theorems

Cauchy’s Integral FormulasCauchy’s Integral Formulas for the n Derivative of a FunctionMorera’s Theorem (the converse of Cauchy’s theorem)

Cauchy’s Inequality

Liouville’s TheoremFoundamental Theorem of Algebra Gauss’ Mean Value TheoremMaximum Modulus Theorem Minimum Modulus Theorem Poisson’s Integral Formulas for a Circle Poisson’s Integral Formulas for a Half Plane

4



Theorems of Convergence of Sequences and Series

Convergence TestsCauchy Root TestD’Alembert or Cauchy Ratio TestMaclaurin or Euler Integral TestKummer’s TestRaabe’s TestGauss’ s Test

Infinite Series, Taylor’s and Laurent SeriesInfinite Series of FunctionsAbsolute Convergence of Series of FunctionsUniformly Convergence of Sequences and SeriesWeierstrass M (Majorant) TestAbel’s TestUniformly Convergent Series of Analytic FunctionsTaylor’s SeriesLaurent’s Series (1843)

5



5

The Argument Theorem Rouché’s Theorem

Foundamental Theorem of Algebra (using Rouché’s Theorem)

Zeros of Holomorphic Functions

Theorem: f(z) Analytic and Nonzero → ln|f(z)| HarmonicPolynomial Theorem

Jensen’s Formula Poisson-Jensen’s Formula for a Disk

6

SOLO Complex VariablesTable of Contents (continue - 4)

Calculation of the Residues

The Residue Theorem, Evaluations of Integral and Series

The Residue Theorem Evaluation of Integrals

Jordan’s LemmaIntegral of the Type Bromwwich-WagnerIntegral of the Type ,F (sin θ, cos θ) is a rational function of sin θ and cos θ

2

0

cos,sin dF

Definite Integrals of the Type .

xdxF

Cauchy’s Principal Value Differentiation Under Integral Sign, Leibnitz’s RuleSummation of Series

Infinite ProductsThe Mittag-Leffler and Weierstrass , Hadamard Theorems

The Weierstrass Factorization TheoremThe Hadamard Factorization Theorem

Mittag-Leffler’s Expansion Theorem

Analytic Continuation

Conformal Mapping

7


Douglas N. Arnold

Gamma FunctionBernoulli Numbers

Fourier Transform

Laplace Transform

Z Transform

Mellin Transform

Hilbert Transform

Zeta Function


Applications of Complex Analysis

References

http://www.ima.umn.edu/~arnold/index.html

8

SOLO Algebra

Set of Numbers – Examples

xnumberrealaisxx ,:R Set of real numbers

yxiyixznumbercomplexaiszzC ,,1,,: Set of complex numbers

,3,2,1,0,1,2,3,,: integeranisiiZSet of integers

,3,2,1,0,0: integernaturalaisnnN Set of positive integers or natural numbers

0,,,/: qZqpwhereqprrQ Set of rational numbers

We have:

CZN R

QxxIR R: Set of irrational numbers

IRQIRQ &R

9


Complex numbers can result by solving algebraic equations

a

cabbx

2

42

1

a

cabbx

2

42

2

042 cab

a

bxx

221

042 cab

042 caby

x

cxbxay 2

a

bcaibx

2

4 2

2,1

y

x

dxcxbxay 23

Three real roots for y = 0

One real & two complexroots for y = 0

kiez 25 1

y

x

5

2

2

i

ez

5

22

2

i

ez

5

23

3

i

ez 5

24

4

i

ez

11 z

72

72

72

72

72

1. Quadratic equations

2. Cubic equations

3. Equation

Examples

02 cxbxa

023 dxcxbxa

015 x

Return to Table of Contents

10


Fundamentals Operations with Complex Numbers z = x + i y

1,sincos

,2

iArgumentModulusi

partImaginaryypartRealxyixz

ieyixz y

x

Division

Addition dbicadicbia

Subtraction dbicadicbia

Multiplication cbdaidbcadbicbidaicadicbia 2

dccdidc

dacbidbca

dic

dic

dic

bia

dic

bia

22

022

2222

dcdc

dacbi

dc

dbca

dic

bia

Conjugate sincos:* iyixz

Absolute Value *22: zzyxz ieyixz :*

ieyixz :

x

y

11



ieyixz y

x

Polar Form of a Complex Number

Multiplication

Division

2121

212121

iii eeezz

*22 zzyxz

sincos: iyixz

xy /tan 1

2121

212121 /// iii eeezz

Euler’s Formula

sincos

!121

!3!1!21

!4!21

!!2!11

sin

123

cos

242

2

i

ki

k

n

iiie

kk

kk

n

i

Leonhard Euler1707- 1783

1,sincos

,2

iArgumentModulusi


12



1,

,2

i

ArgumentModuluse


i ieyixz y

x

Polar Form of a Complex Number ieyixz :

De Moivre Theorem

nine

eiznnin

ninn

sincos

sincos

Roots of a Complex Number

1.2.1.02

sin2

cos

sincos

/1

/1/12/1

nkn

ki

n

k

iez

n

nnkin

kiez 25 1

y

x

5

2

2

i

ez

5

22

2

i

ez

5

23

3

i

ez 5

24

4

i

ez

11 z

72

72

72

72

72

Abraham De Moivre1667 - 1754


13


Axiomatic Foundations of the Complex Number System R babaz ,,Definition of Complex System:

From those relation, for any complex numbers z1,z2,z3 C we obtain:

CzzCzzCzz 212121 ,& Closure Law 1

1221 zzzz Commutative Law of Addition2

312321 zzzzzz Associative Law of Addition3

1221 zzzz Commutative Law of Multiplication4

312321 zzzzzz Associative Law of Multiplication5

3121321 zzzzzzz Distributive Law6

111 00 zzz 111 11 zzz 7

0.. 11 zztsCzuniqueCz zz 18

1..0 11 zztsCzuniqueCz zzz /11

1 9

Equality dbcadcba ,,,A

Sum dbcadcba ,,B

Product cbdadbcadcba ,,, R mbmambam &,,C


14


History of Complex Numbers

Brahmagupta (598-670) writes Khandakhadyaka (665) which solves quadratic equations and allows for the possibility of negative solutions.

Brahmagupta 598 - 670

Brahmagupta also solves quadratic equations of the type a x2 + c = y2 and a x2 - c = y2. For example he solves 8x2 + 1 = y2 obtaining the solutions (x,y) = (1,3), (6,17), (35,99), (204,577), (1189,3363), ... For the equation 11x2+ 1 = y2 Brahmagupta obtained the solutions (x,y) = (3,10), (161/5,534/5), ... He also solves 61x2 + 1 = y2 which is particularly elegant having x = 226153980, y = 1766319049 as its smallest solution.

15



Abraham bar Hiyya Ha-Nasi אברהם בר חייא הנשיא writes the work Hibbur ha-Meshihah ve-ha-Tishboret translated in 1145 into Latin as , חבור המשיחה והתשבורת Liber embadorum, which presents the first complete solution to the quadratic equation.

Abraham bar Hiyya Ha-Nasi (אברהם בר חייא הנשיא Abraham son of [Rabbi] Hiyya "the Prince") (1070 - 1136?) was a Spaish Jewish Mathematician and astronomer, also known as Savasorda (from the Arabic الشرطة Sâhib ash-Shurta "Chief of the Guard"). He lived صاحبin Barcelona.

Abraham bar Ḥiyya ha-Nasi[2] (1070 – 1136 or 1145)

16



Nicolas Chuquet (1445 – 1488)

Chuquet wrote an important text Triparty en la science des nombres. This is the earliest French algebra book .

The Triparty en la science des nombres (1484) covers arithmetic and algebra. It was not printed however until 1880 so was of little influence. The first part deals with arithmetic and includes work on fractions, progressions, perfect numbers, proportion etc. In this work negative numbers, used as coefficients, exponents and solutions, appear for the first time. Zero is used and his rules for arithmetical operations includes zero and negative numbers. He also uses x0 = 1 for any number x. The sections on equations cover quadratic equations where he discusses two solutions.

17



Girolamo Cardano1501 - 1576

Nicolo Fontana Tartaglia1500 - 1557

Solution of cubic equation x3 + b x2 +c x +d = 0

The first person to solve the cubic equation x3 +b x = c wasScipione del Ferro (1465 – 1526), but he told the solution onlyto few people, including his student Antonio Maria Fior.

Nicolo Fontana Tartaglia, prompted by the rumors, manage to solve the cubic equation x3 +b x2 = -d and made no secret ofhis discovery.

Fior challenged Tartaglia, in 1535, to a public contest, each one had to solve 30 problems proposed by the other in 40 to 50 days. Tartaglia managed to solve his problems of type x3 +m x = n in about two hours, and won the contest.

News of Tartaglia victory reached Girolamo Cardan in Milan,where he was preparing to publish Practica Arithmeticae (1539).

Cardan invited Tartaglia to visid him and, after much persuasion,made him to divulge his solution of the cubic equation. Tartagliamade Cartan promise to keep the secret until Tartaglia had published it himself.

18




Nicolo Fontana Tartaglia1500 - 1557


After Tartaglia showed Cardan how to solve cubic equations,Cartan encouraged his student Lodovico Ferrari (1522 – 1565)to use those result and solve quartic equations x4+p x2+q x +r=0.

Since Tartaglia didn’t publish his results and after hearing fromHannibal Della Nave that Scipione del Ferro first solve cubic equations, Cardan pubished in 1545 in Ars Magna (The Great Art) the solutions of the cubic (credit given to Tartaglia) and quartic equations.

This led to another competition between Tartaglia and Cardano, for which the latter did not show up but was

represented by his student Lodovico Ferrari. Ferrari did better than Tartaglia in the competition, and Tartaglia lost both his prestige and income.

19




023 dxcxbx 3/bxy

027

2

33

393

2

273333

332

3

322

3223

23

nymybcb

dyb

cy

dcb

ycb

ybybb

yb

ybydb

ycb

ybb

y

nm

Equivalence between x3 + b x2 +c x +d = 0 and y3+m y = n (depressed cubic equation)

Solutions of y3+m y = n Start from the identity: nymybabababa

nymy

3333 3

027273

33

36

3

3333

mana

a

maban

a

mbbam

3

32

3,2,13,2,1 2742

mnnwa

3

32

3

32

3,2,13,2,1

342

33423

mnn

mmnn

wa

may

2

31,

2

31,13

3,2,1

iiew ki

k

20




3

32

3

32

3,2,1

3

322

3

32

3

32

3,2,1

3

32

3

32

3,2,13,2,1

342342

322

342

3342

342

33423

mnnmnnw

mnn

mnn

mmnnw

mnn

mmnn

wa

may

2

31,

2

31,1

3423423

3,2,13

32

3

32

3,2,13,2,1

iiew

mnnmnnwy ki

k

Solutions of y3+m y = n Note:Tartaglia andCardano knewonly the solutionw1 = 1

22




Viète Solution of y3+m y = n

François Viète1540 - 1603

In 1591 François Viète gave another solution to y3+m y = n

Start with identity CCC

34cos3cos43cos 3cos

3

Substitute in y3+m y = n Cm

y3

2

3

3

3

2

1

3

33

234

32

38

3

m

nCCnC

mmC

m m

Assuming we obtain 323

321

3

2

mn

m

n

cos3

23cos

233

k

m

n

3

1 3

2cos

3

2cos

32

32

m

nkmC

my

23




Comparison of Cardano and Viète Solution of y3+m y = n

François Viète1540 - 1603Cardano solution was

3

1 3

2cos

3

2cos

32

m

nkmssy


32

3

32

342342

mnnmnny

3

1

2

3

323

3

2cos

3

342

33

m

n

em

mnns

ssy ki

ss

or

from which we recover Viète Solution

0s

2

n

3s

3s

0x

3/1s

2s 0s

2s

1s

23

23

nm

3z

120

120

120

24



Rafael Bombelli1526 - 1572

John Wallis1616 - 1703

In 1572 Rafael Bombelli published three of the intended five volumes of “L’Algebra” worked with non-real solutions of the quadratic equation x2+b x+c=0 by using and where and applying addition and multiplication rules.

vu 1 vu 1 11

2

In 1673 John Wallis presented a geometric picture of the complex

numbers resulting from the equation x2+ b x + c=0, that is close with what we sed today.

Wallis's method has the undesirable consequence that is represented by the same point as is

1

1

0,b 0,b

bb c

bb

c1P

2P

2P

1P

Wallis representation of real roots ofquadratics

Wallis representation of non-real roots ofquadratics

25

Vector Analysis History SOLO

Caspar Wessel1745-1818

“On the Analytic Representationof Direction; an Attempt”, 1799

bia

Jean Robert Argand1768-1822

18061i

3 .R.S. Elliott, “Electromagnetics”,pp.564-568

http://www-groups.dcs.st-and.ac.uk/~history/index.html

Wessel's fame as a mathematician rests solely on this paper, which was published in 1799, giving for the first time a geometrical interpretation of complex numbers. Today we call this geometric interpretation the Argand diagram but Wessel's work came first. It was rediscovered by Argandin 1806 and again by Gauss in 1831. (It is worth noting that Gauss redid another part of Wessel's work, for he retriangulated Oldenburg in around 1824.)

26



Leonhard Euler1707- 1783

In 1748 Euler published “Introductio in Analysin Infinitorum” inwhich he introduced the notation and gave the formula1i

xixe ix sincos In 1751 Euler published his full theory of logarithms and complex numbers. Euler discovered the Cauchy-Riemann equations in 1777although d’Alembert had discovered them in 1752 while investigating hydrodynamics.

Johann Karl Friederich Gauss published the first correct proof of the fundamental theorem of algebra in his doctoral thesis of 1797, but still claimed that "the true metaphysics of the square root of -1 is elusive" as late as 1825. By 1831 Gauss overcame some of his uncertainty about complex numbers and published his work on the geometric representation of complex numbers as points in the plane.

Karl Friederich Gauss1777-1855

27



Augustin Louis Cauchy ) 1789-1857(

Cauchy is considered the founder of complex analysis after publishing the Cauchy-Riemann equations in 1814 in his paper “Sur les Intégrales Définies”. He created the Residue Theorem and used it to derive a whole host of most interesting series and integral formulas and was the first to define complex numbers as pairs of

real numbers.

Georg Friedrich BernhardRiemann

1826 - 1866

In 1851 Riemann give a dissertation in the theory of functions.


28


Derivatives

If f (z) is a single-valued in some region C of the z plane, the derivative of f (z) is defined as:

z

zfzzfzf

z

0

lim'

provided that the limit exists independent of the manner in which Δ z→0.

In such case we say that f (z) is differentiable at z.

Analytic Functions

If the derivative of f (z) exists at all points of a region C of the z plane, then f (z) is said to be analytic in C.

analytic = regular = holomorphic

A function f (z) is said to be analytic at a point z0 if there exists a neighborhoodz-z0 < δ in which f ’ (z) exists.

z0

δ

Analytic functions have derivatives of any order which themselves are analytic functions.


29


Derivatives

If f (z) is a single-valued in some region C of the z plane, the derivative of f (z) is defined as:

z

zfzzfzf

z

0

lim'

provided that the limit exists independent of the manner in which Δ z→0.

In such case we say that f (z) is differentiable at z.

Analytic Functions

If the derivative of f (z) exists at all points of a region C of the z plane, then f (z) is said to be analytic in C.

analytic = regular = holomorphic

A function f (z) is said to be analytic at a point z0 if there exists a neighborhoodz-z0 < δ in which f ’ (z) exists.

z0

δ

Analytic functions have derivatives of any order which themselves are analytic functions.

30


Analytic, Holomorphic, MeromorphicFunctions


A Meromorphic Function on an open subset D of the complex plane is a function that is Holomorphic on all D except a set of isolated points, which are poles for the function. (The terminology comes from the Ancient Greek meros (μέρος), meaning “part”, as opposed to holos (ὅλος), meaning “whole”.)

The word “Holomorphic" was introduced by two of Cauchy's students, Briot (1817–1882) and Bouquet (1819–1895), and derives from the Greek ὅλος (holos) meaning "entire", and μορφή (morphē) meaning "form" or "appearance".[2]

Today, the term "holomorphic function" is sometimes preferred to "analytic function", as the latter is a more general concept. This is also because an important result in complex analysis is that every holomorphic function is complex analytic, a fact that does not follow directly from the definitions. The term "analytic" is however also in wide use.

The Gamma Function is Meromorphic in the whole complex plane

Poles

31


Cauchy-Riemann Equations A necessary (but not sufficient) condition that f (z) = u (x,y) +i v (x,y) be analytic in a region C, is that u and v satisfy the Cauchy-Riemann equations:

y

u

x

v

y

v

x

u

&

Proof:



1826 - 1866

z

zfzzfzf

z

0

lim'

Provided that the limit exists independent of the manner in which Δ z→0.

Choose Δ z = Δ x → x

vi

x

uzf

'

Choose Δ z =i Δ y → y

v

y

uizf

'

Equalizing those two expressions we obtain:

y

ui

y

v

x

vi

x

uzf

'y

u

x

v

y

v

x

u

&

The functions u (x,y) and v (x,y) are called conjugate functions,because if one is given we can find the other (with an arbitraryadditive constant). Return to Table of Contents

32


Harmonic Functions If the second partial derivatives of u (x,y) and v (x,y) with respect to x and y existand are continuous in a region C, then using Cauchy-Riemann equations, we obtain:

02

2

2

2

2

22

2

2

2

22

y

u

x

u

y

u

xy

v

y

u

x

v

yx

v

x

u

y

v

x

uxyyx

y

x

02

2

2

2

2

2

2

2

22

22

y

v

x

v

yx

u

x

v

y

u

x

v

y

v

xy

u

y

v

x

uxyyx

x

y

It follows that under those conditions the real and imaginary parts of an analytic function satisfy Laplace’s Equation denoted by:

02

2

2

2

yxor 2

2

2

22 :

yxxi

xxi

x

where02

The functions satisfying Laplace’s Equation are called Harmonic Functions.

Pierre-Simon Laplace(1749-1827)


33


Singular Points A point at which f (z) is not analytic is called a singular point. There are various typesof singular points:

1. Isolated Singularity

The point z0 at which f (z) is not analytic is called an isolated singular point, if we cana neighborhood of z0 in which there are not singular points.

z0

δ If no such a neighborhood of z0 can be found then wecall z0 a non-isolated singular point.

2. Poles

Example 5353

183432

zzzz

zzzf

has a pole of order 2 at z = 3, a pole of order 3 at z = 5, and two simple poles at z = -3 and z = -5.

If we can find a positive integer n such thatand is analytic at z=z0

then z = z0 is called a pole of order n. If n = 1, z is called a simple pole.

0lim 00

Azfzz n

zz zfzzz n

0

34



3. Branch Points

If f (z) is a multiple valued function at z0, then this is a branch point.

Examples:

nzzzf /1

0 has a branch point at z=z0

0201ln zzzzzf has a branch points at z=z01 and z=z02

4. Removable Singularities

The singular point z0 is a removable singularity of f (z) if exists. zfzz 0

lim

Examples: The singular point z = 0 of is a removable singularityz

zsin

1sin

lim0

z

zz

35



5. Essential Singularities

A singularity which is not a pole, branch point or a removable singularity is called an essential singularity.

Example: has an essential singularity at z = z0. 0/1 zzezf

6. Singularities at Infinity

If we say that f (z) has singularities at z →∞. The type of the singularity is the same as that of f (1/w) at w = 0.

0lim

zfz

Example: The function f (z) = z5 has a pole of order 5 at z = ∞, since f (1/w) = 1/w5

has a pole of order 5 at w = 0.


36


Orthogonal Families

If f (z) = u (x,y) + i v (x,y) is analytic, then the one-parameter families of curves

yxvyxu ,,,

where α and β are constant are orthogonal.

Proof:The normal to u (x,y) = α is: y

y

ux

x

uyxu 11,

The normal to v (x,y) = β is: yy

vx

x

vyxv 11,

The scalar product between the normal to u (x,y) = α and the normal to v (x,y) = β is:

y

v

y

u

x

v

x

uyxvyxu

,,

Using the Cauchy-Riemann Equation for the analytic f (z):

y

u

x

v

y

v

x

u

&

0,,

y

v

y

u

y

u

y

vyxvyxu

x

y

yxu ,

yxv ,

planez

u

vplanew


37


Complex Line Integrals

Let f (z) be continuous at all points on a curve C of a finite length L.

n

i

ii

n

i

iiin zfzzfS11

1

C

1z

nzb

2z

0za

1iz

iz

1

2

i

n

Let subdivide C into n parts by n arbitrary pointsz1, z2,…,zn, and call a=z0 and b=zn. On each arc joining zi-1 to zi choose a point ξi. Define the sum:

Let the number of subdivisions n increase in such away that the largest of Δzi approaches zero, then the sum approaches a limit that is called the line integral (also Riemann-Stieltjes integral).

C

b

a

n

i

iiznnzdzfzdzfzfS

i1

0limlim

Properties of Integrals

CCC

zdzgzdzfzdzgzf constantAzdzfAzdzfACC

a

b

b

a

zdzfzdzf b

c

c

a

b

a

zdzfzdzfzdzf

CoflengthLandConMzfLMzdzfzdzfCC


38


Simply and Multiply Connected Regions

A region R is called simply-connected if any simple closed curve Γ, which lies in Rcan be shrunk to a point without leaving R. A region R that is not simply-connectedis called multiply-connected.

C0

x

y

R C1

C0

x

y

RC1

C2

C3

C

x

y

R

C

x

y

Rsimply-connected

multiply-connected.


39


Green’s Theorem in the Plane

C

R

Let P (x,y) and Q (x,y) be continuous and have continuous

partial derivatives in a region R and on the boundary C.

Green’s Theorem states that:

GEORGE STOCKES1819-1903

A more general theorem was given by Stokes

R

dydxy

P

x

QdyQdxP

C

yzxzxy RRR

dzdyz

Q

y

Rdzdx

x

R

z

Pdydx

y

P

x

QdzRdyQdxP

C

or in vector form: S

dAFdrFC

where: zzyxRyzyxQxzyxPzyxF 1,,1,,1,,,,

zdzydyxdxdr 111

zdydxydzdxxdzdydA 111

GEORGE GREEN1793-1841

zz

yy

xx

111

40


Proof of Green’s Theorem in the Plane C

R

P

T

S

Q

a bx

y

xgy 2

xgy 1

Start with a region R and the boundary curve C, definedby S,Q,P,T, where QP and TS are parallel with y axis.

b

a

xgy

Xgy

dyy

Pdxdydx

y

P2

R

By the fundamental lemma of integral calculus:

xgxPxgxPyxPdy

y

yxP xgy

xgy

xgy

Xgy

12 ,,,, 2

1

2

Therefore: b

a

b

a

dxxgxPdxxgxPdydxy

P12 ,,

R

but: a

bSQ

dxxgxPdxxgxP 22 ,, integral along curve SQ

b

aPT

dxxgxPdxxgxP 11 ,, integral along curve PT

If we add to those integrals: 00,, dxsincedxyxPdxyxPQPTS

we obtain:

CTSPTQPSQ

dxyxPdxyxPdxxgxPdxyxPdxxgxPdydxy

P,,,,, 12

R

Assume that PT is defined by the function y = g1 (x) and SQ is defined by the function y = g2 (x), both smooth and

y

P

is continuous in R:

41


Proof of Green’s Theorem in the Plane (continue – 1)

In the same way:

Therefore we obtain:

C

dxyxPdydxy

P,

R

C

dyyxQdydxx

Q,

R

R

dydxy

P

x

QdyQdxP

C

The line integral is evaluated by traveling C counterclockwise.

For a general single connected region, as that described in Figure to the right, can be divided in a finite number of sub-regions Ri, each of each are of the type described in the Figure above. Since the adjacent regions boundaries are traveled in opposite directions, there sum is zero, and we obtain again:

R

dydxy

P

x

QdyQdxP

C

C

R4

x

yR

R3

R1

R2

C

R

P

T

S

Q

a bx

y

xgy 2

xgy 1

42


Proof of Green’s Theorem in the Plane (continue – 2)

The general multiply-connected regions can be transformed in a simply connected region by infinitesimal slits

Since the slits boundaries are traveled in opposite directions, there integral sum is zero:

C0

x

y

R C1

P0

P1

C0

x

y

RC1

C2

C3

R

dydxy

P

x

QdyQdxPdyQdxP

i CC i0

All line integrals are evaluated by traveling Ci i=0,1,… counterclockwise.

00

1

1

0

P

P

P

P

dyQdxPdyQdxP

We obtain:


43


Consequences of Green’s Theorem in the Plane

Let P (x,y) and Q (x,y) be continuous and have continuous first partial

derivative at each point of a simply-connected region R. A necessary and sufficient condition that around every closed path C in

R is that in R. This is synonym to the condition thatis path independent.y

P

x

Q

0C

dyQdxP

Sufficiency: Suppose y

P

x

Q

According to Green’s Theorem 0

R

dydxy

P

x

QdyQdxP

CNecessity: 0

ory

P

x

QSuppose along every path C in R. Assume that

at some point (x0,y0) in R. Since Q/ x and P/ y are continuous existsa region τ around (x0,y0) and boundary Γ for which , therefore:

0C

dyQdxP

0

ory

P

x

Q

0

ordydxy

P

x

QdyQdxP

C

x

y

R

L

dyQdxP

0

y

P

x

QThis is a contradiction to the assumption, therefore q.e.d.


44


Cauchy’s Theorem

C

x

y

R

Proof:

0C

dzzf

If f (z) is analytic with derivative f ‘ (z) which is continuous at all points insideand on a simple closed curve C, then:

yxviyxuzf ,, Since is analytic and has continuous first order derivative

y

ui

y

v

x

vi

x

u

zd

fdzf

iyzxz

'

y

u

x

v

y

v

x

u

& Cauchy - Riemann

0

00

RR

dydxy

v

x

uidydx

y

u

x

v

dyudxvidyvdxudyidxviudzzfCCCC

q.e.d.



45


Cauchy-Goursat Theorem

C

x

y

R

Proof:

0C

dzzf

If f (z) is analytic which is continuous at all points inside and on a simple closed curve C, then:


Goursat removed the Cauchy’s condition

that f ‘ (z) should be continuous in R.

CF

DE

A

B

I

IVII III

Start with a triangle ABC in z in which

f (z) is analytic, Join the midpoints E,D,F to obtain four equal triangles ΔI, ΔII, ΔIII, ΔIV. We have:

IVIIIIII

dzzfdzzfdzzfdzzf

dzzf

DEFDFCDFEBFEDAED

FDEFDEDFFCDFEEBFEDDAE

FCDEBFDAEABCA

Eduard Jean-Baptiste Goursart

1858 - 1936

46


Proof of Cauchy-Goursat Theorem (continue – 1)

If f (z) is analytic which is continuous at all points inside and on a simple closed curve C, then:

CF

DE

A

B

I

IVII III

IVIIIIII

dzzfdzzfdzzfdzzfdzzfABCA

then:

IVIIIIII

dzzfdzzfdzzfdzzfdzzfABCA

Let Δ1 be the triangle in which the absolute value of the integral is maximum.

1

4 dzzfdzzf

Continue this procedure in triangle Δ1 in which Δ2 is the triangle in which theabsolute value of the integral is maximum.

21

244 dzzfdzzfdzzf

n

dzzfdzzf n4

47



CF

DE

A

B

I

IVII III

n

dzzfdzzf n4

For an analytic function f (z) compute 0

0

0

0 ':, zfzz

zfzfzz

0'''lim,lim 000

0

00

00

zfzfzfzz

zfzfzz

zzzz

0000000 ,'&,..,0 zzzzzzzfzfzfzzwheneverzzts

nnnn

dzzzzzdzzzzzdzzzzfzfdzzf

TheoremIntegralCauchy

0000

0

)00 ,,'

n

0z

na

nbnc

z 0zz

0zzcbaP nnnn

2

2

00 2,

nnn

PPdzPdzzzzzdzzf

nnn

But , where Pn the perimeter of Δn and P the perimeter of Δ are related, by construction, by

nPzzzz 00 &,n

n PP 2/

q.e.d. 0

444

02

2

dzzfPP

dzzfdzzfn

nn

n

48



nz1z

2z

1iz

iz

1nz

n1

23

i

C

O

q.e.d.

For the general case of a simple closed curve Cwe take n points on C: z1, z2,…,zn and a pointO inside C. We obtain n triangles Δ1, Δ2,.., Δn,

for each of them we proved Cauchy-Goursat Theorem.

Let define the sum:

n

i zz

iin

ii

zzfS1

1

:

we have: 01

1

i

i

i

ii

z

O

O

z

z

z

dzzfdzzfdzzfdzzf

n

i

i

i

i

i

i

i

ii

S

n

i

z

z

i

n

i

z

z

i

n

i

z

z

ii

n

i

z

z

n

i

dzzfdzzfzfdzzfzfzfdzzfdzzf

11111

1111

0

NnforSdzzfdzzfS n

CC

nn 2

lim

221

1

11111

n

i

ii

n

i

i

n

i

z

z

in

n

i

z

z

in zzL

dzfzfdzzfzfSdzzfzfSi

i

i

i

022

C

nn

CC

dzzfNnforSSdzzfdzzf

Since we proved that , we can write: 0

dzzf


49


Consequences of Cauchy-Goursat Theorem

B

x

y

R

A

C1

C2

D1

D2

a

b If f (z) is analytic in a simply-connected region R, then

is independent of the path in R joining any

two points a and b in R.

b

a

dzzf

Let look at thr closed path AC1D1BD2C2A in R inside which f (z) is analytic.According to Cauchy-Goursat Theorem

BDAcBDAcBDAcBDAcACBDBDAcACBDDAC

dzzfdzzfdzzfdzzfdzzfdzzfdzzf2211221122111122

0

Proof:

If f (z) is analytic in a multiply-connected region R, bounded by two simple closed curves C1 and C2, then:

1

2C1

x

y

R C2

P0

P1

21 CC

dzzfdzzf

The general multiply connected regions can be transformed in a single connected region by an infinitesimal slit P0 to P1.

212

0

1

1

01

0

0CCC

P

P

P

PC

dzzfdzzfdzzfdzzfdzzfdzzf

Proof:


50


Cauchy’s Integral Formulas


If f (z) is analytic inside and on a simple closed curve C and a is any point inside C then

C

dzaz

zf

iaf

2

1

C

x

y

R a

Proof:Let chose a circle Γ with center at a

2,0,: ieazz

Since f (z)/ (z-a) is analytic in the region definedbetween C and the circle Γ we can use:

zd

az

zfzd

az

zf

C

afidafideie

eafzd

az

zf i

i

i

21lim

2

0

2

00

therefore:

C

dzaz

zf

iaf

2

1q.e.d.

Cauchy’s Integral Formulas and Related Theorems Return to Table of Contents

51


Cauchy’s Integral Formulas for the n Derivative of a Function


If f (z) is analytic inside and on a simple closed curve C and a is any point inside C, where the n derivative exists, then

C

n

n dzaz

zf

i

naf

12

!

C

x

y

R a

Proof:

Let prove this by induction.

Assume that this is true for n-1:

Then we can differentiate under the sign of integration:

C

dzaz

zf

iaf

2

1For n = 0 we found

C

n

n dzaz

zf

i

naf

2

!11

C

n

C

n

nn dzaz

nzf

i

ndz

azad

dzf

i

naf

ad

daf

1

1

2

!11

2

!1

q.e.d.Therefore for n we obtain:

C

n

n dzaz

zf

i

naf

12

!

We can see that an analytic function has derivatives of all orders.


52


Morera’s Theorem (the converse of Cauchy’s theorem)

If f (z) is continuous in a simply-connected region R and if

around every simple closed curve C in R then

f (z) ia analytic in R.

0C

dzzf

B

x

y

R

A

C1

C2

D1

D2

a

z

Proof:Since around every closed curve C in

R 0

C

dzzf

BDAcBDAcBDAcBDAcACBDBDAcACBDDAC

dzzfdzzfdzzfdzzfdzzfdzzfdzzf2211221122111122

0

The integral is independent on path

between two points, if the path is in R

z

a

dzzfzF

Let choose a straight path between z and z+Δz

zz

z

z

a

zz

a

udzfufz

zfudufudufz

zfz

zFzzF 11

Since f (z) is continuous zuwheneverzfuf

Therefore zfzd

zFdzudzfuf

zzf

z

zFzzFzz

z

1

C

x

y

R z

z+ z

Since F (z) has a derivative in R, it is analytic, and so are its derivatives, i.e. f (z) Return to Table of Contents

Giacinto Morera1856 - 1907

http://en.wikipedia.org/wiki/File:Giacinto_Morera_.JPG

53


Cauchy’s Inequality


If f (z) is analytic inside and on a circle C of radius r andcenter at z = a, then

,...2,1,0!

nr

nMaf

n

n

where M is a constant such that | f (z) |< M is an upper boundof | f (z) | on C.

Proof:

C

x

y

a

r

Use Cauchy Integral Formula:

,2,1,0

!2

1

2

!1

2

!

2

!1

2

0

11

nr

nMr

r

Mndr

r

Mndz

az

zfnaf

nnn

Mzf

C

n

n

,2,1,02

!1

ndz

az

zf

i

naf

C

n

n

q.e.d.

On the circle C: .ieraz


54


Liouville’s Theorem

Joseph Liouville1809 - 1882

If for all z in the complex plane:

(1) f (z) is analytic

(2) f (z) is bounded, i.e. | f (z) |< M for some constant M

then f (z) must be a constant.

Proof No. 1:

Using Cauchy’s Inequality: ,...2,1,0!

nr

nMaf

n

n

Letting n=1 we obtain: r

Maf '

Since f (z) is analytic in all z plane we can take r → ∞ to obtain

constantafaafr

Maf

r

,00lim ''

q.e.d.

55


Liouville’s Theorem

Joseph Liouville1809 - 1882

If for all z in the complex plane:

(1) f (z) is analytic (2) f (z) is bounded, i.e. | f (z) |< M for some constant M

then f (z) must be a constant.Proof No. 2:

Using Cauchy’s Integral Formula

q.e.d.

Let a and b be any two points in z plane. Draw a circleC with center at a and radius r > 2 | a-b |

C

x

y

a

r

b

2/rba

CCC

dzbzaz

zf

i

abdz

az

zf

idz

bz

zf

iafbf

22

1

2

1

We haveraz 2/rbarbaazbaazbz

r

Mabdr

rr

Mabdz

bzaz

zfabdz

bzaz

zfabafbf

CC

2

2/222

2

0

Since f (z) is analytic in all z plane we can take r → ∞ to obtain

afbfafbf 0 therefore f (z) is constant.


56


Foundamental Theorem of Algebra

Every polynomial equation P (z) = a0 + a1z+a2z2+…+anzn=0 with degreen ≥ 1 and an ≠ 0 (ai are complex constants) has at least one root.

From this it follows that P (z) = 0 has exactly n roots, due attention being paid to multiplicities of roots.

Proof: If P (z) = 0 has no root, then f (z) = 1 / P (z) is analytic for all z. Also | f (z) |= 1 / | P (z) | is bounded. Then by Liouville’s Theorem f (z) andthen P (z) are constant. This is a contradiction to the fact that P (z) is a polynomial in z, therefore P (z) = 0 must have at least one root (zero).

Suppose that z = a is one root of P (z) = 0. Hence P (a) = 0 and

zQazazaazaaza

aaaaaaazazazaaaPzPnn

n

n

n

n

n

22

21

2

210

2

210

Since an ≠ 0, Q (s) is a polynomial of degree n-1.

Applying the same reasoning to the polynomial Q (s) of degree n-1,we conclude that it must also have at least one root. This procedure continuesuntil n = 0, therefore it follows that P (z) has exactly n roots.

q.e.d.Return to Table of Contents

57


Gauss’ Mean Value Theorem

Karl Friederich Gauss1777-1855

C

x

y

a

r

If f (z) is analytic inside and on a circle C with center at a and radius r, then f (a) is the mean of the values of f (z) on C, i.e.,

.2

12

9

derafaf i

Proof:

Use Cauchy Integral Formula:

On the circle C: ii eridzeraz .

C

dzaz

zf

iaf

2

1

derafderier

eraf

idz

az

zf

iaf ii

i

i

C2

1

2

1

2

1

q.e.d.Return to Table of Contents

58


Maximum Modulus Theorem

If f (z) is analytic inside and on a simple closed curve C and is not identicallyequal to a constant, then the maximum value of | f (z) | occurs on C.

Proof:

The proof is based on the continuity of f (z) and on the Gauss’ Mean Value Theorem.

C

x

y

a

r

CC1

C2

C3

R

b

Since f (z) is analytic in C, | f (z) | has a maximum M inside oron C. Suppose that the maximum value is achieved at the point a inside C, i.e. | f (a) |=M =max | f (z) |. Since a is inside C we canfind a circle C1, with center at a that is inside C. Since f (z) is not

constant we can find a point b in C1 such that | f (b)|=M – ε< | f (a)|. Using the continuity of f (z) we can find a circle around b, C2,

bzforbfzfzC 2/:2

2/2/2/ MMbfzf

Now apply the Gauss’ Mean Value Theorem for point a and the circle with center at apassing trough b, C3. Define by α the arc of C3 inside C2. .

2

12

9

derafaf i

42

222/.

2

12

9 2/

2

9

MM

MderafderafderafMaf

M

i

M

ii

59


Maximum Modulus Theorem

If f (z) is analytic inside and on a simple closed curve C and is not identicallyequal to a constant, then the maximum value of | f (z) | occurs on C.

Proof (continue):

42

222/.

2

12

9 2/

2

9

MM

MderafderafderafMaf

M

i

M

ii

We obtained that is impossible, therefore

a for which |f (z)| is maximum cannot be inside C, but on C.

4

?

MMafC

x

y

a

r

CC1

C2

C3

R

b

q.e.d.


60


Minimum Modulus Theorem

If f (z) is analytic inside and on a simple closed curve C and f (z) ≠ 0inside C then | f (z) | assumes its minimum value on C.

Proof:

Since f (z) is analytic inside and on a simple closed curve C and f (z) ≠ 0inside C it follows that 1/ f (z) is analytic inside and on C. Then according to Maximum Modulus Theorem 1/| f (z) | assumes its maximum vale on C and

therefore | f (z) | assumes its minimum value on C.

q.e.d.

x

y C

R


61


Poisson’s Integral Formulas for a Circle

Siméon Denis Poisson1781-1840

Let f (z) be analytic inside and on the circle C defined by |z| = R, and let z = r e iθ be any point inside C, then:

2

0

2

02

22

2

22

2

022

22

2

1

2

1

cos22

1

deRfzeR

zRdeRf

ereR

rR

deRfrrRR

rRerzf

i

i

i

ii

ii

C

x

y

R

R

z'

rR2/r

z

z

r

zRz /2

1

Proof: Since f (z) is analytic in C we can apply Cauchy’s IntegralFormula:

C

i dzzz

zf

ierfzf '

'

'

2

1

C

dzzRz

zf

i'

/'

'

2

10

2

If we subtract those equations we obtain:

The inverse of the point z with respect to C is andlies outside C, therefore by Cauchy’s Theorem:

zRz /2

1

CC

i dzzfzRzzz

zRz

idzzf

zRzzzierfzf ''

/''

/

2

1''

/'

1

'

1

2

12

2

2

62


Poisson’s Integral Formulas for a Circle


Proof (continue):

2

022

22

2

0

22

2

0

22

2

02

2

2

2

cos22

1

2

1

2

1

/

/

2

1

''/''

/

2

1

deRfRrRr

rR

deRfereRereR

rR

deRfeRerereR

eRr

deRieRferReRereR

erRer

i

dzzfzRzzz

zRz

ierfzf

i

iiiii

iiiii

i

iiiiii

ii

C

i

Writing we have: ,, rviruerf i

2

022

22

,cos22

1, dRu

RrRr

rRru

2

022

22

,cos22

1, dRv

RrRr

rRrv

q.e.d.

C

x

y

R

R

z'

rR2/r

z

z

r

zRz /2

1


63


Poisson’s Integral Formulas for a Half Plane


C

x

y

R

z

zR

Let f (z) be analytic in the upper half y ≥ 0 of the z plane and let z = (x + i y) any point in this upper half plane, then:

dw

yxw

wfyzf

22

Proof: Let C be the boundary of a a semicircle of radius Rcontaining as an interior point, but doesnot contain

yixz yixz

Using Cauchy’s Integral Formula we have:

C

dwzw

wf

izf

2

1

By subtraction we obtain:

C

dwzw

wf

i2

10

CC

CC

dwwfyxw

yi

idwwf

yixwyixw

yixyix

i

dwwfzwzw

zz

idwwf

zwzwizf

22

2

2

1

2

1

2

111

2

1

64


Poisson’s Integral Formulas for a Half Plane


Proof (continue):

Where Γ is the upper a semicircle of radius R.

dwwfyxw

ydwwf

yxw

y

dwwfyxw

yi

izf

R

R

C

2222

22

11

2

2

1

If we take R→∞ we obtain: 0lim

122

dwwfyxw

yR

dw

yxw

wfyzf

22

Therefore:

dw

yxw

wufyyxu

22

0,,

dw

yxw

wvfyyxv

22

0,,

Writing and since w varies on x axis , and we have:

yxviyxuzf ,,

0,0, wviwuwf

q.e.d.

C

x

y

R

z

zR


65

SOLOInfinite Series

Given a series:


n

iin uS

1Convergence Definition:The series Sn converges to S as n →∞ if for all ε > 0 there exists an positive integer N such that

If no such N exists then we say that the series diverges.

NnallforSuSSn

iin

1

Convergence Theorem:The series Sn converges as n →∞ if and only if there exists an positive integer M such that

If no such M exists then we say that the series diverges.

11

NallforMuSN

iiN

If S is unknown we can use the Cauchy Criterion for convergence: for all ε > 0 there exists an positive integer N such that

NmnallforuuSSm

jj

n

iimn

,11


A necessary (but not sufficient) condition for convergence is that lim i→∞ ui = 0

Return to the Table of Content

Cauchy Convergence Criterion

66

SOLOInfinite Series

Given a series:


n

iin uS

1Convergence Tests

Cauchy Root Kummer, anEuler-Maclurin

Integral

D’AlembertCauchy Ratio Raabe

Gauss

(Comparison withGeometric Series)

(Also by Comparison withGeometric Series)

(Comparison withIntegral)

1na nan

nnan ln

In term by term a series of terms 0 ≤ un ≤ an, in which the an form a convergent series,then is also convergent.n nu


67

SOLO

The Geometric Series

r

ra

r

rrarararaa

r

rS nnnG

1

1

1

1

1

1 1321

Multiply and divide by (1 – r)

1

0

1321

n

i

innG rararararaaS

We can see that

diverger

convergerr

a

r

raS

n

nnG

n1

11

1

1limlim 1

Given a series:


n

iin uS

1

Infinite Series

68

SOLO

Convergence Tests

Cauchy Root Test


If (an) 1/n ≤ r < 1 for all sufficiently large n, with r independent of n, then is convergent. If (an) 1/n ≥ 1 for all sufficiently large n, then is divergent.

n na

n na

The first part of this test is verified easily by raising (an) 1/n ≤ r to the nth power. We get:

1 nn ra

n na Since rn is just the nth term in a Convergent Geometric Series, is convergent by the Comparison Test. Conversely, if (an) 1/n ≥ 1, the an ≥ 1 and the series diverge. This Root Test is particularly useful in establishing the properties of Power Series.

Given a series:


n

iin uS

1

Infinite Series


69

SOLO

Convergence Tests

D’Alembert or Cauchy Ratio Test

Jean Le Rond D’Alembert1717 - 1783

If (an+1/an) ≤ r < 1 for all sufficiently large n, with r independent of n, then is convergent. If (an+1/an) ≥ 1 for all sufficiently large n, then is divergent.

n na

n na

Convergence is proved by direct comparison with the geometric series (1+r+r2+ …)

ateindetermin,1

,1

,1

lim 1 divergence

econvergenc

a

a

n

n

n

n

i

nn nS

1

2/Example: convergentn

n

a

a n

nnn

n

n 2

12

2

1limlim

11

Given a series:


n

iin uS

1

Infinite Series


70

SOLO

Convergence Tests

Maclaurin or Euler Integral Test

xf

x1 2 3 4

11 af

22 af

Comparison of Integral and Sum-Blocks Leading

xf

x1 2 3 4

11 af

Comparison of Integral and Sum-Blocks Lagging

Is geometrically obvious that:

1111

fxdxfaxdxfn n

Given a series:


n

iin uS

1

Infinite Series


SOLO

Convergence Tests

Kummer’s Test

Ernst Eduard Kummer(1810 – 1893)

Consider a Series of positive terms ui and a sequence of positive constants ai.

The two tests can be written as:

divergea

convergeCa

u

ua

i in

n

nn

n 111 &0

0lim

Given a series:


n

iin uS

1

Infinite Series

http://en.wikipedia.org/wiki/File:ErnstKummer.jpg

72

SOLO

Convergence Tests

Kummer’s Test (continue – 1)


If

for all n ≥ N, where N is some fixed number, then converges.

011

Cau

ua n

n

nn

1i iuErnst Eduard Kummer

(1810 – 1893)

nnnnn

NNNNN

NNNNN

uauauC

uauauC

uauauC

11

22112

111

Proof:

Add and divide by CC

ua

C

uau nnNNn

Ni i 1

C

uau

C

ua

C

uauuuuS NNN

i innNNN

i i

n

Ni i

N

i i

n

i in 11111

The partial sums Sn have an upper bound. Since the lower bound is zero the sum must converge. iu

q.e.d.

Given a series:


n

iin uS

1

Infinite Series


73

SOLO

Convergence Tests

Kummer’s Test (continue – 2)

Ernst Eduard Kummer(1810 – 1893)


Proof: Nnuauaua NNnnnn ,11

Since an > 0n

NNn a

uau

and

1

1

1 NiiNN

Nii auau

If diverges, then by comparison test diverges.

1

1

iia

1iiu

q.e.d.

Given a series:


n

iin uS

1

Infinite Series



74

SOLO

Convergence Tests

Raabe’s Test

Proof:

In Kummer’s Test choose an = n and P = C + 1.

Given a series:


n

iin uS

1

Infinite Series


75

SOLO

Convergence Tests

Gauss’ s Test

Carl Friedrich Gauss(1777 – 1855)

If un > 0 for all finite n and

in which B (n) is a bounded function of n for n → ∞, thenconverges for h > 1 and diverges for h ≤ 1. There is no indeterminate case here.

2

1

1n

nB

n

h

u

u

n

n

n nu

Proof:

For h > 1 and h < 1 the proof follows directly from Raabe’s Test:

h

n

nBh

n

nB

n

hn

u

un

nnn

n

n

lim11lim1lim

21

If h = 1, Raabe’s Test fails. However if we return to Kummer’s Test and use an=n ln n:

nnnnn

n

nnn

nnn

nB

n

hnn

nn

h

n

11lnlnln1lim1ln1

1lnlim

1ln11lnlim

1

2

Given a series:


n

iin uS

1

Infinite Series

76

SOLO

Convergence Tests

Gauss’ s Test


If un > 0 for all finite n and

in which B (n) is a bounded function of n for n → ∞, thenconverges for h > 1 and diverges for h ≤ 1. There is no indeterminate case here.

2

1

1n

nB

n

h

u

u

n

n

n nu

Proof (continue – 1):

Kummer’s withan=n ln n:

nnnn

n

nB

n

hnn

n

h

n

11ln1lim1ln11lnlim

1

2

013

1

2

111lim

11ln1lim

32

nnnn

nn

nn

Hence we have a divergence for h = 1. This is an example of a successful application of Kremmer’s Test in which Raabe’s Test failed.

Given a series:


n

iin uS

1

Infinite Series


77


Infinite Series, Taylor’s and Laurent Series

Let {un} :=u1 (z), u2 (z),…,un (z),…, be a sequence of single-valued functions of z insome region of z plane.

We call U (z) the limit of {un} ,if given any positive number ε we can find a number N (ε,z) such that and we write this: zNnzUzu n ,

zUzuorzUzun

nnn

lim

x

y C

R

If a sequence converges for all values z in a region R, we call Rthe region of convergence of the sequence. A sequence that is notconvergent at some point z is called divergent at z.

Infinite Series of Functions

78

SOLO Complex VariablesInfinite Series, Taylor’s and Laurent Series

Infinite Series of Functions From the sequence of functions {un} let form a new sequence {Sn} defined by:

n

i

inn zuzuzuzuzS

zuzuzS

zuzS

1

21

212

11

If , the series is called convergent and S (z) is its sum. zSzS nn

lim

A necessary (but not sufficient) condition for convergence is that lim n→∞ un(z) = 0

Example: The Harmonic Series

nnn

1

4

1

3

1

2

11

1

1

01

limlim n

un

nn

By grouping the terms in the sum as

2

1

22

1

2

1

1

2

1

1

1

8

1

7

1

6

1

5

1

4

1

3

1

2

11

p

p

pppp


79


Absolute Convergence of Series of Functions

Given a series of functions:

n

i

in zuzS1

If is convergent the series is called absolutely convergent.

n

i

i zu1

If is convergent but is not, the series is called conditionally convergent.

n

i

i zu1

n

i

i zu1


80


Uniformly Convergence of Sequences and Series

If for the sequence of functions {un(z)} we can find for each ε>0 a number N (ε)

such that for all zR we say that {un} uniformlyconverges to U (z). ( N is a function only of ε and not of z)

NnzUzu n

If the series of functions {Sn(z)} converges to S (z) for all zR we define the remainder

1

:nz

inn zuzSzSzR

The series of functions {Sn(z)} is uniformly convergent to S (z)

if for all for all ε>0 and for all zR we can find a number N (ε)

such that NnzSzS n x

y C

R


81


Weierstrass M (Majorant) Test

Karl Theodor Wilhelm Weierstrass

(1815 – 11897)

The most commonly encountered test for Uniform Convergence is the Weierstrass M Test.

Proof:

Since converges, some number N exists such that for n + 1 ≥ N,

If we can construct a series of numbers , in which Mi ≥ |ui(x)| for all x in the interval [a,b] and is convergent, the series ui(x) will be uniformly convergent in [a,b].

1 iM

1 iM

1 iM

1niiM

This follows from our definition of convergence. Then, with |ui(x)| ≤ Mi for all x in the interval a ≤ x ≤ b,

1nii xu

Hence

1niin xuxsxS


http://en.wikipedia.org/wiki/File:Karl_Weierstrass.jpg


Abel’s Test

Niels Henrik Abel ( 1802 – 1829)

If

and the functions fn(x) are monotonic decreasing |fn+1(x) ≤ fn(x)| and bounded, 0 ≤ fn(x) ≤ M, for all x in [a,b], then Converges Uniformly in [a,b].

convergentAa

xfaxu

n

nnn

,

n n xu

bainconvergentuniformlyisxuxd

dbaincontinuousarexu

xd

dandxu

n nnn ,&,1


http://en.wikipedia.org/wiki/File:Niels_Henrik_Abel.jpg


Uniformly Convergent Series of Analytic FunctionsSuppose that(i)Each number of a sequence of functions u1(z), u2(z),…,un(z),… is Analytic inside a Region D,(ii)The Series

is Uniformly Convergent through Every Region D’ interior to D.Then the function is Analytic inside D, and all its Derivatives can be calculated by term-by-termDifferentiation.

1n n zu

1n n zuzf

C

x

y

R

z0

z

w

r

P0D

Proof:Let C be a simple closed contour entirely inside D, and let z a Point inside D. Since un(z) is Analytic inside D, we have:

C

nn wd

zw

wu

izu

2

1

for each function un(z). Hence

11 2

1n

C

nn n wd

zw

wu

izuzf


Uniformly Convergent Series of Analytic FunctionsC

x

y

R

z0

z

w

r

P0D


Since is Uniformly Convergent on C, we may multiply by 1/(w-z) and integrate term-by-term:

and we obtain

11 2

1n

C

nn n wd

zw

wu

izuzf

1n n zu

11 nC

n

Cn

n wdzw

wuwd

zw

wu

CC

nn wd

zw

wf

iwd

zw

wu

izf

2

1

2

11

The last integral proves that f(z) is Analytic inside C, and since C is an arbitrary closed contour inside D, f(z) is Analytic inside D.



x

y

R

z0

z

w

r

P0D


Since f(z) is Analytic in D, the same is true for f’(z), therefore we can write

C

wdzw

wf

izf 22

1'

Therefore

q.e.d.

11 2

1 22

'2

1

2

1

2

1'

n nnC

neConvergenc

Uniform

Cn n

C

zuwdzw

wu

i

zw

wdwu

iwd

zw

wf

izf

Hence the Series can be Differentiate term-by-term



x

y

R

z0

z

w

r

P0D

Remarks on the above Theorem

(i)The contrast between the conditions for term-by-term differentiation of Real Series, and of Series of Analytic Functions is that - In the case of Real Series we have to assume that the Differentiated Series is Uniformly Convergent. - In the case of Series of Analytic Series the Theorem proved that the Differentiated Series is Uniformly Convergent.

(ii)If we merely assumed that the given Series is Uniformly Convergent on a certain Closed Curve C, we could prove as before that f(z) is Analytic at all points inside C.

(iii) Even if we assume that each un(z) is Analytic on the Boundary of the Domain D, and the Series is Uniformly Convergent on the Boundary, we can not prove that f(z) is Analytic on the Boundary, or the Differentiated Series Converges on the Boundary.

(iv) The Theorem may be stated as a Theorem on Sequences of Functions: If fn(z) is Analytic in D for each value of n, and tends to f(z) Uniformly in any Region interior to D, then f(z) is Analytic inside D, and fn’(z) tends to f’(z) Uniformly in any Region interior to D.


87


Let f (z) be analytic at all points within a circle C0 with center at z0 and radius r0.Then at each point z inside C0:

Taylor’s Series

nn

zzn

zfzz

zfzzzfzfzf 0

02

00

000 !!2

'''

Power Series

Brook Taylor1685 - 1731

a convergent power series for some |z-z0|<R (radius of convergence).

C

x

y

R z0

C0

C1

zz'

r0

r1

r

Proof:

Start with the Cauchy’s Integral Formula:

C

zdzz

zf

izf '

'

'

2

1

Use the identity:

11

1

1 12n

n

for:

nn

zz

zz

zz

zzzz

zz

zz

zz

zz

zz

zzzzzzzzzz

0

0

0

0

1

0

0

0

0

0

0

0000

'

'1

1

''1

'

1

'1

1

'

1

'

1

'

1

Since z inside C0 |z-z0|=r < r0. For z’ is on C1 we have |z’-z0|=r1<r0

88


Taylor’s Series (continue - 1)

Power Series

C

x

y

R z0

C0

C1

zz'

r0

r1

r

Proof (continue - 1):

Using the Cauchy’s Integral Formula:

C

zdzz

zf

izf '

'

'

2

1

nn

zz

zz

zz

zzzz

zz

zz

zz

zz

zf

zz

zf

0

0

0

0

1

0

0

0

0

0 '

'1

1

''1

'

'

'

'

n

nn

R

C

n

n

n

nzf

C

n

zf

C

zf

C

Rzzn

zfzzzfzf

zzzz

zdzf

i

zz

zzzz

zdzf

izz

zz

zdzf

izz

zdzf

i

n

n

00

000

0

0

1

0

!/

0

0

!1/'

2

00

!'

''

''

2

'

''

2

1

'

''

2

1

'

''

2

1

0

0

0

0

0

0

0

We have:

n

i

n

n

C

n

n

n r

r

rr

Mrder

rrr

Mr

zzzz

zdzfzzR

11

1

2

0

1

110

0

2''

''

20

where |f (z)|<M in C0 and r/r1< 1, therefore: 0

n

nR q.e.d.

0100 ' rrzzrzz

89


Let f (z) be analytic at all points within a circle C0 with center at z0 and radius r0.Then at each point z inside C0:

Taylor’s Series (continue – 2)

nn

zzn

zfzz

zfzzzfzfzf 0

02

00

000 !!2

'''

Power Series


a convergent power series for some |z-z0|<R (radius of convergence).

C

x

y

R z0

C0

C1

zz'

r0

r1

r


0

010

1 !k

kk

zzk

zfzfSuppose the series converges for z=z1:

01

0

1

001

0

0

01

0

0

0

0

1!!

zz

zz

MaM

zz

zzzz

k

zfzz

k

zfzf

a

k

k

k

k

kk

k

kk

Since the series converges all its terms are bounded

,2,1,0! 01

0 nMzz

k

zf kk

Define:01

0:

zz

zza

Therefore the series f (z) converges for all 010 zzzz

The region of convergence of a Taylor series of f (z) around a point z0 is a circle centered at z0 and radius of convergence R that extends until f (z) stops to be analytic.

90


Taylor’s Series (continue – 3)

n

n

zn

fz

fzffzf

!

0

!2

0''0'0 2

Power Series


When z0 = 0 the series is called Maclaurin’s series after Colin Maclaurin a contemporary of Brook Taylor.

Colin Maclaurin1698 - 1746

Examples of Taylor’s Series

zn

ze

n

nz

0 !

z

n

zz

n

nn

0

121

!121sin

zn

zz

n

nn

0

2

!21cos

zn

zz

n

n

0

12

!12sinh

zn

zz

n

n

0

2

!2cosh

111

1

0

zzz n

nn


91


Laurent’s Series (1843)

Power Series

If f (z) is analytic inside and on the boundary of the ringshaped region R bounded by two concentric circles C1 andC2 with center at z0 and respective radii r1 and r2 (r1 > r2),

then for all z in R:

Pierre Alphonse Laurent1813 - 1854

C1

x

y

RC2R2

R1

z0

z

z'

r

P1

P0

z'

1 00

0

nn

n

n

n

nzz

azzazf

,2,1,0''

'

2

1

2

1

0

nzdzz

zf

ia

C

nn

,2,1,0''

'

2

1

1

1

0

nzd

zz

zf

ia

C

nn

Proof:

Since z is inside R we have R1 <|z-z0|=r < R2 , and |z’-z0|= R1 on C1 and R2 on C2.

Start with the Cauchy’s Integral Formula:

212

0

1

1

01

''

''

'

''

'

''

'

''

'

''

'

'

0

CCC

P

P

P

PC

dzzz

zfdz

zz

zfzfdzdz

zz

zfdz

zz

zfdz

zz

zfdz

zz

zfzf

92


Laurent’s Series (continue - 1)

Power SeriesPierre Alphonse Laurent

1813 - 1854

C1

x

y

RC2R2

R1

z0

z z'

r


Since z and z’ are inside R we have R1 >|z-z0|=r >R2, |z’-z0|=R1.

From Cauchy’s Integral Formula:

21

''

''

'

'

CC

dzzz

zfdz

zz

zfzf

Use the identity:

11

1

1 12n

n

For I integral:

nn

zz

zz

zz

zzzz

zz

zz

zz

zz

zz

zzzzzz 0

0

0

0

1

0

0

0

0

0

0

00 '

'1

1

''1

'

1

'1

1

'

1

'

1

n

n

n

R

C

n

n

n

zs

C

n

za

C

za

C

Rzzzazzzazazzzz

zdzf

i

zz

zzzz

zdzf

izz

zz

zdzf

izz

zdzf

i

n

n

0000100

0

0

1

0

0

02

00

2

0

2

01

2

00

2

''

''

2

'

''

2

1

'

''

2

1

'

''

2

1

1

''

'

2

1

C

zdzz

zf

i

We have:

n

n

n

C

n

n

n R

r

rR

MRdR

rRR

Mr

zzzz

zdzfzzR

11

1

2

0

1

110

0

2''

''

20

where |f (z)|<M in R and r/R1< 1, therefore: 0

n

nR

93


Laurent’s Series (continue - 2)

Power SeriesPierre Alphonse Laurent

1813 - 1854

C1

x

y

RC2R2

R1

z0

z z'r


Since z and z’ are inside R we have R1 >|z-z0|=r > R2, |z’-z0|=R2.

From Cauchy’s Integral Formula:

21

''

''

'

'

CC

dzzz

zfdz

zz

zfzf

Use the identity:

11

1

1 12n

n

For II integral:

nn

zz

zz

zz

zzzz

zz

zz

zz

zz

zz

zzzzzz 0

0

0

0

1

0

0

0

0

0

0

00

'

'1

1''1

1'

1

11

'

1

n

n

n

R

C

n

n

n

za

C

n

za

CC

Rzzzazzzazzzz

zdzfzz

i

zzzz

zdzf

izzzz

zdzf

izdzf

i

n

n

1

001

1

001

0

0

1

0

1

00

2

0

0

0

01

0

01

00

'

'''

2

1

1

'

''

2

11

'

''

2

1''

2

1

C

zdzz

zf

i'

'

'

2

1

We have:

n

n

n

C

n

n

n r

R

rR

RMdR

rRr

MR

zzzz

zdzfzzR

2

2

2

2

0

2

2

2

0

0

2'

'''

2

1

0

where |f (z)|<M in R and R2/r< 1, therefore: 0

n

nR


94


Zeros of Holomorphic Functions

00 001

01

0 zfandzfzfzf kk

We say that Holomorphic Function f (z) has a Zero of Order k at z = z0 if

If f (z) has a Zero of Order k at z = z0, by Taylor expansion, we can write

with Holomorphic and nonzero.

zfzzzf kk

0

kk

zz

zfzf

0

:

Note:

(1) For g (z) = 1/ f (z) the Order k Zeros of f (z) are Order k Poles of g (z)

zfzzzfzg

kk

111

0

(2) For

zf

zf

zz

k

zf

zfzf

zd

d

k

k ''ln

0

z = z0, is a Simple Pole.


95


Theorem: f(z) Analytic and Nonzero → ln|f(z)| Harmonic

If f (z) is analytic for in an Open Set Ω, and has no zeros in Ω, then ln |f(z)| is Harmonic in Ω.

Proof :

Since f (z) is analytic and has no zeros the logarithm of f(z) is also Analyticg (z) := ln f (z) is Analytic

Therefore

q.e.d.

zgizgzgizgzg eeeezf ImReImRe and

Harmoniczgizgzg )(Im)(Re

zgzgizg eeezf Re

1

ImRe

Harmoniczgezf zg Relnln Re

meaning

Harmonicyixgyixg

yixgy

yixgx

yixgy

yixgx

)(Re),(Re

0)(Im)(Im&0)(Re)(Re2

2

2

2

2

2

2

2


96


Polynomial Theorem

If f (z) is analytic for all finite values of z, and as |z| → ∞, and

then f (z) is a polynomial of degree ≤ k.Proof :

Integrating this result we obtain

q.e.d.

kgivenAsomeforzforzAzfk

&0,

Using Taylor Series around any analytic point z = a

afn

azafazafzf n

n

!1

Aafazz

az

naf

z

az

kaf

z

az

z

af

z

zf nkn

k

k

kk

k

kkk

!

1

!

11

01

1 azazazf kk

kk

Continuing to Integrate we obtain

11

kk

k azazf


97

SOLO Complex VariablesThe Argument Theorem

If f (z) is analytic inside and on a simple closed curve C except for a finite number of poles inside C(this is called a Meromorphic Function), then

PNdzzf

zf

iC

'

2

1

where N and P are respectively the number of zeros and poles inside C.

Proof:

Let write f (z) as:

zGz

z

zf

j

p

j

k

n

k

j

k

where: , j

j

k

k pPnN &

zGzpznzf jjk

k

k lnlnlnln

Differentiate this equation:

Cinanalytic

j

j

k k

k

zG

zG

z

p

z

n

zf

zf

'

'

PNpn

zG

zG

iz

p

iz

n

idz

zf

zf

i

j

k

k

C

p

C j

j

k

n

C k

k

C

jk

0

'

2

1

2

1

2

1'

2

1

and G (z) ≠ 0 and analytic in C (G’ (z) exists).

k

C k

k

C k

k ndzz

n

idz

z

n

ik

2

1

2

1

j

C j

j

C j

j pdzz

p

idz

z

p

ij

2

1

2

1

x

y C

R

1

3

2 k1

2j

kC

3C

2C

2C

1C

jC

q.e.d.Return to the Table of Content

98


Rouché’s Theorem Eugène Rouché

1832 - 1910

If f (z) and g (z) are analytic inside and on a simple closed curve Cand if |g (z)| < |f (z)| on C, then f (z) + g (z) and f (z) have the same number of zeros inside C.

Proof:

Let F (z):= g (z)/f (z)

If N1 and N2 are the number of zeros inside C of f (z) + g (z) and f (z) respectively, and using the fact that those functions are analytic and C, therefore they have no poles

inside C, using the Argument Theorem we have

C

dzzf

zf

iN

'

2

12

C

dzzgzf

zgzf

iN

''

2

11

0

1'2

1

1

'

2

1'

2

1

1

''

2

1

'

2

1

1

'1'

2

1'

2

1'''

2

1

32

21

CCCC

CCCC

dzFFFFi

dzF

F

idz

f

f

idz

F

F

f

f

i

dzf

f

idz

Ff

FfFf

idz

f

f

idz

Fff

FfFff

iNN

We used the fact that |F|=|g/f|<1 on C, so the series 1-F+F2+… is uniformlyconvergent on C and integration term by term yields the value zero. Thus N1=N2


99


Foundamental Theorem of Algebra (using Rouché’s Theorem)

Every polynomial equation P (z) = a0 + a1z+a2z2+…+anzn=0 with degreen ≥ 1 and an ≠ 0 (ai are complex constants) has at exactely n zeros.

Proof:

Define:

Take C as the circle with the center at the origin and radius r > 1.

q.e.d.

n

n zazf : 1

1

2

210: n

n zazazaazg

ra

aaaa

ra

rararara

ra

rararaa

za

zazazaa

zf

zg

n

n

n

n

n

n

nnn

n

n

n

n

n

n

n

n

1210

1

1

1

2

1

1

1

0

1

1

2

210

1

1

2

210

By choosing r large enough we can make |g (z)|/|f (z)|<1, and using Rouché’s Theorem

n

n

n

n zazazazaazgzfzP

1

1

2

210

and (n zeros at the origin z = 0) have the same number of zeros, i.e. P (z) has exactly n zeros.

n

n zazf :


100


Jensen’s Formula

Johan Ludwig William Valdemar Jensen

(1859 – 5 1925)

C

x

y

r

1a

na

D

which is the Mean-Value Property of the Harmonic Function ln |f(z)|.

Suppose that ƒ is an Analytic Function in a region in the complex plane which contains the closed disk D of radius r about the origin, a1, a2, ..., an are the zeros of ƒ in the interior of D repeated according to multiplicity, and ƒ(0) ≠ 0. Jensen's formula states that

This formula establishes a connection between the Moduli of the zeros of the function ƒ(z) inside the disk D and the average of log |f(z)| on the boundary circle |z| = r, and can be seen as a generalization of the Mean Value Property of Harmonic Functions. Namely, if f(z) has no zeros in D, then Jensen's formula reduces to

2

01

ln2

1ln0ln derf

r

af j

n

k

k

2

0

ln2

10ln derff j

101


Jensen’s Formula (continue – 1)

Proof

If f has no Zeros in D, then we can use Gauss’ Main Value Theorem to ln f(z) that is Harmonic in D

2

0

ln2

10ln derff j

Since f has Zeros a1, a2,…, an inside D, ( |z| < r) let define the Holomorphic Function F (z) without Zeros in D :

n

k k

k

raz

rzazfzF

1

2

/

/1:

Apply the Gauss’ Main Value theorem for |z|=r

2

0

2

01

ln2

1ln

2

1ln0ln0ln derfderF

r

afF ii

n

k

k

and:

11

1

1

/1

1

/1

/

/1

/

/1

2

2

1

2

22

1

22

22

rz

a

rz

a

rzz

za

rza

za

rza

z

r

raz

rza

raz

rzarzzz

k

k

k

k

k

k

k

k

k

k

The Zeros of f(z) are cancelled.

q.e.d.

We have: i.e. zk1 is outside the Disk D.

rzarzforrza k

ra

kkkk

k

12

12

1 /0/1

C

x

y

D

r

z'

r

ka

ka

r

kk arz /2

102




(1859 – 5 1925)

zh

zgzzf l

Jensen's formula may be generalized for functions which are merely meromorphic on D. Namely, assume that

where g and h are analytic functions in D having zeros at

respectively, then Jensen's formula for meromorphic functions states that

Jensen's formula can be used to estimate the number of zeros of analytic function in a circle. Namely, if f is a function analytic in a disk of radius R centered at z0 and if |f(z)| is bounded by M on the boundary of that disk, then the number of zeros of f(z) in a circle of radius r < R centered at the same point z0 does not exceed

0\,,0\,, 11 DbbandDaa mn

2

01

1 ln2

1ln

0

0ln i

m

nnm erfbb

aar

h

g

0

ln/ln

1

zf

M

rR

C

x

y

r

1a

na

D

103




(1859 – 5 1925)

Jensen's formula may be put in an other way. If n (r) denotes the number of Zeros, including multiplicity, and p (r) denotes the number of Poles, including multiplicity, for |z| < r, then the Jensen’s Formula can be written as

C

x

y

r

1a

na

D

0lnln2

1lnln

2

0110

fderfb

r

a

rdx

x

xpxn jn

j j

m

k k

r

Proof

n

jj

m

kk

n

j j

m

k k

brnarmb

r

a

r

1111

lnlnlnlnlnln

n

n

jjjm

m

kkk brnbbjarnaak lnlnlnlnlnlnlnln

1

11

1

11

r

r

n

j

r

r

r

r

m

k

r

r n

j

jn

k

kx

xdn

x

xdj

x

xdm

x

xdk

1

1

1

1

11

104



Johan Ludwig William Valdemar Jensen (1859 – 5 1925)

Jensen's formula may be put in an other way. If n (r) denotes the number of Zeros, including multiplicity, and p (r) denotes the number of Poles, including multiplicity, for |z| < r, then the Jensen’s Formula can be written as

C

x

y

r

1a

na

D

0lnln2

1lnln

2

0110

fderfb

r

a

rdx

x

xpxn jn

j j

m

k k

r

Proof (continue – 1)

r

r

n

j

r

r

r

r

m

k

r

r

n

j j

m

k k n

j

jn

k

kx

xdn

x

xdj

x

xdm

x

xdk

b

r

a

r 1

1

1

111

11

lnln

But k = n (x) for rm ≤ x ≤ rm+1, m = n (x) for rn ≤ x ≤ r, andj = p (x) for rj ≤ x ≤ rj+1, n = p (x) for rn ≤ x ≤ r Hence

xd

x

xpxd

x

xn

b

r

a

r rrn

j j

m

k k

0011

lnlnq.e.d.


105


Poisson-Jensen’s Formula for a Disk


(1859 – 5 1925)

Poisson Formula states:


C

x

y

R

1a

na

D

Let g (z) be analytic inside and on the circle C defined by |z| = R, and let z = r e iθ be any point inside C, then:

2

02

22

2

1deRg

ereR

rRerg i

ii

i

In our case ƒ is an analytic function in a region in the complex plane which contains the closed disk D of radius R about the origin, a1, a2, ..., am are the Zeros, and b1, b2, ..., bn are the Poles of ƒ in the interior of D repeated according to multiplicity.

Since f has Zeros a1, a2,…, an inside D, ( |z| < r) let define the Holomorphic Function F (z) without Zeros in D :

RzRzb

Rbz

Raz

RzazfzF

m

k

n

j j

j

k

k

1 12

2

/1

/

/

/1:

Zeros and Poles of f(z) are cancelled, and new ones are outside D.

106




(1859 – 5 1925)

Let apply the Poisson Formula to g (z) = ln |F (z)|:

n

j j

jm

k k

k

Rzb

Rbz

Raz

RzazfzFzg

12

1

2

/1

/ln

/

/1lnlnln:

1/1

/

/

/12

2 Rz

j

jRz

k

k

Rzb

Rbz

Raz

Rza


C

x

y

R

1a

na

D

We proved that:

ii eRfeRg ln

The Poisson Formula to g (z) = ln F (z) is:

RzdeRfereR

rR

Rzb

Rbz

Raz

Rzazf

i

ii

n

j j

jm

k k

k

2

02

22

12

1

2

ln2

1

/1

/ln

/

/1lnln

RzRzbandRzRza jjjkkk 12

112

1 0/1,0/1

107




(1859 – 5 1925)


C

x

y

R

1a

na

D

The Poisson-Jensen’s Formula for a Disk is:

RzdeRfereR

rR

Rzb

Rbz

Raz

Rzazf

i

ii

n

j j

jm

k k

k

2

02

22

12

1

2

ln2

1

/1

/ln

/

/1lnln

For z = r = 0 we obtain the Jensen’s Formula:

2

021

21 ln2

1ln0ln deRfR

aaa

bbbf inm

m

n

If there are no Zeros or Poles in D, it reduces to Poisson’s Formula:

RzdeRfereR

rRzf i

ii

2

02

22

ln2

1ln


108



If f (z) is analytic inside and on the boundary of a circle C, except it’s center z0, thenaccording to Laurent’s Series:

C

x

y

R

R

z0

z

z'

r

P0

1 00

0

nn

n

n

n

nzz

azzazf

,2,1,0'

'

'

2

11

0

nzdzz

zf

ia

C

nn

,2,1,0'

'

'

2

11

0

nzdzz

zf

ia

C

nn

Let compute

,2,1,0'

'

'''''

1 00

0

nzdzz

zfazdzzazdzf

n C

nn

C n C

n

n

12

10'

'

'&,2,1,00''

0

0 ni

nzd

zz

zfnzdzz

C

n

C

n

Therefore: 12'' aizdzfC

Because only a-1 is involved in the integral above, it is called the residue of f (z) at z = z0.


109



According to residue definition the residue of f (z) at z = z0 can be computed as follows:C

x

y

R

R

z0

z

z'

r

P0

If z = z0 is a pole of order k, i.e. the Laurent series atz0 is

Then:

C

zdzfi

a ''2

11

Calculation of the Residues

2

02010

0

2

0

2

0

1 zzazzaazz

a

zz

a

zz

ak

k

zfzzzd

d

ka k

k

k

zz 01

1

1 !1

1lim

0

2

02

1

0100

2

02

1

010

kkk

k

kkk zzazzazzaazzazzazfzz

and:

If z = z0 is a pole of order k=1, then:

zfzzazz 01

0

lim

k

n

nn

n

n

nzz

azzazf

1 00

0


110


The Residue Theorem, Evaluations of Integral and Series The Residue Theorem

If f (z) is analytic inside and on the boundary of a closed curve C, except at the singularitiesz01, z02,…,z0n, which have residues Re1, Re2,…,Ren, then:

Proof:

n

C

izdzf ReReRe2'' 21

x

y C

R

01z

nzC

2zC

02z1zC

nz0Surround every singularity z0i by a small closed curveCzi, that enclosed only this singularity. Connect thoseCurves to C by a small corridor (the width of which shrinks to zero, so that the integration along the oppositedirections will cancel out)

0''''''''21

znzz CCCC

zdzfzdzfzdzfzdzf

We have , therefore: i

C

izdzfzi

Re2''

n

CCCC

izdzfzdzfzdzfzdzfznzz

ReReRe2'''''''' 21

21


111



Evaluation of Integrals

Theorem 1If |F (z)| ≤ M/Rk for z = R e iθ where k > 1 and M are constants, then

where Γ is the semicircle arc of radius R, center at origin, in theupper part of z plane.

0lim

zdzF

R

x

y

R

Proof:

1

0

1

0

kk

i

k

eRz

k R

Md

R

MdeRi

R

Mzd

R

MzdzF

i

Therefore: 0limlim1

1

0

1

k

kRkR R

Md

R

MzdzF

0lim

zdzF

R

and: 0limlimlim0

zdzFzdzFzdzF

RRR


112


The Residue Theorem, Evaluations of Integral and Series Evaluation of Integrals

Jordan’s LemmaIf |F (z)| ≤ M/Rk for z = R e iθ where k > 0 and M are constants, then

where Γ is the semicircle arc of radius R, center at origin, in theupper part of z plane, and m is a positive constant.

0lim

zdzFe zmi

R

x

y

R

Proof:

0lim

zdzFe zmi

R

using:

q.e.d.

0

deRieRFezdzFe iieRmieRz

zmi i

i

2/

0

sin

1

0

sin

1

0

sin

0

sincos

00

2

dReR

MdRe

R

MdReRFe

deRieRFedeRieRFedeRieRFe

Rm

k

Rm

k

iRm

iiRmRmiiieRmiiieRmi ii

2/0/2sin for2/

1sin

/2

Rm

k

Rm

k

Rm

k

iieRmi eR

Mde

R

Mde

R

MdeRieRFe

i

1

2222/

0

/2

1

2/

0

sin

1

0

012

limlim0

Rm

kR

iieRmi

Re

R

MdeRieRFe

i

Marie Ennemond Camille Jordan1838 - 1922

113



Jordan’s Lemma GeneralizationIf |F (z)| ≤ M/Rk for z = R e iθ where k > 0 and M are constants, then

for Γ a semicircle arc of radius R, and center at origin:

00lim

mzdzFe zmi

R

x

y

R

where Γ is the semicircle, in the upper part of z plane.

1

00lim

mzdzFe zmi

R

xy

R

where Γ is the semicircle, in the down part of z plane.

2

00lim

mzdzFe zm

R x

y

R

where Γ is the semicircle, in the right part of z plane.

3

00lim

mzdzFe zm

R

where Γ is the semicircle, in the left part of z plane.

4x

y

R


114



Integral of the Type Bromwwich-Wagner

jc

jc

ts sdsFei2

1

The contour from c - i ∞ to c + i ∞ is called Bromwich Contour

Thomas Bromwich1875 - 1929

x

y

0t

R

c

x

y0t

R c

0

0

2

1

lim2

1

2

1

tzFeRes

tzFeReszdzF

i

sdsFesdsFei

sdsFei

tf

tz

planezRight

tz

planezLeft

ts

ic

ic

ts

R

ic

ic

ts

where Γ is the semicircle, in the right part of z plane, for t < 0.

where Γ is the semicircle, in the left part of z plane, for t > 0.

This integral is also the Inverse Laplace Transform.


115



Integral of the Type ,F (sin θ, cos θ) is a rational function ofsin θ and cos θ

2

0

cos,sin dF

Let z = e iθ

22cos,

22sin

11

zzee

i

zz

i

ee iiii

zizdddzideizd i /

Czi

zdzz

i

zzFdF

2,

2cos,sin

112

0

where C is the unit circle with center at the origin.

C

x

y

R=1


116



Definite Integrals of the Type .

xdxF

If the conditions of Theorem 1, i.e.:if |F (z)| ≤ M/Rk for z = R e iθ where k > 1 and M are constants, thenand we can write

0lim

zdzF

R

x

y

R

zFResizdzFzdzFxdxFxdxFplanezUpper

R

RR

2lim

Example: Heaviside Step Function x

e

ixF

txi

2

1:

x

y

R0t

xy

R0t

This function has a single pole at z = 0.

For t > 0 Γ is the semicircle, in the upper part of z plane.We also include on the path x = - ∞ to x = + ∞ a small semicircle such that the pole z = 0 is included. For t < 0 Γ is the semicircle, in the lower part of z plane.We also include on the path x = - ∞ to x = + ∞ a small semicircle such that the pole z = 0 is excluded.

00/

01/

2

1

tzeRes

tzeResxd

x

e

i tzi

planezLower

tzi

planezUppertxi


117



Cauchy’s Principal Value

Cauchy’s Principal value deals with integrals that have singularities along theintegration paths. Start with the following:

Theorem 1:

If f (z) is analytic on and inside a positive-sensed circle C of radius ε, centered at z = z0, then 0

00

lim zfizz

zdzf

C

where Cψ is every arc on C of angle ψ.

Proof:

Since f (z) is analytic inside and on C we can usethe Taylor series expansion to write

1

00

0 !n

nn

zzn

zfzfzf

zg

n

nn

zzn

zf

zz

zf

zz

zf

1

1

00

0

0

0 !

Consider the integral on Cψ defined by z=z0 + ε e iθ θ0 ≤ θ ≤ θ0 + ψ

C

x

yC

0z0

0

O

118



Cauchy’s Principal Value (continue – 1)


Since g (z) is bounded inside and on C , there is a positive number M such that|g (z)| < M for all z such that |z – z0| < ε

CCC

zdzgzz

zdzf

zz

zdzf

0

0

0

00

0

0

0

0

0

0

0

zfidizfzz

zdzf

zz

zdzf iezz

CC

MLMzdzgzdzgCC

Where L = ψ ε is the length of Cψ.

0lim0limlim000

CC

zdzgMzdzg

0lim0limlimlim00000

CCCC

zdzgzdzgzdzgzdzg

Therefore 0

00

lim zfizz

zdzf

C

q.e.d.

Note: For ψ = 2 π we recover the Cauchy’s Integral result.

C

x

yC

0z0

0

O

119




Theorem 2:

If F (z) is analytic on and inside a positive-sensed circle C of radius ε, except at thecenter of C, z = z0, that is a simple pole of F (z), then

C

x

yC

0z0

0

O

00lim zFResizdzF

C

where Cψ is every arc on C of angle ψ.

Proof:

Since f (z) is analytic inside and on C by using Theorem 1 we obtain the desired result.

The function:

000

00

0

lim zzzFzzzFRes

zzzFzzzf

zz

120




Theorem 3:

If F (z) is analytic on and inside a positive-sensed curve C, except at the interior poleszint1, zint2,…, zint n and the simple poles on the curve C, zcont1, zcont2,…, zcont m,then

Proof:x

y

1intz

2intz

nz int

1scontextz

2czckz

1c

2c

0C

2scontextz

1in tcontz2intcontz

mcontextz

scontzin t

scontin t

mcontext

2intcont

2scontext

1scontext

2in tcontCz

2scontextCz

1scontextCz

1in tcontCz

mcontextCz

scontCzint

m

k

kcontCCC1

0

s

k

kcont

n

j

j

m

sk C

s

k CC

zFReszFResizdzFzdzFzdzFkextcontkcont

1

int

1

int

11

2int0

m

k

kcont

n

j

j

C

zFReszFResizdzF11

int 2

12

Let encircle the simple poles zcont k on the C contour bysemicircles Ccont k of radiuses ε cont k such that,randomly, zcont int 1,…, zcont int s, are inside theintegration contour, and zcont ext s+1,…,zcont ext m areoutside the integration contour. We have:

where

121


The Residue Theorem, Evaluations of Integral and Series Evaluation of Integrals Cauchy’s Principal Value (continue – 4) Proof (continue – 1):

x

y

1intz

2intz

nz int

1scontextz

2czckz

1c

2c

0C

2scontextz

1in tcontz2intcontz

mcontextz

scontzin t

scontin t

mcontext

2intcont

2scontext

1scontext

2in tcontCz

2scontextCz

1scontextCz

1in tcontCz

mcontextCz

scontCzint

s

k

kcont

s

k C

zFResizdzFkcont

kcont1

int

10

int

int

lim

The integrals along the semicircles Ccont k of thesingularities zcont ext s+1,…,zcont ext m that areoutside the integration contour, are in the negativedirection, and we have, according to Theorem 2:

m

k

kcont

n

j

j

CC

zFReszFResizdzFzdzFc 11

int0 2

12lim

0

therefore:

Since the integrals along the semicircles Ccont k of thesingularities zcont int 1,…, zcont int s, that are inside theintegration contour, are in the positive direction we have,according to Theorem 2:

m

sk

kextcont

m

sk C

zFResizdzFkextcont

kcont11

0int

lim

s

k

kcont

n

j

j

m

sk C

s

k CC

zFReszFResizdzFzdzFzdzFkextcontkcont

1

int

1

int

11

2int0

Note: This result is independent on the way that we encircled the simple poles on the curve C.

q.e.d.

122




If F (x) is continuous in a ≤ x ≤ b except at a point x0 such that a < x0 < b, then if

ε1 and ε2 are positive then the integral exists if and only if the limit:

b

x

x

a

b

a

xdxFxdxFxdxF10

10

2

100

lim

b

a

xdxF

b

x

x

a

xdxFxdxF10

10

2

100

lim

exists. If the limit does exist, the integral is equal to the value of this limit:

For this limit to exist, it must always have the same definite value regardless of howthe quantities ε1 and ε2 approach zero.

Cauchy’s Principal Value is defined as:

b

x

x

a

b

a

xdxFxdxFxdxFPV

0

0

0lim:

Clearly, if the integral exists, then PV exists and is equal to the integral,

but the opposite is not true.

b

a

xdxF

123




Example:x

y

x

1

aa

a

a

xdx

1

a

a

xdx

xdx

1

1

2

1

11lim

00

doe’s not exists

011

lim11

lim

11lim

1

00

0

aa

xua

a

vu

a

a

a

a

xdx

xdx

udu

xdx

vdv

xdx

xdx

PV

Cauchy’s PV does exist, in this case, but has no meaning.


124

SOLO

Example

0

sindk

k

krLet compute:

x

y

R

A

B

C

D

E

F

G

H

Rx Rx

For this use the integral: 0ABCDEFGHA

zi

dzz

e

Since z = 0 is outside the region of integration

0

BCDEF

ziR xi

GHA

zi

R

xi

ABCDEFGHA

zi

dzz

edx

x

edz

z

edx

x

edz

z

e

00

0000

sin2

sin2

sinlim2limlimlim dk

k

rkidx

x

xidx

x

xidx

x

eedx

x

edx

x

eR

R

R xixi

R

R xi

RR

xi

R

idideideie

edz

z

e i

ii

eii

i

eiez

GHA

zi

00

1

0

0

00limlimlim

012

2

0

/2/2sin

0

sin

00

R

RRReRii

i

eRieRz

BCDEF

zi

eR

dedededeRieR

edz

z

e i

ii

Therefore: 0sin

20

idkk

rkidz

z

e

ABCDEFGHA

zi 2

sin

0

dkk

kr

Complex Variables

125

SOLO

Example 2

0,1

21

2

1

RRxdx

R

R

kLet compute: for k integer and positive

x

y

R

A

B

C

D

E

F

G

H

Rx Rx

This integral has a singularity on the path ofintegration on x = 0:

Complex Variables

1

100lim

lim11

lim1

1

2

1

2

1

1

1

1

00

11

00

00

2

1

2

2

1

1

2

1

2

2

1

12

1

2

1

kdefinednot

kkRkRkk

xkxkxdx

xdx

xdx

kkkk

Rk

R

k

R

k

R

k

R

R

k

Let compute:

oddk

ke

kkRkRkk

xkeR

deRixkxd

xzd

zxd

xxd

x

kik

kkkk

Rk

kik

i

R

k

R

k

C

k

R

k

R

R

k

&10

1001

1lim

lim111

lim1

21

11

2

1

1

1

0

1

0

1

00

2

2

1

1

2

1

2

1

126



Differentiation Under Integral Sign, Leibnitz’s Rule

constantbaxd

xFxdxF

d

db

a

b

a

,

,,

This is true if a and b is constant, α is real and α1 ≤ α ≤ α2 whereα1 and α2 are constants, and F (x,α) is continuous and hascontinuous partial derivative with respect to α for a ≤ x ≤ b,α1 ≤ α ≤ α2.

Gottfried Wilhelmvon Leibniz(1667-1748)

b

a

d

bd

d

ad

b

a

xdxF

xdxFxdxFd

d

,,,

When a and/or b are functions of α, then:


127


The Residue Theorem, Evaluations of Integral and Series Summation of Series

Proof:

Start with the following contour of integration CN, that is a square with vertices at (N+1/2) (±1±i): x

y

iN

1

2

1

1 2 N 1N1 N N 12

iN

1

2

1

iN

1

2

1 iN

1

2

1

NC

We want to show that on CN:

e

ez

1

1cot

For y > 1/2

12

2

:1

1

1

1

cot

Ae

e

e

e

ee

ee

ee

ee

ee

ee

ee

eez

y

y

yy

yy

yxiyxi

yxiyxi

yxiyxi

yxiyxi

zizi

zizi

For y < - 1/2

12

2

1

1

1

1cot A

e

e

e

e

ee

ee

ee

eez

y

y

yy

yy

yxiyxi

yxiyxi

zfzsnfzfpoles

cotRe

128


The Residue Theorem, Evaluations of Integral and Series Summation of Series (continue – 1)


Start with the following contour of integration CN, that is a square with vertices at (N+1/2) (±1±i):

x

y

iN

1

2

1

1 2 N 1N1 N N 12

iN

1

2

1

iN

1

2

1 iN

1

2

1

NC

We want to show that on CN:

e

ez

1

1cot

For y > ½ and y < - 1/21:

1

1cot A

e

ez

For - ½ ≤ y ≤ ½ consider, first, z = N +1/2 + i y

e

eAAy

yiyiNz

1

12/tanhtanh

2/1cot2/1cotcot

12

For - ½ ≤ y ≤ ½ and z = -N -1/2 + i y

122/tanhtanh2/1cotcot AAyyiNz

y

e

e

1

1

2

1

zcot

2

1

2tanh

129


The Residue Theorem, Evaluations of Integral and Series Summation of Series (continue -2)

Proof (continue -2) :

x

y

iN

1

2

1

1 2 N 1N1 N N 12

iN

1

2

1

iN

1

2

1 iN

1

2

1

NC

We proved that on CN: Ae

ez

1

1cot

Residue of π cot (π z) f (z) at the poles of cot (π z), i.e.z = n, n = 0, ±1, ±2, …

Case 1: f (z) has finite number of poles

nfnfnz

zfzz

nz

zfznzzfzRes

nz

HopitalL

nz

nznz

coscos

limcossin

lim

cotlimcot

'

zfzszfzResdzzfz

NNCin

zfpoles

n

nnf

nzC

cotRecotcot

048limlimcotlim48

NN

MALM

N

Adzzfz

kN

NL

CkNC

N

NC

N

N

zfzsnfNCin

zfpoles cotRe

130


The Residue Theorem, Evaluations of Integral and Series Summation of Series (continue -3)

zfzsnfNCin

zfpoles cotRe

Proof (continue -3) :

x

y

iN

1

2

1

1 2 N 1N1 N N 12

iN

1

2

1

iN

1

2

1 iN

1

2

1

NC

We proved that on CN: Ae

ez

1

1cot

Case 1: f (z) has finite number of poles

0cotlimcotlim NN C

NC

Ndzzfzdzzfz

zfzsnfdzzfzNN

Cinzfpoles

n

nC

cotRecot

Therefore

zfzsnfzfpoles

cotRe

q.e.d.

Case 2: f (z) has infinite number of poles

Since CN is expanding to include all s plane, when N → ∞, it will encircle, at thelimit all the poles of f (z).


131



Proof:

Start with the same contour of integration CN, that is a square with vertices at (N+1/2) (±1±i): x

y

iN

1

2

1

1 2 N 1N1 N N 12

iN

1

2

1

iN

1

2

1 iN

1

2

1

NC

On CN:2csc Az

zfzsnfzfpoles

n cscRe1

Residue of π csc (π z) f (z) at the poles of csc (π z), i.e.z = n, n = 0, ±1, ±2, …

nfnfz

zfz

nz

zfznzzfzRes

n

nz

HopitalL

nz

nznz

1cos

limsin

lim

csclimcsc

'

zfzszfzResdzzfz

NnN

Cinzfpoles

n

nnf

nzC

N cscRecsccsclim0

1

zfzsnfzfpoles

n cscRe1

q.e.d.


132



Proof:


y

iN

1

2

1

1 2 N 1N1 N N 12

iN

1

2

1

iN

1

2

1 iN

1

2

1

NC

On CN:3tan Az

zfzs

nf

zfpoles tanRe

2

12

Residue of π tan (π z) f (z) at the poles of tan(π z), i.e.z = (2n+1)/2, n = 0, ±1, ±2, …

2

12

2

12

coslimsin

cos

212

lim

tan2

12limtan

2/12

'

2/12

2/122/12

nf

nf

zzfz

z

nz

zfzn

zzfzRes

nz

HopitalL

nz

nznz

zfzszfzResdzzfzNN

Cinzfpoles

n

nn

f

nzC

N tanRetancsclim0

2

12

zfzs

nf

zfpoles tanRe

2

12

q.e.d.


133



Proof:


y

iN

1

2

1

1 2 N 1N1 N N 12

iN

1

2

1

iN

1

2

1 iN

1

2

1

NC

On CN:4sec Az

zfzsn

fzfpoles

n secRe2

121

Residue of π sec (π z) f (z) at the poles of sec (π z), i.e.z = (2n+1)/2, n = 0, ±1, ±2, …

2

121

2

12

sinlim

cos

212

lim

2

12limsec

2/12

'

2/12

2/122/12

nf

nf

zzf

z

nz

zfzescn

zzfzRes

n

nz

HopitalL

nz

nznz

zfzszfzResdzzfz

N

nN

Cinzfpoles

n

nn

f

nzC

N secResecseclim0

2

121

zfzsn

fzfpoles

n secRe2

121

q.e.d.


134

SOLO

Perron’s Formula

11

10

2

1

2:Re aif

aifds

s

a

i ss

s

Oskar Perron

( 1880 – 1975)

1 32

t

10 adss

a

LC

s

10 adss

a

RC

s

LC

RC

R

eRsCj

L

0cos2

2:

R

eRsCj

R

0cos

2:

R

Proof

Define the two semi-circular paths CL (left side), CR (right side) with s=2 as the common origin., and R → ∞.

RLRL

RLRL

C

R

C

R

C

iiRR

C

s

dadRR

a

deRiiRR

ads

s

a

,,

,,

coscos

sincos

sincos

LR

LR

RLRL

CC

CC

C

R

RC

s

R aora

aora

dadss

a

)0cos&1()0cos&1(

)0cos&1()0cos&1(0

limlim,,

cos

Complex Variables

135

SOLO

Perron’s Formula

11

10

2

1

2:Re aif

aifds

s

a

i ss

s

1 32

t

10 adss

a

LC

s

10 adss

a

RC

s

LC

RC

R

eRsCj

L

0cos2

2:

R

eRsCj

R

0cos

2:

R

Proof (continue)

We can see tat

10Residue

11lim

1Residue

1Residue

2Re

0

2Re

2Re

as

a

as

as

as

a

as

a

s

Cs

s

s

s

Cs

s

Cs

RR

L

q.e.d.

1Residue

1Residue

1

1

1

1

2

1

2Re

2Re

2Re

2Re

0

2

22:Re as

a

as

a

adss

a

adss

a

adss

a

adss

a

dss

ads

s

a

i s

Cs

s

Cs

Cs

s

Cs

s

C

s

C

s

i

i

s

ss

s

R

L

R

L

R

L

Complex Variables


136

SOLO

z

zofzerosn

n

z

z

z

n

sin

,2,11sin

12

2

zofzerosn

enz

ez

zte

n

n

zz

zt

1

,2,1

1

1

1

0

1

Euler’s Product

2/1

1

2/2/

1

1

2

010

1

211

11

zzz

n

n

z

zeroszTrivialzerosztrivialNon

z

zofpole

zba

primep

z en

ze

z

z

epz

Weierstrass Product

Hadamard Product

1

22 sin1

1

n

zz

n

zz

zz

0Im0

2/12ln

12/112

s

es

ss

es

Infinite ProductsComplex Variables

137

SOLO

In 1735 Euler solved the problem, named “Basel Problem” , posed by Mengoli in 1650, by showing that

6

1

4

1

3

1

2

11

2

12232

n n

122

2

2

2

2

2

2

2

19

14

11sin

k k

xxxx

x

x

He did this by developing an Infinite Product for sin x /x:

The roots of sin x are x =0, ±π, ±2π, ±3π,…. However sin x/x is not a polynomial, but Euler assumed (and check it by numerical computation) that it can be factorized using its roots as

We now that if p (x) and q (x) are two polynomials, then using the roots of the two polynomials we have:

m

n

qqqq

pppp

xxxxxxa

xxxxxxa

xq

xp

21

21

We want to show how to express a general solution for complex function f (x) using the zeros and the poles (finite or infinite) of f (x).


138

SOLO

Definition 1:We say that the Infinite Product converges, if for any N0 > iN the limit

exists and is nonzero.

If this is satisfied then we can compute

N

Nj iN

N

Nj iN

N

Ni iN

N0000lnlimlnlimlimlnln

We transformed the Infinite Product in an Infinite Series, and we know that a necessary (but not sufficient) condition for an Infinite Series to converge is

1lim0lnlim j

jj

j

For simplicity we will define

0lim1 j

jjj aa

1j j

00lim N

N

Nj jN


139

SOLO

Lemma 2:Let aj ϵ C be such that |aj| < 1. Let . Then

N

j jN aQ1

1:

Nj j

Nj j a

N

aeQe 112

1

Proof:

Since 1 + |aj| ≤ e |aj|

N

j j

N

a

Q

N eaa 111 1

On the other hand, since ex ≤ 1 + 2 x for 0 ≤ x ≤ 1,

NN

aaa

QaaeeeNj

jNj

jNNj j

2/212/21 122

22

111

11 q.e.d.

Proof: Suppose . Then, by the previous Lemma, QN ≤ eM, for all N. Since Q1 ≤Q2 ≤ …., the sequence of “partial products” {QN} converges.Conversely, if the Infinit Product converges to Q, then Q ≥ 1 andfor all N. Then converges.

Maj j

11

1j ja

QaN

j j ln21

Lemma 3:Let aj ϵ C be such that |aj| < 1. Then converges if and only if converges.

11

j ja

1j ja

q.e.d.


140

SOLO

Proof: Since the product converges, then |aj| → 1, so that aj ≠ 0 for j ≥ j0. Let assume j0 = 1, and define

11

j ja

N

j jN

N

j jN aQandaP11

1:,1:

Note that for a suitable choice of indexes ajk

N

n

n

k j

N

j jN kaaP

1 1111:

Then 111 11 1

N

N

n

n

k j

N

n

n

k jN QaaPkk

and for N, M > 1, N > M

MN

n

Mj jM

n

Mj j

M

j j

NMN

j

M

j jjMN

QQaQ

aaaaPP

11

11111

1

111 1

Hence, {PN} is a Cauchy Sequence, since {QN} is, and it converges.


141

SOLO


We need to prove that {PN} does not converge to zero. By Lemma 2

2

31

NMj jaN

Mj j ea

for M ≥ j0, and N > M. Then using

2

11

2

31111

N

Mj j

N

Mj j aa

for M ≥ j0, and N > M. Hence 2

11

N

Mj ja

so that

012

111limlim 0

11

j

j j

N

Mj j

M

j jj

Nj

aaaP

q.e.d.

11 NN QP


142

SOLO

The Mittag-Leffler and Weierstrass , Hadamard Theorems

Magnus Gösta Mittag-Leffler1846 - 1927

Karl Theodor Wilhelm

Weierstrass (1815 – 11897)

We want to answer the following questions:

• Can we find f ϵ M (C) so that f has poles exactly a prescribed sequence {zn} that does not cluster in C, and such that f has prescribed principal parts (residiu) at these poles (this refers to fixing the entire portion of the Laurent Series with negative powers at each pole)?

A positive answer to this question was given by Mittag-Leffler

• Can we find f ϵ H (C) so that f has zeros exactly at a given sequence {zn} ?

A positive answer to this question was given by Weierstrassand improved by Hadamard


Jacques Salomon Hadamard

) 1865– 1963 (

143


The Residue Theorem, Evaluations of Integral and Series Mittag-Leffler’s Expansion Theorem

Magnus Gösta Mittag-Leffler1846 - 1927

1

110

n nn

n aazafResfzf

Suppose that the only singularities of f (z) in the z-plane are thesimple poles a1, a2,…, arranged in order of increasing absolutevalues. The respective residues of f are Res { f (a1)}, Res { f (a2)}, …

C

x

y

RN

1ana

CN

Proof:

Assume ξ is not a pole of f (z), then has simple polesat a1, a2,…, and ξ.

z

zf

Residue of at an, n = 1,2,… is z

zf

n

nnaz a

afRes

z

zfaz

n

lim

Residue of at ξ is z

zf

fz

zfz

naz

lim

Let take a circle CN at the origin with a radius RN → ∞

By the Residue Theorem

NNCinn n

n

Ca

afResfdz

z

zf

Assume f (z) is analytic at z = 0, then NN

Cinn n

n

Ca

afResfdz

z

zf0

144


The Residue Theorem, Evaluations of Integral and Series Mittag-Leffler’s Expansion Theorem (continue – 1)

C

x

y

RN

1ana

CN


Let take a circle CN at the origin with a radius RN → ∞

NN

Cinn n

n

Ca

afResfdz

z

zf

i

2

1

NN

Cinn n

n

Ca

afResfdz

z

zf

i0

2

1

NN

N

CC

Cinn nn

n

dzzz

zf

idz

zzzf

i

aaafResff

2

11

2

1

110

Since | z-ξ | ≥ | z | - | ξ |=RN - | ξ | for z on CN, we have if | f(z) | ≤ M

0

2limlim

NN

N

RC

R RR

RMdz

zz

zfN

N

N

0lim

N

N

CR

dzzz

zf

1

110

n nn

n aazafResfzf

Therefore using this result and ξ → z, we obtain

q.e.d.

145


The Residue Theorem, Evaluations of Integral and Series Mittag-Leffler’s Expansion Theorem (continue – 1)

Example: Expand 1/sinz

Define zz

zf1

sin

1:

0cossincos

sinlim

cossin

cos1lim

sin

sinlim

1

sin

1lim0

0

'

0

'

00

zzzz

z

zzz

z

zz

zz

zzf

z

HopitalL

z

HopitalL

zz

f (z) has Simple Poles at n π, n=±1, ±2,… with Residue

n

nz

HopitalL

nznznz

z

z

nz

zznzzf

1cos

1lim

sinlim

1

sin

1limRes

'

1222

11

10

112

111

111

11Res0

1

sin

1

n

n

n

n

n

n

n nnn

nzz

nnznnz

aazaff

zzzf

146


The Residue Theorem, Evaluations of Integral and Series Generalization of Mittag-Leffler’s Expansion Theorem

q.e.d.

Suppose that the only singularities of f (z) in the z-plane are the poles a1, a2,…, arranged in order of increasing absolute values, and having Higher Order then One. The respective residues of f are Res { f (a1)}, Res { f (a2)}, … Suppose that exists a Positive Integer p such that for |z| = RN

|f (z)| < RNp+1

and the poles a1, a2,…, an are all inside the Circle of Radius RN around the origin (|a1|≤ |a2|≤…≤ |an | < RN). Then

p

i jp

j

p

jjjj

ii

j jp

j

p

ip

p

a

z

a

z

aazaf

i

zf

aza

zaff

p

zf

zfzf

1 112

11

11

11Res

!

0

Res0!

0!1

0

Proof: Start with the Integral

Nj

N

Cina jp

j

j

pwpzw

Cp

zaa

af

zww

wf

zww

wf

dwzww

wf

iI

1101

1

ResResRes

2

1

C

x

y

RN

1ana

CN

Return to Infinite Product

147




but

111limRes

ppzwpzw z

zf

zww

wfzw

zww

wf

C

x

y

RN

1ana

CN

i

i

ip

pp

iw

pp

p

p

wpw

wd

wfd

zwwd

d

ipi

p

p

zww

wfw

wd

d

pzww

wf

1

!!

!

!

1lim

!

1limRes

1

00

11

010

1

1 !11

ip

ip

ip

p

zw

ip

zwwd

d

p

iip

i

p

ii

i

ip

ip

wpw

zi

f

wd

wfd

zw

ip

ipi

p

pzww

wf

01

01010

!

0

!1

!!

!

!

1limRes

Therefore

Leibnitz Formula for RepeatedDifferentiation of a Product

148




but

Nj Cina j

pj

jp

iip

i

p zaa

af

zi

f

z

zf1

011

Res

!

0

C

x

y

RN

1ana

CN

Therefore

Nj

N

Cina jp

j

j

pwpzw

Cp

zaa

af

zww

wf

zww

wf

dwzww

wf

iI

1101

1

ResResRes

2

1

0max2

2

1

2

11max

11

N

pN

NCN

N

Rn

RwfCN

pN

N

Cp

wfzRR

Rdw

zww

wf

iI

0

Res

!

0

11

011

j jp

j

jp

iip

i

p zaa

af

zi

f

z

zf

11

1

0

Res

!

0

j jp

j

pj

p

i

ii

aza

zaf

i

zfzf

149



We can see that for p = 0 we get

C

x

y

RN

1ana

CN

11

11Res0

Res0

n nnn

n nn

n

aazaff

aza

zaffzf

We recovered the Mittag-Leffler’s Expansion Theorem

11

1

0

Res

!

0

j jp

j

pj

p

i

ii

aza

zaf

i

zfzf

112

11

111

ResRes

jp

j

p

jjjj

j jp

j

pj

a

z

a

z

aazaf

aza

zaf

q.e.d.


150

SOLO

Start with some introductory results:

TheoremLet f (z) be entire holomorphic (analytic for all z ϵ C) and f (z) ≠ 0 everywhere. There is an entire function g (s) for which f = eg.

CorollaryIf f (z) is entire Holomorphic (analytic) with finitely many zeros {ai≠0}(with multiplicity) and m zeros at z=0, then there exists an entire g (z) such that

nzgm azezzf /1Proof:

Since is entire with no zeros we can apply the previous Theorem

nm azzzf /1/

q.e.d.

The Weierstrass Factorization Theorem


(1815 – 11897)


zf

zfzf

zd

d 'ln

Proof :

Since f (z) ≠ 0 and entire, f’ (s) is also entire, and so is f’(z)/ f (z), therefore

entireiszf

zd

d

zf

zfzg

zd

dln

':

and taking g (0)= 1 we obtain zgezf q.e.d.

151

SOLO


DefinitionWe define the Weierstrass Elementary Factors as

,2,11

01,

2

2

nez

nznzE

n

zzz

n

LemmaFor |z| ≤ 1, |1 – E (z,n)| ≤ |z|n+1.

Proof: The case n = 0 is trivial. Let n ≥ 1. Let differentiate E (z,n)

n

zzz

nn

zzz

nn

zzz

n

zzz

nn

zzz

nnnnn

ezezeezzzenzEzd

d

222212

22222

111,

By developing in a Taylor series

0, 2

2

knk

kk

n

zzz

n bzbeznzEzd

dn

0

10

1,,k

kk

k

kk zaknzE

sd

dzanzE

,2,10

0

1,0

21

0

jjn

ba

aaa

nEa

jnjn

n


(1815 – 11897)

Inspired by the fact that

321

1ln&11

321

1ln zz

zz

ez z w have the following


152

SOLO


DefinitionWe define the Weierstrass Elementary Factors as

,2,11

01,

2

2

nez

nznzE

n

zzz

n

LemmaFor |z| ≤ 1, |1 – E (z,n)| ≤ |z|n+1.


01,1

k

nk

kk azanzE

So for |z| ≤ 1

1

0

1

1

1

1

11

1

11

1

11

1

,11

,1

nn

nkk

n

nkk

ns

nk

nk

k

n

nk

nkk

n

nk

kk

znEzazaz

zazzazzanzE

q.e.d.


153

SOLO

The Weierstrass Product

Let {zj} be a sequence of complex numbers such that limj→∞ |zj|=+∞. We may assume that 0 < |z1| ≤ |z2| ≤… Let {pj} be integers. Then the Weierstrass Product defined as

converges uniformly on every set {|z|≤r}, to a holomorphic entire function F. The zeros of F are precisely the points {zj} counted with the corresponding multiplicity.

1

1

2

1

1

2

1,j

z

z

pz

z

z

z

jj j

j

jp

jjjjez

zp

z

zE

Proof:Let r > 0 be fixed. Let j0 be such that |zj| > r for j ≥ j0. Thus,

11

1,

jj

p

j

p

jj

j z

r

z

zp

z

zE

By the hypothesis on the pj’s,

00

1

1,jj

p

jjj j

j

j

z

rp

z

zE

By Weierstrass M (Majorant Test) it follows that converges uniformly on {|z| ≤ r}, for any r > 0. Then exist C > 0 such that

01,/

jj jj pzzE

C

jj

pz

zEp

z

zE

jj jj

eeeCpz

zE

jj

jj jj

0

0

0

1,1,

1,

C

jj

pz

zE

jj jj

jj jj

jj

pz

zE

eepz

zEp

z

zEe

jj

Taylorj

j

0000

1,,

1,1,


154

SOLO

Genus of the Canonical Product


155

SOLO


LemmaLet {zj} be a sequence of complex numbers such that limj→∞ |zj|=+∞. Then there exists an entire function F whose zeros are precisely the {z j}, counting multiplicity.This function is

This is a Generalization of the Fundamental Theorem of Algebra

11,:

kjj

k jz

zEssF


156

SOLO

The Hadamard Factorization Theorem

Examples:

(1) Polynomials have Order 0. Let N be the degree of p (z)

for all ε > 0 and a suitable constant Cε .

raseCrCazazazp rNnn

01

(2) The exponential ex has order 1, and more generally, have order n

and no smaller power of r would suffice.(3) sin z, cos z, sinh z, cosh z have order 1.(4) exp {exp z} has infinite order.

nnnn rzzr eeee Re

nxe



) 1865– 1963 (

157

SOLO


Relation berween Order ρ and Integer Genus p of an Entire Function



) 1865– 1963 (

Paul Garrett, “Weirstrass and Hadamard Products”, March 17, 2012, http://www.math.umn.edu/~garrett/

158



Expansion of an Integral Function as an Infinite Product

An Integral Function is a function which is Analytic for all finite values of z.For example ez, sin z, cos z are Integral Functions. An Integral Function may be regarded as a generalization of a Polynomial.

Let f (z) be an Integral Function (no Poles) with Simple/Non-simple Zeros at a1, a2,…,an,.., arranged in increasing order (|a1|≤ |a2|≤…≤|an|≤…. ). Suppose that exists a Positive Integer p such that for |z| = RN

|f (z)| < RNp+1

and the Zeros a1, a2,…, an are all inside the Circle of Radius RN around the origin (|a1|≤ |a2|≤…≤ |an | < RN).Then f (z) can be expanded as an Infinite Product (Hadamard):

pii

ff

zdd

c

ea

zefzf

z

i

i

i

j

a

z

pa

z

a

z

j

zczc pj

p

jjp

p

,,1,0!1

:

10

0

1

1

1

1

1

2

11

1

2

2

111

C

x

y

RN

1ana

CN

Note:1.The minimum p for which |f(z)|<RN

p+1 is called the Order of f(z)2.If f(z) has no poles or zeros then the previous relation reduces to

1110

pp zczcefzf


Return to Gamma F.

Return to Zeta F.

159




Proof:

C

x

y

RN

1ana

CN

Let compute: zf

zfzf

zd

d 1

ln

1limlimRes1

21'11

f

fazf

f

faz

f

f j

az

HopitalLj

azaz

jjj

Define

pii

ff

zdd

c z

i

i

i ,,1,0!1

: 0

1

1

112

01

1 111

jp

j

p

jjj

p

i

ii

a

z

a

z

aazzic

zf

zf

All Zeros of f (z) (a1, a2,…,an,..) are Simple Poles of f(1)(z)/f(z), therefore we can apply the previous result and write:

1

12

1

0

0

1

1 11Res

! jp

j

p

jjjaz

p

i

i

z

i

i

a

z

a

z

aazf

f

i

zf

fzd

d

zf

zf

j


160





C

x

y

RN

1ana

CN

Integrating from 0 to z along a path not passing through any of aj, we obtain

1

1

1

2

2

0

11 1

1

2

1ln

0ln

jp

j

p

jjj

jp

i

ii

a

z

pa

z

a

z

a

azzc

f

zf

The values of the logarithms will depend on the path chosen, but when we take exponentials all the ambiguities disappear,

112

01

1 111

jp

j

p

jjj

p

i

ii

a

z

a

z

aazzic

zf

zf

1

1

1

2

11

1

2

2

0

11

10 j

a

z

pa

z

a

z

j

zc pj

p

jj

p

i

ii

ea

ze

f

zf

q.e.d.

If |f(z)| < RNp+1 it will be true for all q > p. If we choose the ρ = min p for which

the inequality holds, then we obtain the Hadamard’s Factorization .


161



Example: Expand sinz/z

Define z

zzf

sin:

11

coslim

sinlim0

0

'

0

z

z

zf

z

HopitalL

z

f (z) has Simple Zeros at n π, n=±1, ±2,…


122

2

11

111sin

0 nnn n

z

n

z

n

z

z

z

f

zf

01sin 0 pR

z

zzf N

Leonhard Euler)1707 – 1783 (

We recovered the Euler Product Formula 1735

162

SOLO

Analytic continuation (sometimes called simply "continuation") provides a way of extending the domain over which a complex function is defined. The most common application is to a complex analytic function determined near a point by a power series

0

0k

kk zzazf

Such a power series expansion is in general valid only within its radius of convergence. However, under fortunate circumstances (that are very fortunately also rather common!), the function will have a power series expansion that is valid within a larger-than-expected radius of convergence, and this power series can be used to define the function outside its original domain of definition. This allows, for example, the natural extension of the definition trigonometric, exponential, logarithmic, power, and hyperbolic functions from the real line to the entire complex plane . Similarly, analytic continuation can be used to extend the values of an analytic function across a branch cut in the complex plane.


Analytic continuation of natural logarithm (imaginary part)

Complex Variables

163

SOLO


Complex Variables

164


Conformal Mapping Transformations or Mappings

x

y

u

v

r

xd

yd

r

ud

vdA B

CD

'A

'B

'C'DThe set of equations

yxvv

yxuu

,

,

define a general transformation or mapping between (x,y) plane to (u,v) plane.

If for each point in (x,y) plane there corresponds one and only one point in (u,v)plane, we say that the transformation is one to one.

vdv

rud

u

rvdy

v

yx

v

xudy

u

yx

u

x

yvdv

yud

u

yxvd

v

xud

u

xyydxxdrd

u

r

u

r

1111

1111

If is a vector that defines a point A in (x,y) plane, we have: vuryxr ,,

r

The area dx dy of a region A,B,C,D, in (x,y) plane is mapped in the area A’,B’,C’,D’, du dv in the (u,v) plane. We have

zvdudu

y

v

x

v

y

u

xvdudy

v

yx

v

xy

u

yx

u

x

vdudv

r

u

rzydxdydxd

y

r

x

rSd

yx

11111

1

11

If x and y are differentiable

165


Conformal Mapping Transformations or Mappings

yxvv

yxuu

,

,

The transformation is one to one if and only if, for distinct points A, B, C, D, in (x,y)we obtain distinct points A’,B’,C’,D’, in (u,v). For this a necessary (but not sufficient)condition:

''''det1det

11

DCBA

ABCD

Sd

v

y

u

y

v

x

u

x

zvdud

v

y

u

y

v

x

u

x

zvdudu

y

v

x

v

y

u

xzydxdSd

Transformation is one to one 00 '''' DCBAABCD SdSd

0det:

,

,

v

y

u

y

v

x

u

x

vu

yxJacobian of theTransformation

By symmetry (change x,y to u,v) we obtain:

ABCDDCBA Sd

y

v

x

v

y

u

x

u

Sd

det''''

1detdet

v

y

u

y

v

x

u

x

y

v

x

v

y

u

x

u

one to one

transformation

1

,

,

,

,

vu

yx

yx

vu

x

y

u

v

r

xd

yd

r

ud

vdA B

CD

'A

'B

'C'D

166


Conformal Mapping Complex Mapping

In the case that the mapping is done by a complex function, i.e.

yixfzfviuw

we say that f is a complex mapping.If f (z) is analytic, then according to Cauchy-Riemann equation:

2222

det,

,

zd

zfd

y

ui

x

u

y

u

x

u

x

v

y

u

y

v

x

u

y

v

x

v

y

u

x

u

yx

vu

x

v

y

u

y

v

x

u

&

If follows that a complex mapping f (z) is one to one in regions where df/dz ≠ 0.

Points where df/dz = 0 are called critical points.

167


Conformal Mapping Complex Mapping – Riemann’s Mapping Theorem

In the case that the mapping is done by a complex function, i.e. yixfzfviuw


1826 - 1866

we have:

x

y

u

vC 'C

1

RR' Let C be the boundary of a region R in the z plane,

and C’ a unit circle, centered at the origin of thew plane, enclosing a region R’.

The Riemann Mapping Theorem states that for each pointin R, there exists a function w = f (z) that performs aone to one transformation to each point in R’.

Riemann’s Mapping Theorem demonstrates the existence of theone to one transformation to region R onto R’, but it not providesthis transformation.

168


Conformal Mapping Complex Mapping (continue – 1)

yxvv

yxuu

,

,

x

y

u

v

r

2zd

1zd

r

2wd

1wdA

B

C

'A

'B

'C

yixfzfviuw

Consider a point A in (x,y) plane mapped to pointA’ in (u,v) plane

Consider a small displacement from A to Bdefined as dz1, that is mapped to a displacementfrom A’ to B’ defined as dw1

1

1

argarg

11

arg

11

zdzd

zfdi

AA

wdi Aezdzd

zfdzd

zd

zfdewdwd

Consider also a small displacement from A to C defined as dz2, that is mapped to a displacement from A’ to C’ defined as dw2

2

2

argarg

22

arg

22

zdzd

zfdi

AA

wdi Aezdzd

zfdzd

zd

zfdewdwd

We can see that dw ≠ 0 if dz ≠ 0, i.e. a one-to-one transformation, if and only if

0

Azd

zfd

169


Conformal Mapping Complex Mapping (continue – 2)

yxvv

yxuu

,

,

x

y

u

v

r

2zd

1zd

r

2wd

1wdA

B

C

'A

'B

'C

yixfzfviuw

Consider a point A in (x,y) plane mapped to pointA’ in (u,v) plane

1

1

argarg

11

arg

11

zdzd

zfdi

AA

wdi Aezdzd

zfdzd

zd

zfdewdwd

2

2

argarg

22

arg

22

zdzd

zfdi

AA

wdi Aezdzd

zfdzd

zd

zfdewdwd

We can see that:

12

1212

argarg

argargargargargarg

zdzd

zdzd

zfdzd

zd

zfdwdwd

AA

Consider two small displacements from A to BAnd from A to C, defined as dz1 and dz2, that are mapped to displacements from A’ to B’ and from A’ to C’, defined as dw1 and dw2

Therefore the angular magnitude and sense between dz1 to dz2 is equal to that between dw1 to dw2. Because of this the transformation or mapping is called aConformal Mapping.

170


Conformal Mapping

RzRzzzf 2/122ln

z

Rzz

RzzRRzz

Rzz

RRzz

w

Rw 2

222

2/12222/122

2/122

22/122

2

Define RzRzzzgw 2/122

w

Rw

2

1z

2

ww

Rw

w

Rwwiww

2

1

wwiww

Rwwiww

2

1yix

argsinargcosargsinargcos


22

2

w

Rww

2

1y

w

Rww

2

1x

22

argsin&argcos

171


Conformal Mapping

RzRzzzf 2/122ln

Define RzRzzzgw 2/122

w

Rww

2

1y

w

Rww

2

1x

22

argsin&argcos

From those equations we have:

2

22

22

22

4

1

4

1

argsinargcosR

w

Rw

w

Rw

w

y

w

x

4

1

w

Rw

y

w

Rw

x

2

2

2

2x

y warg

wln

wiwzf argln

172


Conformal Mapping

2/tt eex

0122 tt exe

2/cosh tt eet

2/12 1ln xxxacosh

http://www.mathworks.com/company/newsletters/news_notes/clevescorner/sum98cleve.html

173


Conformal Mapping

2/122ln Rzzzf

z

Rzz

RzzRRzz

Rzz

RRzz

w

Rw 2

222

2/12222/122

2/122

22/122

2

Define 2/122 Rzzzgw

w

Rw

2

1z

2

ww

Rw

w

Rwwiww

2

1

wwiww

Rwwiww

2

1yix



22

2

w

Rww

2

1y

w

Rww

2

1x

22

argsin&argcos

174


Conformal Mapping

2/122ln Rzzzf

Define 2/122 Rzzzgw

w

Rww

2

1y

w

Rww

2

1x

22

argsin&argcos

From those equations we have:

2

22

22

22

4

1

4

1

argsinargcosR

w

Rw

w

Rw

w

y

w

x

4

1

w

Rw

y

w

Rw

x

2

2

2

2x

y warg

wln

wiwzf argln

175


Conformal Mapping


2/tt eex

0122 tt exe

2/sinh tt eet

2/12 1ln zzzasinh:zf

176


Conformal Mapping

dz

dzkviuw

ln

kueydx

ydx

dz

dz /2

22

222

dx

dyx

ydxydx

ydxydx

e

e

k

uku

ku

21

1coth

222

2222

2222

/2

/2

kudkudykudx /sinh/1/coth/coth 222222

kvikukviku eedz

dzee

dz

dz ////

dyi

dyxi

dzdz

dzzi

dz

dz

dz

dzdz

dz

dz

dz

iee

eei

k

uctg

kvikvi

kvikvi

2

2222

//

//

kvdkvctgdkvctgdyx /sin/1// 222222

kviku eedz

dz //

177


Conformal Mapping

dz

dzkviuw

ln

kudykudx /sinh//coth 2222

kvdkvctgdyx /sin// 2222

ddx

y kvdkvctgdyx /sin// 2222

kudykudx /sinh//coth 2222

1v

2v3v

3u

2u

1u

We have two families of orthogonal circles.

All those circle passe through (-d,0) and (d,0)

v

u

178


Conformal Mapping


1

12

2

w

w

e

ez zze w 112

1

1tanh

2

2

w

w

ww

ww

e

e

ee

eew

z

zzatanhw

1

1ln

2

1

179


Conformal Mapping

The complex squaring map (on left half square)

The complex squaring map (on right half square)

The complex squaring map (on entire square)

2zzf -3/2

-3/2

+3/2

+3/2

x

y

Transform the square under the map

Douglas N. Arnold

180


Conformal Mapping

The complex exponential map

zezf Transform the strip ± i π under the exponential map

Douglas N. Arnold

+i π

x

y

-i π

181


Conformal Mapping

The complex cosine map

Douglas N. Arnold

zzf sin

-π

-1

+π+1

x

y

Transform the square under the maps

The complex sine map

zzf cos

182


Conformal Mapping

Douglas N. Arnold

An important property of analytic functions is that they are conformal maps everywhere they are defined, except where the derivative vanishes. A conformal map is one that preserves angles. More precisely, if two curves meet at a point and their tangents make a certain angle there, then the angle between the images curves under any analytic function (with non-vanishing derivative) will be the same in both sense and magnitude

zzf

183


Mobius Transformation

Douglas N. Arnold

10/11 4

t

zit

itzzf t

August Ferdinand Möbius1790 - 1868

184


Schwarz-Christoffel Mappings

Hermann Amandus Schwarz

1843 - 1921

Elwin Bruno Cristoffel 1829 - 1900

1

23

45

61w

2w

3w

4w

5w

6w

u

v

x

y

1x 2x 3x 4x 5x 6x

A Schwarz – Christoffel transformation is an analytic mapping ofThe upper half-plane (x,y) onto a polygon in (u,v) plane.

Let take n points on x axis:nxxx 21

Define the derivative of the mapping as:

11

2

1

1

21

n

nxzxzxzAzd

fd

zd

wd

or BdzxzxzxzAzfw

n

n 11

2

1

1

21

where A and B are complex constants.

185


Schwarz-Christoffel Mappings1

23

45

61w

2w

3w

4w

5w

6w

u

v

x

y

1x 2x 3x 4x 5x 6x

11

2

1

1

21

n

nxzxzxzAzd

fd

zd

wd

Since for xi-1 < x < xi the slope of d w/ d z is constant, i.e. the real axis is mapped instraight lines.

We can see that for x > xn: Azd

wdargarg

1

1

1

1 arg1arg1argarg

ni

xzxzAzd

wd ni For xi-1 < x < xi:

For x > xi: 1

1

1

11 arg1arg1argarg

1

ni

xzxzAzd

wd ni

(1) Any three of the points can be chosen at will.nxxx 21

(2) The constants A and B determine the size, orientation and position of the polygon.

(3) If we choose xn at infinity, the last term that includes xn is not present.

(4) Infinite open polygons are limiting cases of closed polygons.

186



Douglas N. Arnold

According to the Riemann mapping theorem, there exists a conformal map from the unit disk to any simply connected planar region (except the whole plane). However, finding such a map for a specific region is generally difficult. An important special case where a formula is known is when the target region is polygonal. In that case we have the Schwarz-Christoffel formula, written as

z n

jj dzcfzf j

0 1

10

Here the polygon has n vertices, the interior angles at the vertices are , , in counterclockwise order, and c is a complex constant. The numbers , , are the pre-images of the polygon's vertices, or prevertices, which lie in order on the unit circle.

n ,,1

nzz ,,1

The first animation illustrates the effect of the prevertices. The prevertices start in a random configuration, and the resulting image polygon is shown. Then the prevertices are moved (linearly in argument) into a configuration leading to a symmetric "X." Notice how the angles remain fixed, but the side lengths vary nonlinearly into the final configuration.

187



Douglas N. Arnold

z n

jj dzcfzf j

0 1

10

Here the polygon has n vertices, the interior angles at the vertices are , , in counterclockwise order, and c is a complex constant. The numbers , , are the pre-images of the polygon's vertices, or prevertices, which lie in order on the unit circle.

n ,,1

nzz ,,1

A variation on the first animation is to leave the prevertices fixed and vary the angles assigned to them. Here we "square" the ends of the X into right angles. The color of a (pre)vertex indicates its distance from being a right angle.

The last sequence The color indicates the radius of a point's image in the disk. Notice how the arms of the X originate from points quite close to the boundary of the disk.

188


Applications of Complex Analysis

Douglas N. Arnold

Gamma FunctionBernoulli Numbers

Fourier Transform

Laplace Transform

Z Transform

Mellin Transform

Hilbert Transform

Zeta Function

189

SOLO Primes

t

tt

z

tde

tz

0

1

Proof:

Gamma Function

0& xyixz

t

tt

zt

tt

zt

tt

z

tde

ttd

e

ttd

e

t

1

11

0

1

0

1

For the first part:

xt

xxt

xtdttd

e

ttd

e

t x

t

t

t

xt

t

xet

tt

yixt

tt

z t 1lim

1110

1

0

1

0

111

0

11

0

1

The first integral converges for any x ≥ δ > 0.

For the second integral, using integration by parts:

t

tt

x

e

t

t

txedv

tu

t

tt

x

e

t

t

txedv

tu

t

tt

xt

tt

yixt

tt

z

tde

txxetx

e

tde

txettd

e

ttd

e

ttd

e

t

t

x

t

x

1

3

/1

1

2

1

2

/1

1

1

1

1

1

1

1

1

2111

1

2

1

Euler’s Second IntegralGamma integral is defined, and

converges uniformly for x > 0.

190

SOLO Primes

t

tt

z

tde

tz

0

1

Proof (continue):

Gamma Function

0& xyixz

For the second integral, using integration by parts:

t

tt

x

e

t

t

txedv

tu

t

tt

x

e

t

t

txedv

tu

t

tt

xt

tt

yixt

tt

z

tde

txxetx

e

tde

txettd

e

ttd

e

ttd

e

t

t

x

t

x

1

3

/1

1

2

1

2

/1

1

1

1

1

1

1

1

1

2111

1

2

1

After [x] (the integer defined such that x-[x] < 1) such integration the power of t in the integrand becomes x-[x]-1 < 0. and we have:

t

tt

t

ttxx

tde

xxxxtdet

xxxx11

1

121

121

Therefore the Gamma integral is defined, and converges uniformly for x > 0.

Gamma integral is defined, and converges uniformly for x > 0.

q.e.d.

191

SOLO

t

tt

z

tde

tz

0

1

Proof:

Gamma Function

0& xyixz

zzz 1

zztdetztdtzeettdetzt

t

tzt

t ud

z

v

t

v

t

u

zdtedvtu

partsby

t

t

tztz

0

1

0

1

0

,

nintegratio0

01

Properties of Gamma Function: 1

Note that for the evaluation of Gamma Function for a Positive Real Number we need to know only the value of Γ (x) for 0 < x < 1

xxxnxnxnx 121

121

nxnxxx

nxx

For x < 0 with –n < x < -n+1 or 0 < x+n < 1, we define

We can see that for x = 0 or a negative integer the denominator of the right side is zero, and so Γ (x) is undefined (goes to infinity)

Gamma Function

,2,1,0!1 nnn

192

SOLO Primes

t

tt

z

tde

tz

0

1

Proof:

Gamma Function

!1

1Residue

1

1

nz

n

nzResidues of Gamma Function at x = 0,-1, -2,---,-n:..,

121

nxnxxx

nxx

q.e.d.

,2,1!1

1

121

1

1211limResidue

11

11

nnnn

nxnxxx

nxnxx

n

nxnx

193

SOLO Primes

t

tt

z

tde

tz

0

1

Gamma Function

Absolute value |Γ (z)|

Real value ReΓ (z)

Imaginary value ImΓ (z)

194

SOLO Primes

t

tt

z

tde

tz

0

1

Gamma Function

zzz 1

Let compute

110

0

tt

t

t etde

Therefore for any n positive integer:

!1122112111 nnnnnnnnn

Properties of Gamma Function : 1

2

q.e.d.

195

SOLO Primes

Second definition identical to First

bayxallyfxfyxf ,,1,011

xa by yx 1

yxf 1

yfxf 1

Convex Function:

A Function f (x) is called Convex in an interval (a,b) if for every x,y ϵ (a,b) we have

A Function f (x), defined for x > 0, is called Convex, if the corresponding function

y

xfyxfy

defined for all y > -x, y ≠ 0, is monotonic Increasing throughout the range of definition.

x yx y

yxf

xf

If 0 < x1 < x < x2, are given by choosing y1 = x1 – x < 0, y2 = x2 – x > 0, we express the condition of convexity as

xx

xfxfy

xx

xfxfy

2

22

1

11

xxxfxfxxxfxf 1221

1

12

12

12

21 xx

xxxf

xx

xxxfxf

One other equivalent definition:

196

SOLO Primes

1,0ln1ln1ln yfxfyxf

Logarithmic Convex Function :

A Function f (x)>0 is called logarithmic-convex or simply log-convex if ln (f (x) ) is convex or

This is equivalent to 1ln1ln yfxfyxf

Since the logarithm is a momotonic increasing function we obtain

yxyfxfyxf ,1,01 1

197

SOLO Primes

t

tt

z

tde

tz

0

1

Proof :

Gamma Function

0& xyixz

1,0ln1ln1ln baba

Properties of Gamma Function :

3Gamma is a Log Convex Function

1

1

0

1

0

1

0

111

0

111

badtetdtet

dtetetdtetba

tbtaInequalityHolder

tbtatba

q.e.d.

198

SOLO Primes

t

tt

z

tde

tz

0

1

Proof :

Gamma Function

Other Gamma Function Definitios:

nxxx

nnx

x

n

1

!limGauss’ Formula

Since the Gamma Function is monotonically increasing the logarithm of Gamma Function is also monotonic increasing and for 0 < x < 1 and any n > 2 we have

nnx

nnx

lnln

!1

!ln

!2

!1ln

1

!1ln!ln!1lnln

1

!1ln!2ln

n

n

n

n

nn

x

nnxnn

n

xn

nx

n ln!1

ln1ln

x1 1

yln

0

1

1

ln1ln

x

nn

nn nn

nnx

1

ln1ln1


199

SOLO Primes

t

tt

z

tde

tz

0

1

Proof (continue - 1) :

Gamma Function


Since the Gamma Function is monotonically increasing the logarithm of Gamma Function is also monotonic increasing and for 0 < x < 1 and any n > 2 we have

n

xn

nx

n ln!1

ln1ln

xx nn

nxn ln

!1ln1ln

10 x

!1!11 nnnxnn xx

Use xxxnxnxnx

0

121

xxnxnx

nnx

xxnxnx

nn xx

121

!1

121

!11

nxxx

nnx

x

n

1


Euler 1729Gauss 1811

200

SOLO Primes

t

tt

z

tde

tz

0

1


Gamma Function


xxnxnx

nnx

xxnxnx

nn xx

121

!1

121

!11

xxnxnx

nnx

xxnxnx

nn xx

11

!1

11

!

Take the limit n → ∞

xxnxnx

nn

nx

xxnxnx

nn x

n

x

n

x

n 11

!lim

11lim

11

!lim

1

1,011

!lim

x

xxnxnx

nnx

x

n

Substitute n+1 for n

nxxx

nnx

x

n

1


201

SOLO Primes

t

tt

z

tde

tz

0

1

Let substitute x + 1 for x

Gamma Function


1,011

!lim

x

xxnxnx

nnx

x

x

n

n

q.e.d

nxxx

nnx

x

n

1



1,011

!lim

1lim

11

!lim1

1

1

xxxxxnxnx

nn

nx

nx

xnxnx

nnx

x

x

nn

x

n

The right side is defined for 0 < x <1. The left side extend the definition for(1 , 2). Therefore the result is true for all x , but 0 and negative integers.

202

SOLO Primes

t

tt

z

tde

tz

0

1

Gamma Function


Start from Gauss Formula xx nn

lim

q.e.d

constantMascheroni-Euler57721566.0ln1

2

11lim

11

nn

kx

e

x

ex

n

k

k

xx

Weierstrass’ Factorization Formula for Gamma Function

Proof :

nx

nxx

x

eeee

xx

nx

nx

n

xxnxnx

nnx

n

xxx

nnx

xx

n

11

11

1

11

111

11

!:

211

2

11ln

11

1

2

11ln

11

1limlim

k

k

xxn

k

k

x

nnx

nn

n

kx

e

x

e

kx

e

xexx


(1815 – 11897)

203

SOLO Primes

t

tt

z

tde

tz

0

1

Gamma Function

Other Gamma Function Definitios:Weierstrass’ Factorization Formula for Gamma Function (continue)

Karl Theodor Wilhelm Weierstrass) 1815 – 11897(

1

11

k

k

zz e

k

zez

z

Γ (z) has Poles at zk = - k, k=0,1,2,…, and no Zeros therefore 1/ Γ (z) has Zeros at zk = - k, k=0,1,2,…, and no Poles, andFrom previous development we obtain Weierestrass Factorization

Return to ζ (z)

204

SOLO

t

tt

z

tde

tz

0

1

Gamma Function Gamma integral is defined, and converges uniformly for x > 0.

Differentiation of Gamma Function:

q.e.d

0,2!11'

ln

01'''

ln

constantMascheroni-Euler57721566.0111'

ln

11

1

122

2

2

2

1

xnkx

n

x

x

xd

dx

xd

d

kxx

xxxx

xd

d

kxkxx

xx

xd

d

kn

n

n

n

n

n

k

k

Proof :

Start from Weierstrass Formula

1 1k

k

xx

kx

e

x

ex

11

1lnlnlnkk k

x

k

xxxx

11 1

111

lnkk

kx

kkx

xxd

d

0111111

ln0

21

221

2

2

kkk kxkxxkxkxxd

dx

xd

d

0

1

1 !11'ln

kn

n

n

n

n

n

kx

n

x

x

xd

dx

xd

d

Gamma Function

We can see that

1

11

1 1

11lim

1

1

1

1'1ln

n

n

kn kk

xxd

d

205

SOLO Primes

t

tt

z

tde

tz

0

1

Gamma Function


1

11

1

1

k

k

zz e

k

ze

zzz

Return to ζ (z)

Hadamard Infinite Product Expansion of Gamma Function

1

1

1

2

11

1

2

2

111 10

j

a

z

pa

z

a

z

j

zczc pj

p

jjp

p ea

zefzf

Since 1/z Γ (z)=1/Γ (1+z) has Zeros at zk = - k, k=1,2,…, and no Poles we can use the Hadamard Infinite Product Expansion

pii

ff

zdd

c z

i

i

i ,,1,0!1

: 0

1

1

Gamma Function Γ (1+z) has Order p=0, and

1

1':

0

1

1

zf

fc

1101 fzzfDefine

We recovered the Weierstrass Formulausing Hadamard Expansion

206

SOLO Primes

1

0

11 1,s

s

zy sdsszyBBeta Function

Beta Function is related to Gamma Function:

u

u

uy

duudt

utt

t

ty udeutdety0

12

20

1 22

2

zy

zyzyB

,

Proof:

In the same way:

v

v

vz vdevz0

12 2

2

u

u

v

v

vuuzy vdudevuzy0 0

1212 22

4

Use polar coordinates:

drdrdrdr

rdrd

vrv

uruvdud

rv

ru

cossin

sincos

//

//

sin

cos

2/

0

1212

0

12

0

2/

0

121212

sincos22

sincos4

2

2

drder

drderzy

zy

zy

r

r

rzy

r

r

rzyzy

Euler’s First Integral

207

SOLO Primes

1

0

11 1,s

s

zy sdsszyBBeta Function Euler’s First Integral

Beta Function is related to Gamma Function: zy

zyzyB

,

Proof (continue):

2/

0

1212 sincos2

dzyzy zy

Change variables in the integral using dsds cossin2sin 2

zyBsdssds

s

yzzy ,1sincos21

0

112/

0

1212

zyBzyzy ,Therefore q.e.d.

Use z→y and y → 1 - z

u

u

zu

u

z

z

zu

us

u

udsd

s

s

zz

udu

u

u

ud

u

u

u

u

dssszzBzz

0

1

021

11

1

1

0

1

1111

1

11,11

2 q.e.d.

208

SOLO Primes

Proof

yzBzyzyBzyyzyzBzyB

,,,,

Use y → 1 - z

u

u

zu

u

z

z

zu

us

u

udsd

s

s

zz

udu

u

u

ud

u

u

u

u

dssszzBzz

0

1

021

11

1

1

0

1

1111

1

11,11

2

t

tt

z

tde

tz

0

1

Gamma Function

Other Gamma Function Properties: zzz

sin1 Euler

Reflection Formula

209

SOLO Primes

Proof (continue - 1)

u

u

x

udu

uxx

0

1

11

t

tt

z

tde

tz

0

1

Gamma Function

Other Gamma Function Properties:

R

RC

C

1

planeu

uRe

uIm

Replace the path from 0 to ∞ by the Hankel contour Hε

in the Figure, described by four paths, traveled in counterclockwise direction :

1 .going counterclockwise above the real axis, (u = |u|)2 .along the circular path CR ,

3 .bellow the real axis, (u= |u|e -2πi )4 .along the circular path Cε.

C

yR yyi

C

yR y

udu

uud

u

ueud

u

uud

u

u

R1111

2

Define y = 1 – x, and assume x,y ϵ (0,1)

zzz

sin1 Euler

Reflection Formula

210

SOLO Primes


u

u

x

udu

uxx

0

1

11

t

tt

z

tde

tz

0

1

Gamma Function


R

RC

C

1

planeu

uRe

uIm

This path encloses the pole u=-1 of that has the residue1

u

u y

yi

eu

yy

euu

ui

11

Residue

By the Residue Theorem

For z ≠ 0 we have

yzyzyzyy zeeez lnlnReln

zzz

sin1 Euler

Reflection Formula

yiy

eu

y

C

yR yiy

C

yR y

eiu

uui

u

uizd

z

zud

u

uezd

z

zud

u

u

i

R

21

1lim2

1Residue2

1111

1

2

211

SOLO Primes


t

tt

z

tde

tz

0

1

Gamma Function


R

RC

C

1

planeu

uRe

uIm

yi

C

yR yiy

C

yR y

eizdz

zud

u

uezd

z

zud

u

u

R

2

11112

For the second and forth integral we have

0lnlnReln zzeeezyzyzyzyy

z

z

z

z

z

zyyy

111

Hence for small ε we have :

and for large R we have :

01

21

01

y

C

y

zdz

z

01

21

1

Ry

C

y

R

Rzd

z

z

R

Therefore the integrals on the circular paths are zero for ε→0 and R∞→

zzz

sin1 Euler

Reflection Formula

212

SOLO Primes


t

tt

z

tde

tz

0

1

Gamma Function


R

RC

C

1

planeu

uRe

uIm

yiy

iyy

eiudu

ueud

u

u

2

11 0

2

0

We obtain

Multiply both sides by yie

iudu

uee

yiyiy 2

10

yee

iud

u

uiyiy

y

sin

2

10

Rearranging we obtain

Since both sides of this equation are Holomorphic (analytic) in x ϵ (0,1) we can extend the result for all analytic parts of z ϵ C (complex plane).

1,0sin1sin11

10

1

0

1

xxx

udu

uud

u

uxx

u

u

yxyu

u

x

Substituting y = 1 – x we obtain

zzz

sin1 Euler

Reflection Formula

213

SOLO Primes

Onother Proof

t

tt

z

tde

tz

0

1

Gamma Function


Start with Weierstrass Gamma Formula

zzz

sin1 Euler

Reflection Formula

1 1k

k

xx

kx

e

x

ex

12

22

1

2 1111

kk k

x

k

xxx

k

xx

e

kx

e

kx

eexxx

Use the fact that Γ (-x)=- Γ (1-x)/x to obtain

12

2

11

1

k k

xx

xx

Now use the well-known infinite product

12

2

1sink k

xxx

q.e.d.

214

SOLO Primes

Proof

t

tt

z

tde

tz

0

1

Gamma Function

Other Gamma Function Properties: zzz

cos2

1

2

1

Start from

Substitute ½ +z instead of z

zzz

sin1

zz

zz

cos

21

sin2

1

2

1

q.e.d.

215

SOLO Primes

t

tt

z

tde

tz

0

1

Gamma Function

Duplication and Multiplication Formula:

0Re222

112

zzzz

z

Legendre Duplication Formula1809

Adrien-Marie Legendre )1752 – 1833 (

Proof:

2/1,2sin22sin2

2sin22sincos2,

212/

0

1221

0

1221

2/

0

12212/

0

1212

zBdd

ddzzB

zzzzz

zzzz

0Re2/1

2/122/1,2,

22121

zz

zzBzzB

z

zz zzWe have

therefore

q.e.d

0Re222

112

2

1

zzzzz

216

SOLO

t

tt

z

tde

tz

0

1

Gamma Function

Duplication and Multiplication Formula:

znnn

nz

nz

nzz znn

2/12/12

121

Gauss Multiplication

Formula

nz

1

Carl Friedrich Gauss)1777 – 1855 (

nn

n

nn

n 2/12121

Euler

Multiplication Formula

Gamma Function

217

SOLO Primes

t

tt

z

tde

tz

0

1

Gamma Function

Some Special Values of Gamma Function:

q.e.d

2222/1

02

0

22 t

t

uut

duudt

t

t

t

udetdt

e

n

nnnnnn

2

125312/12/112/32/12/12/12/1

12531

21

2/12/32/1

2/1

2/1

2/32/1

nnnn

nn

nn

2/1

n

nn

2

125312/1

12531

212/1

n

nnn

Proof:

218

Jacob Bernoulli1654-1705

The Bernoulli numbers are among the most interesting and important number sequences in mathematics. They first appeared in the posthumous work "Ars Conjectandi" (1713) by Jakob Bernoulli (1654-1705) in connection with sums of powers of consecutive integers. Bernoulli numbers are particularly important in number theory, especially in connection with Fermat's last theorem (see, e.g., Ribenboim (1979)). They also appear in the calculus of finite differences (Nörlund (1924)), in combinatorics (Comtet (1970, 1974)), and in other fields.

Bernoulli Numbers

The Bernoulli numbers Bn play an important role in several topics of mathematics. These numbers can be defined by the power series

SOLO

0 !1 n

n

nz n

zB

e

z

Complex Variables

219

SOLO

Bernoulli Numbers

0 !1 n

n

n

seriesTaylor

z n

zB

e

z

Let compute the Bernoulli number using

1Residue2

2

!

12

!

11

z

n

eCnzn e

zi

i

n

z

zd

e

z

i

nB

z

R

RC

C

planez

zRe

zIm

The zeros of e z = 1 are at z = ± 2 π i k

1 1

1 12

1'

22

1'

2

1

2

1

2

1!

1lim

1

2lim

1lim

1

2lim2

2

!

1Residue2

2

!

k knn

k knkiz

Hopitall

zkiznkiz

Hopitall

zkiz

z

n

en

kikin

ze

kiz

ze

kizi

i

n

e

zi

i

nB

z

01

x

xn

n

n e

x

xd

dB

Complex Variables

220

SOLO

Bernoulli Numbers

0 !1 n

n

nz n

zB

e

z

Let compute the Bernoulli number using

1Residue2

2

!

12

!

11

z

n

eCnzn e

zi

i

n

z

zd

e

z

i

nB

z

R

RC

C

planez

zRe

zIm

The zeros of e z -1 = 1 are at z = ± 2 π i k

1 11 2

1

2

1!

1Residue2

2

!

k knnz

n

en

kikin

e

zi

i

nB

z

oddn

evennk

oddn

evennki

kikik

nn

k

n

kn

kn

0

12

0

2111

2/

111

oddn

evennnn

k

nB n

n

knn

n

n

0

2

!12

1

2

!12

2/

1

2/

,2,1,0

120

222

!212 2

m

mn

mnmm

B m

m

n

Complex Variables

1

1

knk

nwhere is the Zeta Function

SOLO

Euler Zeta Function and the Prime History

232 4

1

3

1

2

11

In 1650 Mengoli asked if a solution exists for

P. Mengoli1626 - 1686

The problem was tackled by Wallis, Leibniz, Bernoulli family, without success .The solution was given by the young Euler in 1735. The problem was named “Basel Problem” for Basel the town of Bernoulli and Euler.

Euler started from Taylor series expansion of the sine function

!7!5!3

sin753 xxx

xx

Dividing by x, he obtained

!7!5!3

1sin 642 xxx

x

x

The roots of the left side are x =±π, ±2π, ±3π,…. However sinx/x is not a polynomial, but Euler assumed (and check it by numerical computation) that it can be factorized using its roots as

2

2

2

2

2

2

91

411

21

2111

sin

xxxxxxx

x

x

Leonhard Euler)1707 – 1783 (

Return to Euler

Riemann's Zeta Function

SOLO

!7!5!3

1sin 642 xxx

x

x

2

2

2

2

2

2

91

411

sin

xxx

x

x

Leonhard Euler)1707 – 1783 (If we formally multiply out this product and collect all the x2 terms, we

see that the x2 coefficient of sin(x)/x is

122222

11

9

1

4

11

n n

But from the original infinite series expansion of sin(x)/x, the coefficient of x2 is −1/(3!) = −1/6. These two coefficients must be equal; thus,

1

22

11

6

1

n n 6

1 2

12

n n

Euler extend this to a general function, Euler Zeta Function

,4,3,24

1

3

1

2

11: nn

nnn The sum diverges for n ≤ 1 and

converges for n > 1.

Euler computed the sum for n up to n = 26. Some of the values are given here

,9450

8,945

6,90

4,6

28642

Euler checked the sum for a finite number of terms.

EulerZeta Function and the Prime History (continue – 1)


SOLO

Euler Product Formula for the Zeta Function

Leonhard Euler proved the Euler product formula for the Riemann zeta function in his thesis Variae observationes circa series infinitas (Various Observations about Infinite Series), published by St Petersburg Academy in 1737

primepx

nx pn 1

11

1

where the left hand side equals the Euler Zeta Function

Euler Proof of the Product Formula

xxxxx

s8

1

6

1

4

1

2

1

2

1

xxxxxxxx

13

1

11

1

9

1

7

1

5

1

3

11

2

11

xxxxxxxxx

33

1

27

1

21

1

15

1

9

1

3

1

2

11

3

1

xxxxxxxx

17

1

13

1

11

1

7

1

5

11

2

11

3

11

all elements having a factor of 3 or 2 (or both) are removed

xxxx

nxn

x5

1

4

1

3

1

2

11

1

1

converges for integer x > 1

all elements having a factor of 2 are removed

Leonhard Euler)1707 – 1`783 (

EulerZeta Function and the Prime History (continue – 2)Riemann's Zeta Function

SOLO

Leonhard Euler)1707 – 1`783 (

Euler Product Formula for the Zeta Function

primepx

nx pn

x1

11

1

Euler Proof of the Product Formula (continue)

xxxxxxxx

17

1

13

1

11

1

7

1

5

11

2

11

3

11

Repeating infinitely, all the non-prime elements are removed, and we get:

12

11

3

11

5

11

7

11

11

11

13

11

17

11

x

xxxxxxx

Dividing both sides by everything but the ζ(s) we obtain

xxxxxx

x

131

1111

171

151

131

121

1

1

Therefore

primepx

nx pn

x1

11

1

EulerZeta Function and the Prime History (continue – 3)


225

SOLO Riemann's Zeta Function

The Riemann Zeta Function or Euler–Riemann Zeta Function, ζ(z), is a function of a complex variable z that analytically continues the sum of the infinite series

yixzn

zn

z

1

1

“On the Number of Primes Less Than a Given Magnitude”, 7 page paper offered to the Monatsberichte der Berliner Akademie on October 19, 1859. The exact publication date is unknown.

zz

zz zz

1

2sin12 1

To construct the analytic Continuation of the Zeta Function, Riemann established the relation (see proof ).

where Γ(s) is the Gamma Function, which is an equality of Meromorphic Functions valid on the whole complex plane. This equation relates values of the Riemann Zeta Function at the points z and 1 − z. The functional equation (owing to the properties of sin) implies that ζ(z) has a simple zero at each even negative integer z = −2n — these are known as the trivial zeros of ζ(z). For s an even positive integer, the product sin(πz/2)Γ(1−z) is regular and the functional equation relates the values of the Riemann Zeta Function at odd negative integers and even positive integers.

Georg Friedrich Bernhard Riemann )1826– 1866(

SOLO

,2,11

1 1

nn

Bn nn

Bn are the Bernoulli numbers

Those roots are called the Trivial Zeros of the Zeta Function. The remaining zeros of ζ (z) are called Nontrivial Zeros or Critical Roots of the Zeta Function.

The Nontrivial Zeros are located on a Critical Strip defined by 0 < x < 1.

Since Bn+1 = 0 for n + 1 odd (n even) we also have ,2,102 mm

xyixzpn

zprimep

zn

s

Re1

11

1

Riemann Zeta Function Zeros

Since the product contains no zero factors we see that ζ (z) ≠ 0 for Re {z} >1.

Graph showing the Trivial Zeros, the Critical Strip and the Critical Line of ζ (z) zeros.

We shall prove that

227


228

Re ζ (z) in the original domain, Re z > 1.

Re ζ (z) after Riemann’s extension.


229

SOLO

The position of the complex zeros can be seen slightly more easily by plotting the contours of zero real (red) and imaginary (blue) parts, as illustrated above. The zeros (indicated as black dots) occur where the curves intersect

The figures bellow highlight the zeros in the complex plane by plotting |ζ(z)|) where the zeros are dips) and 1/|ζ(z)) where the

zeros are peaks(.


230


The Riemann Hypothesis

The Non-Trivial Zeros ρ of ζ (z) has Re ρ½ = This Hypothesis was never proved.

1z

231

SOLO

1Re10

1

zxfordte

tzz

t

tt

z

u

uu

z

due

uz

0

1

Proof:

Gamma Function

Change of variables u=nt

t

tnt

zz

t

tnt

z

tde

tntdn

e

ntz

0

1

0

1

Thus for n=1,2,3,…,N

t

tNt

z

z

t

tt

z

z

t

tt

z

z

tde

t

Nz

tde

tz

tde

tz

0

1

02

1

0

1

1

2

1

1

1

0& xyixz

Summing those equationsfor x > 0

t

t

zNtttzzz

tdteeeN

z0

12

1111

2

1

1

1

_________________________________________________


232

SOLO

Proof (continue – 1): 0& xyixz

Since converges only for Re (z)= x > 1, then letting N → ∞, we obtain for x > 1

1n

zn

Uniform convergence of

t

t

zNtttNzz

tdteee

z0

12

111lim

2

1

1

1

01

1

1

111

1

2

2

tqeeee t

q

q

t

q

t

q

t

allows to interchange between limit and the integral :

RatioGoldentde

ttd

e

ttd

e

tz

t

tt

zt

tt

zt

tt

z

zz

2

51

1112

1

1

1

ln2

1ln2

0

1

0

1

ln2

02

1ln2

0

1ln2

0

1 11

11

t

ttt

xt

tt

xyixzt

tt

z

tdee

ttde

ttd

e

t

The first integral gives

The integral diverges for 0 < x ≤ 1, and converges only for x > 1

1Re10

1

zxfordte

tzz

t

tt

z


233

SOLO

Proof (continue – 2): 0& xyixz

t

tt

zt

tt

zt

tt

z

zztd

e

ttd

e

ttd

e

tz

ln2

1ln2

0

1

0

1

1112

1

1

1

In the second integral we have

This integral converges only for x > 1, therefore we proved that

ln21 2/ tforee tt

since RatioGoldeneforee ttt

2

5101 2/2/

t

tt

x

termfinite

t

txtu

dtedv

t

tt

xt

tt

xiyxzt

tt

z

tde

txettd

e

ttd

e

ttd

e

t x

t

ln22/

2

ln2

2/1

ln22/

1

ln2

1

ln2

1

12211

1

2/

finite

t

ttxx

xxt

tt

x

tdet

xxxxtermsfinitetde

t

ln22/1

ln22/

1 1212

1Re12

1

1

1

0

1

zxfortde

tzzz

t

tt

z

zz

1Re10

1

zxfordte

tzz

t

tt

z

After [x] (the integer defined such that x-[x] < 1) such integration the power of t in the integrand becomes x-[x]-1 < 0. and we have:

q.e.d.


234

SOLO

Proof

The integral can be rewritten as

00

1sin2

1 0

0

1

itoreturnsandzeroencirclesiatstartspaththe

de

izz

zi

i

z

i

iy

x

i

i

2

IntegralIII

i

i

z

originaroundCircleIntegralII

i

i

z

IntegralI

i

i

z

i

i

z

de

de

de

de

1lim

1lim

1lim

11

0

1

0

1

0

0

0

1


235

SOLO


The first integral can be written as

i

iy

x

i

i

2

t

tt

zzi

et

tt

zzi

t

tit

ziiti

i

z

tde

tetd

e

tetd

e

eitd

e

i

0

110 11

1

0

0 1

0 111lim

1lim

The second integral can be written as

0

1

2lim2

1

2lim

21

2lim

1lim

2

020

2

02

1

0

2

02

1

0

21

0

de

deie

e

deie

ed

e

ii

i

i

e

x

i

e

iyxi

i

e

iyxie

originaroundCircle

i

i

z

00

1sin2

1 0

0

1


de

izz

zi

i

z


236

SOLO


The third integral can be written as

i

iy

x

i

i

2

t

tt

zzi

et

tt

zzi

t

tit

ziiti

i

z

tde

tetd

e

tetd

e

eitd

e

i

0

11

0

11

1

0

1

0 111lim

1lim

Therefore

t

tt

zt

tt

zzizit

tt

zzizi

i

i

z

tde

tzitd

e

t

i

eeitd

e

teed

e 0

1

0

1

0

10

0

1

1sin2

122

11

But we found that 1Re10

1

zxfordte

tzz

t

tt

z

00

1sin2

1 0

0

1


de

izz

zi

i

z

Therefore

0

0

1

1sin2

1 i

i

z

de

izz

z

The right hand is analytic for any z ≠ 1. Since it equals Zeta Function in the half plane x > 1, it is the Analytic Continuation of Zeta to the complax plane for any z ≠ 1 .

0

0

1

1sin2

1 i

i

z

de

izz

z

q.e.d.


237

SOLO

Proof

0

0

1

12

sin221

i

i

zz

zz

ide

R

RC

C

plane

Re

Im

Let add a circular path of radius R → ∞. On this path

0

1lim

1

2

0

1

d

e

eRd

ei

i

i

R

eR

zi

R

eR

deRdC

z

Therefore we have

Since the integral is over a closed path in the complex λ plane, we can use the Residue Theorem to calculate it. The residues are given by

,2,121 nnie

1

1

1

110

0

1

222211 n

z

n

zzi

i

z

niiniide

de

d

ed

ed

ed

e

z

C

zi

i

zi

i

z

R1111

110

0

10

0

1


238

SOLO

Proof (continue)

0

0

1

12

sin221

i

i

zz

zz

ide

R

RC

C

plane

Re

Im

11

11

1

1

1

10

0

1 122222

1 nz

zzz

n

z

n

zi

i

z

niiiniiniid

e

2sin22/2/lnln11 z

eeieeiiiiiii izizizizzzzz

znn

z

1

1

11

zz

ide

zi

i

z

12

sin221

0

0

1

q.e.d.


239

SOLO

i

iy

x

i

i

2

00

12

1 0

0

1


dei

zz

i

i

z

We also found

zz

ide

zi

i

z

12

sin221

0

0

1

Has zeros for

,...4,2,002

sin

zforz

,7,5,301 zforz

z 1 Has no zeros, but has simple poles for z = 1,2,3,4.…,

If we return to ζ (z) equation we can see that the zeros of are cancelled by the poles of Γ (1-z). Only the simple pole

at z = 1 remain and is the single pole of ζ (s) .

0

0

1

1

i

i

z

de

Let find the Residue of this pole:

0

0

1

111 12

1lim11lim1lim

i

i

z

zzzd

eizzzz

1cos

limsin

1lim11lim

1

'

1

1

sin1

1

zzz

zzz

z

HopitalL

z

zzz

z

0

0

1

1 12

1lim

i

i

z

zd

ei


240

SOLO

Proof

zz

zzz z

1

2sin22sin2

zz

ide

zi

i

z

12

sin221

0

0

1

We found

0

0

1

1sin2

1 i

i

z

de

izz

z

Combining those two relations, we get

zz

zzz z

1

2sin22sin2

q.e.d.


241

SOLO

Proof

zzzz zz 112/sin2 1

Start from

use

zz

zzz z

1

2sin22sin2

zzz

sin1 z

zz

1

sin

or

zz

zz

z

1

2sin2

1

zz

zz zz

1

2sin12 1

q.e.d.Return to Riemann Zeta Function


242

SOLO

Proof

Start from

use

zz

zzz z

1

2sin22sin2

zzz

sin1

zzz

1sin

21

22

sinzz

z z

z

2

zz

zzz

zz

1

2sin

2/12/

12

1

1

or

zz

zzz zzz

12/1

1122/ 2/12/12/

z

z

z

z zzzz

1

2/12/ 12/12/


243

SOLO

Proof (continue)

z

z

z

z zzzz

1

2/12/ 12/12/

or

zz

zzz zzz

12/1

1122/ 2/12/12/

2

1

22 12/1 zz

z z

2/12

121 2/1 z

zz z

z

z

1

2

1

2/1

1122/1 z

zzz

therefore

q.e.d. z

zzz zz

12

12/ 2/12/

Use LegendreDuplication Formula: 0Re2

22

112

zzzz

z

2/z

z


244

SOLO

Proof

i

iy

x

i

i

2

00

1sin2

1 0

0

1


de

izz

zi

i

z

We found

and zzz

sin1 z

zz

sin

1

0

0

12

1 0

0

1

itoreturnsand

zeroencirclesiatstartspaththe

dei

zz

i

i

z

0 1

21 12

!1

12

!1 i

i

n

Cnzn d

ei

n

z

zd

e

z

i

nB

i

iy

x

i

i

2

therefore 1!1

11

n

n Bn

zz

0

0

12

1 0

0

1

itoreturnsand

zeroencirclesiatstartspaththe

dei

zz

i

i

z

zz

nz 11 1

n

Bn nn

,2,1,01

1 1

nn

Bn nn Bn are the Bernoulli numbers

q.e.d.

We found

Zeta-Function Values and the Bernoulli Numbers

Return to Riemann Zeta Function Zeros


245

SOLO

Zeta Function Values and the Bernoulli Numbers

,3,2,1

!22

212 2

2

mBm

m m

mm

zz

zz zz

12

12/ 2/12/ Let use

with z = 2 m mm

mm mm 212

212 2/21

m

mm

m

mmm

m

m

Bmm

mB

mm

m

mm

mm

2

2/112

2

22/12

2/1

2/1!2

!121

!22

21

2/1

!1

22/1

!121

We found

,3,2,1

2/1!2

!12121 2

2/112

mBmm

mm m

mm


246

SOLO

Zeta Function Values and the Bernoulli Numbers

We found

2/112

2/11

2/1

!12

!121

224212531

121221

12531

212/1

m

m

mm

m

mm

mm

mmmmm

,3,2,1

2/1!2

!12121 2

2/112

mBmm

mm m

mm Therefore

Finally

,2,1,01

1 1

nn

Bn nn

,3,2,12

121 2 mB

mm m

We also found The two expressionsAgree .

Riemann Zeta Function


247

SOLO

Hadamard Infinite Product Expansion of Zeta Function


1

1

1

2

11

1

2

2

111 10

j

a

z

pa

z

a

z

j

zczc pj

p

jjp

p ea

zefzf

Since (z-1) ζ (z) is Analytic and has only Zeros we can use the Hadamard Infinite Product Expansion

Zeta Function ζ (z) has Order p=0, and

2

1

2

10 1 B

The Zero of the Zeta Function ζ (z) are-Trivial Zeros at z = -2n, n=1,2,…- Nontrivial Zeros ρ on the Critical Zone 0 < Re ρ < 1

1

2

0102/10

21101 1

n

n

z

zofzerostrivial

z

zofzerosnontrivial

zc

fzf

en

ze

zezz

Hadamard Infinite Product Expansion of (z-1) ζ (z) is:

pii

ff

zdd

c z

i

i

i ,,1,0!1

: 0

1

1

12ln2/1

2/2ln2/1

0

0'0:

1

0

1

1

zzzf

zf

fc

2

10&

2

2ln0'


248

SOLO

Hadamard Infinite Product Expansion of Zeta Function (continue)


1

2

010

12ln

211

21

n

n

z

zofzerostrivial

z

zofzerosnontrivial

z

en

ze

zezz

1

22

21

22/

1

n

n

zz

en

ze

z

z

Hadamard Infinite Product Expansion of (z-1) ζ (z) is:

We found the Weierstrass Expansion for the Gamma Function:

100

2/12ln

12/112

Re

zz

ez

zz

ez

2

122

21

22

1

2

ze

zze

en

zzz

n

n

z

Hadamard (1893) used the Weierstrass product theorem to derive this result. The plot above shows the convergence of the formula along the real axis using the first 100 (red), 500 (yellow), 1000 (green), and 2000 (blue) Riemann zeta function zeros.


Fourier Transform

dttjtftfF exp:F

SOLO

Jean Baptiste JosephFourier

1768 - 1830

F (ω) is known as Fourier Integral or Fourier Transformand is in general complex

jAFjFF expImRe

Using the identities

tdtj

2exp

we can find the Inverse Fourier Transform Ftf -1F

002

1

2exp

2expexp

2exp

tftfdtfdd

tjf

dtjdjf

dtjF

2exp:

dtjFFtf -1F

002

1

tftfdtf

If f (t) is continuous at t, i.e. f (t-0) = f (t+0)

This is true if (sufficient not necessary)f (t) and f ’ (t) are piecewise continue in every finite interval1

2 and converge, i.e. f (t) is absolute integrable in (-∞,∞)

dttf

Fourier TransformSOLO

tf-1F

F FProperties of Fourier Transform

Linearity 1

221122112211 exp: FFdttjtftftftf

F

Symmetry 2 tF

-1FF f2

tFdttjtFfdt

tjtFfd

tjFtft

F

exp22

exp2

exp

Proof:

Conjugate Functions3 tf *

-1FF *F

Proof:

tfd

tjFd

tjFtf ****

2exp

2exp 1-F

Fourier Transform

a

Faa

d

ajfdttjtaftaf

ta 1

expexp:F

FjdttjjtfFd

ddttjtftfF nn

n

n

FF expexp:

SOLO

tf-1F


Scaling4

Derivatives5

Proof:

taf-1F

F

aF

a

1

Proof:

Corollary: for a = -1 tf

-1FF F

tftj n-1F

F

Fd

dn

n

tftd

dn

n

-1FF Fj n

Fj

dtjjFtf

td

ddtjFFtf nn

n

n1-1- FF

2

exp2

exp


tf-1F


Convolution6

Proof:

212121

212121

expexpexp

expexpexp:

FFFdjfdduujufjf

ddttjtfjfdtdtfftjdtff

ut

F

tftf 21-1F

F 21 * FF

dtfftftf 2121 :*-1F

F 21 FF

The animations above graphically illustrate the convolution of two rectangle functions (left) and two Gaussians (right). In the plots, the green curve shows the convolution of the blue and red curves as a function of t, the position indicated by the vertical green line. The gray region indicates the product as a function of g (τ) f (t-τ) , so its area as a function of t is precisely the convolution.

http://mathworld.wolfram.com/Convolution.html


tf-1F


dFFdttftf 2*

12*

1 2

1Parseval’s Formula7

Proof:

dttjtfF exp11

22exp

2exp 2

*

112*

2*

12*

1

dFF

ddttjtfFdt

dtjFtfdttftf

22exp

2exp 21122121

dFF

ddttjtfFdt

dtjFtfdttftf

2exp*

2

*

2

dtjFtf

dttjtfFdttjtfF expexp 1111

dFFdFFdttftf 212121 2

1

2

1

Signal Duration and BandwidthSOLO

tf-1F

F FRelationships from Parseval’s Formula

dFFdttftf 2*

12*

1 2

1Parseval’s Formula7

Choose tstjtftf m 21

,2,1,0

2

12

22

ndd

Sddttst

m

mm

tftj n-1F

F

Fd

dn

n

and use 5a

Choose n

n

td

tsdtftf 21 and use 5b tf

td

dn

n

-1FF Fj n

,2,1,02

1 22

2

ndSdttd

tsd mn

n

Choosec

,2,1,0,,2,1,0

2*

mndd

SdS

jdt

td

tsdtstj

m

mn

n

n

nmm

n

n

td

tsdtf 1

tstjtf m2


tf-1F


Modulation9

Shifting: for any a real 8

Proof:

ttf 0cos -1F

F 002

1 FF

Proof:

tjtjt 000 expexp2

1cos

atf -1F

F ajF exp tajtf exp-1F

F aF

Fajdajfdttjatfatfat

expexpexp:F

aFdttajtfdttjtajtftajtf

expexpexp:expF

use shifting property with a=±ω0

atf -1F

F ajF exp

Fourier TransformSOLO tf

-1FF FProperties of Fourier Transform (Summary)

Linearity 1 221122112211 exp: FFdttjtftftftf

F

Symmetry 2

tF-1F

F f2

Conjugate Functions3 tf *

-1FF *F

Scaling4 taf-1F

F

aF

a

1

Derivatives5 tftj n-1F

F

Fd

dn

n

tftd

dn

n

-1FF Fj n

Convolution6

tftf 21-1F

F 21 * FF

dtfftftf 2121 :*-1F

F 21 FF

dFFdttftf 2*

12*

1 2

1

Parseval’s Formula7

Shifting: for any a real 8 tajtf exp

-1FF aF

Modulation9 ttf 0cos -1F

F 002

1 FF

dFFdFFdttftf 212121 2

1

2

1

Laplace’s Transform

C2

f

a

0t

00

t

js s - plane

SOLO

Laplace L-Transform (continue – 1)

The Inverse Laplace’s Transform (L -1) is given by:

j

j

ts dsesFj

tf2

1

Using Jordan’s Lemma (see “Complex Variables” presentation or the end of this one)

Jordan’s Lemma GeneralizationIf |F (z)| ≤ M/Rk for z = R e iθ where k > 0 and M are constants, then

for Γ a semicircle arc of radius R, and center at origin:

00lim

mzdzFe zm

R

where Γ is the semicircle, in the left part of z plane.

x

y

R

we can write

j

j

tstsf

f

dsesFj

dsesFj

sFtf

2

1

2

11-L

dsesFj

dsesFj

dsesFj

sFtf ts

C

tsj

j

ts

2

1

2

1

2

1

0

1-L

If the F (s) has no poles for σ > σf+, according to Cauchy’s Theoremwe can use a closed infinite region to the left of σf+, to obtain

Laplace’s TransformSOLO

Properties of Laplace L-Transform

s - Domaint - Domain

tf

f

st sdtetfsF Re0

1 if

M

iii zsFc maxRe

1

Linearity

M

iii tfc

1

3 000 1121 nnnn ffsfssFs Differentiation n

n

td

tfd

4

t

tdf

ss

sF 0

lim1Integration

t

df

5 s

sFReal DefiniteIntegration

t

df0

t

ddf0 0

2s

sF

2

a

sF

a

1Scaling taf


Properties of Laplace L-Transform (continue – 1)


tf

f

st sdtetfsF Re0

6 n

n

sd

sFdMuliplicity by tn tft n

7

0

dssFDivision by t t

tf

8 sFe sTime shifting tutf

9 asF Complex Translations

tfe ta

10 sHsF Convolutiont - plane

0

dthfthtf

11

j

j

dsHFj

sHsFj

2

1

2

1Convolution s - plane

thtf


Properties of Laplace L-Transform (continue – 2)


tf

f

st sdtetfsF Re0

12 Initial Value Theorem sFstfst

limlim0

13 Final Value Theorem sFstfst 0limlim

14 Parseval’s Theorem

j

j

j

j

ts

j

j

ts

dssGsFj

dsdtetgsFj

dttgdsesFj

dttgtf

2

1

2

1

2

1

0

00

SOLO

Z- Transform and Discrete Functions

Z Transform

The Z- Transform (one-sided) of a sequence { f (nT); n=0,1,… } is defined as :

0

:n

nzTnfzFTnfZ

where T, the sampling time, is a positive number.

tf

0n

T Tntt

0

*

n

T TntTnfttftf

tf *

tfT t

tf

0n

T Tntt

0

*

n

T TntTnfttftf

tf *

tfT t

f

ts dtetftfsF0

L

SOLO

Sampling and z-Transform

0

1

1

00sT

n

sTn

n

T eeTnttsS LL

0

00**

1

1

2

1

f

j

j

tsT

n

sTn

n

de

Fj

ttf

eTnfTntTnf

tfsF

L

LL

tse

ofPoleststs

FofPoles

tsts

n

nsT

e

FResd

e

F

j

e

FResd

e

F

j

eTnf

sF

1

1

0

*

112

1

112

1

2

1

Poles of

Tse 1

1

Poles of

F

planes

Tnsn

2

j

j

0s

Laplace Transforms

The signal f (t) is sampled at a time period T.

12

R

R

Poles of

Tse 1

1

Poles of

F

plane

Tnsn

2

j

j

0s

Z Transform

tf

0n

T Tntt

0

*

n

T TntTnfttftf

tf *

tfT t

SOLO

Sampling and z-Transform (continue – 1)

nnTse

nts

T

njs

T

njs

e

ofPolests

T

njsF

TeT

Tn

jsF

T

njsF

eT

njs

e

FRessF

ts

n

ts

212

lim

2

1

2

lim1

1

2

21

1

*

Poles of

F

j

0s

T

2

T

2

T

2

Poles of

*F plane

js


The poles of are given by tse 1

1

T

njsnjTsee n

njTs 221 2

n T

njsF

TsF

21*

Z Transform

SOLO

F F-1

frequency-B/2 B/2B

F F-1

-B/2 B/2

B

1/Ts-1/Ts frequency

Sample

Sampling a function at an interval Ts (in time domain)

Anti-aliasing filters is used to enforce band-limited assumption.

causes it to be replicated at 1/ Ts intervals in the other (frequency) domain.


Bandlimited Continuous Time Signal

1/B sec

ampl

itud

e

time (sec)

-0.4

-0.2

0.2

0

0.4

0.6

0.8

1

0 5 10 15-15 -10 -5

Discrete-Time (Sampled) Signal

ampl

itud

e

sample

-0.4

-0.2

0.2

0

0.4

0.6

0.8

1

0 10 20-20 -10

Z Transform

tf

0n

T Tntt

0

*

n

T TntTnfttftf

tf *

tfT t

SOLO


0z

planez

Poles of

zF

C


The z-Transform is defined as:

iF

iF

iiF

Ts

FofPoles

T

F

n

n

ze

ze

F

zTnf

zFsFtf

1

0*

1

lim:Z

00

02

1 1

n

RzndzzzFjTnf

fCC

n

Z Transform

SOLO


0

* 21

n

nsT

n

eTnfT

njsF

TsF

We found

The δ (t) function we have:

1

dtt fdtttf

The following series is a periodic function: n

Tnttd :

therefore it can be developed in a Fourier series:

n

n

n T

tnjCTnttd 2exp:

where: T

dtT

tnjt

TC

T

T

n

12exp

12/

2/

Therefore we obtain the following identity:

nn

TntTT

tnj 2exp

Second Way

Z Transform

dttjtftfF 2exp:2 F

0

* 21

n

nsT

n

eTnfT

njsF

TsF

dtjFFtf 2exp2:2-1F

SOLOSampling and z-Transform (continue – 5)

We found

Using the definition of the Fourier Transform and it’s inverse:

we obtain

dTnjFTnf 2exp2

0

111

0

* exp2exp2expnn

n sTndTnjFsTTnfsF

111

* 2exp22 dTnjFjsFn

nn T

nF

Td

T

n

TFjsF 2

1122 111

*

We recovered (with –n instead of n)

n T

njsF

TsF

21*

Second Way (continue)

Making use of the identity: with 1/T instead of T

and ν - ν 1 instead of t we obtain:

nn T

n

TTnj 11

12exp

nn

TntTT

tnj 2exp

Z Transform

Z TransformSOLO

Properties of Z-Transform Functions

Z - Domaink - Domain

kf

ffk

k rzrzkfzF0

1

ii ff

M

iii rzrzFc minmax

1

Linearity

M

iii kfc

1

2 ,2,10 kkfmkf zFz mShifting

mkf

m

k

kmm zkfzFz1

mkf

m

k

kmm zkfzFz1

1kf 0fzFz

3 Scaling kfak

ffk

krazrazakfzaF

0

11

Z TransformSOLO

Properties of Z-Transform Functions (continue – 1)

4 Periodic Sequence kf

1111 ffN

N

rzrzFz

z

N = number of units in a period

Rf1- ,+ = radiuses of convergence in F(1) (z)

F(1) (z) = Z -Transform of the first period

5 Multiplication by k kfk

ff rzrzd

zFdz

6 Convolution

0

:m

mkhmfkhkf hfhf rrzrrzHzF ,min,max

7 Initial Value zFfz

lim0

8 Final Value existsfifzFzkfzk

1limlim1


kf

ffk

k rzrzkfzF0

Z TransformSOLO

Properties of Z-Transform Functions (continue – 2)

9 Complex Conjugate kf * ff rzrzF **

10 Product khkf hfhf

C

rrzrrz

zdzHzF

j,1,

2

1 1

12 Correlation

1,1,2

1 11

0

krrzrr

z

zdzzHzF

jkmhmfkhkf hfhf

C

k

m

11 Parceval’s Theorem

hfhf

Ck

rrzrrz

zdzHzF

jkhkf ,1,

2

1 1

0


kf

ffk

k rzrzkfzF0

Z TransformSOLO

Table of Z-Transform Functions


kf f

k

k RzzkfzF

0

1

mkf 110 11 mfzfzfzFz mm 2

mkf zFz m3

kfkfkf 1: 01 fzzFz 4

kfkfkfkf 122:2 1021 2 fzfzzzFz 5

kf3 2130331 23 fzfzzfzzzzFz 6

272

SOLOMellin Transform

0

1 xdxfxsFxf sMM

We can get the Mellin Transform from the two side Laplace Transform

Robert Hjalmar Mellin ( 1854 – 1933)

xdxfesFxf sx

2LL2

10

11

0

1

sFxdxfxxdxfxxxfx ssMM

ic

ic

s sdsFxi

x M1- fsfM

2

1

Example:

sxdexe xsx

0

1M

xexf

273

SOLO

Mellin Transform (continue – 1)

0

1 xdxfxsFxf sMM

Relation to Two-Sided Laplace Transformation

Robert Hjalmar Mellin ( 1854 – 1933)

tdexdex tt ,

Let perform the coordinate transformation

tdeeftdeeftdeefesF tsttstttst

0

1M

After the change of functions teftg :

tdetgsGtdeefsF tstst

2LM

Inversion Formula

xfefsdxsFi

sdesGi

tgxe

tic

ic

sexic

ic

tstt

ML

L

2

12

2

1

2

1

c

t

cxdssFxRC

s 0M

RC

R

Mellin Transform

274

SOLO

Properties of Mellin Transform (continue – 2)

fkk

k

k

fkk

k

k

fz

fk

kk

fa

f

f

s

SszsFstftd

dt

sksksks

SkszsFkstftd

d

SzszsFCztft

SssFsd

dtft

SsasFaRatf

SsFaataf

SsFtf

HolomorphyofStriptdtftsFtftf

M

M

M

M

M

M

M

MM0t,

1

11:

1

,

ln

0,,

0,11

1

0

1

Original Function Mellin Transform Strip of Convergence

Mellin Transform

275

SOLO

Properties of Mellin Transform (continue – 3)

21

0

21

1

0

1

0

1

//

1

1

11:

1

11:

1

ff

t

t

k

fkk

k

kk

k

fkkk

k

k

f

s

SSssFsFxxdxtfxf

sFsxdxf

sFsxdxf

kssss

SssFstftd

dt

sksksks

SssFkstfttd

d

SsFtf

HolomorphyofStriptdtftsFtftf

M2M1

M

M

M

M

M

MM0t,

Original Function Mellin Transform Strip of Convergence

Mellin Transform

276

Hilbert Transform SOLO

F (u) a analytical function in the right half u plan including infinity.

According to Cauchy theorem:

C

dzsz

zF

jsF

2

1s

R

1C

2C

s

*s

*s

complexplane

complexplane

Let take the point –s* , where s* is the complex conjugate of s. Since –s* is outside thecontour C, we have

C

dzsz

zF

j *2

10

By adding and subtracting those two relations we obtain:

CC

dzzFszszj

dzzFszszj

sF*

11

2

1

*

11

2

1

where C = C1 + C2 is a closed curve composed by- C1 a semicircle in the right half plane- C2 a straight line on the imaginary axis of the complex plane


277

SOLO

Let compute the integrals:s

R

1C

2C

s

*s

*s

complexplane

complexplane

along C1, assuming R → ∞ we have

jRszsz exp

1

*

11

CC

dzzFszszj

dzzFszszj

sF*

11

2

1

*

11

2

1

djRjzd exp

0exp22

1lim

*

11

2

1 2

2

1

1

atanalyticF

RC

FdjRFjj

dzzFszszj

I

0*

11

2

1

1

3

C

dzzFszszj

I

along C2, assuming R → ∞ we have

j

j

vjz

jsjs

j

jC

vdvjFv

vjzdzF

szsz

ssz

jdzzF

szszjI

22*

2

1

*

*2

2

1

*

11

2

1

2

j

j

vjz

jsjs

j

jC

vdvjFv

dzzFszsz

ss

jdzzF

szszjI

22*

4

1

*

*

2

1

*

11

2

1

2

j

j

j

j

vdvjFv

vdvjFv

vjsF

2222

11

Hilbert Transform

278

SOLO

Let write

j

j

j

j

vdvjFv

FvdvjFv

vjsF

2222

11

jFjjFjF ImRe

j

j

j

j

vdjFjjFv

vdjFjjFv

vjjFjjF

ImRe1

ImRe1

ImRe

22

22

We obtain

By equaling the real and imaginary parts we obtain

j

j

j

j

vdvjFv

vdvjFv

vjF Re

1Im

1Re

2222

j

j

j

j

vdvjFv

vdvjFv

vjF Im

1Re

1Im

2222

From those relation we can see that if F (s) is analytic in the right half plane, then it is enough to know it’s value on the imaginary axis to compute F (s) in the entire right half plane.

Hilbert Transform

279

SOLO

We are interested in cases when σ = 0, i.e. points on the imaginary axis. In this case:

j

j

j

j

vdvjFv

vdvjFv

vjF Re

1Im

1Re

2222

j

j

j

j

vdvjFv

vdvjFv

vjF Im

1Re

1Im

2222

It seams that we have a singular point ν = ω, on the path of integration, but we will see how this can be taken care.

j

j

vdv

vjFjF

Im1

Re

j

j

vdv

vjFjF

Re1

Im

Hilbert Transform

280

SOLO

Suppose that F (z) is an analytic functionon the lower (or upper) half complex plane . Re

Im

R

RR

'C

C

We can write

R

CRC

djF

djF

djF

djF

djF

'

0

0lim

'

C

djF

Now

jFjdjFje

edjF

djFd

jF

C

j

j

CC

2

Therefore

R

RR

djF

djFj

jF

0limlim

Define Cauchy Principal Value

R

R

R

R

dqdqdqPV

0lim:

d

jFPV

jjF

From the development we can see that the limit exist and are finite, therefore we removed the singularity at ν = ω. Augustin Louis Cauchy

) 1789-1857(

Hilbert Transform

281

SOLO

Suppose that F (z) is an analytic functionon the lower (or upper) half complex plane .

We can write

dvjF

PVj

dvjF

PVjFjjFjFReIm1

ImRe

Comparing real and imaginary parts we obtain

jFdvjF

PVjF

jFdvjF

PVjF

ReRe1

Im

ImIm1

Re

H

H

Where H stands for Hilbert Transform.

d

jFPV

jjF

Re Im

R

RR

'C

C

David Hilbert1862 - 1943


Hilbert Transform

282

SOLOReferences

[2] Churchill, R.V., “Complex Variables and Applications”, McGraw-Hill, Kõgakusha’ 1960

[3] Spiegel, M.R., “Complex Variables with an introduction to Conformal Mapping

and its applications”, Schaum’s Outline Series, McGraw-Hill, 1964

[4] Hauser, A.A., “Complex Variables with Physical Applications”, Simon & Schuster, 1971

[5] Fisher, S.D., “Complex Variables”, Wadsworth & Brooks/Cole Mathematics Series, 1986

Complex Variables

[6] Tristan, N., “Visual Complex Analysis”, Clarendon Press, Oxford, 1997

[1] E.C. Titchmarsh, “The Theory of Functions”, Oxford University Press, 2nd Ed., 1939

http://www.ima.umn.edu/~arnold/complex.html

SOLO

References (continue - 1)

Complex Variables

F.B. Hildebrand, “Advanced Calculus for Applications”, 2nd Ed., Prentice Hall, 1976, Ch.10, “Functions of a Complex Variable”

http://en.wikipedia.org/wiki/

G.B. Arfken, H.J. Weber, “Mathematical Methods for Physicists”, Academic Press, Fifth Ed., 2001

Sokolnikoff, I.S., Redheffer, R.M., “Mathematics of Physics and Modern Engineering”, 2nd Ed., McGraw Hill, Kõgakusha, 1966

Walter Rudin, “Real and Complex Analysis”, 2nd Ed., McGraw Hill,1974

Marco M. Peloso, “Complex Analysis”, January 21, 2011, University of Milano

http://www.math.umn.edu/~garrett/m/complex/

Wilhelm Schlag, “A Concise Course in Complex Analysis and Riemann Surfaces”, University of Chicago

Alexander D. Poularikas, Ed., “Transforms and Applications Handbook”, 3th Edition, CRC Press, 2010

SOLO

References (continue -2)

Complex Variables

S. Hermelin, “Fourier Transform”

S. Hermelin, “Gamma Function”

S. Hermelin, “Primes”

S. Hermelin, “Hilbert Transform”

S. Hermelin, “Z Transform”

April 9, 2023 285

SOLO

TechnionIsraeli Institute of Technology

1964 – 1968 BSc EE1968 – 1971 MSc EE

Israeli Air Force1970 – 1974

RAFAELIsraeli Armament Development Authority

1974 – 2013

Stanford University1983 – 1986 PhD AA

Complex Variables

286

SOLO

Laplace Fields (general three dimensional)

Vector Analysis

A vector field is said to be a Laplace Field if rAA

0 rA

In this case we have

and

022

00 2

AAAAA

0 rA

Harmonic Functions

A continuous function φ with continuous first and second partial derivatives is saidto be harmonic if it satisfies Laplace’s Equation 02 Properties of Harmonic Functions

Pierre-Simon Laplace1749-1827

022 2

1 0

S

dSn

n

iiSS

1

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

SS

dSn

dSn

General Three Dimensional Complex function

If φ ‘(z) is analytical inside and on C

0C

dzzd

zd Cauchy’s Th.

C

R

If φ ,φ’, ψ, ψ’ are analytical inside and on C

CCC

dzzd

ddz

zd

ddz

zd

d 0

seeVector Analysis.ppt

287

SOLO Vector Analysis

Harmonic Functions (continue 1)

A continuous function φ with continuous first and second partial derivatives is saidto be harmonic if it satisfies Laplace’s Equation 02 Properties of Harmonic Functions (continue – 1)

n

iiSS

1

iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

3 A function φ harmonic in a volume V can be expressed in terms of the function and its normal derivative on the surface S bounding V.

S SFSF

F dSrrnnrr

Tr

11

4

where

VoutsidendSndSSonF

VinFT

11

2

1

1

General Three Dimensional Complex function

If φ (z) is analytic inside and on a simple closed curve C and a is any point inside C then

C

dzaz

z

ia

2

1

Cauchy’s Integral Formula

C

x

y

R a


288




RS

dSn

1

V

Fr

Sr

FSF rrR

4 If the surface S is a sphere SR of radius R with center at then

RS

RF dSR

r

24

1

Fr If f (z) is analytic inside and

on a circle C of radius r andcenter at z = a, then

Complex functionGeneral Three Dimensional

2

02

1

2

1

dera

dzaz

z

ia

ieraz

C

i

C

x

y

a

r


289




RS

dSn

1

V

Fr

Sr

FSF rrR

5 If φ is harmonic in a volume V bounded by the surface S and if φ = c = constantat every point on S, then φ = c at every point of V.


Gauss’ Mean Value Theorem

If φ (z) is analytic inside and on a closed curve C and φ (z) =c =constant at every point on C, then φ (a) = c at every point inside C, i.e.,

Cinsidezcdc

dzazi

cdz

az

z

ia

ieraz

CC

2

02

1

22

1

C

x

y

R a

If φ is harmonic in a region V bounded by a surface S and ∂ φ/∂ n = 0 at every point of S, then φ = constant at every point of V.

6


290




A non-constant function φ harmonic in a region V can have neither a maximum nor a minimum in V. S

dSn

1

V

Fr

Sr

SF rrr

SF

7 Maximum Modulus Theorem

If f (z) is analytic inside and on a simple closed curve C and is not identically equal to a constant, then the maximum value of | f (z) | occurs on C.


Minimum Modulus Theorem

If f (z) is analytic inside and on a simple closed curve C and f (z) ≠ 0inside C then | f (z) | assumes its minimum value on C.


291




8 If φ1 and φ2 are two solutions of Laplace’s equation in a volume V whose normal derivatives take the same value ∂ φ1/∂ n = ∂ φ2/∂ n on the surface S bounding V, then φ1 and φ2 can differ only by a constant.

S

dSn

1

V

Fr

Sr

FSF rrr

0

Sn

Table of Contents


If φ1 and φ2 are two analytic functions inside a curve C whose derivatives takethe same value ∂ φ1/∂ n = ∂ φ2/∂ n onC, then φ1 and φ2 can differ only by a constant.

C

C

dzaz

z

ia

dzaz

z

ia

'

2

1'

'

2

1'

22

11

Cinsideaaa '' 21

Cinsideaconstaa 21

Cona

zz

'' 21


292


Blaschke Products

Wilhelm Johann Eugen Blaschke

(1885 - 1962,)

A sequence of points (an) inside the unit disk is said to satisfy the Blaschke Condition when

Given a sequence obeying the Blaschke Condition, the Blaschke Product is defined as

provided a n≠ 0. Here an* is the complex conjugate of an. When an = 0 take B(0,z) = z.

The Blaschke Product B(z) defines a function analytic in the open unit disc, and zero exactly at the an (with multiplicity counted): furthermore it is in the Hardy class H∞.

The sequence of an satisfying the convergence criterion above is sometimes called a Blaschke Sequence.

n

na1

n

znB

n

n

n

n

za

za

a

azB

,

*1

mathematics and history of complex variables

Science

c b d i b c

q q q multiplicationdivisionz

dc d i d c c d c

r cos q

r q p r q q2z

c b d

zr q q r qy z

complex numbers z