﻿ Complex number study notes tute0065 Complex number study notes tute0065
video list , update 2015-04-05

Complex number: 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 11, 12, 13, 14,
15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33,
Principal argument of z -π<θ≦π
Zeta function: tute0057, tute0058, tute0059, tute0060,
Integration by parts: tute0061
Gamma function: tute0062 tute0063
Taylor Series: tute0064
Why gamma function this way? Γ(s) = ∫[t=0,t=∞]{ts-1*e-t*dt} eq.CG01
Integral Test, Euler's formula,
How to prove eq.CG16? Taylor series expansion equation

```<a name="cplx104a_index">
2015-03-13-17-07 start build tute0065.htm page
and build cplx104a index
save as cplx104a\JacobBains_utube_a40207a.htm
2015-02-07-14-27 done

<a name="cplx104a_index1">
Analysis of a Complex Kind
•by Jacob Bains
•28 videos
•7,887 views
•Last updated on Jun 16, 2014

<a name="cplx104a_index2">
This course provides an introduction to
complex analysis, that is the theory of
complex functions of a complex variable
. We'll start by introducing the complex
plane along with the algebra and geometry
of complex numbers and make our way via
differentiation, integration, complex
dynamics and power series representation
into territories at the edge of what's
known today.
<a name="cplx104a_index3">
This was a MOOC Coursera.org Internet
Course from Winter 2013, through Wesleyan
University.
https://class.coursera.org/complexanalysis-001/wiki/syllabus
This course provides an introduction to
complex analysis, that is the theory of
complex functions of a complex variable
. We'll start by introducing the complex
plane along with the algebra and geometry
of complex numbers and make our way via
d... more

Lecturer is Prof. Petra Bonfert-Taylor
1.1 History of Complex Numbers ; 19:26
1.2 Algebra and Geometry in the Complex Plane ; 30:40
1.3 Polar Representation of Complex Numbers ; 32:32
1.4 Roots of Complex Numbers ; 14:10
1.5 Topology in the Plane ; 20:56

<a name="cplx104aIndex2">
Lecturer is Prof. Petra Bonfert-Taylor
2.1 Complex Functions ; 26:38
2.2 Sequences and Limits of Complex Numbers ; 30:47
2.3 Iteration of Quadratic Polynomials, Julia Sets ; 25:42
2.4 How to Find Julia Sets ; 20:32
2.5 The Mandelbrot Set ; 18:33

<a name="cplx104aIndex3">
Lecturer is Prof. Petra Bonfert-Taylor
3.1 The Complex Derivative ; 34:25
3.2 The Cauchy-Riemann Equations ; 28:49
3.3 The Complex Exponential Function ; 24:44
3.4 Complex Trigonometric Functions ; 21:23
3.5 First Properties of Analytic Functions ; 24:54

<a name="cplx104aIndex4">
Lecturer is Prof. Petra Bonfert-Taylor
4.1 Inverse Functions of Analytic Functions ; 25:28
4.2 Conformal Mappings ; 26:16
4.3 Möbius transformations, Part 1 ; 27:02
4.4 Möbius Transformations, Part 2 ; 17:35
4.5 The Riemann Mapping Theorem ; 15:15

<a name="cplx104aIndex5">
Lecturer is Prof. Petra Bonfert-Taylor
5.1 Complex Integration ; 27:21
5.2 Complex Integration: Examples and First Facts ; 32:29

[[
19:10
…5.3 The Fundamental Theorem of Calculus for Analytic Functions
by Jacob Bains
•10 months ago
•542 views
]]
5.3 The Fundamental Theorem of Calculus for Analytic Functions ; Petra Bonfert-Taylor 19:10

5.4 Cauchy's Theorem and Integral Formula ; 32:36
5.5 Consequences of Cauchy's Theorem and Integral Formula ; 28:11

<a name="cplx104aIndex6">
Lecturer is Prof. Petra Bonfert-Taylor
6.1 Infinite Series of Complex Numbers ; 22:03
6.2 Power Series ; 25:22
6.3 The Radius of Convergence of a Power Series ; 27:53
6.4 The Riemann Zeta Function And The Riemann Hypothesis ; 22:53

Above total 29 files.

<a name="cplx104aIndex7">
Lecturer is Prof. Petra Bonfert-Taylor
Residue theorem, 22:08
branches of inverse function, 19:57
ArgumentPrinciple, 22:53
Differentiability in the Real Sense, 21:21

Above total 29+4=33 files.

<a name="cplx104aIndex8">
…An Invitation to Complex Analysis
by Petra Bonfert-Taylor
•1 year ago
•2,786 views
…In this video we describe a course on complex analysis, including geometry, complex dynamics, and perhaps most importantly, ...
An Invitation to Complex Analysis ;
Petra Bonfert-Taylor 3:00
2015-03-13-18-08 done build cplx104a list

<a name="Taylor_Index">
2015-02-08-21-30 LiuHH access
Taylor Series
•by MrYouMath
•9 videos
•2,037 views
•Last updated on Apr 22, 2014

Taylor Series - 1 - Motivation and Derivation.mp4
●●●

2015-01-01-15-18 start
On
2014-03-25-20-36 Liu,Hsinhan access

Gamma Function
•by MrYouMath
•12 videos
•5,940 views
•1 hour, 48 minutes
Gamma Function - Part 1 - Functional Equation
●●●

Gamma Function - Part 7 - Euler Integral I
●●●
2015-01-01-15-38 stop

2014-10-29-07-56
2014-10-29-08-24
MrYouMath_17_files_Riemann.htm
"Zeta Function - Part 1 - Convergence"
●●●

"Zeta Function - Part 5 - Prime Zeta Function"
●●●

"Zeta Function - Part 8 - Zeta of 2n - Part 1"
●●●

"Zeta Function - Part 11 - Riemann Functional Equation I"
●●●

2015-03-13-20-38 start
Next is study notes of
Analysis of a Complex Kind
1.1 History of Complex Numbers ; 19:26

Lecturer is Prof. Petra Bonfert-Taylor.
Notes recorder is Liu,Hsinhan.
Recorded notes is about 60% match, not exact.
Liu,Hsinhan inserted text is blue Italic.
LiuHH notes may contain error!

<a name="docA002"> Begin video 1 of 33
Welcome to the course "Analysis of a Complex Kind".
This is the first lecture in this course. I will
introduce the course to you tell you some of the
topics which will be covering and give you the
history of complex numbers.

<a name="docA003">
● Prof. Petra Bonfert-Taylor
● Born, raised and educated in Germany (Berlin)
● Ph.D. 1996, Technical University of Berlin.
● Postdoc at University of Michigan.
● Professor at Wesleyan University since 1999

<a name="docA004">
● Complex numbers, their geometry and algebra.

Complex numbers have a real part x and an
imaginary part y.
z=x+i*y --- eq.AC01
In a complex plane, mark x value on horizontal
x axis and mark y value on vertical y axis.
From x value draw a line parallel to y axis,
from y value draw a line parallel to x axis.
Two auxiliary lines meet at point x+i*y.

<a name="docA005">
● Historical explorations
We will talk about some important people who
were important in developing complex analysis.

● Complex dynamics: Mandelbrot set, Julia sets.
We will learn complex dynamics. You probably
have seen this beautiful Mandelbrot set or
zoom into Mandelbrot set. We will be able to
learn how to construct Mandelbrot set and how
to calculate.

This is a Julia set. We will study both of these
in this course.

<a name="docA007">
● Complex functions, continuity, complex
differentiation.
After study complex number, we will study complex
functions. Complex function input a complex number
output to another complex number. We study their
continuity and complex differentiation.
If we talk about one function it takes a complex
argument in the first picture we saw. We know a
complex number can be displayed as a point in
complex real_x and complex imag_y coordinate system.
z=x+i*y. Complex function map a portion of input
complex plane to an output complex plane. We need
two pictures, one for domain (input) and one for
the image (output) We study how to visualize
complex function
<a name="complex2.gif">
ivory background section is Liu,Hsinhan work.
ivory background section is not in Bonfert video.

http://freeman2.com/complex2.gif
2015-03-14-22-18 include complex2.gif
This purple circle is input curve.
Red  curve is output complex sin(z)
Blue curve is output complex cos(z)
This graph draw input and output in one coordinate system.
Following files are in http://freeman2.com/
complex1.gif ; complex2.gif ; complex3.gif ;
complex4.gif ; complex5.gif ; complex6.gif ;
cos_cplx.gif
<a name="docA008">
● Conformal mapping, Möbius transformations and the
Riemann mapping theorem.
We will be able to talk about the most important
theorems in complex analysis. One of them being the
Riemann mapping theorem. You will be able to prove
theorem like that. We will be able to study it.
Riemann mapping theorem says, no matter what shape
a shoe has that does not have any holes. You will
be able find a mapping, analytic function map this
shoe on to the unit disk. We will explain some
details, and learn why this is important theorem.

<a name="docA009">
● Complex integration, Cauchy theory and consequences.
We will learn complex integration and make a formula
like this
```
 f(a) = 1 2πi ∫ γ f(z) z－a dz
--- eq.AC02
eq.AC02 make sense after the course.
width of above equation a403132152
```<a name="docA010">
2015-03-13-22-01 here
We will be able to prove
Fundamental theorem of Algebra
Given any polynomial of degree n, eq.AC03 left side
equation. We will be able to factor this polynomial
into n factors, eq.AC03 right side equation.
This is the Fundamental theorem of Algebra.
```
<a name="docA011">
Fundamental theorem of Algebra
 anzn+an-1zn-1+●●●+a1z+a0 = an(z-z1)(z-z2)●●●(z-zn)
--- eq.AC03
width of above equation a403132158
```2015-03-13-22-06 stop at video 4:29/19:25
2015-03-14-14-48 start
● Power series representation of analytic functions.
Riemann hypothesis.
Prof. Petra Bonfert-Taylor lecture screen show
eq.cz01 and show eq_cz01b below. Liu,Hsinhan include
following equations from tute0057.htm to tute0065.htm
2015-03-14-15-18
```
<a name="RiemannZeta">
2015-03-14-14-56 include next three equations
s be a complex number, example s='1.1+2.2i'
Riemann Zeta function ζ(s), if Real(s)>1
ζ(s)=∑[n=1,∞]{1/n^s} = ∏[p=prime]{p^s/(p^s-1)} --- eq_cz01a
Next is same equation in better form.
2014-10-08-00-19
<a name="czeta_1"> Bernhard Riemann (1859)
if Real(s)>1 , ζ(s) is defined by next equation.

 ζ(s) = n=∞ ∑ n=1 1 ns = p=∞ ∏ p=prime ps ps-1
--- eq.cz01
width of above equation

<a name="docZ011a">
2014-11-05-12-25 next is for entry level reader.
s be a complex number, example s='1.1+2.2i'
<a name="czeta_1a">
complex number s real part MUST BE > 1.0

 n=∞ ∑ n=1 1 ns = 1 1s + 1 2s + 1 3s + 1 4s + 1 5s + ...
--- eq_cz01b
width of above equation
<a name="czeta_1b">

 p=∞ ∏ p=prime ps ps-1 = 2s 2s-1 * 3s 3s-1 * 5s 5s-1 * 7s 7s-1 * 11s 11s-1 * ...
--- eq_cz01c
width of above equation
In eq_cz01b all denominator are positive integers.
In eq_cz01c all denominator are prime numbers.
Both eq_cz01b and eq_cz01c extend to infinity.
2014-11-05-12-42 above is for entry level reader.

<a name="docZ012">
s be a complex number, example s='0.5-3.4i'
complex number s real part MUST BE 0<Real(s)<=1

Riemann Zeta function ζ(s), if 0<Real(s)<=1
ζ(s)=∑[n=1,∞]{-1n+1/n^s}/(1-21-s) --- eq_cz02a
or
ζ(s)={[1-2^(1-s)]^(-1)}*∑[n=1,∞]{-1n+1/n^s} --- eq_cz02b
Next is same equation in better form.
2014-10-08-00-38
do NOT have '=' . Next line, LiuHH added '=' ;

<a name="czeta_2">
if 0<Real(s)<=1 , ζ(s) is defined by next equation.

 ζ(s) = 1 1-21-s * n=∞ ∑ n=1 (-1)n+1 ns
--- eq.cz02
width of above equation

<a name="docZ013">
s be a complex number, example s='-0.78+1.23i'
complex number s real part MUST BE <=0

Riemann Zeta function ζ(s), if Real(s)<=0
ζ(s)=2s*PIs-1*sin(PI*s/2)*Γ(1-s)*ζ(1-s) --- eq_cz03a
or
ζ(s)=2^s*PI^(s-1)*sin(PI*s/2)*Γ(1-s)*ζ(1-s) --- eq_cz03b
here PI=π=3.14159265358979323846264338327950288419716939937510 ...
Next is same equation in better form.
2014-10-08-00-48
do NOT have '=' . Next line, LiuHH added '=' ;

<a name="czeta_3">
if Real(s)<=0 , ζ(s) is defined by next equation.

 ζ(s) = 2s * πs-1 * sin ( π*s 2 ) * Γ(1-s) * ζ(1-s)
--- eq.cz03
width of above equation
above ζ(1-s) is Riemann Zeta function with input 1-s
above Γ(1-s) is complex Gamma function with input 1-s
Because if Real(s)<=0 , then Real(1-s)>=1

<a name="docZ014">
Liu,Hsinhan access next video and find above three equations.
MrYouMath: "Zeta Function - Part 12 - Riemann Functional Equation II"

eq.cz01 , eq.cz02 , eq.cz03 from next URL
singingbanana "The Riemann Hypothesis"
From time 11:40 to 13:05 screen has three equations .
total minute/second is 19:35 .

2014-10-29-19-29 done include and modify in tute0057.htm.
2015-03-14-15-14 done include and modify in tute0065.htm.
```<a name="docA012">
2015-03-14-15-20 here
We will study power series representation of
analytic functions. and we even be able to talk
about the Riemann hypothesis which is an open
conjecture by Bernhard Riemann (1859). To this
day it has not been proved. Zeta Function is
related to prime numbers. We will study how this
is related to prime numbers. Riemann hypothesis
is related to distribution to prime numbers.

<a name="docA013">
Brief History of Complex Numbers
x2=m*x+b --- eq.AC04

● Solutions are
```

 x = m 2 ± √ [ m2 4 + b ]
--- eq.AC05
width of above equation a403141548
```<a name="docA014">
2015-03-14-15-48 here
eq.AC05 represent two points, intersection of
y=x2 --- eq.AC06
and
y=m*x+b --- eq.AC07
In y=x2, x2 come from eq.AC04 equality left side.
In y=m*x+b , m*x+b come from eq.AC04 right side.
y=m*x+b is x first power straight line.
y=x2 is x second power curve. 2015-03-14-15-53

<a name="docA015">
Let us start talk about Brief History of Complex
x2=m*x+b --- eq.AC04
Its solution is
x=(m/2) ± √[(m2/4)+b] --- eq.AC05
If you look at eq.AC04 by itself, it really
represent the intersection of graph y=x2 and
another graph y=m*x+b this is just a line.
eq.AC04 says curve y=x2 equal line y=m*x+b
Where curve y=x2 and line y=m*x+b intersect
is the equality points.
Two intersection x values are given by eq.AC05.
In eq.AC05 take '±' as '+' that is one answer,
take '±' as '-' that is second answer.
That was discovered long time ago. Greek knew
this formula. video 6:14/19:25
<a name="docA017">
Solutions: x=(m/2) ± √[(m2/4)+b] --- eq.AC05
● What if (m2/4)+b < 0 ?
● In particular
x2=-1 --- eq.AC08
has no real solution.
● It is often argued that x2=-1 led to i=√-1 .
● But ... Historically, no interest in non-real
solutions. Since the graphs of y=x2 and y=m*x+b
simply do not intersect in that case.

<a name="docA018">
Here again I wrote down the solution eq.AC05.
What happen if the number under the square root was
negative? Then you would not be able get a solution.
In particular if your equation use m=0 and b=-1
then eq.AC04 become x2=-1 it has no real solution.
It is often argued that this x2=-1 led to invention
of complex number i=√-1 . But, historically, there
is no interest in non-real solutions.
<a name="docA019">
Because in x2=-1 case, simply curve y=x2 and
line y=m*x+b do not intersect.
y=x2 is a parabola pass (0,0) and above x axis.
y=-1 is a horizontal line pass (0,-1) below x axis.
These two graphs simply do not intersect. They have
nothing to do with each other. We do not need to
make up a complex solution x=√-1 . There is simply
historically no interest.

So, where complex number come really important?
It is with cubic equations those were the real
reason for the importance of complex number.

<a name="docA020">
History
● Cubic equations were the real reason. Consider
x3=p*x+q --- eq.AC09
● Represents intersection of
y=x3 --- eq.AC10
and
y=p*x+q --- eq.AC11
● There always must be a solution.

Let us look at cubic equation eq.AC09. In eq.AC09
there is no quadratic x2 term here. It is just an
example of cubic equation. For eq.AC09 we look at
y=x3 --- eq.AC10
and
y=p*x+q --- eq.AC11
Solution of eq.AC09 represent the intersection of
cubic curve y=x3 and line y=p*x+q . You can see
that cubic curve y=x3 go all the way from plus
infinity to minus infinity. No matter what line
you draw, y=p*x+q always intersect with y=x3.
There always must be a solution for cubic eq.AC09

<a name="docA022">
Solution to cubic equation
● Del Ferro (1465-1526) and Tartaglia (1499-1577)
followed by Cardano (1501-1576) showed that
x3=p*x+q --- eq.AC09
has a solution given by
```

 x = ∛ { √ [ q2 4 － p3 27 ] ＋ q 2 } － ∛ { √ [ q2 4 － p3 27 ] － q 2 }
--- eq.AC12
width of above equation a403141726
```2015-03-14-17-28 here video 9:51/19:25
<a name="docA023">
2015-03-14-19-00 start
● Try it out for
x3=-6*x+20 --- eq.AC13

Del Ferro and Tartaglia followed by Cardano showed
x3=p*x+q --- eq.AC09
has solution given by the complicated formula
eq.AC12 . Let us try eq.AC13 First we notice
<a name="docA024">
what is p? what is q? Compare eq.AC13 with eq.AC09
get p=-6 and q=20. Plug p and q into eq.AC12 get
q2/4 = 20*20/4=100
p3/27 = (-6)*(-6)*(-6)/27=-8
q/2=20/2=10 then eq.AC12 has
x = ∛{√[100-(-8)]+10} - ∛{√[100-(-8)]-10}
x = ∛{√[108]+10} - ∛{√[108]-10}
<a name="docA025">
You get your calculator out, find the solution is
x=2
in http://freeman2.com/complex4.htm#Box3JS
copy next seven lines to "Box3, JS command"
[[
aa=csqrt(108)
ab=aa+10
ac=aa-10
b1=cpowf(ab,1/3)
b2=cpowf(ac,1/3)
cc=b1-b2
cc
]]
then click [eval Box3] Box4 output next number
1.9999999999999995,0
this number is 2+0i or just 2.

Indeed, in eq.AC13, 23 is 8. -6*x is -6*2=-12
8 = -12 + 20 is true.
2 is indeed the solution of eq.AC13.

Bombelli's Problem
● About 30 years after the discovery of this
formula. Bombelli (1526-1572) considered the
equation
x3=15*x+4 --- eq.AC14
● Plug in p=15 and q=4 into the formula get
x = ∛{2+√-121} + ∛{2-√-121} --- eq.AC15
● Bombelli had a "wild thought"

<a name="docA027">
But thirty years after the discovery of eq.AC12
Another Italian mathematician by the name Bombelli
considered the following cubic equation
x3=15*x+4 --- eq.AC14
So p=15 and q=4 So now plug in to eq.AC12 you
find
x = ∛{2+√-121} + ∛{2-√-121} --- eq.AC15
The square root of minus 121 is a problem. You
cannot take square root of a negative number.
Then eq.AC15 does not give you a solution. However
we know a cubic equation MUST have a solution,
which you can not find it using eq.AC12 Bombelli
had a really wild thought. He think if take √-121
as a regular number and carry calculation, may be
√-121 will cancel out and get a real solution.

<a name="docA028">
Bombelli's Idea
● Bombelli discovered that
∛{2+√-121} = 2+√-1  --- eq.AC16
and
∛{2-√-121} = 2-√-1  --- eq.AC17
● These clearly add up to 4, the desired solution.
● Check it out
(2+√-1)3 = 2+√-121 --- eq.AC18
and
(2-√-1)3 = 2-√-121 --- eq.AC19

<a name="docA029">
∛{2+√-121} = 2+√-1  --- eq.AC16
and
∛{2-√-121} = 2-√-1  --- eq.AC17
then you add up eq.AC16 and eq.AC17 as you have
to in the formula eq.AC15 No matter what √-1 is,
plus √-1 and minus √-1 cancel out. What left is
2+2=4 and 4 is the solution of eq.AC14 Now let
us check it out
(2+√-1)3 = 2+√-121 --- eq.AC18
Is eq.AC18 true?

<a name="docA030">
How to cube one equation?
(a+b)3 = a3+3*a2*b+3*a*b2+b3 --- eq.AC20
To use eq.AC20 right here in eq.AC18
(2+√-1)3 = 2+√-121 --- eq.AC18
(2+√-1)3 = 23+3*22*√-1+3*2*√-12+√-13 --- eq.AC21
Because
√-12=-1 --- eq.AC22
and
√-13 = √-12*√-1 = -√-1 --- eq.AC23
eq.AC21 change to eq.AC24 below
(2+√-1)3 = 23+3*22*√-1+3*2*(-1)+(-√-1) --- eq.AC24
Simplify eq.AC24 to eq.AC25 below
(2+√-1)3 = 8+12*√-1-6-√-1 = 2+11*√-1 --- eq.AC25
What is 11*√-1? Square 11 to 121 and move 121 in
to square root sign, get 11*√-1=√-121
eq.AC25 is same as eq.AC26
(2+√-1)3 = 2+√-121 --- eq.AC26
Bombelli is indeed correct.
∛{2+√-121} = 2+√-1  --- eq.AC16
and
∛{2-√-121} = 2-√-1  --- eq.AC17
are both correct.
<a name="docA032">
Problem
x3=15*x+4 --- eq.AC14
has solution
x = ∛{2+√-121} + ∛{2-√-121} --- eq.AC15
solution can be written as
x = 2+√-1 + 2-√-1 = 4  --- eq.AC27
All of a sudden, Bombelli's Idea solved a
perfect real problem of cubic equation.
("perfect real problem" say eq.AC14
To solve eq.AC14 required accepting √-1 as an
important object that can do calculation with
them as if they were object behave according to
the rule of real numbers. This is considered as
the birth of complex analysis.

<a name="docA033">
The Birth of Complex Analysis
● Bombelli's discovery is considered "The
Birth of Complex Analysis"
● It showed that perfectly real problem require
complex arithmetic for their solution.
● Note: Need to be able to manipulate complex
numbers according to the same rules we are used
to from real numbers (distributive law etc.)
● We will study this next.
2015-03-14-20-36 stop

2015-03-16-23-00 start
Next is study notes of
Analysis of a Complex Kind
1.2 Algebra and Geometry in the Complex Plane ; 30:40

Lecturer is Prof. Petra Bonfert-Taylor.
Notes recorder is Liu,Hsinhan.
Recorded notes is about 60% match, not exact.
Liu,Hsinhan inserted text is blue Italic.
LiuHH notes may contain error!

<a name="docA052"> Begin video 2 of 33
Welcome to the second lecture of our course
"Analysis of a Complex Kind".
Today we will be talking about Algebra and
Geometry in the Complex Plane. Learn what complex
numbers are, how to add them, how to multiply
them, how to subtract them, how to divide them,
and some of their properties.

<a name="docA053">
Complex numbers: expressions of the form
z=x+iy --- eq.AC28
where
x is called real part of z;
x=Re z --- eq.AC29
and
y is called the imaginary part of z;
y=Im z --- eq.AC30
<a name="docA054">
2015-03-16-23-08
One point is very important. In z=x+iy
only i=√-1 is imaginary number.
x MUST be real and y MUST BE REAL.
i*y together is imaginary of z. On the other
hand. If x=1+2i and y=3+4i then the expression
z=x+iy is z1=1+2i + i*(3+4i) We must convert
from z1 to z2=1+2i + i*3+i*4*i=(1-4)+i*(2+3)
for further discussion. This practice avoid
future distortion. (future discussion all
treat x,y both be real) 2015-03-16-23-16

<a name="docA055">
Complex numbers is simply the expressions of
the form z=x+iy where x and y are real numbers.
For example z=3+i*5 would be a complex number.
Write in the form z=3+5i is same.
In z=x+iy , x is called real part, because there
is no i=√-1 attach to x.
In z=x+iy , y is called imaginary part, because
there is an i=√-1 attach to y.
The set of all complex number is noted by 'ℂ'.
If reader write document/web page and need 'ℂ'.
You can find related hollow characters at here.
You can find a tool at
http://freeman2.com/utility4.htm#showUnicode
This too allow you start from in hand 'ΓΔ' and
find related 'ΣΤΥΦΧΨΩ' From in hand '∫' find
Unicode ∬　∭　∮∯　∰　∱　∲　∳　∴　∵　∶　∷
∸ ∹　∺　∻　∼　∽　∾　∿　≀　≁　≂
utility4.htm is a handy tool. 2015-03-16-23-33

<a name="docA057">
Set of complex numbers: 'ℂ' (the complex plane)
It is a 'C' with double bar in it.

Real numbers: subset of the complex numbers
(those whose imaginary part is zero)
The number z=x+0*i is considered a real number.

The complex plane can be identified with ℝ2
<a name="docA058">
Set of real numbers: 'ℝ', both x-axis and
y-axis are real is ℝ2 mean two dimensional
real. Draw a x-y plane coordinate system.
Use x axis for real part of z, x
Use y axis for imaginary part of z, y
With that convention we can draw a complex
number z=x+i*y Mark number x on x-axis, and
mark number y on y-axis. From number x draw
an auxiliary line parallel to y-axis. From
number y draw an auxiliary line parallel to
x-axis. Two auxiliary lines meet at point
z=x+i*y.
2015-03-16-23-47 stop at video 2:47/30:39
<a name="docA059">
2015-03-17-11-17 start
z=x+i*y --- eq.AC31
w=u+i*v --- eq.AC32
z+w = (x+u) + i*(y+v) --- eq.AC33
Real(z+w)=x+u=Real(z)+Real(w) --- eq.AC34
Imag(z+w)=y+v=Imag(z)+Imag(w) --- eq.AC35

<a name="docA060">
How do we add complex numbers?
Suppose we have two numbers z and w.
z is defined in eq.AC31.
w is defined in eq.AC32.
How do we add them? Here we need finding the
Real part result Real(z+w) = Real(z)+Real(w)
x and u both no i=√-1 attach to them. x+u is
and v, put i=√-1 in front of (y+v): i*(y+v).
The addition result complex number has real
part x+u which is sum of two real part, and
addition result complex number has imaginary
part y+v which is sum of two imaginary part.
In other words
Real(z+w)=x+u=Real(z)+Real(w) --- eq.AC34
Imag(z+w)=y+v=Imag(z)+Imag(w) --- eq.AC35

<a name="docA062">
Let me give you an example.
Suppose we are adding  (3+5i) + (-1+2i) The
-1 get 2 for addition real part result.
part result. Get (3+5i) + (-1+2i) = 2+7i

<a name="docA063">
to two vector addition. Let us go back to our
coordinate system (Please view video 4:24/30:39)
Draw a z vector from (0,0) to (5,1)
Draw a w vector from (0,0) to (4,4)
To add these two vectors, all we have to do
is parallel move z vector from start at (0,0)
to start at (4,4) [new z from (4,4) to (9,5)]
The diagonal vector from (0,0) to (9,5) is
the sum of (5,1) + (4,4) = (9,5)
Addition of two complex numbers correspond to
2015-03-17-11-56 stop at video 5:49/30:39

<a name="docA064">
2015-03-17-15-31 start
The Modulus of a Complex Number
Definition
The modulus of a complex number z=x+i*y is
the length of the vector z
|z|=√x*x+y*y --- eq.AC36

<a name="docA065">
The modulus of a complex number z=x+i*y is
the distance of the number z from the origin
(0,0) or the length of the vector from origin
to the point z. In x-y complex plane, draw z
point and draw line from (0,0) to z=(x,y) .
Call this line the modulus of complex number.
I can see the drawing has a right triangle.
In order to find the length of hypotenuse of
a right triangle, all I need to do is square
the length of two legs. How long are those
legs? This leg along x-axis is x, the leg
along y-axis is y. Square the length of two
legs are x*x and y*y Together x*x+y*y gives
the length of hypotenuse squared
|z|2=x2+y2 --- eq.AC37
Take square root of |z|2 get |z|, see eq.AC36.

Multiplication of Complex Numbers
Motivation:
(x+iy)*(u+iv)=xu+ixv+iyu+i2yv --- eq.AC38

Let us pretend for a second, they behave just
like normal real numbers we know from elementary
school. Suppose all the regular rules held.
From (x+iy)*(u+iv) I get x*u+x*iv+iy*u+iy*iv
We get exactly four terms in eq.AC38 right side.
If i2 has the property -1, then i2 is just -1.

<a name="docA067">
"Suppose all the regular rules held" is the
motivation and i2=-1 is the complex rule
(i2=-1 is NOT the regular rule)
This motivation is the thought and goto the
following definition
(regular rule + complex rule together.
In eq.AC39 '-' come from complex rule.)
Definition:
(x+iy)*(u+iv)=(xu-yv)+i*(xv+yu) ∈ℂ --- eq.AC39
<a name="docA068">
iy*iv become i2yv and it is -yv. Here, -yv
is a real number.
Multiplication real part is (xu-yv)
Multiplication imaginary part is i*(xv+yu)
Indeed, (xu-yv)+i*(xv+yu) is how we define
multiplications.
Let us look at one example.
(3+4i)*(-1+7i)=[3*(-1)-4*7]+i*[3*7+4*(-1)]
(3+4i)*(-1+7i)=[-3-28]+i*[21-4]=-31+17i --- eq.AC40
Follow eq.AC39, get eq.AC40 a correct result.

<a name="docA069">
One can check with the definition of usual
properties (for real number)
still hold (for complex number) :
associative
(z1z2)z3=z1(z2z3) --- eq.AC41
commutative
z1z2=z2z1 --- eq.AC42
distributive
z1(z2+z3)=z1z2+z1z3 --- eq.AC43
2015-03-17-16-28 stop at 10:10/30:39
2015-03-17-18-12 start
Commutative says
(3+4i)*(-1+7i)=(-1+7i)*(3+4i)
You should check this out.

<a name="docA070">
Now what exactly is i?
i=0+1*i --- eq.AC44
so
i2=(0+1*i)*(0+1*i)
=(0*0-1*1)+i*(0*1+1*0)=-1 --- eq.AC45
So far, i=√-1 is just a symbol sitting there.
i is same as 0+1*i. Now I can calculate i2
Use the rule of multiplication find i2=-1 as
shown in eq.AC45 . i squared is indeed -1.
Multiplication we defined has the property
that we wanted it to have i2=-1.

What is i cubed?
i3=i2*i=-1*i=-i --- eq.AC46
From associative law
i3=i*i2=i*(-1)=-i --- eq.AC47
i to the fourth power
i4=i2*i2=(-1)*(-1)=+1 --- eq.AC48
i to the fifth power
i5=i4*i=1*i=i --- eq.AC49
i to the sixth power
i6=i4*i2=1*(-1)=-1 --- eq.AC50
.....
keeps going, i=i, i2=-1, i3=-i, i4=1
these four numbers i,-1,-i,+1 keep repeating.
There is really four different values the power
of i can have.
2015-03-17-18-39 stop at video 14:12/30:39
<a name="docA072">
2015-03-17-23-50 start
How do you divide complex numbers?
Suppose that z=x+iy and w=u+iv What is z/w ?
Require |w| not= zero. Here is the great trick.
```

 z w = x+iy u+iv = (x+iy)*(u-iv) (u+iv)*(u-iv)
--- eq.AC51

 = (xu+yv)+i*(-xv+yu) u2+v2+i(-uv+vu) = xu+yv u2+v2 + i* yu-xv u2+v2
--- eq.AC52
u2+v2 must NOT = 0
width of above equation a403180007
```<a name="docA073">
2015-03-18-00-07 here
See eq.AC51, z/w equal to (x+iy)/(u+iv). Let us
use actual number example. Let z=3+4i , w=-1+7i
I want to calculate (3+4i)/(-1+7i) How do I do
that? The idea is to multiply numerator and
denominator by this number (u-iv) In the actual
number example multiply numerator and denominator
by (-1-7i) get
```
<a name="docA074">

 z w = 3+4i -1+7i = (3+4i)*(-1-7i) (-1+7i)*(-1-7i)
--- eq.AC53

 = [3*(-1)+4*7]+i*[-3*7+4*(-1)] (-1)2+72+i*[-(-1)*7+7*(-1)]
--- eq.AC54

 = 3*(-1)+4*7 (-1)2+72 + i* -3*7+4*(-1) (-1)2+72
--- eq.AC55

 = 25 50 + i* -25 50 = 1 2 － i* 1 2
--- eq.AC56
width of above equation a403180032
```<a name="docA075">
2015-03-18-00-33 here
The trick we use is to make denominator REAL.
(see eq.AC53 red terms that is inserted "*1")
We make denominator u+iv REAL by multiply u-iv
to denominator and to numerator.
Multiply u-iv to u+iv this make imaginary cancel
out. So the denominator become pure real. Then
we can separate real from imaginary. See from
eq.AC54 to eq.AC55 has this "denominator become
pure real" process.
eq.AC52 is the formula of z/w
In particular, if we want just one over complex z.
In eq.AC52 write z as 1+0i (set x=1 and set y=0)
and write w as z (set u=x and set v=y) We get
```
Reciprocal of a complex number

 1 z = 1 x+iy = x-iy x2+y2
--- eq.AC57
width of above equation a403180047
```
2015-03-18-00-50 stop at video 19:10/30:39
<a name="docA077">
2015-03-18-15-00 start
The complex conjugate
Note the importance of the quantity x-iy in
the previous calculation!
This quantity x-iy will just replace '+' with
'-' that was really important in the previous
calculation. It is so important, we give it a
special name "complex conjugate".
<a name="docA078">
Definition
If z=x+iy, then z is the complex conjugate of z.
Notice the z with a bar on top of it. Let us
look at some of its properties.
● z with double bar on top = z --- eq.AC58
● z+w=z+w ------------------------ eq.AC59
● |z|=|z| ------------------------ eq.AC60
● zz=(x+iy)*(x-iy)=x2+y2=|z|2 ---- eq.AC61
● 1/z = z/[z*z] = z/|z|2 --------- eq.AC62

<a name="docA079">
If you take conjugate twice, you end up with
original number. What take conjugate mean
graphically? Let us go back to our coordinate
system.
Suppose you have a complex z point here which
is x+iy. (see video 20:28/30:39) Then where is
x-iy? Make y negative, take minus y instead of
y. In other words, reflect z with respect to x
axis, get complex conjugate z.

<a name="docA080">
Continue properties of complex conjugate.
If you take complex conjugate of sum of z+w
I might just as well take complex conjugate
of z first and take complex conjugate of w
and add these two numbers. (see eq.AC59)

The absolute of z is same as the absolute of
z (see eq.AC60) You may see the picture.
The length of z and the length of z are same.

Here is a really neat property
zz=(x+iy)*(x-iy)=x2+y2=|z|2 ---- eq.AC61
we use this property in the quotient rule.
(see eq.AC51 inserted red w/w which is one)

We can use eq.AC61 re-write 1/z (see eq.AC62)
1/z = z/[z*z] = z/|z|2 --------- eq.AC62
In 1/z, multiply top and bottom by z make
denominator a real number.

<a name="docA082">
More Properties of the Complex Conjugate
● When is z=z?
We learn complex conjugate z is the reflection
of z with respect to x-axis. When is the
reflection with respect to x-axis is same as
number itself? Only if the number is on the
real axis to begin with.
z=z is the case if z is a real number. z∈ℝ

<a name="docA083">
Here is another interesting thing to note.
● z+z=(x+iy)+(x-iy)=2x. So
```

 Real z = z+z 2
--- eq.AC63
Similarly

 Imag z = z－z 2*i
--- eq.AC64
width of above equation a403181621
```<a name="docA084">
2015-03-18-16-22 here
If you add z+z you get something real 2*x.
Divide the equation by 2, get x which is real
part of z . We find a nice formula to find the
real part of z (see eq.AC63)

Similarly, if you subtract z-z, then you get
2*i*y . If you want imaginary by itself, use
eq.AC64 above.

<a name="docA085">
● |z*w|=|z|*|w| --- eq.AC65
Now you can verify that length of z*w is same as
length of z multiply length of w (see eq.AC65)

You can also verify
```

 ● z ─ w = z w
--- eq.AC66
w≠0
width of above equation a403181641
```<a name="docA087">
2015-03-18-16-42 here
Finally
● |z|=0 if and only if z=0 --- eq.AC67
This is very easy to verify. Let us look at
the square of length.
|z|2=x2+y2 --- eq.AC68
x and y both are real numbers. (video 25:05/30:39)
x2+y2 is zero only if x=y=0 both be zero.
2015-03-18-16-51 stop at video 25:25/30:39
<a name="docA088">
2015-03-18-18-00 start
Some Inequalities
● -|z| ≦ Real z ≦ |z| --- eq.AC69
● -|z| ≦ Imag z ≦ |z| --- eq.AC70
One of them is obvious looking at a picture.
Let me draw a complex plane here.
|z| is hypotenuse,
Real z=x is |z| projection on x-axis.
Imag z=y is |z| projection on y-axis.
Right triangle hypotenuse≧leg is obvious.

<a name="docA089">
● |z+w| ≦ |z| + |w| --- eq.AC71
eq.AC71 is triangle inequality.
Let me show you why that is the case.
|z+w| and |z| and |w| are three sides of a
triangle. Side |z+w| has two end points.
Between these two end points,
|z+w| is a straight line go shortest distance.
|z| and |w| are two broken lines go longer
distance. Therefore |z+w| ≦ |z| + |w| is true.

<a name="docA090">
● |z-w| ≧ |z| - |w| --- eq.AC72
eq.AC72 is reverse triangle inequality.
This reverse triangle inequality eq.AC72 follows
triangle inequality.
2015-03-18-20-26 insert start
In real number |-3|＞|2| ok, (-3)2＞22 is OK.
In complex |2i|＞|1| ok, (2i)2＞12 is ERROR.
You cannot compare complex numbers.
Above complex inequality ALL COMPARE LENGTH !
2015-03-18-20-36 insert stop

Finally I will look at
The Fundamental Theorem of Algebra
Theorem
If a0, a1 ●●● an are complex numbers with an≠0,
then the polynomial
p(z)=anzn+an-1zn-1+●●●+a1z+a0 --- eq.AC73
has n roots z1, z2 ●●●, zn in ℂ.
It can be factored as
p(z)=an(z-z1)(z-z2)●●●(z-zn) --- eq.AC74
<a name="docA092">
For example
p(z)=5*z3+4*z2-2*z+14 --- eq.AC75
eq.AC75 is called degree of 3 polynomial.
The Fundamental Theorem of Algebra say eq.AC75
has three roots. Three complex numbers can make
eq.AC75 equal to zero. eq.AC75 can be factored
as eq.AC76 below
p(z)=a3(z-z1)(z-z2)(z-z3) --- eq.AC76
We will be able to prove this theorem using
complex analysis later on in this course.

<a name="docA093">
Consider the polynomial
p(x)=x2+1 --- eq.AC77
in ℝ. It has no real roots. eq.AC77 is degree
of two polynomial. We consider real answer.
eq.AC77 never ever cross the x axis. Cruve
y=x2+1 is always above x-axis. Curve never
equal to zero. We cannot factor either. It
has no real roots.
<a name="docA094">
If you look eq.AC77 in ℂ, eq.AC77
can be factored
z2+1=(z+i)*(z-i) --- eq.AC78
i and -i let z2+1 become zero.
We hope to prove this theorem later on in
this course.
2015-03-18-19-03 stop

<a name="docA095">
2015-03-18-19-03 start
To solve
p(z)=5*z3+4*z2-2*z+14 --- eq.AC75
http://freeman2.com/polyroot.htm#begin0
In "max.poly.order" box fill in 3 for 3rd degree.
In C00:[  ] C01:[  ] C02:[  ] C03:[  ] fill in
C00:[14] C01:[-2] C02:[ 4] C03:[ 5]
Then click [fast get root] button.

Box 1, answer output to next
[[
Below is answer, complex polynomial root ([1,2] same as 1+2i)
1st root is [-1.8421948711135218,0]
2nd root is [0.5210974329859563,-1.1173108393094164]
3rd root is [0.5210974381275649,1.117310836878581]

Below is answer too, complex number only, no text.
-1.8421948711135218+0i
0.5210974329859563-1.1173108393094164i
0.5210974381275649+1.117310836878581i
]]

<a name="docA097">
Box 2, debug, verify output to next
[[
Below substitute root into polynomial to evaluate.
If result is zero, root is correct, otherwise wrong.
1st root -1.8421948711135218,0
evaluation is 1.6548324133225378e-7,0
Absolute value is 1.6548324133225378e-7 should be zero.
2nd root 0.5210974329859563,-1.1173108393094164
evaluation is 1.6548326620124953e-7,8.305598608870923e-8
Absolute value is 1.8515671259179744e-7 should be zero.
3rd root 0.5210974381275649,1.117310836878581
evaluation is 1.654832644248927e-7,8.305599052960133e-8
Absolute value is 1.851567129962405e-7 should be zero.
]]
Please verify substitute root into polynomial
to evaluate. See whether absolute value is zero.
2015-03-18-19-17 stop

<a name="a40319a">
2015-03-19-14-13 start
http://freeman2.com/tute0065.htm
2015-03-18-21-49 Liu,Hsinhan access
http://freeman2.com/polyroot.htm#begin0
It is a surprise, polyroot.htm not work!?
LiuHH has two Acer computers.
Acer1 off line more than one year.
Acer2 go online, update file and search other's
web page.
<a name="a40319b">
Acer1 open local polyroot.htm get answer Box 1
Acer2 open online polyroot.htm but no answer.
Both local polyroot.htm and online polyroot.htm
have same file size.
Acer1 accept xygraph code and draw graph.
Acer2 NOT draw graph.

<a name="a40319c">
2015-03-19-14-28 use Acer2 open
http://freeman2.com/2014util.htm
click [Yr] Acer2 get
Thu Mar 19 2015 14:28:50 GMT-0700 (Pacific Daylight Time)
click [Yr] Acer1 get
Thu Mar 19 14:39:00 PDT 2015
click [Y0] Acer2 get NaN03191429
click [Y0] Acer1 get a403191429
click [Y1] Acer2 get -1796,03,19,14,29
click [Y0] Acer1 get 104,03,19,14,29
Acer2 start trouble from 2014-Jan
2015-03-19-14-41 stop

2015-03-19-17-15 start
Next is study notes of
Analysis of a Complex Kind
1.3 Polar Representation of Complex Numbers ; 32:32

Lecturer is Prof. Petra Bonfert-Taylor.
Notes recorder is Liu,Hsinhan.
Recorded notes is about 60% match, not exact.
Liu,Hsinhan inserted text is blue Italic.
LiuHH notes may contain error!

<a name="docA102"> Begin video 3 of 33
Welcome to lecture three of our course
"Analysis of a Complex Kind". Today we
Polar Representation of Complex Numbers.
Another representation which is different
from the Cartesian representation we have
been talk about so far. Polar Representation
what multiplication means for example.

<a name="docA103">
● Consider
z=x+iy  --- eq.AC79
z∈ℂ z≠0+0i
● z can be also described by the distance r
from the origin (r=|z|) and the angle θ
between the positive x-axis and the line
segment from 0+0i to z=x+iy .
<a name="docA104">
Draw a picture (video 0:50/32:31) point z
located at z=x+iy=4+3i . x component is x
y component is y. Distance from the origin
r is called modulus of z.
r=|z| --- eq.AC80
Angle between line from 0+0i to z=x+iy and
positive x-axis is called θ. θ is a Greek
symbol. Often use α, β, γ, θ, φ for angle in
mathematics. Together, radius r and angle θ
describe exactly what point we are talking
<a name="docA105">
● (r,θ) are the polar coordinates of z.
If I give you radius r and angle θ, you can
re-construct the Cartesian form of z=x+iy.
How are they related?
● Relation between Cartesian and polar coordinates:
x=r*cos(θ) --- eq.AC81
y=r*sin(θ) --- eq.AC82
Cartesian coordinates use (x,y)
Polar coordinates use (r,θ)
You see the picture (video 2:40/32:31) right
triangle Sine of this angle θ is y/r
sin(θ)=y/r --- eq.AC83
Solve eq.AC83 for y, find eq.AC82 y=r*sin(θ)
Cosine of θ is x/r
cos(θ)=x/r --- eq.AC84
Solve eq.AC84 for x, find eq.AC81 x=r*cos(θ)
So z,x,y are related from each other.

<a name="docA107">
Instead of writing z=x+iy (eq.AC79) I could
replace x by r*cos(θ), replace y by r*sin(θ)
Change eq.AC79 to eq.AC85.
z=x+iy --- eq.AC79
z=r*cos(θ)+i*r*sin(θ) --- eq.AC85
Take common factor r out, get eq.AC86
z=r*[cos(θ)+i*sin(θ)] --- eq.AC86
eq.AC86 is called the Polar representation of z.
2015-03-19-18-25 stop at video 03:55/32:31

<a name="docA108">
2015-03-19-19-05 start
The argument of a complex number
● z=x+iy=r*[cos(θ)+i*sin(θ)]
● r=|z| is easy to find, but how to find θ?
Note: θ is not unique!
If giving x and y, how to find θ? First of all.
angle θ is not unique. There are lots of way to
find angle θ. For example draw z=2+2i the angle
is π/4 (45 degree) Sine and cosine have period
2*π. Now angle θ=π/4, angle 2π+π/4 or 4π+π/4 or
6π+π/4 or 8π+π/4 their sin() and cos() values
are the same as angle θ=π/4 . Angle is not unique.
2015-03-19-19-15 here guess
<a name="a40322a">
2015-03-22-15-46 LiuHH notes start
"If giving x and y, how to find θ?"
θ=arctan(y/x) but this formula make (-1,-1) and
(1,1) same answer; also make (-1,1) and (1,-1)
In javascript, use the function theta=atan2(y,x)
Goto http://freeman2.com/complex4.htm#Box3JS
In [Box3, JS command] input next four lines.
[[
x=-1
y=-1
a=atan2(y,x)*180/PI
a
]]
then click [test box3] (NOT click [eval Box3])
Box4 output to
[[
a
-135
]]
In Box3 change x, y values and test few other
complex points. x, y value represent z=x+i*y
2015-03-22-15-55 LiuHH notes stop
2015-03-19-19-15 here guess
<a name="docA109">
Definition //table
The principal argument of z, called Arg z,
(Arg use capital A) is the value of θ for which
-π<θ≦π //see notes at a40324a
In this case z=2+2i , we call principal argument
of z is π/4 (2π+π/4, 4π+π/4 etc are NOT principal)
● arg z={Arg z+2*π*k: k=0,±1,±2 ...} z≠0
(arg use lowercase a)
Uppercase Arg limit θ in the range -π<θ≦π
Lowercase arg allow θ be out of Arg range
k is 0 or -1, or 1 or -2 or 2 any integer.
By the way, when you talk about the polar
representation, complex number z cannot be
zero. If z=0+0i there is no angle there.

<a name="docA110">
Let us look at some example.
Arg i = π/2  --- eq.AC87
Draw coordinate axis. Point 'i' is 0+1*i
a point on y axis at y=1. You need to find
the angle. Line segment from 0+0*i to 0+1*i
is on y-axis. This is a 90 degree angle from
x-axis to y-axis. We have to use radian.
90 degree in radian is π/2. θ=π/2 is indeed
in -π<θ≦π. So the principal argument of i is
π/2.

Let us look at the complex number z=1+0i
Arg 1 = 0  --- eq.AC88
One is a real number, but we can also look at
one as complex number 1+0i. Number z=1+0i is
at (1,0) it has no imaginary part. The line
segment from (0,0) to (1,0) is on x-axis.
The angle between (0,0) to (1,0) and x-axis
is zero. θ=0 is indeed in -π<θ≦π. So the
principal argument of 1 is zero (radian).

<a name="docA112">
Arg(-1) = π  --- eq.AC89
The line segment from (0,0) to (-1,0) make
an angle with positive x-axis is π. θ=π is
in -π<θ≦π. The principal argument of -1 is π.

<a name="docA113">
What is the argument of one minus i? z=1-i
Arg(1-i) = -π/4  --- eq.AC90
In Cartesian coordinate z=1-i has coordinate
(1,-1) (video 8:31/32:31) What is the angle
line segment from (0,0) to (1,-1) make with
positive x-axis ? From positive x-axis go
around counter-clockwise direction all the
way to (1,-1).
<a name="docA114">
This angle is 315 degree (7*π/4) which is
more than π. Angle 7*π/4>π violate principal
argument range -π<θ≦π. We have to add multiple
of 2π to bring 7*π/4 into -π<θ≦π.
7*π/4 + (-2π) = -π/4 ; -2π is multiple of 2π
a403221615 Another way to measure is start
from positive x-axis go negative direction
(clockwise) π/4 reach (1,-1) Call that angle
minus π/4. θ=-π/4 is in -π<θ≦π.
The principal argument of z=1-i is -π/4.

<a name="docA115">
Finally, another example.
Arg(-i) = -π/2  --- eq.AC91
The argument of negative i. Negative i is z=-i
Cartesian coordinate of negative i is (0,-1).
Again, start from positive x-axis go counter
clockwise to z=-i. The angle is 3π/2. This
value 3π/2 is out of principal argument range
-π<θ≦π. Measure in other direction, start from
positive x-axis go clockwise to z=-i. The angle
is -π/2. θ=-π/2 is in -π<θ≦π.
The principal argument of z=-i is -π/2.
(video 10:01/32:31)

<a name="expcosin">
Exponential Notation
● Convenient notation: //short hand notation
eiθ=cos(θ)+i*sin(θ) --- eq.Eul1
Euler's formula eq.Eul1
Please see Tom M. Apostol prove eq.Eul1
2015-03-19-20-17 here
Shortcut notation eq.Eul1 we are going to
introduce. For now, eiθ is a shortcut notation.
eiθ is a place holder for cos(θ)+i*sin(θ) Later
on we will understand why we choose eiθ for
cos(θ)+i*sin(θ) Indeed it has to do with the
number e. e is Euler's number e=2.7182818284590...
<a name="docA117">
For now you can look at eiθ as a symbol.
● z=r*[cos(θ)+i*sin(θ)] --- eq.AC86
become
z=r*eiθ --- eq.AC92
z=r*eiθ is the polar form of z=x+iy.
z=r*eiθ is much shorter than z=r*[cos(θ)+i*sin(θ)]
● Note:
ei(θ+2π)=eiθ=ei(θ+4π)=...=ei(θ+2kπ) --- eq.AC93
k∈ℤ (k is an integer -∞...-3,-2,-1,0,1,2,3...∞)
In eq.AC93 add 2kπ to θ, not change cos(), sin()
value.

<a name="docA118">
Example
eiπ/2=i --- eq.AC94
What eiπ/2 really stand for?
eiπ/2 stand for cos(π/2)+i*sin(π/2)
Since cos(π/2)=0 and sin(π/2)=1, then
eiπ/2=cos(π/2)+i*sin(π/2)=0+1*i=i
Another way to look at eiπ/2 is a complex
number with radius one and angle π/2 We know
what number that is. Real=0, imaginary=1
this is point (0,1) or 0+1*i=i

<a name="docA119">
What is eiπ ?
eiπ=-1 --- eq.AC95
Again, I can write eiπ as cos(π)+i*sin(π)
cos(π)=-1 and sin(π)=0,
cos(π)+i*sin(π)=-1+0*i=-1
Or I can see cos(π)+i*sin(π) as
1*[cos(π)+i*sin(π)]
radius is one. Angle is π. This is (-1,0)
point in complex plane.

<a name="docA120">
Similarly
eiπ/4=(1+i)/√2 --- eq.AC96
radius is one. Angle is π/4. This is (1/√2,1/√2)
point in complex plane.
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From 1*[cos(π/4)+i*sin(π/4)] I can find cos(π/4)
and sin(π/4) values to get eiπ/4 or
On unit circle find π/4=45 degree point. Must
be this point (1/√2,1/√2) What is the coordinate
of this number? x and y must have same number.
Since hypotenuse is one (unit circle) two legs
equal length, must each length be 1/√2
This complex number eiπ/4 is 1/√2+i/√2
Another form is  √2/2+i*√2/2 .

<a name="docA122">
Finally,
e2πi=1 --- eq.AC97
What is that. By definition
e2πi=1*[cos(2π)+i*sin(2π)]=1*[1+0i]=1 --- eq.AC98
You possibly see e2πi=1 before.
One equation incorporated many important
constants e,2,π,i,1
1 is the most important number, it is unit.
2 is smallest prime number.
i=√-1 is imaginary unit
π is 3.141592653589793
e is 2.718281828459045
e2πi=1 combine five most important constants
in mathematics into one equation.
2015-03-19-21-50 stop at video 15:30/32:31

<a name="docA123">
2015-03-20-14-57 start
Properties of the Exponential Notation
● |eiθ|=1 --- eq.AC99
From eq.Eul1  eiθ=cos(θ)+i*sin(θ) --- eq.Eul1
Therefore
|eiθ|=|cos(θ)+i*sin(θ)| --- eq.ACa0
How do you find the length of a complex number
cos(θ)+i*sin(θ) ? You take the square root of
real squared plus imaginary squared. That is
|eiθ|=√cos(θ)*cos(θ)+sin(θ)*sin(θ)=√1=1 --- eq.ACa1
eiθ can be viewed as a complex number whose
length r=1 and angle from positive x-axis is θ.
<a name="docA124">
___
● eiθ=e-iθ --- eq.ACa2
What is the complex conjugate of eiθ ?
We have to look at cos(θ)+i*sin(θ) and take
complex conjugate to it. How do you take the
complex conjugate? You flip the sign of imaginary
part and get
cos(θ)+i*sin(θ)=cos(θ)-i*sin(θ) --- eq.ACa3
How do you write cos(θ)-i*sin(θ) in exponential
form? See  eiθ=cos(θ)+i*sin(θ) --- eq.Eul1
<a name="docA125">
Need write cosine something PLUS sine something.
How do you do that? Sine property allow us pull
-i*sin(θ)=+i*sin(-θ) --- eq.ACa4
In order to use exponential notation we have to
use SAME angle for both sin() and cos()
See  eiθ=cos(θ)+i*sin(θ) --- eq.Eul1
[eiθ ≠ cos(+θ)+i*sin(-θ) ]
cosine is an even function,
cos(θ)=cos(-θ) --- eq.ACa5
write cos(θ)-i*sin(θ) in exponential form as
cos(-θ)+i*sin(-θ)=e-iθ --- eq.ACa6
Here we have it. cos(-θ)+i*sin(-θ) is short
hand for ei(-θ)=e-iθ
Here we showed eq.ACa3 is true.

<a name="docA127">
● 1/eiθ=e-iθ --- eq.ACa7
What is one over eiθ ?
We have talked about the reciprocal of a complex
number. see eq.AC57
```
<a name="docA128">
Reciprocal of a complex number e

 1 eiθ = 1 cos(θ)+i*sin(θ) = cos(θ)－i*sin(θ) [cos(θ)]2+[sin(θ)]2 = e－iθ 1 = e－iθ
--- eq.ACa8
width of above equation a403201617
```<a name="docA129">
2015-03-20-16-18 here
Multiply top and bottom with complex conjugate
of eiθ We just calculated, it is e-iθ
"top and bottom with complex conjugate" see
red term in eq.AC51. Here eq.ACa8 skipped
"top and bottom with complex conjugate"
eq.ACa8 third term denominator is
eiθ*e-iθ=[cos(θ)]2+[sin(θ)]2=1 --- eq.ACa9
This calculation end up to
1/eiθ=e-iθ --- eq.ACa7
2015-03-20-16-33 stop at video 18:55/32:31

<a name="docA130">
2015-03-20-18-47 start
Finally
● ei(θ+φ)=eiθ*eiφ --- eq.ACb0
Here we meet another Greek symbol φ for angle.
From eq.Eul1  eiθ=cos(θ)+i*sin(θ) --- eq.Eul1
Change θ to θ+φ we have
ei(θ+φ)=cos(θ+φ)+i*sin(θ+φ) --- eq.ACb1
Remind
cos(θ+φ)=cos(θ)*cos(φ)-sin(θ)*sin(φ) --- eq.ACb2
sin(θ+φ)=sin(θ)*cos(φ)+cos(θ)*sin(φ) --- eq.ACb3
Then eq.ACb1 become
ei(θ+φ)=[cos(θ)*cos(φ)-sin(θ)*sin(φ)]
+i*[sin(θ)*cos(φ)+cos(θ)*sin(φ)] --- eq.ACb4
We can re-arrange eq.ACb4 as next
ei(θ+φ)= //attention i*i=-1
[cos(θ)*cos(φ)+i*i*sin(θ)*sin(φ)]
+i*sin(θ)*cos(φ)+i*cos(θ)*sin(φ)
=cos(θ)*cos(φ)+i*cos(θ)*sin(φ)
+i*sin(θ)*cos(φ)+i*sin(θ)*i*sin(φ)
=cos(θ)*[cos(φ)+i*sin(φ)]
+i*sin(θ)*[cos(φ)+i*sin(φ)]
=[cos(θ)+i*sin(θ)]*[cos(φ)+i*sin(φ)]
=eiθ*eiφ --- eq.ACb5
eq.ACb5 verify eq.ACb0 is true.

<a name="docA132">
Conclusions for the Argument Function
● arg(z)=-arg(z) --- eq.ACb6
Equation we just established eq.ACb5 has
consequences for argument function eq.ACb6 .
Notice in eq.ACb6 I use the lowercase arg()
[not uppercase Arg() It must be -π＜Arg(z)≦π]
Mainly the argument of conjugate of z agrees
with minus of argument of z and possibly
Draw z=1+i and draw z=1-i
arg(z)=π/4 and arg(z)=-π/4

<a name="docA133">
● arg(1/z)=-arg(z) --- eq.ACb7
These equation follow directly from short hand
notation eq.Eul1

<a name="docA134">
● arg(z1z2)=arg(z1)+arg(z2) --- eq.ACb8
eq.ACb8 can have differ by 2π. Let me show you
one example.
arg(i*i)=arg(-1)=π --- eq.ACb9
arg(i)+arg(i)=π/2+π/2=π --- eq.ACc0
In above example, eq.ACb9 and eq.ACc0 both are
in principal angle range -π<θ≦π In this case
eq.ACb9 and eq.ACc0 get same answer π.
<a name="docA135">
Next example is different.
arg((-1)*(-1))=arg(1)=0 --- eq.ACc1
arg(-1)+arg(-1)=π+π=2*π --- eq.ACc2
In second example, eq.ACc1 is in principal angle
range -π<θ≦π. But eq.ACc2 2*π is out of range
-π<θ≦π. In this case eq.ACc1 and eq.ACc2 get
different answer 0 and 2*π. In
arg(z1z2)=arg(z1)+arg(z2) --- eq.ACb8
Left side and right side may differ by multiple
of 2*π.

● Consider
z1=r1*eiφ1 --- eq.ACc3
z2=r2*eiφ2 --- eq.ACc4
What is the polar form of z1z2?
Just plug in get
z1z2=r1*eiφ1 * r2*eiφ2 --- eq.ACc5
z1z2=(r1r2)*ei(φ1+φ2) --- eq.ACc6
Video 26:16/32:31 has picture.

<a name="docA137">
De Moivre's Formula
● eiθ*eiθ=ei(θ+θ)=ei*2θ --- eq.ACc7
We can use eq.ACb0
ei(θ+φ)=eiθ*eiφ --- eq.ACb0
set θ=φ to get eq.ACc7
eq.ACc7 relate to De Moivre's Formula
eq.ACc7 say eiθ squared is same as ei*2θ.
[eiθ]3=[eiθ]2 * [eiθ] --- eq.ACc8

<a name="docA138">
In eq.ACc8 eiθ cubed apply eiθ square eq.ACc7
get
● [eiθ]3=ei*2θ * eiθ --- eq.ACc9
By the rule we already established, find
[eiθ]3=ei*3θ --- eq.ACd0
Similarly, you find eiθ to power n is
● [eiθ]n=ei*nθ --- eq.ACd1
eq.ACd1 is also true for negative n.
[eiθ]-1=1/eiθ=e-iθ --- eq.ACd2

<a name="docA139">
Recall eiθ is simply short for cos(θ)+i*sin(θ)
Thus the last formula eq.ACd1 means
● [cos(θ)+i*sin(θ)]n=cos(nθ)+i*sin(nθ) --- eq.ACd3
De Moivre's Formula
● [eiθ]n=ei*nθ --- eq.ACd1
● [cos(θ)+i*sin(θ)]n=cos(nθ)+i*sin(nθ) --- eq.ACd3

<a name="docA140">
De Moivre's Formula let us get cosine and sine
multiple angle formula. Example n=3
cos(3θ)=cos3(θ)-3*cos(θ)*sin2(θ) --- eq.ACd4
sin(3θ)=3*cos2(θ)*sin(θ)-sin3(θ) --- eq.ACd5
[cos(θ)+i*sin(θ)]3=cos(3θ)+i*sin(3θ) --- eq.ACd6
On the left hand side, I can multiply through.
Remember our formula eq.AC20
(a+b)3 = a3+3*a2*b+3*a*b2+b3 --- eq.AC20
Let
a=cos(θ)   --- eq.ACd7
b=i*sin(θ) --- eq.ACd8
We can use eq.AC20 to find [cos(θ)+i*sin(θ)]3 value
[cos(θ)+i*sin(θ)]3=
cos3(θ)+3*cos2(θ)*[i*sin(θ)]
+3*cos(θ)*[i*sin(θ)]2+[i*sin(θ)]3 --- eq.ACd9

[cos(θ)+i*sin(θ)]3=
cos3(θ)+3*i*cos2(θ)*[sin(θ)]
-3*cos(θ)*[sin2(θ)]-[i*sin3(θ)] --- eq.ACe0
//remind: i*i=-1 ; i*i*i=-i ;

<a name="docA142">
[cos(θ)+i*sin(θ)]3=
cos3(θ)-3*cos(θ)*[sin2(θ)]
3*i*cos2(θ)*[sin(θ)]-[i*sin3(θ)] --- eq.ACe1

[cos(θ)+i*sin(θ)]3=
cos3(θ)-3*cos(θ)*[sin2(θ)]
i*[3*cos2(θ)*sin(θ)-sin3(θ)] --- eq.ACe2
Compare eq.ACe2 with eq.ACd6
[cos(θ)+i*sin(θ)]3=cos(3θ)+i*sin(3θ) --- eq.ACd6
Both are equation for [cos(θ)+i*sin(θ)]3

<a name="docA143">
real(eq.ACe2) = real(eq.ACd6) get
cos(3θ)=cos3(θ)-3*cos(θ)*sin2(θ) --- eq.ACe3
eq.ACe3 and eq.ACd4 are identical
cos(3θ)=cos3(θ)-3*cos(θ)*sin2(θ) --- eq.ACd4

<a name="docA144">
imag(eq.ACe2) = imag(eq.ACd6) get
sin(3θ)=3*cos2(θ)*sin(θ)-sin3(θ) --- eq.ACe4
eq.ACe4 and eq.ACd5 are identical
sin(3θ)=3*cos2(θ)*sin(θ)-sin3(θ) --- eq.ACd5

<a name="docA145">
De Moivre's Formula helped us find eq.ACd4
cos(3θ) formula in terms of cos(θ) and sin(θ)

De Moivre's Formula helped us find eq.ACd5
sin(3θ) formula in terms of cos(θ) and sin(θ)
2015-03-20-21-24 stop

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2015-03-22-19-38 start
Next is study notes of
Analysis of a Complex Kind
1.4 Roots of Complex Numbers ; 14:10

Lecturer is Prof. Petra Bonfert-Taylor.
Notes recorder is Liu,Hsinhan.
Recorded notes is about 60% match, not exact.
Liu,Hsinhan inserted text is blue Italic.
LiuHH notes may contain error!

<a name="docA152"> Begin video 4 of 33
Welcome to lecture four of our course
"Analysis of a Complex Kind". Today we will
talk about Roots of Complex Numbers. You
all know that square root of nine is three,
or the square root of four is two, cubic
root of twenty seven is three. How to take
square root of 3+4i for example, or fifth
root of negative i? That is what we are

<a name="docA153">
The n-th root. Definition
Let w be a complex number. An n-th root of w
is a complex number z such that
zn=w --- eq.ACe5

Let us begin with definition (above).
We will see: if w≠0, there are exactly n
distinct n-th roots. So, there are two distinct
square roots for any number and three distinct
cubic roots etc.
<a name="docA154">
Use the polar form for w and z:
w=ρ*eiφ --- eq.ACe6
z=r*eiθ --- eq.ACe7
w=ρ*eiφ is given, one value.
Here introduce another Greek symbol ρ for
The equation
zn=w --- eq.ACe5
becomes
<a name="docA155">
rn*einθ=ρ*eiφ --- eq.ACe8
Compare eq.ACe8 left side with right side.
Both side are the same, then
both modulus are equal
rn=ρ --- eq.ACe9
and both arguments are equal
einθ=eiφ --- eq.ACf0
eq.ACf0 gives
nθ=φ+2kπ --- eq.ACf1
sin(α)=sin(α+2kπ) and cos(α)=cos(α+2kπ)
Thus, from eq.ACe9 get
r = n√ρ --- eq.ACf2
ρ is radius which is a positive real number,
therefore n√ρ is taking regular n-th root of rho.
No complex calculation involved in eq.ACf2.

<a name="docA157">
from eq.ACf1 nθ=φ+2kπ , k∈ℤ (ℤ =integer set) so
θ = φ/n + 2kπ/n --- eq.ACf3
In eq.ACf3 look k values from 0,1,2,... up to n-1
I have exhausted all possibilities. After k=n-1
k values start repeat (k=0 and k=n are identical)
When I look at k=n for example. When k=n, what
would I get? I get θ = φ/n + 2kπ/n = φ/n + 2nπ/n
n cancel out, so get θ = φ/n + 2π. But I already
look at θ = φ/n when k=0. φ/n and φ/n + 2π are
same angle. I get different angle if k=0,1,2,...
up to k=n-1 . I get all of my different solutions
We write  //2015-03-22-20-36 here
```
<a name="docA158">
w=ρ*e --- eq.ACe6

nw
=
w1/n
=
nρ
e
 i* [ φ n + 2kπ n ]

--- eq.ACf4
k=0,1,2,... up to k=n-1
width of above equation a403222100
exp power use three <br> to push higher. 2015-03-22-21-01
```<a name="docA159">
2015-03-22-21-05 here
w1/n is how I denote my solutions eq.ACf4 .
eq.ACf4 is n-th roots of complex number w. Next
we do a bunch examples.
2015-03-22-21-09 stop at 04:48/14:10

<a name="docA160">
2015-03-23-14-57 start
Let us find
● Square roots of 4i:
First of all, bring 4i into this form eq.ACe6
w=ρ*eiφ --- eq.ACe6
We need find out what is ρ and what is φ.
4i=4*eiπ/2 --- eq.ACf5
so ρ=4, φ=π/2 and n=2.
φ is argument, 4i=(0,4) on positive y-axis
φ=angle from positive x-axis to positive y-axis
n=the root number; square root has n=2.
2015-03-23-15-09
Now apply ρ=4, φ=π/2 and n=2 to eq.ACf4.
We find √4i is equal to
```

4i
=

4
e
 i* [ π 2*2 + 2kπ 2 ]

--- eq.ACf6
k=0,1
π/(2*2) is φ/n = (π/2)/2
width of above equation a403231522
```<a name="docA162">
2015-03-23-15-22 here
If k=0, 2kπ/2=0, eq.ACf6 become first answer:
√4i = 2*eiπ/4 = √2+i*√2 --- eq.ACf7

If k=1, 2kπ/2=π, eq.ACf6 become second answer:
√4i = 2*ei(π+π/4) =-√2-i*√2 --- eq.ACf8
In eq.ACf8 factor eiπ rotate 180 degree
change sign to both real and imaginary.
2*ei(π+π/4) is actually negative of 2*ei(π/4)
<a name="docA163">
Square root has only two answers.

Remember eiπ/4 is √2/2 + i*√2/2 on unit circle.
Present problem Square roots of 4i has modulus
ρ=4 and √ρ=√4=2. Now multiply √2/2 + i*√2/2 by
that number √ρ=2 get √2+i*√2.
Draw picture see video 07:50/14:10

<a name="docA164">
2015-03-23-16-12 here video 08:40/14:10
Let us do same thing for
● Cubed roots of -8
First step again we have to write -8 in polar
form.
w=ρ*eiφ --- eq.ACe6
We cannot write modulus as ρ=-8, because polar
form must have a positive radius ρ. Minus sign
in -8 must be carried by argument (phase angle)
<a name="docA165">
We write
-8=8*eiπ --- eq.ACf9
Here eiπ=-1 carry minus sign.
so ρ=8, φ=π and n=3.
modulus must be positive, modulus cannot be zero,
(z=0+0i a dull case) modulus cannot be negative.
φ is argument, -8=(-8,0) on negative x-axis
φ=angle from positive x-axis to negative x-axis
n=the root number; Cubed roots has n=3.
2015-03-23-16-29
```
<a name="docA166">

-8
=
8
e
 i* [ π 3 + 2kπ 3 ]

--- eq.ACg0
k=0,1,2
π/3 is φ/n ; φ=π and n=3
width of above equation a403231643
```<a name="docA167">
2015-03-23-16-43 here
If k=0, 2kπ/3=0, eq.ACg0 become first answer:
∛-8=∛8*eiπ/3=2*[cos(π/3)+i*sin(π/3)] --- eq.ACg1

If k=1, 2kπ/3=2π/3, 2π/3 + π/3=π ;
∛-8=∛8*eiπ=2*[cos(π)+i*sin(π)]=2*[-1+i*0]
∛-8=-2 --- eq.ACg2

<a name="docA168">
If k=2, 2kπ/3=4π/3, 4π/3 + π/3=5π/3 ;
∛-8=∛8*ei5π/3=2*[cos(5π/3)+i*sin(5π/3)] --- eq.ACg3
2015-03-23-16-56 here
Draw picture video 11:02/14:10

<a name="docA169">
2015-03-23-17-00 start at video 12:00/14:10
Definition
The n-th roots of 1 are called the n-th root
of unity.
Since
1=1*ei0 --- eq.ACg4
Phase angle φ is zero Complex 1 = (1,0) on x-axis
so ρ=1, φ=0 and n=unspecified, still use n.
```
<a name="docA170">

n1
=
n1
e
 i* [ 0 n + 2kπ n ]

--- eq.ACg5

n1
=
e
 i* [ 2kπ n ]

--- eq.ACg6
k=0,1,2,... n-1
width of above equation a403231724
```<a name="docA171"> t065link()
2015-03-23-17-25 here
Draw picture video 12:57/14:10
For example, find eighth root of one, n=8.
If k=1, 2kπ/8=π/4, eq.ACg6 become first answer:
11/8=eiπ/4=cos(π/4)+i*sin(π/4) --- eq.ACg7

If k=2, 2kπ/8=π/2, eq.ACg6 become second answer:
11/8=eiπ/2=cos(π/2)+i*sin(π/2)=i --- eq.ACg8

<a name="docA172">
If k=3, 2kπ/8=3π/4, eq.ACg6 become third answer:
11/8=ei3π/4=cos(3π/4)+i*sin(3π/4) --- eq.ACg9

If k=4, 2kπ/8=π, eq.ACg6 become fourth answer:
11/8=eiπ=cos(π)+i*sin(π)=-1 --- eq.ACh0

<a name="docA173">
If k=5, 2kπ/8=5π/4, eq.ACg6 become fifth answer:
11/8=ei5π/4=cos(5π/4)+i*sin(5π/4) --- eq.ACh1

If k=6, 2kπ/8=3π/2, eq.ACg6 become sixth answer:
11/8=ei3π/2=cos(3π/2)+i*sin(3π/2)=-i --- eq.ACh2

<a name="docA174">
If k=7, 2kπ/8=7π/4, eq.ACg6 become seventh answer:
11/8=ei7π/4=cos(7π/4)+i*sin(7π/4) --- eq.ACh3

If k=0, 2kπ/8=0, eq.ACg6 become zeroth answer:
11/8=e0=cos(0)+i*sin(0)=1 --- eq.ACh4

Because eighth root operation, zeroth answer and
2015-03-23-17-47 stop

<a name="a40324a">
Cartesian complex ≡ Polar complex
2015-03-24-14-56 start
Given Cartesian coordinate complex z
z=x+i*y --- eq.AC01
Given polar coordinate complex z
z=r*eiθ --- eq.AC92
eq.AC01 and eq.AC92 are ONE complex number
two different expressions. Therefore
<a name="a40324b">
Cartesian component x,y and polar component
r,θ are related as next
Next is formula from r,θ to z=x+i*y .
x=r*cos(θ) --- eq.AC81
y=r*sin(θ) --- eq.AC82
Next is formula from z=x+i*y to r and θ .
r=|z| --- eq.AC80
r=|z|=√x*x+y*y --- eq.AC36
θ=atan(y/x)=tan-1(y/x) --- eq.LB01
//do NOT use eq.LB01, use eq.LB02, see table
eq.LB01 = EQuation used by Liuhh, B sequence
equation number 01. a403241528
<a name="a40324c">
θ1=atan(y/x)=atan((-1)/(+1))
θ2=atan(y/x)=atan((+1)/(-1))
2015-03-24-15-38 stop
2015-03-24-18-08 start
But atan(y/x) output both to fourth quadrant.
Similarly atan((negative)/(negative)) third

<a name="a40324d">
atan(y/x) take only one parameter y/x.
atan2(y,x) take two parameters y and x.

Earlier notes LiuHH suggest reader use atan2()
θ=theta=atan2(y,x) --- eq.LB02
to find argument angle when z=x+i*y is given.
atan2(y,x) return angle in RADIAN, not degree.
atan2(y,x) is correct, atan2(x,y) is wrong.
atan2(imag,real) is correct.

<a name="a40324e">
Base on atan(y/x) and x/y positive/negative
Principal argument of z -π<θ≦π
positive y axis x＝0, y＞0: arg(z)=+π/2
negative y axis x＝0, y＜0: arg(z)=-π/2
positive x axis x＞0, y＝0: arg(z)=0
negative x axis x＜0, y＝0: arg(z)=-π
<a name="a40324f">
Above table adjust atan(y/x) to atan2(y,x)
atan(y/x) output range from -π/2 to +π/2
atan2(y,x) output range from -π to +π
atan2(y,x) NOT output to -π, report -π as +π.
function atan2(y,x) need two parameters y,x
2009-03-17-11-15 Liu,Hsinhan accessed
http://personal.maths.surrey.ac.uk/st/C.Wulff/Modules/MS224/cvch1.pdf
save as cvch1.pdf ... cvch5.pdf
cvch1.pdf page 11/18 has above table.
2015-03-24-18-55 stop

2015-03-25-14-19 start
Next is study notes of
Analysis of a Complex Kind
1.5 Topology in the Plane ; 20:56

Lecturer is Prof. Petra Bonfert-Taylor.
Notes recorder is Liu,Hsinhan.
Recorded notes is about 60% match, not exact.
Liu,Hsinhan inserted text is blue Italic.
LiuHH notes may contain error!

<a name="docA202"> Begin video 5 of 33
Welcome back and welcome to lecture five in
our course "Analysis of a Complex Kind". Today
we will talk a little bit about topology in the
plane. Topology is study of shapes. We need to
name some shapes and find out how to write down
complex notation and learn some concept about
topology. In order to move to some really needed
application next week.

<a name="docA203">
Sets in complex plane
Let us start something very simple. A circle
need center and radius. Center z0 see eq.ACh5
Radius is r. See graph at video 0:42/20:56
Inside of a circle is called disk. Circle itself
is called circle of boundary.

<a name="docA204">
● Circles and disks: radius r and center at
z0=x0+i*y0 --- eq.ACh5
Br(z0)={z∈ℂ: z has distance less than r from z0} --- eq.ACh6
Br(z0) is disk of radius r centered at z0.
Kr(z0)={z∈ℂ: z has distance r from z0} --- eq.ACh7
Kr(z0) is circle of radius r centered at z0.
Br(z0) use inequality distance less than r
Kr(z0) use equality distance r a403261512

<a name="docA205">
How do we measure distance in complex plane?
Draw a picture. See graph at video 1:40/20:56
Suppose center z0=x0+i*y0 LiuHH guess z0=3+4i
Another point z=x+i*y LiuHH guess z=7+7i
On x-axis, draw x0 and x two points.
On y-axis, draw y0 and y two points.
Draw a line from z0=x0+i*y0 to z=x+i*y
Draw a right triangle from z0 to z as hypotenuse.
Draw horizontal side x-x0 and vertical side y-y0
● How do we measure distance?
Horizontal side, vertical side form a right angle.
Call hypotenuse as distance d. In order to find d,
I can just as well find two legs length and use
Pythagoras theorem to get d. How long are these
two legs?
The length of leg parallel to x-axis is x-x0
The length of leg parallel to y-axis is y-y0
<a name="docA207">
I can use Pythagoras theorem to find
● d=√[(x-x0)2+(y-y0)2] --- eq.ACh8
d=|(x-x0)+i*(y-y0)|  --- eq.ACh9
d=|z-z0|             --- eq.ACi0
Pythagoras theorem give us eq.ACh8 . eq.ACh8 is
same as eq.ACh9 the length of (x-x0)+i*(y-y0).
Complex number in eq.ACh9 is really z-z0.
<a name="docA208">
This distance d is actually the modulus of
complex number z-z0. With that notation I can
write down disk expression
Br(z0)={z∈ℂ: |z-z0|＜r} --- eq.ACi1
write down circle expression
Kr(z0)={z∈ℂ: |z-z0|=r} --- eq.ACi2
View from geometry, Br(z0) use inequality and
Br(z0) has two-dimension area. On the other
hand, Kr(z0) use equality, Kr(z0) is a one
dimensional curve. a403261520

<a name="docA209">
Definition (interior point)
Let E⊂ℂ. A point z0 is an interior point of E
if there is some r＞0 such that Br(z0)⊂E
E⊂ℂ say set E is within set ℂ.
c∈ℂ say element c is within set ℂ.
Let E be a complex number set //E⊂ℂ
A point z0 is interior point of E ? or not an
interior point of E ? This is determined by
next condition.
Take point z0 as center, if we can find a disk
Br(z0) centered at z0 and with radius r＞0 such
that disk Br(z0) is completely stay in set E,
then this point is an interior point of E .
<a name="docA210">
Counter example is
If set E has boundary point, then boundary point
is not interior point. Since disk centered at
boundary point always has partial in E and partial
outside of E for any radius r＞0.
If set E has NO boundary point, then limit point
is not interior point. Because limit point itself
do not belong to set E.

For example, if set E is (0,1] that is 0<x≦1
Point x=0.999 is interior point. If radius r=0.02
Consider disk center at 0.999 and radius=0.02
point 0.999-0.01=0.989 is in disk and in set (0,1]
point 0.999+0.01=1.009 is in disk but NOT in (0,1]
point 0.999-1.e-5 is in disk and in set (0,1]
point 0.999+1.e-5 is in disk and in set (0,1]
<a name="docA212">
Although radius r=0.02 find "NOT in set (0,1]"
BUT definition use "there is some r＞0",
r=2.e-5 determined point x=0.999 is interior
point. Because 0.999+1.e-5=0.99901 in set (0,1]
and 0.999+1.999e-5=0.99901999 in set (0,1]
2015-03-25-16-09 video 4:07/20:56

<a name="docA213">
2015-03-25-16-45 start
Definition (boundary point)
Let E⊂ℂ. A point b is a boundary point of E
if every disk around b contains a point in E
and a point not in E.
The boundary of the set E⊂ℂ, ∂E, is the set
of all boundary points of E.
∂E is a symbol of boundary of set E.

<a name="docA214">
Interior point definition mentioned "some r＞0"
Boundary point definition NOT mention "some r＞0"
Boundary point definition mentioned "every disk"
Disk Br(b) has center b and radius r. Change
center b or change radius r then change disk.
The word "every disk" is same as "every r＞0"

<a name="docA215">
For a given center b, change radius r change
disk. For "every disk" is for every radius r＞0.
Boundary point definition still mention for any
radius r＞0, this disk contain both point in E
and point not in E. Key point is for any r＞0.
No matter how small radius r is, boundary point
disk always contain both point in E and not in E.

For example, if set E is (0,1] that is 0<x≦1
Point x=1 is boundary point. If radius r=2.e-99
point 1-1.e-99 is in disk and in set (0,1]
point 1+1.e-99 is in disk but not in (0,1]
point 1+1.e-99999 is still not in (0,1]
video 5:12/20:56  2015-03-25-17-32

<a name="docA217">
Definition (open and closed sets) video 5:55/20:56
A set U⊂ℂ is open if every one of its points is
an interior point.
A set A⊂ℂ is closed if it contains all of its
boundary points.
Set contain partial of its boundary points
is neither open nor closed. a403261602

<a name="docA218">
Examples:
● {z∈ℂ :|z-z0|＜r} and
{z∈ℂ :|z-z0|＞r} are open
A ball centered at z0 with radius r are open.
Points on the edge are excluded, on-edge-points
are not in these set. because definition use
strictly less than r (strictly greater than r )
<a name="docA219">
● ℂ and ∅ are open
Whole complex plane is open, that is trivial.
Because in ℂ any center z any radius r always
contains complex points, no non-complex points.
Empty set ∅ is open, that is a little bit funky.
Empty set ∅ does not have interior points. Then
no center point, no disk can be used.
We define empty set ∅ is open. (please verify)
2015-03-25-18-01 stop at video 7:50/20:56

<a name="docA220">
2015-03-25-19-05 start
● {z∈ℂ :|z-z0|≤r} and
{z∈ℂ :|z-z0|=r} are closed set
They are closed set because these set definition
include all of its boundary points.
A circle {z∈ℂ :|z-z0|=r} is also closed set.
A circle has only "boundary points". A circle
satisfy closed set definition.
● ℂ and ∅ are closed // video 8:35/20:56
It is kind of unsettling.
ℂ does not have any boundary points, you could
say ℂ contain all boundary points.
∅ is closed, ∅ does not have anything, you could
say ∅ contain them all.
ℂ and ∅ two sets are both open and both closed.
open and closed in mathematics is different from
open or close for a door. A door can be open or
can be closed. A door cannot be both open and
closed. There are sets neither open nor closed.
<a name="docA222">
● {z∈ℂ :|z-z0|＜r}∪
{z∈ℂ :|z-z0|＝r and imag(z-z0)＞0}
is neither open nor closed.
Above definition, if r=1, it is a unit circle
upper half circle has solid line, lower half
circle has dash line. 2015-03-25-19-27
// video 10:50/20:56

<a name="docA223">
Definition Closure and interior of a set
Let E be a set in ℂ (complex number set)
The closure of E (E) is the set E together with
all of its boundary points.
E=E∪∂E --- eq.ACi3
。
The interior of E (E)is the set of all interior
points of E. // video 11:36/20:56
2015-03-25-19-48 stop

<a name="docA224">
2015-03-25-22-28 start
An example again.
______
● Br(z0)=Br(z0)∪Kr(z0) --- eq.ACi4
______
Br(z0)={z∈ℂ:|z-z0|≦r} --- eq.ACi5
The closure of disk of radius r centered at z0
is disk of radius r centered at z0 together
with its circle. Adding all the boundary points.
<a name="docA225">
______
● Kr(z0)=Kr(z0) --- eq.ACi6
The closure of a circle is the circle itself.
Because circle has only boundary points. No
extra boundary points could be added to it.
___________
● Br(z0)\{z0}={z∈ℂ:|z-z0|≦r} --- eq.ACi7
If a disk Br(z0) take its center away "\{z0}"
then take its closure. The closure will take
its center back. Because removed center is a
boundary point. Taking closure will take all
boundary points back, that is take center back.
// video 12:18/20:56
<a name="docA227">
● With E={z∈ℂ:|z-z0|≦r} --- eq.ACi8
。
E=Br(z0) --- eq.ACi9
E is disk together with its circle (include all
boundary points) If I want to take interior of
E the result is disk Br(z0) only and throw away
all boundary points.
<a name="docA228">
On the other hand, if E is circle Kr(z0) only
● With E=Kr(z0) --- eq.ACi9
。
E=∅ --- eq.ACj0
Circle has only boundary points.
Interior operation, throw away all boundary points.
What remain is an empty set ∅. // video 13:05/20:56

<a name="docA229">
Connectedness
Intuitively: A set is connected if it is "in one
piece" How do we make this precise?

We also need a notion of Connectedness.
Definition Connectedness
Two sets X,Y in ℂ are separated if there are
disjoint open set U,V so that X⊂U and Y⊂V.
A set W in ℂ is connected if it is impossible
to find two separated non-empty sets whose
union equals W.

<a name="docA230">
Disjoint mean two sets U,V have nothing in
common. That is if
U∩V=∅ --- eq.ACj1
is true, then two sets U,V are disjoint.
Example:
Suppose X=[0,1) and Y=(1,2] then X,Y two sets
are disjoint. To see this choose
U=B1(0) a disk center at 0 and radius=1
and
V=B1(2) a disk center at 2 and radius=1
X⊂B1(0) and Y⊂B1(2). We also know
B1(0)∩B1(2)=∅.
Both disk has no common point. Then verified
X,Y two sets are disjoint. // video 14:40/20:56
2015-03-25-23-20 stop

<a name="docA232">
2015-03-26-12-08 start at video 15:40/20:56
choose U=B1(0) and choose V=B1(2), thus
X∪Y=[0,2]\{1} is not connected.
It is easier to check two sets are not connected.
It is harder to check one set  is connected.

For open set, there is a much easier criterion
to check whether or not a set is connected. This
is true only in complex plane though.

<a name="docA233">
Theorem  //video 16:08/20:56
Let G be an open set in C. Then G is connected
if and only if any two points in G can be joined
in G by successive line segments.
We are not going to prove this theorem.
On the other hand if G is formed by two parts.
Choose one point in one part and choose other
point in second part. It is impossible to
connect two points without crossing undefined
section. Then set G is not connected.

<a name="docA234">
Definition Bounded sets //video 17:50/20:56
A set A in ℂ is bounded if there exists a number
R＞0 such that A⊂BR(0). If no such R exists then
A is called unbounded.
If there exist a disk BR(0) center at (0,0) with
finite radius R this disk contain whole set A.
Then set A is called bounded. If no such disk
can be found, this set is unbounded.

<a name="docA235">
Unbounded set example
If first ray from (0,0) to infinity at
phase angle θ1=30 degree.
If second ray from (0,0) to infinity at
phase angle θ2=60 degree.
The set from angle 30 degree to 60 degree is
unbounded. Because its radius extend to
infinity.

Another example of unbounded set is whole
half plane. All complex for which their real
part is positive. This is an unbounded set.

The whole complex plane itself is unbounded set.
There are plenty of unbounded set.

<a name="docA237">
//video 19:21/20:56
In ℝ (real number set), there are two directions
that give rise to ±∞, 1,2,3,4,5,... → ∞ and
-1,-2,-3,-4,-5,... → -∞

<a name="docA238">
In ℂ (complex number set) there is only one ∞
1, 2, 3, ...              → ∞
-1,-2,-3, ...              → ∞
i,2i,3i, ...              → ∞
1,2i,-3,-4i,5,6i,-7 ...   → ∞
In complex plane, no matter which direction
you go, always reach one infinity.
2015-03-26-12-55 stop

2015-03-27-13-57 start
Next is study notes of
Analysis of a Complex Kind
2.1 Complex Functions ; 26:38

Lecturer is Prof. Petra Bonfert-Taylor.
Notes recorder is Liu,Hsinhan.
Recorded notes is about 60% match, not exact.
Liu,Hsinhan inserted text is blue Italic.
LiuHH notes may contain error!

<a name="docA252"> Begin video 6 of 33
Welcome to week two of our course
"Analysis of a Complex Kind". This is first
lecture of complex functions.

<a name="docA253">
This week:
● Julia sets for quadratic polynomials.
● The Mandelbrot set.
● We laid the ground last week!
● Just a little more preparation
Complex functions (Lecture 1)
Sequences and limits (Lecture 2)
● We will need to study quadratic polynomial
of this form
f(z)=z2+c --- eq.ACj2

<a name="docA254">
Functions
● Recall: A function f:A→B is a rule that assigns
each element of A exactly one element of B.
● Example f:ℝ→ℝ f(x)=x2+1 --- eq.ACj3
● The graph helps us understand the function.

<a name="docA255">
A function f:A→B is a rule that assigns each
element of A exactly one element of B. For example
f(x)=x2+1 --- eq.ACj3
If input x=+2, eq.ACj3 output f( 2)=22+1=5
If input x=-3, eq.ACj3 output f(-3)=(-3)2+1=10
We often graph function in order to see them
better. Here we see graph //video 01:40/26:38
For example if x=0, eq.ACj3 output f(0)=02+1=1

Now imagine we study function not only plug in
real numbers into function. But possibly input
complex numbers. Graph work from real number to
real number, x axis is input, y axis is output.
We are able to plot real number function graph.
But for a complex function, input complex number
which has real component and imaginary component
real and imaginary are independent. In other
words, complex input graph need 2-dimensional
area. Similarly, graph complex function, output
complex number need another 2-dimensional area.
<a name="a40328a">
If input is a single complex curve and output
is another single complex curve. LiuHH merge
input complex plane with output complex plane.
Please see complex2.gif In this special case,
it is easy to see |sin(complex)|＜＝＞|complex|

<a name="docA257">
Complex Functions
● Now: f:ℂ→ℂ f(z)=z2+1 --- eq.ACj4
● How do we graph this? Need 4 dimensions?
● Writing
z=x+i*y  --- eq.ACj5
we see
w=f(z)=(x+i*y)2+1 --- eq.ACj6
w=f(z)=(x2-y2+1)+i*2xy --- eq.ACj7
w=f(z)=u(x,y)+i*v(x,y) --- eq.ACj8
where u and v are ℝ2→ℝ

<a name="docA258">
To graph complex functions need complex input
and complex output. For example
f(z)=z2+1 --- eq.ACj4
In eq.ACj4, input z is a complex, and output
f(z) is a complex too. How do we graph this?
We need four dimension to graph complex input
and complex output.
<a name="docA259">
Write input as   z=x+i*y  --- eq.ACj5
For complex function f(z)=z2+1 --- eq.ACj4
write it as w=f(z) see eq.ACj6.
Multiply out eq.ACj6 get eq.ACj7. In eq.ACj7,
assign real part as u(x,y)=(x2-y2+1) --- eq.ACj9
assign imag part as v(x,y)=2*x*y  --- eq.ACk0
Both u(x,y) and v(x,y) are ℝ2→ℝ that is
both u(x,y) and v(x,y) need two real numbers as
input and both u(x,y) and v(x,y) output one real
number. (this is ℝ2→ℝ )
u(x,y) output real number which is eq.ACj7 real part
v(x,y) output real number which is eq.ACj7 imag part
2015-03-27-14-54 stop

<a name="docA260">
2015-03-27-15-18 start at video 05:20/26:38
Graphing Complex Functions
● Idea: Consider TWO complex planes:
one for the domain (input points/curves)
one for the range (output points/curves)
input variable is z, input plane is z-plane.
output variable is w,output plane is w-plane.
● Analyze how geometric configurations in the
z-plane are mapped under f(z) to the w-plane

Example //at video 06:12/26:38
f(z)=z2 --- eq.ACk1
w=(x+i*y)2=(x2-y2)+2ixy --- eq.ACk2
Not so useful?
More useful in this case: polar coordinates!
z=r*eiθ --- eq.ACe7
then
<a name="docA262">
w=r2*e2iθ --- eq.ACk3
so
|w|=|z|2 --- eq.ACk4
and
arg(w)=2*arg(z) --- eq.ACk5
eq.ACk4 and eq.ACk5 really help us understanding
the geometrical meaning of f(z)=z2 --- eq.ACk1

<a name="docA263">
see video 07:28/26:38 for f(z)=z2 graph.
As z moves around a circle of radius r once,
w moves around a circle of radius r2 at double
speed twice.
see video 09:55/26:38 for f(z)=z2 movie.
see video 10:08/26:38 for f(z)=z2 mesh mapping.
see video 11:25/26:38 for f(z)=z2+1 graph.
<a name="docA264">
More complicated Functions
● How do we understand more complicated functions
such as
f(z)=z2+1 --- eq.ACk6
● Same idea!
Video graph show first step change from z to z2.
First step is square operation.
Second step change from z2 to z2+1
Second step is parallel moving operation (add 1).

2015-03-27-15-56 external mouse stop work
2015-03-27-15-58 turn off computer
2015-03-27-16-10 turn on  computer

<a name="docA265">
2015-03-27-16-19 start
at video 11:25/26:38 for f(z)=z2+1 graph.
for f(z)=z2+1 idea is the same, except take
this function apart into f(z)=z2 and separate
function just add one to the result. We first
look at w1 plane for f(z)=z2 (not add one)
We already know how this function work. See
|w|=|z|2 --- eq.ACk4
and
arg(w)=2*arg(z) --- eq.ACk5
Then we compose f(z)=z2 with adding one
is shift graph to the right one unit. The circle
is no longer centered at 0+0i, but centered at
1+0i. All together from z plane to w1 plane to
w plane get the final result f(z)=z2+1

<a name="docA267">
f(z)=z2+c --- eq.ACk7
c∈ℂ //c is a complex number.
● Same idea (again)!
In graph, complex c=1+i, then parallel move in
real direction +1 and parallel move in imaginary

<a name="docA268">
Iteration of Functions //video 13:35/26:38
Let
f(z)=z+1 --- eq.ACk8
then
● f2(z)=f(f(z))=(z+1)+1=z+2 --- eq.ACk9
● f3(z)=f(f2(z))=f(z+2)=(z+2)+1=z+3 --- eq.ACL0
● ●●●
● fn(z)=z+n --- eq.ACL1
● fn (read "Eff n") is called the n-th iterate
of f. (Not to be confused with n-th power of f)
<a name="docA269">
When we talk about Julia sets, Mandelbrot set we
talk about iteration of function. Let us try to
understand what that means. We are interested in
function of type f(f(z)), we plug in f(z) into
f(z).
2015-03-27-16-54 external mouse freeze one minute.
This period computer auto start external storage
hard drice e:\.  //video 13:56/26:38
2015-03-27-16-56 external mouse freeze on and off
on and off not in control
2015-03-27-16-58 turn off computer
2015-03-27-17-00 turn on  computer
<a name="docA270">
2015-03-27-17-05 start at video 13:51/26:38
Suppose
f(z)=z+1 --- eq.ACk8
This function move a point to the right by one.
What happen if apply this function again? It
will map that point one more time, goto z+2.
Then if apply f(z)=z+1 again, it goto z+3.
Look at number. Suppose z=i, what is f(z)=z+1?
f(z) is simply i+1 in that case. Apply one more
time. What is f(f(z)) ?
f(f(z))=f(i+1)=(i+1)+1=i+2 --- eq.ACL2
If apply f(z) one more time to that
f(f(f(z)))=f(i+2)=(i+2)+1=i+3 --- eq.ACL3
In general
● f2(z)=f(f(z))=(z+1)+1=z+2 --- eq.ACk9
● f3(z)=f(f2(z))=f(z+2)=(z+2)+1=z+3 --- eq.ACL0
You get the idea keep doing this n times, get
● fn(z)=z+n --- eq.ACL1

<a name="docA272">
f two is also written as f compose with f
f2 is also written as f◦f
f3 is also written as f◦f◦f
and so forth. In other words,
fn=f◦f◦f...f◦f◦f n times.
We read fn as "Eff n". Some time people write
fn as f○n with a little circle before n.
//2015-03-27-17-30 here video 18:10/26:38
f○n is not n-th power of f(z)
n-th power of f(z) is something entirely different.
If   f(z)=z+1 --- eq.ACk8
f○n(z)=z+n
[f(z)]n=[z+1]n

<a name="docA273">
Let me demonstrate to you the n-th power of f(z)
just as a comparison. // video 18:21/26:38
n-th power of f(z) is calculated f(z) then rise
the result to n-th power.
The notation of n-th iteration f○n(z) and
the notation of n-th power fn(z) or [f(z)]n
are easily confused.
<a name="docA274">
For now we use fn(z) as n-th iteration.
If we are interested at n-th power of f(z) we
write as (z+1)n that is zn+n*zn-1+... and so forth.
A complicated formula. It is not at all the same
thing as n-th iterated formula f○n(z)=z+n.
We are talking about n-th iterate of f(z).
2015-03-27-17-47 stop at video 19:32/26:38

http://freeman2.com/calcity0.pdf
http://freeman2.com/calcity1.jpg

<a name="docA275">
2015-03-27-21-57 start at video 19:35/26:38
Another Example
Let
f(z)=3z --- eq.ACL4
Then
● f2(z)=f(f(z))=f(3z)=3*3z=32z --- eq.ACL5
● f3(z)=f(f2(z))=f(32z)=3*32z=33z --- eq.ACL6
● ●●●
● fn(z)=3nz --- eq.ACL7

Two more examples
Let
f(z)=zd --- eq.ACL8
Then
● f2(z)=f(f(z))=f(zd)=(zd)d=zd*d --- eq.ACL9
● f3(z)=f(f2(z))=f(zd*d)=(zd*d)d=zd*d*d --- eq.ACm0
● ●●●
● fn(z)=z(dn) --- eq.ACm1

<a name="docA277">
Last example. Now let
f(z)=z2+2 --- eq.ACm2
Then
● f2(z)=f(f(z))=f(z2+2)=(z2+2)2+2=z4+4z2+6 --- eq.ACm3
● f3(z)=(z4+4z2+6)2+2=z8+... --- eq.ACm4
● fn(z) is a polynomial of degree 2n.

2015-03-27-22-17 here at video 23:47/26:38
<a name="docA278">
Julia Sets
● To study the Julia set of the polynomial
f(z)=z2+c --- eq.ACm5
We will study the behavior of the iterates
f,f2,f3,f4 ... fn ... of this function.
● The Julia set of f(z) is the set of points z
in the complex plane at which this sequence
of iterates behaves "chaotically".
● We thus need one more preparation: we need
to study sequences of complex numbers.
That is next!
<a name="docA279">
To study the Julia set we study quadratic
polynomial z2+c We do not need other type
quadratic polynomial. For example we do
NOT study
3*z2-15z+4 --- eq.ACm6
z2+c does not allow any linear term and
not allow any number before z2 Later we study
Julia set we will see why z2+c is enough.

<a name="docA280">
We need to study sequences of complex numbers.
For example, when //video 25:10/26:38
z=0 --- eq.ACm7
f(z) iteration is
0, f(0), f(f(0)), f(f(f(0))) ●●●
For f(z)=z2+c , the sequence is
0, c, c2+c, [c2+c]2+c ●●●
it keeps going like that. You can multiply
through to get a sequence of complex numbers.
Question we are going to ask is what that
sequence do when it keeps going? Does it run
wildly in complex plane? Does it go circle?
It goes to infinity? It goes to zero? It go
closer closer to a point? We will study
complex sequence. That is next topic.
2015-03-27-22-43 stop

<a name="a40328b">
2015-03-28-13-12 start
This afternoon, LiuHH and sister, brother,
sister-in-law will goto California City house.
Study notes may be on/off during next one month
because LiuHH and sister will move there. See
03/14/2015  08:15 PM 6,658,797 calcity0.pdf
03/27/2015  07:30 PM   611,555 calcity1.jpg
2015-03-28-13-15 stop

<a name="a40405">
Liu,Hsinhan and sister will move from
3727 West Ave. K11, Lancaster CA 93536 rent house
to 7325 Dogwood Ave California City self own house.
2015-April study notes page will stop a month.
Next is an interesting problem copied from online.

<a name="a40405a">
sin(α)+sin(α+β)+sin(α+2β)+...+sin(α+(n-1)*β)
=
sin(α + (n-1)β/2) * sin(n*β/2) / sin(β/2)

<a name="a40405b">

Trigonometry + Complex Numbers Sin,
Cos Alpha + Beta varying in AP Sum
of series
2015-04-02-16-18 no proof, just formula

2015-04-03-01-09 copy his equation below
```
<a name="a40405c">
screen mark next URL
http://zookeepersblog.wordpress.com/iit-jee-ipho-apho-physics-mathematics-video-for-std-9-to-12/
Above URL is invalid. 2015-04-05-14-14 "Oops! That page can’t be found"

<a name="a40405d">
2015-04-03-01-15 start build equation
sin(α)+sin(α+β)+sin(α+2β)+...+sin(α+(n-1)*β)

=
 sin ( α+(n-1)* β 2 )

 sin ( β 2 )
*
sin
(
n*β

2
)
--- eq.LAm1
Qnk7lTM20mI
2015-04-03-01-33 here

<a name="a40405e">
cos(α)+cos(α+β)+cos(α+2β)+...+cos(α+(n-1)*β)

=
 cos ( α+(n-1)* β 2 )

 sin ( β 2 )
*
sin
(
n*β

2
)
--- eq.LAm2
Qnk7lTM20mI
2015-04-03-01-42 here ; no proof, self go find proof.
width of above equation Two table a404030142
```<a name="a40405f">
Go online find how to proof eq.LAm1 and eq.LAm2

2015-04-05-14-21
sin(α+(n-1)*β)

Doubt in a formula cos(α)+cos(α+β)+cos(α+2β)+___+cos(α+{n-1}β)  Yahoo Answers
2015-04-05-14-23
eq.LAm3 to eq.LAn3 are reference.

<a name="a40405g">
sin(A+B)=sin(A)*cos(B)+cos(A)*sin(B) --- eq.LAm3
cos(A+B)=cos(A)*cos(B)-sin(A)*sin(B) --- eq.LAm4

sin(A-B)=sin(A)*cos(-B)+cos(A)*sin(-B)
=sin(A)*cos(B) -cos(A)*sin(B)  --- eq.LAm5
cos(A-B)=cos(A)*cos(-B)-sin(A)*sin(-B)
=cos(A)*cos(B) +sin(A)*sin(B)  --- eq.LAm6

sin(A+B)+sin(A-B)=2*sin(A)*cos(B)  --- eq.LAm7
cos(A+B)+cos(A-B)=2*cos(A)*cos(B)  --- eq.LAm8
cos(A-B)-cos(A+B)=2*sin(A)*sin(B)  --- eq.LAm9
2015-04-05-15-31 numerically correct.

<a name="a40405h">
sin(α)+sin(β)=2*sin((α+β)/2)*cos((α-β)/2) --- eq.LAn0
sin(α)-sin(β)=2*cos((α+β)/2)*sin((α-β)/2) --- eq.LAn1
cos(α)+cos(β)=2*cos((α+β)/2)*cos((α-β)/2) --- eq.LAn2
cos(α)-cos(β)=-2*sin((α+β)/2)*sin((α-β)/2) --- eq.LAn3
CRC Standard Math Table 27th ed. page 138
line 7 to 10.

<a name="a40405i">
Let S be the sum
S=sin(α)+sin(α+β)+sin(α+2β)+...+sin(α+(n-1)*β)  --- eq.LAn4
Let
t0=sin(α)    --- eq.LAn5
t1=sin(α+β)  --- eq.LAn6
t2=sin(α+2β) --- eq.LAn7
t3=sin(α+3β) --- eq.LAn8
.....
tn-1=sin(α+(n-1)β) --- eq.LAn9
therefore
S=∑[i=0,n-1]ti  --- eq.LAo0
<a name="a40405j">
Consider
2*t0*sin(β/2)=2*sin(α)*sin(β/2)=cos(α-β/2)-cos(α+β/2) --- eq.LAo1
2*t1*sin(β/2)=2*sin(α+β)*sin(β/2)=cos(α+β-β/2)-cos(α+β+β/2) --- eq.LAo2
2*t2*sin(β/2)=2*sin(α+2β)*sin(β/2)=cos(α+2β-β/2)-cos(α+2β+β/2) --- eq.LAo3
2*t3*sin(β/2)=2*sin(α+3β)*sin(β/2)=cos(α+3β-β/2)-cos(α+3β+β/2) --- eq.LAo4
2*t4*sin(β/2)=2*sin(α+4β)*sin(β/2)=cos(α+4β-β/2)-cos(α+4β+β/2) --- eq.LAo5
.....
2*tn-2*sin(β/2)=2*sin(α+(n-2)β)*sin(β/2)=cos(α+(n-2)β-β/2)-cos(α+(n-2)β+β/2) --- eq.LAo6
2*tn-1*sin(β/2)=2*sin(α+(n-1)β)*sin(β/2)=cos(α+(n-1)β-β/2)-cos(α+(n-1)β+β/2) --- eq.LAo7
<a name="a40405k">
Now sum above equations
Left side is
2*∑[i=0,n-1]ti*sin(β/2)  --- eq.LAo8
which is
2*S*sin(β/2)  --- eq.LAo9

<a name="a40405l">
Sum above equations right side,
red term cancel red term
blue term cancel blue term
purple term cancel purple term
After cancellation, right side is
cos(α-β/2)-cos(α+(n-1)β+β/2) --- eq.LAp1
Left side = right side get
2*S*sin(β/2)=cos(α-β/2)-cos(α+(n-1)β+β/2) --- eq.LAp2
<a name="a40405m">
Because
cos(A-B)-cos(A+B)=2*sin(A)*sin(B)  --- eq.LAm9
let (A-B)=M --- eq.LAp3
(A+B)=N --- eq.LAp4
solve for A, B get
A=(M+N)/2 --- eq.LAp5
B=(N-M)/2 --- eq.LAp6
cos(A-B)-cos(A+B)=2*sin(A)*sin(B) --- eq.LAp7
can be re-written as
<a name="a40405n">
cos(M)-cos(N)=2*sin((M+N)/2)*sin((N-M)/2) --- eq.LAp8
Then, cos(α-β/2)-cos(α+(n-1)β+β/2) is next
2*sin((α-β/2 + α+(n-1)β+β/2)/2)
*sin((α+(n-1)β+β/2 - [α-β/2])/2) --- eq.LAp9
Above line must have (square) bracket to guard sign.
<a name="a40405o">
Simplify to
cos(α-β/2)-cos(α+(n-1)β+β/2)
= 2*sin(α + (n-1)β/2)
*sin(n*β/2) --- eq.LAq0
Whole equation is
2*S*sin(β/2)=2*sin(α + (n-1)β/2)*sin(n*β/2) --- eq.LAq1
Since S is target, find
S=sin(α)+sin(α+β)+sin(α+2β)+...+sin(α+(n-1)*β)
=sin(α + (n-1)β/2) * sin(n*β/2) / sin(β/2) --- eq.LAq2
Equation eq.LAq2 is same as eq.LAm1.
<a name="a40405p">
discussion.
LiuHH notes may contain error!
To prove eq.LAm2 that is reader's home work.
2015-04-05-16-22 stop

<a name="a40405q">
The importance of eq.LAm1 and eq.LAm2 is that
if sin(α)+sin(α+β)+sin(α+2β)+...+sin(α+(n-1)*β)
has big n value, original problem has too many
calculation, but eq.LAm1 and eq.LAm2 give user
a much easier formula to get answer.
2015-04-05-20-32

Following is frequently needed strings ∞ z0
z=x+iy ∈ℂ z≠0+0i Cartesian coordinate π -π<θ≦π ∅
z∈ℂ z∈ℝ k∈ℤ z complex conjugate coordinate system
● √ ∛ ∜ mathematician ℂ ℍ　ℕ ℙ　ℚ　ℝ ℤ x2
z=r*eiθ --- eq.ACe7 ; z=r*eiπ/4
i=√-1 parabola parabolic
z = complex number, z=complex conjugate
<a style='text-decoration:overline;'>z</a>=complex conjugate
double overline NOT work z=z 2015-03-13-18-19
overequality NOT work z=z 2015-03-13-18-21
√π √p ε α → ∞ negligible θ0 auxiliary
an an+1 Taylor series representation
trigonometric Pythagoras Cartesian criterion criteria
implicity differentiation ┌ │ ┐ ┘ └ | ⇍　⇎　⇏
Α Β Γ Δ Ε Ζ Η Θ Ι Κ Λ Μ Ν Ξ Ο Π Ρ ΢ Σ Τ Υ Φ Χ Ψ Ω
α β γ δ ε ζ η θ ι κ λ μ ν ξ ο π ρ ς σ τ υ φ χ ψ ω
¬ ~ ∨ ⊻ ∧ → ← ↔ ⇒ ⇐ ⇔ ↑ ↓ ⇓ ⇑ ⇕　⇖　⇗　⇘　⇙
≦ ≠ ≧ ＜ ＝ ＞ ± ≡ ≈ ≌ ≒ ∏ ∑ √ ∛ ∜ ∝ → ∞ ⊕ ⊙ ⊗
〈v,w〉 ∈∀ ∂ ⊥ ∃ ∋ ∆ ∇ ∟ ∠ ∫ ∬ ∭ ∮ ∥ ○ ● ◎
∧ ∨ ∩ ∪ ∴ ∵ ∶ ∷ ⊂ ⊃ ⊄ ⊅ ⊆ ⊇ ⊿ ＋ － ＊ ／
ℂ ℍ　ℕ ℙ　ℚ　ℝ ℤ ℀　℁　ℂ
℃　℄　℅　℆　ℇ　℈　℉　ℊ　ℋ　ℌ
ℍ　ℎ　ℏ　ℐ　ℑ　ℒ　ℓ　℔　ℕ　№
℗　℘　ℙ　ℚ　ℛ　ℜ　ℝ　℞　℟　℠
℡　™　℣　ℤ　℥　Ω　℧　ℨ　℩　K
Å　ℬ　ℭ　℮　ℯ　ℰ　ℱ　Ⅎ　ℳ　ℴ
ℵ　ℶ　ℷ　ℸ　　⅍　ⅎ
■□　▢▣▤▥▦▧▨▩▪▫ × ÷ ° ◦ º ¹ ² ³
⇽　⇾　⇿　∀　∁　∂　∃　∄　∅　∆
∇　∈　∉　∊　∋　∌　∍　∎　∏　∐
∑　−　∓　∔　∕　∖　∗　∘　∙　√
∛　∜　∝　∞　∟　∠　∡　∢　∣　∤
∥　∦　∧　∨　∩　∪　∫　∬　∭　∮
∯　∰　∱　∲　∳　∴　∵　∶　∷　∸
∹　∺　∻　∼　∽　∾　∿　≀　≁　≂
≃　≄　≅　≆　≇　≈　≉　≊　≋　≌
≍　≎　≏　≐　≑　≒　≓　≔　≕　≖
≗　≘　≙　≚　≛　≜　≝　≞　≟　≠
≡　≢　≣　≤　≥　≦　≧　≨　≩　≪
≫　≬　≭　≮　≯　≰　≱　≲　≳　≴
≵　≶　≷　≸　≹　≺　≻　≼　≽　≾
≿　⊀　⊁　⊂　⊃　⊄　⊅　⊆　⊇　⊈
⊉　⊊　⊋　⊌　⊍　⊎　⊏　⊐　⊑　⊒
⊓　⊔　⊕　⊖　⊗　⊘　⊙　⊚　⊛　⊜
⊝　⊞　⊟　⊠　⊡　⊢　⊣　⊤　⊥　⊦
⊧　⊨　⊩　⊪　⊫　⊬　⊭　⊮　⊯　⊰
⊱　⊲　⊳　⊴　⊵　⊶　⊷　⊸　⊹　⊺
⊻　⊼　⊽　⊾　⊿　⋀　⋁　⋂　⋃　⋄
⋅　⋆　⋇　⋈　⋉　⋊　⋋　⋌　⋍　⋎
⋏　⋐　⋑　⋒　⋓　⋔　⋕　⋖　⋗　⋘
⋙　⋚　⋛　⋜　⋝　⋞　⋟　⋠　⋡　⋢
⋣　⋤　⋥　⋦　⋧　⋨　⋩　⋪　⋫　⋬
⋭　⋮　⋯　⋰　⋱　⋲　⋳　⋴　⋵　⋶
⋷　⋸　⋹　⋺　⋻　⋼　⋽　⋾　⋿　⌀
⌁　⌂　⌃　⌄　⌅　⌆　⌇　⌈　⌉　⌊

ts-1*e-t*dt ●●●
rigorously 嚴謹的 negligible 可忽略的
[直角三角形的]斜邊	hypotenuse (of a right triangle) a403171542

&#197;=Å angstrom A top circle.
Ů　ů has top circle.
Ǻ　ǻ　Ȅ　ȅ　Ȇ　ȇ Ӗ　ӗ
no E top circle a403281006

<a name="NumberSetsChar">
ℂ　Complex numbers ; 複數
ℍ　Hello ;
ℕ　Natural numbers ; 自然數（正整數及零）
ℙ　Prime numbers ; 素數
ℚ　Quotient, Rational numbers ; 有理數
ℝ　Real numbers ; 實數
ℤ　Zahl, Integers ; (from Zahl, German for integer) ;
ℤ　整數（正整數及零及負整數）
2015-03-13-18-52

```

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TUTor, English, 65 th .htm
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