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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
SOLO Complex Variables
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
SOLO Complex Variables
Table of Contents (continue - 2)
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
SOLO Complex Variables
Table of Contents (continue - 3)
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
SOLO Complex Variables
Douglas N. Arnold
Gamma FunctionBernoulli Numbers
Fourier Transform
Laplace Transform
Z Transform
Mellin Transform
Hilbert Transform
Zeta Function
Table of Contents (continue - 5)
Applications of Complex Analysis
References
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
SOLO Complex Variables
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
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10
SOLO Complex Variables
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
SOLO Complex Variables
Fundamentals Operations with Complex Numbers z = x + i y
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
partImaginaryypartRealxyixz
12
SOLO Complex Variables
Fundamentals Operations with Complex Numbers z = x + i y
1,
,2
i
ArgumentModuluse
partImaginaryypartRealxyixz
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
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13
SOLO Complex Variables
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
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14
SOLO Complex Variables
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
SOLO Complex Variables
History of Complex Numbers
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
SOLO Complex Variables
History of Complex Numbers
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
SOLO Complex Variables
History of Complex Numbers
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
SOLO Complex Variables
History of Complex Numbers
Girolamo Cardano1501 - 1576
Nicolo Fontana Tartaglia1500 - 1557
Solution of cubic equation x3 + b x2 +c x +d = 0
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
SOLO Complex Variables
History of Complex Numbers
Solution of cubic equation x3 + b x2 +c x +d = 0
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
SOLO Complex Variables
History of Complex Numbers
Solution of cubic equation x3 + b x2 +c x +d = 0
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
SOLO Complex Variables
History of Complex Numbers
Solution of cubic equation x3 + b x2 +c x +d = 0
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
SOLO Complex Variables
History of Complex Numbers
Solution of cubic equation x3 + b x2 +c x +d = 0
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
Girolamo Cardano1501 - 15763
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
SOLO Complex Variables
History of Complex Numbers
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
SOLO Complex Variables
History of Complex Numbers
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
SOLO Complex Variables
History of Complex Numbers
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.
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28
SOLO Complex Variables
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.
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29
SOLO Complex Variables
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
SOLO Complex Variables
Analytic, Holomorphic, MeromorphicFunctions
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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
SOLO Complex Variables
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:
Augustin Louis Cauchy ) 1789-1857(
Georg Friedrich BernhardRiemann
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
SOLO Complex Variables
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)
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33
SOLO Complex Variables
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
SOLO Complex Variables
Singular Points A point at which f (z) is not analytic is called a singular point. There are various typesof singular points:
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
SOLO Complex Variables
Singular Points A point at which f (z) is not analytic is called a singular point. There are various typesof singular points:
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.
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36
SOLO Complex Variables
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
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37
SOLO Complex Variables
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
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38
SOLO Complex Variables
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.
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39
SOLO Complex Variables
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
SOLO Complex Variables
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
SOLO Complex Variables
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
SOLO Complex Variables
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:
Return to Table of Contents
43
SOLO Complex Variables
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.
Return to Table of Contents
44
SOLO Complex Variables
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.
Augustin Louis Cauchy ) 1789-1857(
Return to Table of Contents
45
SOLO Complex Variables
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:
Augustin Louis Cauchy ) 1789-1857(
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
SOLO Complex Variables
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
SOLO Complex Variables
Proof of Cauchy-Goursat Theorem (continue – 2)
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
SOLO Complex Variables
Proof of Cauchy-Goursat Theorem (continue – 3)
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
Return to Table of Contents
49
SOLO Complex Variables
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:
Return to Table of Contents
50
SOLO Complex Variables
Cauchy’s Integral Formulas
Augustin Louis Cauchy ) 1789-1857(
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
SOLO Complex Variables
Cauchy’s Integral Formulas for the n Derivative of a Function
Augustin Louis Cauchy ) 1789-1857(
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.
Return to Table of Contents
52
SOLO Complex Variables
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
53
SOLO Complex Variables
Cauchy’s Inequality
Augustin Louis Cauchy ) 1789-1857(
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
Return to Table of Contents
54
SOLO Complex Variables
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
SOLO Complex Variables
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.
Return to Table of Contents
56
SOLO Complex Variables
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
SOLO Complex Variables
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
SOLO Complex Variables
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
SOLO Complex Variables
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.
Return to Table of Contents
60
SOLO Complex Variables
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
Return to Table of Contents
61
SOLO Complex Variables
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
SOLO Complex Variables
Poisson’s Integral Formulas for a Circle
Siméon Denis Poisson1781-1840
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
Return to Table of Contents
63
SOLO Complex Variables
Poisson’s Integral Formulas for a Half Plane
Siméon Denis Poisson1781-1840
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
SOLO Complex Variables
Poisson’s Integral Formulas for a Half Plane
Siméon Denis Poisson1781-1840
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
Return to Table of Contents
65
SOLOInfinite Series
Given a series:
Theorems of Convergence of Sequences and 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
Augustin Louis Cauchy ) 1789-1857(
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:
Theorems of Convergence of Sequences and 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
Return to the Table of Content
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:
Theorems of Convergence of Sequences and Series
n
iin uS
1
Infinite Series
68
SOLO
Convergence Tests
Cauchy Root Test
Augustin Louis Cauchy ) 1789-1857(
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:
Theorems of Convergence of Sequences and Series
n
iin uS
1
Infinite Series
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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:
Theorems of Convergence of Sequences and Series
n
iin uS
1
Infinite Series
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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:
Theorems of Convergence of Sequences and Series
n
iin uS
1
Infinite Series
Return to the Table of Content
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:
Theorems of Convergence of Sequences and Series
n
iin uS
1
Infinite Series
72
SOLO
Convergence Tests
Kummer’s Test (continue – 1)
Consider a Series of positive terms ui and a sequence of positive constants ai.
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:
Theorems of Convergence of Sequences and Series
n
iin uS
1
Infinite Series
73
SOLO
Convergence Tests
Kummer’s Test (continue – 2)
Ernst Eduard Kummer(1810 – 1893)
Consider a Series of positive terms ui and a sequence of positive constants ai.
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:
Theorems of Convergence of Sequences and Series
n
iin uS
1
Infinite Series
Return to the Table of Content
74
SOLO
Convergence Tests
Raabe’s Test
Proof:
In Kummer’s Test choose an = n and P = C + 1.
Given a series:
Theorems of Convergence of Sequences and Series
n
iin uS
1
Infinite Series
Return to the Table of Content
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:
Theorems of Convergence of Sequences and Series
n
iin uS
1
Infinite Series
76
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 (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:
Theorems of Convergence of Sequences and Series
n
iin uS
1
Infinite Series
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77
SOLO Complex Variables
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
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79
SOLO Complex Variables
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
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80
SOLO Complex Variables
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
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81
SOLO Complex Variables
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
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SOLO Complex Variables
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
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SOLO Complex Variables
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
SOLO Complex Variables
Uniformly Convergent Series of Analytic FunctionsC
x
y
R
z0
z
w
r
P0D
Proof (continue – 1):
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.
SOLO Complex Variables
Uniformly Convergent Series of Analytic FunctionsC
x
y
R
z0
z
w
r
P0D
Proof (continue – 2):
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
SOLO Complex Variables
Uniformly Convergent Series of Analytic FunctionsC
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.
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87
SOLO Complex Variables
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
SOLO Complex Variables
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
SOLO Complex Variables
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
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 (continue – 2):
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
SOLO Complex Variables
Taylor’s Series (continue – 3)
n
n
zn
fz
fzffzf
!
0
!2
0''0'0 2
Power Series
Brook Taylor1685 - 1731
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
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91
SOLO Complex Variables
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
SOLO Complex Variables
Laurent’s Series (continue - 1)
Power SeriesPierre Alphonse Laurent
1813 - 1854
C1
x
y
RC2R2
R1
z0
z z'
r
Proof (continue – 1):
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
SOLO Complex Variables
Laurent’s Series (continue - 2)
Power SeriesPierre Alphonse Laurent
1813 - 1854
C1
x
y
RC2R2
R1
z0
z z'r
Proof (continue – 1):
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
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94
SOLO Complex Variables
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.
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95
SOLO Complex Variables
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
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96
SOLO Complex Variables
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
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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
SOLO Complex Variables
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
q.e.d.Return to the Table of Content
99
SOLO Complex Variables
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 :
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100
SOLO Complex Variables
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
SOLO Complex Variables
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
SOLO Complex Variables
Jensen’s Formula (continue – 2)
Johan Ludwig William Valdemar Jensen
(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
SOLO Complex Variables
Jensen’s Formula (continue – 3)
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
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
SOLO Complex Variables
Jensen’s Formula (continue – 4)
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.
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105
SOLO Complex Variables
Poisson-Jensen’s Formula for a Disk
Johan Ludwig William Valdemar Jensen
(1859 – 5 1925)
Poisson Formula states:
Siméon Denis Poisson1781-1840
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
SOLO Complex Variables
Poisson-Jensen’s Formula for a Disk
Johan Ludwig William Valdemar Jensen
(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
Siméon Denis Poisson1781-1840
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
SOLO Complex Variables
Poisson-Jensen’s Formula for a Disk
Johan Ludwig William Valdemar Jensen
(1859 – 5 1925)
Siméon Denis Poisson1781-1840
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
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108
SOLO Complex Variables
The Residue Theorem, Evaluations of Integral and Series
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.
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109
SOLO Complex Variables
The Residue Theorem, Evaluations of Integral and Series
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
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110
SOLO Complex Variables
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
q.e.d.Return to the Table of Content
111
SOLO Complex Variables
The Residue Theorem, Evaluations of Integral and Series
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
q.e.d.Return to the Table of Content
112
SOLO Complex Variables
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
SOLO Complex Variables
The Residue Theorem, Evaluations of Integral and Series Evaluation of Integrals
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
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114
SOLO Complex Variables
The Residue Theorem, Evaluations of Integral and Series Evaluation of Integrals
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.
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115
SOLO Complex Variables
The Residue Theorem, Evaluations of Integral and Series Evaluation of Integrals
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
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116
SOLO Complex Variables
The Residue Theorem, Evaluations of Integral and Series Evaluation of Integrals
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
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117
SOLO Complex Variables
The Residue Theorem, Evaluations of Integral and Series Evaluation of Integrals
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
SOLO Complex Variables
The Residue Theorem, Evaluations of Integral and Series Evaluation of Integrals
Cauchy’s Principal Value (continue – 1)
Proof (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
SOLO Complex Variables
The Residue Theorem, Evaluations of Integral and Series Evaluation of Integrals
Cauchy’s Principal Value (continue – 2)
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
SOLO Complex Variables
The Residue Theorem, Evaluations of Integral and Series Evaluation of Integrals
Cauchy’s Principal Value (continue – 3)
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
SOLO Complex Variables
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
SOLO Complex Variables
The Residue Theorem, Evaluations of Integral and Series Evaluation of Integrals
Cauchy’s Principal Value (continue – 5)
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
SOLO Complex Variables
The Residue Theorem, Evaluations of Integral and Series Evaluation of Integrals
Cauchy’s Principal Value (continue – 6)
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.
Return to the Table of Content
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
SOLO Complex Variables
The Residue Theorem, Evaluations of Integral and Series Evaluation of Integrals
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:
Return to the Table of Content
127
SOLO Complex Variables
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
SOLO Complex Variables
The Residue Theorem, Evaluations of Integral and Series Summation of Series (continue – 1)
Proof (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
SOLO Complex Variables
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
SOLO Complex Variables
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).
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131
SOLO Complex Variables
The Residue Theorem, Evaluations of Integral and Series Summation of Series
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.
Return to the Table of Content
132
SOLO Complex Variables
The Residue Theorem, Evaluations of Integral and Series Summation of Series
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: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.
Return to the Table of Content
133
SOLO Complex Variables
The Residue Theorem, Evaluations of Integral and Series Summation of Series
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: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.
Return to the Table of Content
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
Return to the Table of Content
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).
Infinite ProductsComplex Variables
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
Infinite ProductsComplex Variables
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.
Infinite ProductsComplex Variables
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.
Infinite ProductsComplex Variables
141
SOLO
Proof (continue – 1):
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
Infinite ProductsComplex Variables
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
Infinite ProductsComplex Variables
Jacques Salomon Hadamard
) 1865– 1963 (
143
SOLO Complex Variables
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
SOLO Complex Variables
The Residue Theorem, Evaluations of Integral and Series Mittag-Leffler’s Expansion Theorem (continue – 1)
C
x
y
RN
1ana
CN
Proof (continue – 1):
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
SOLO Complex Variables
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
SOLO Complex Variables
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
SOLO Complex Variables
The Residue Theorem, Evaluations of Integral and Series Generalization of Mittag-Leffler’s Expansion Theorem
Proof (continue – 1):
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
SOLO Complex Variables
The Residue Theorem, Evaluations of Integral and Series Generalization of Mittag-Leffler’s Expansion Theorem
Proof (continue – 2):
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
SOLO Complex Variables
The Residue Theorem, Evaluations of Integral and Series Generalization of Mittag-Leffler’s Expansion Theorem
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.
Proof (continue – 3):
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
Karl Theodor Wilhelm Weierstrass
(1815 – 11897)
Infinite ProductsComplex Variables
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
The Weierstrass Factorization Theorem
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
Karl Theodor Wilhelm Weierstrass
(1815 – 11897)
Inspired by the fact that
321
1ln&11
321
1ln zz
zz
ez z w have the following
Infinite ProductsComplex Variables
152
SOLO
The Weierstrass Factorization Theorem
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 (continue – 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.
Infinite ProductsComplex Variables
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,
Infinite ProductsComplex Variables
154
SOLO
Genus of the Canonical Product
Infinite ProductsComplex Variables
155
SOLO
The Weierstrass Factorization Theorem
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
Infinite ProductsComplex Variables
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
Infinite ProductsComplex Variables
Jacques Salomon Hadamard
) 1865– 1963 (
157
SOLO
The Hadamard Factorization Theorem
Relation berween Order ρ and Integer Genus p of an Entire Function
Infinite ProductsComplex Variables
Jacques Salomon Hadamard
) 1865– 1963 (
Paul Garrett, “Weirstrass and Hadamard Products”, March 17, 2012, http://www.math.umn.edu/~garrett/
158
SOLO Complex Variables
The Residue Theorem, Evaluations of Integral and Series
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
The Hadamard Factorization Theorem
Return to Gamma F.
Return to Zeta F.
159
SOLO Complex Variables
The Residue Theorem, Evaluations of Integral and Series
Expansion of an Integral Function as an Infinite Product
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
The Hadamard Factorization Theorem
160
SOLO Complex Variables
The Residue Theorem, Evaluations of Integral and Series
Expansion of an Integral Function as an Infinite Product
Proof (continue – 1):
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 .
The Hadamard Factorization Theorem
161
SOLO Complex Variables
The Residue Theorem, Evaluations of Integral and Series
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,…
Expansion of an Integral Function as an Infinite Product
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
Analytic continuation of natural logarithm (imaginary part)
Complex Variables
163
SOLO
Analytic Continuation
Complex Variables
164
SOLO Complex Variables
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
SOLO Complex Variables
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
SOLO Complex Variables
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
SOLO Complex Variables
Conformal Mapping Complex Mapping – Riemann’s Mapping Theorem
In the case that the mapping is done by a complex function, i.e. yixfzfviuw
Georg Friedrich BernhardRiemann
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
SOLO Complex Variables
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
SOLO Complex Variables
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
SOLO Complex Variables
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
argsinargcosargsinargcos
22
2
w
Rww
2
1y
w
Rww
2
1x
22
argsin&argcos
171
SOLO Complex Variables
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
SOLO Complex Variables
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
SOLO Complex Variables
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
argsinargcosargsinargcos
argsinargcosargsinargcos
22
2
w
Rww
2
1y
w
Rww
2
1x
22
argsin&argcos
174
SOLO Complex Variables
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
SOLO Complex Variables
Conformal Mapping
http://www.mathworks.com/company/newsletters/news_notes/clevescorner/sum98cleve.html
2/tt eex
0122 tt exe
2/sinh tt eet
2/12 1ln zzzasinh:zf
176
SOLO Complex Variables
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
SOLO Complex Variables
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
SOLO Complex Variables
Conformal Mapping
http://www.mathworks.com/company/newsletters/news_notes/clevescorner/sum98cleve.html
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
SOLO Complex Variables
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
SOLO Complex Variables
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
SOLO Complex Variables
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
SOLO Complex Variables
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
SOLO Complex Variables
Mobius Transformation
Douglas N. Arnold
10/11 4
t
zit
itzzf t
August Ferdinand Möbius1790 - 1868
184
SOLO Complex Variables
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
SOLO Complex Variables
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
SOLO Complex Variables
Schwarz-Christoffel Mappings
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
SOLO Complex Variables
Schwarz-Christoffel Mappings
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
SOLO Complex Variables
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
Carl Friedrich Gauss(1777 – 1855)
199
SOLO Primes
t
tt
z
tde
tz
0
1
Proof (continue - 1) :
Gamma Function
Other Gamma Function Definitios:
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
!limGauss’ Formula
Euler 1729Gauss 1811
200
SOLO Primes
t
tt
z
tde
tz
0
1
Proof (continue - 2) :
Gamma Function
Other Gamma Function Definitios:
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
!limGauss’ Formula
201
SOLO Primes
t
tt
z
tde
tz
0
1
Let substitute x + 1 for x
Gamma Function
Other Gamma Function Definitios:
1,011
!lim
x
xxnxnx
nnx
x
x
n
n
q.e.d
nxxx
nnx
x
n
1
!limGauss’ Formula
Proof (continue - 3) :
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
Other Gamma Function Definitios:
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
Karl Theodor Wilhelm Weierstrass
(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
Other Gamma Function Definitios:
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
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
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
Proof (continue - 2)
t
tt
z
tde
tz
0
1
Gamma Function
Other Gamma Function Properties:
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
Proof (continue - 3)
t
tt
z
tde
tz
0
1
Gamma Function
Other Gamma Function Properties:
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
Other Gamma Function Properties:
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)
Riemann's Zeta Function
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)
Riemann's Zeta Function
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
Riemann's Zeta Function
228
Re ζ (z) in the original domain, Re z > 1.
Re ζ (z) after Riemann’s extension.
Riemann's Zeta Function
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(.
Riemann's Zeta Function
230
Riemann's Zeta Function
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
_________________________________________________
Riemann's Zeta Function
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
Riemann's Zeta Function
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.
Riemann's Zeta Function
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
Riemann's Zeta Function
235
SOLO
Proof (continue – 1)
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
itoreturnsandzeroencirclesiatstartspaththe
de
izz
zi
i
z
Riemann's Zeta Function
236
SOLO
Proof (continue – 2)
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
itoreturnsandzeroencirclesiatstartspaththe
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.
Riemann's Zeta Function
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
Riemann's Zeta Function
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.
Riemann's Zeta Function
239
SOLO
i
iy
x
i
i
2
00
12
1 0
0
1
itoreturnsandzeroencirclesiatstartspaththe
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
Riemann's Zeta Function
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.
Riemann's Zeta Function
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
Riemann's 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/
Riemann's Zeta Function
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
Riemann's Zeta Function
244
SOLO
Proof
i
iy
x
i
i
2
00
1sin2
1 0
0
1
itoreturnsandzeroencirclesiatstartspaththe
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
Riemann's Zeta Function
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
Riemann's Zeta Function
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
Riemann's Zeta Function
247
SOLO
Hadamard Infinite Product Expansion of Zeta Function
Graph showing the Trivial Zeros, the Critical Strip and the Critical Line of ζ (z) zeros.
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'
Riemann's Zeta Function
248
SOLO
Hadamard Infinite Product Expansion of Zeta Function (continue)
Graph showing the Trivial Zeros, the Critical Strip and the Critical Line of ζ (z) zeros.
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.
Riemann's Zeta Function
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
F FProperties of Fourier Transform
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
Fourier TransformSOLO
tf-1F
F FProperties of Fourier Transform
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
Fourier TransformSOLO
tf-1F
F FProperties of Fourier Transform
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
Fourier TransformSOLO
tf-1F
F FProperties of Fourier Transform
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
Laplace’s TransformSOLO
Properties of Laplace L-Transform (continue – 1)
s - Domaint - Domain
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
Laplace’s TransformSOLO
Properties of Laplace L-Transform (continue – 2)
s - Domaint - Domain
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 signal f (t) is sampled at a time period T.
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.
Sampling and z-Transform (continue – 2)
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
Sampling and z-Transform (continue – 3)
0z
planez
Poles of
zF
C
The signal f (t) is sampled at a time period T.
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
Sampling and z-Transform (continue – 4)
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
Z - Domaink - Domain
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
Z - Domaink - Domain
kf
ffk
k rzrzkfzF0
Z TransformSOLO
Table of Z-Transform Functions
Z - Domaink - Domain
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
Augustin Louis Cauchy ) 1789-1857(
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
Return to Table of Contents
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
seeVector Analysis.ppt
288
SOLO Vector Analysis
Harmonic Functions (continue 2)
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 – 2)
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
seeVector Analysis.ppt
289
SOLO Vector Analysis
Harmonic Functions (continue 3)
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 – 3)
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.
Complex functionGeneral Three Dimensional
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
seeVector Analysis.ppt
290
SOLO Vector Analysis
Harmonic Functions (continue 4)
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 – 4)
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.
Complex functionGeneral Three Dimensional
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.
seeVector Analysis.ppt
291
SOLO Vector Analysis
Harmonic Functions (continue 5)
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 – 5)
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
Complex functionGeneral Three Dimensional
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
seeVector Analysis.ppt
292
SOLO Complex Variables
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