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 
2015-02-07-14-05 LiuHH use YTD download 
https://www.youtube.com/playlist?list=PLX7wL1Q2SMr-gtmj-V-qrZqmS7WKSnrHY
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

<a name="cplx104aIndex"> 
Lecturer is Prof. Petra Bonfert-Taylor
1.1 History of Complex Numbers ; 19:26
http://www.youtube.com/watch?v=4TfWZjn6330
1.2 Algebra and Geometry in the Complex Plane ; 30:40
http://www.youtube.com/watch?v=6miBS7qRZEs
1.3 Polar Representation of Complex Numbers ; 32:32
http://www.youtube.com/watch?v=CZpUIPJHwMI
1.4 Roots of Complex Numbers ; 14:10
http://www.youtube.com/watch?v=DB_n9WvnjU0
1.5 Topology in the Plane ; 20:56
http://www.youtube.com/watch?v=2kg6bYa9gTc

<a name="cplx104aIndex2">
Lecturer is Prof. Petra Bonfert-Taylor
2.1 Complex Functions ; 26:38
http://www.youtube.com/watch?v=zWZApXIqXZw
2.2 Sequences and Limits of Complex Numbers ; 30:47
http://www.youtube.com/watch?v=cnDnnPG2j88
2.3 Iteration of Quadratic Polynomials, Julia Sets ; 25:42
http://www.youtube.com/watch?v=M9he1yxUgXk
2.4 How to Find Julia Sets ; 20:32
http://www.youtube.com/watch?v=d_5hfFAGZSg
2.5 The Mandelbrot Set ; 18:33
http://www.youtube.com/watch?v=c6rn2Oj_ErE

<a name="cplx104aIndex3">
Lecturer is Prof. Petra Bonfert-Taylor
3.1 The Complex Derivative ; 34:25
http://www.youtube.com/watch?v=XWVx0upRqu8
3.2 The Cauchy-Riemann Equations ; 28:49
http://www.youtube.com/watch?v=yszStQLd0p8
3.3 The Complex Exponential Function ; 24:44
http://www.youtube.com/watch?v=7mPmcr634lA
3.4 Complex Trigonometric Functions ; 21:23
http://www.youtube.com/watch?v=JYfAKA4gYE4
3.5 First Properties of Analytic Functions ; 24:54
http://www.youtube.com/watch?v=KhMCydBax3w

<a name="cplx104aIndex4">
Lecturer is Prof. Petra Bonfert-Taylor
4.1 Inverse Functions of Analytic Functions ; 25:28
http://www.youtube.com/watch?v=A5sNSHwZi8k
4.2 Conformal Mappings ; 26:16
http://www.youtube.com/watch?v=kZtH8H3G6rQ
4.3 Möbius transformations, Part 1 ; 27:02
http://www.youtube.com/watch?v=eIBU2FOrhIs
4.4 Möbius Transformations, Part 2 ; 17:35
http://www.youtube.com/watch?v=4nhJaCQODx0
4.5 The Riemann Mapping Theorem ; 15:15
http://www.youtube.com/watch?v=Wdi_Ybg2EL4

<a name="cplx104aIndex5">
Lecturer is Prof. Petra Bonfert-Taylor
5.1 Complex Integration ; 27:21
http://www.youtube.com/watch?v=A3XB1XHqxiQ
5.2 Complex Integration: Examples and First Facts ; 32:29
http://www.youtube.com/watch?v=62tTuv2d84Q

[[
19:10
…5.3 The Fundamental Theorem of Calculus for Analytic Functions
by Jacob Bains
•10 months ago
•542 views
]]
2015-02-07-14-37 use YTD download 
5.3 The Fundamental Theorem of Calculus for Analytic Functions ; Petra Bonfert-Taylor 19:10
https://www.youtube.com/watch?v=JZkKpUAFspU

5.4 Cauchy's Theorem and Integral Formula ; 32:36
http://www.youtube.com/watch?v=E_hscM-DZl8
5.5 Consequences of Cauchy's Theorem and Integral Formula ; 28:11
http://www.youtube.com/watch?v=6h-E7TVqFEA

<a name="cplx104aIndex6">
Lecturer is Prof. Petra Bonfert-Taylor
6.1 Infinite Series of Complex Numbers ; 22:03
http://www.youtube.com/watch?v=G_kjUZaRS0M
6.2 Power Series ; 25:22
http://www.youtube.com/watch?v=JjYjoKwHIv0
6.3 The Radius of Convergence of a Power Series ; 27:53
http://www.youtube.com/watch?v=NSEG3_GUhGo
6.4 The Riemann Zeta Function And The Riemann Hypothesis ; 22:53
http://www.youtube.com/watch?v=OXrXHypVbWw

Above total 29 files. 

<a name="cplx104aIndex7">
Lecturer is Prof. Petra Bonfert-Taylor
2015-02-07-14-49 use YTD download 4 videos
Residue theorem, 22:08
https://www.youtube.com/watch?v=LIj12IyhJ7o
branches of inverse function, 19:57
https://www.youtube.com/watch?v=dN2-iUFZWOQ
ArgumentPrinciple, 22:53
https://www.youtube.com/watch?v=oz9nyExEh_E
Differentiability in the Real Sense, 21:21
https://www.youtube.com/watch?v=aA9kczRVjIM

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, ...
2015-02-07-14-52 use YTD download 1 video
An Invitation to Complex Analysis ; 
Petra Bonfert-Taylor 3:00
https://www.youtube.com/watch?v=zQ1IxVLi8SA
2015-03-13-18-08 done build cplx104a list 


<a name="Taylor_Index"> 2015-02-08-21-30 LiuHH access https://www.youtube.com/playlist?list=PL68EAB0099AFEAAA5 2015-02-08-23-51 use YTD download Taylor Series •by MrYouMath •9 videos •2,037 views •Last updated on Apr 22, 2014 <a name="TaylorIndex"> Taylor Series - 1 - Motivation and Derivation.mp4 http://www.youtube.com/watch?v=ttnl671QBcE ●●● <a name="Gamma_Indx"> 2015-01-01-15-18 start On 2014-03-25-20-36 Liu,Hsinhan access http://www.youtube.com/playlist?list=PL3E4136E122545FBE find and download next 12 video files. Gamma Function •by MrYouMath •12 videos •5,940 views •1 hour, 48 minutes <a name="GammaIndex"> Gamma Function - Part 1 - Functional Equation http://www.youtube.com/watch?v=2iBNo4j3vRo ●●● Gamma Function - Part 7 - Euler Integral I http://www.youtube.com/watch?v=VF7ud3Al6d8 ●●● 2015-01-01-15-38 stop <a name="ZetaIndex"> 2014-10-29-07-56 https://www.youtube.com/playlist?list=PL32446FDD4DA932C9 2014-10-29-08-24 MrYouMath_17_files_Riemann.htm "Zeta Function - Part 1 - Convergence" http://www.youtube.com/watch?v=ZlYfEqdlhk0 ●●● "Zeta Function - Part 5 - Prime Zeta Function" http://www.youtube.com/watch?v=3eN9tQX3JJ4 ●●● "Zeta Function - Part 8 - Zeta of 2n - Part 1" http://www.youtube.com/watch?v=axQqExF7NsU ●●● "Zeta Function - Part 11 - Riemann Functional Equation I" http://www.youtube.com/watch?v=K6L4Ez4ZVZc 13/17 ●●●
<a name="docA001"> 2015-03-13-20-38 start Next is study notes of Analysis of a Complex Kind 1.1 History of Complex Numbers ; 19:26 http://www.youtube.com/watch?v=4TfWZjn6330 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"> About the Lecturer ● 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"> About this course ● 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. We talk about Riemann We talk about Cauchy We talk about Weierstrass We talk about i=√-1 We talk about limit We talk about e=2.718281828459045... We talk about open set We talk about connectedness <a name="docA006"> ● 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
jump to lecture docA012 (skip Riemann Zeta function) 
<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
http://www.youtube.com/watch?v=rGo2hsoJSbo
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
http://www.youtube.com/watch?v=rGo2hsoJSbo
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.
2014-03-25-18-28 use YTD download (YouTube Downloader)
http://www.youtube.com/watch?v=TnRnlJBecRg
MrYouMath: "Zeta Function - Part 12 - Riemann Functional Equation II"

eq.cz01 , eq.cz02 , eq.cz03 from next URL
2014-03-26-10-20 use YTD download
http://www.youtube.com/watch?v=rGo2hsoJSbo
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 
● Consider a quadratic equation 
  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 
Numbers. Look at quadratic equation 
  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 
<a name="docA016"> 
Where curve y=x2 and line y=m*x+b intersect 
is the equality points. 
Please watch video 6:00/19:25
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. 

<a name="docA021"> 
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.

<a name="docA026"> 
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">
Bombelli came about a thought 
  ∛{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 
<a name="docA031"> 
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 
 no imaginary number to start with.) 
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 
2015-03-14-23-01 done first proofread 
2015-03-15-12-34 done second proofread 


<a name="docA051"> 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 http://www.youtube.com/watch?v=6miBS7qRZEs 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'. <a name="docA056"> 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 Adding complex numbers 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 addition of two complex numbers. Real part result Real(z+w) = Real(z)+Real(w) x and u both no i=√-1 attach to them. x+u is the real part addition result. Then we add y and v, put i=√-1 in front of (y+v): i*(y+v). <a name="docA061"> 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 way to add two numbers is simply add 3 with -1 get 2 for addition real part result. add 5 with +2 get 7 for addition imaginary part result. Get (3+5i) + (-1+2i) = 2+7i <a name="docA063"> Graphically, complex number addition corresponds 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 two vector addition. 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. <a name="docA066"> 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. <a name="docA071"> 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 
<a name="docA076"> 
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
ask |z| > 0
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. 

<a name="docA081"> 
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)

<a name="docA086"> 
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.
(Please see video 25:35/30:39)
|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.
(Please see video 26:45/30:39)
|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

<a name="docA091"> 
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 2015-03-18-20-42 done first proofread 2015-03-18-21-09 done second proofread
<a name="docA095"> 2015-03-18-19-03 start To solve p(z)=5*z3+4*z2-2*z+14 --- eq.AC75 Please goto 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. <a name="docA096"> 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 ]] Please verify roots. <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 2015-03-18-21-47 upload tute0065.htm to 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
<a name="docA101"> 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 http://www.youtube.com/watch?v=CZpUIPJHwMI 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 will learn about the Polar Representation of Complex Numbers. Another representation which is different from the Cartesian representation we have been talk about so far. Polar Representation is helpful in understanding geometrically 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 about. <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) <a name="docA106"> 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) same answer. 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. <a name="docA111"> 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"> How about negative one? z=-1+0i 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 e=cos(θ)+i*sin(θ) --- eq.Eul1 Euler's formula eq.Eul1 Please see Tom M. Apostol prove eq.Eul1 Please see MrYouMath prove eq.Eul1 2015-03-19-20-17 here <a name="docA116"> Shortcut notation eq.Eul1 we are going to introduce. For now, e is a shortcut notation. e is a place holder for cos(θ)+i*sin(θ) Later on we will understand why we choose e 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 e as a symbol. ● z=r*[cos(θ)+i*sin(θ)] --- eq.AC86 become z=r*e --- eq.AC92 z=r*e is the polar form of z=x+iy. z=r*e is much shorter than z=r*[cos(θ)+i*sin(θ)] ● Note: ei(θ+2π)=e=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 e ? e=-1 --- eq.AC95 Again, I can write e 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. 2015-03-19-20-56 external mouse stop work video 13:35/32:31 2015-03-19-20-57 turn off computer 2015-03-19-21-20 turn on computer <a name="docA121"> 2015-03-19-21-25 start 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 ● |e|=1 --- eq.AC99 From eq.Eul1 e=cos(θ)+i*sin(θ) --- eq.Eul1 Therefore |e|=|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 |e|=√cos(θ)*cos(θ)+sin(θ)*sin(θ)=√1=1 --- eq.ACa1 e can be viewed as a complex number whose length r=1 and angle from positive x-axis is θ. <a name="docA124"> ___ ● e=e-iθ --- eq.ACa2 What is the complex conjugate of e ? 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 e=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 minus sign inside. So -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 e=cos(θ)+i*sin(θ) --- eq.Eul1 [e ≠ cos(+θ)+i*sin(-θ) ] cosine is an even function, cos(θ)=cos(-θ) --- eq.ACa5 <a name="docA126"> 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/e=e-iθ --- eq.ACa7 What is one over e ? We have talked about the reciprocal of a complex number. see eq.AC57
<a name="docA128">
Reciprocal of a complex number e
 
1

e
=
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 e 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 
  e*e-iθ=[cos(θ)]2+[sin(θ)]2=1 --- eq.ACa9 
This calculation end up to 
  1/e=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)=e*e --- eq.ACb0 
Here we meet another Greek symbol φ for angle. 
From eq.Eul1  e=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 
<a name="docA131"> 
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)] 
     =e*e --- 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 
adding 2π. An example 
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. 

<a name="docA136"> 
● Consider 
  z1=r1*e1 --- eq.ACc3
  z2=r2*e2 --- eq.ACc4
What is the polar form of z1z2?
Just plug in get 
  z1z2=r1*e1 * r2*e2 --- eq.ACc5 
  z1z2=(r1r2)*ei(φ12) --- eq.ACc6 
eq.ACc6 is the answer. 
Video 26:16/32:31 has picture. 

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

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

<a name="docA139">
Recall e 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 ● [e]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 For n=3 , eq.ACd3 reads [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 <a name="docA141"> 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 2015-03-21-22-22 done first proofread 2015-03-22-16-38 done second proofread 2015-03-20-21-34 external mouse stop work 2015-03-20-21-36 restart computer
<a name="docA151"> 2015-03-22-19-38 start Next is study notes of Analysis of a Complex Kind 1.4 Roots of Complex Numbers ; 14:10 http://www.youtube.com/watch?v=DB_n9WvnjU0 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 going to talk about today. <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*e --- eq.ACe6 z=r*e --- eq.ACe7 w=ρ*e is given, one value. z=r*e is answer, n values. Here introduce another Greek symbol ρ for radius of w. Because r is used for radius of z. The equation zn=w --- eq.ACe5 becomes <a name="docA155"> rn*einθ=ρ*e --- 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=e --- eq.ACf0 <a name="docA156"> eq.ACf0 gives nθ=φ+2kπ --- eq.ACf1 eq.ACf1 adding 2kπ, because sin(α)=sin(α+2kπ) and cos(α)=cos(α+2kπ) Thus, from eq.ACe9 get r = nρ --- eq.ACf2 ρ is radius which is a positive real number, therefore nρ 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*e --- 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. 
ρ is modulus (radius), ρ=|4i|=|4|*|i|=4*1=4 
φ is argument, 4i=(0,4) on positive y-axis
φ=angle from positive x-axis to positive y-axis
φ=90 degree=π/2 radian. 
n=the root number; square root has n=2.
2015-03-23-15-09
<a name="docA161"> 
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 e 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">
Second answer is negative of first answer.
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*e --- 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*e --- eq.ACf9
Here e=-1 carry minus sign.
so ρ=8, φ=π and n=3. 
ρ is modulus (radius), ρ=|-8|=8 
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 
φ=180 degree=π radian
n=the root number; Cubed roots has n=3.
2015-03-23-16-29
I can go ahead and find answer. 
<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 ; 
eq.ACg0 become second answer: 
  ∛-8=∛8*e=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 ; 
eq.ACg0 become third answer: 
  ∛-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"> 
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=e=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

zeroth answer is same as eighth answer of unity.
Because eighth root operation, zeroth answer and 
eighth answer repeat. 
2015-03-23-17-47 stop 
2015-03-23-19-33 done first proofread
2015-03-23-19-55 done second proofread

<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*e --- 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">
eq.LB01 has disadvantage. 
  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 
θ1 is in fourth quadrant
θ2 is in second quadrant
But atan(y/x) output both to fourth quadrant.
second quadrant answer lost. 
Similarly atan((negative)/(negative)) third 
quadrant answer become first quadrant answer.

<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 
atan2(y,x) output follow next table 

Principal argument of z -π<θ≦π first quadrant x>0, y>0: arg(z)=tan-1(y/x) second quadrant x<0, y>0: arg(z)=π+tan-1(y/x) third quadrant x>0, y<0: arg(z)=tan-1(y/x) fourth quadrant x<0, y<0: arg(z)=-π+tan-1(y/x) 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 to find answer in correct quadrant. 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
<a name="docA201"> 2015-03-25-14-19 start Next is study notes of Analysis of a Complex Kind 1.5 Topology in the Plane ; 20:56 http://www.youtube.com/watch?v=2kg6bYa9gTc 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 <a name="docA206"> ● 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 Eℂ say set E is within set ℂ. cℂ 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. <a name="docA211"> 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] Next change radius to r=2.e-5 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. <a name="docA216"> 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] Even if radius r=2.e-99999 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. <a name="docA221"> ● 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. <a name="docA226"> ___________ ● 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. <a name="docA231"> 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. <a name="docA236"> 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 2015-03-26-16-51 done first proofread 2015-03-26-19-02 done second proofread
<a name="docA251"> 2015-03-27-13-57 start Next is study notes of Analysis of a Complex Kind 2.1 Complex Functions ; 26:38 http://www.youtube.com/watch?v=zWZApXIqXZw 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 <a name="docA256"> 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| Recall: |sin(real)|<=|real| a403281031 proofread <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 <a name="docA261"> 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*e --- 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 <a name="docA266"> Then we compose f(z)=z2 with adding one operation. In w plane just add one. 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"> How about ... //video 12:45/26:38 ● How about 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 direction +i get final answer. <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 <a name="docA271"> 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 2015-03-27-20-02 upload file to 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 <a name="docA276"> 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. <a name="docA281"> 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 2015-03-28-13-03 done first proofread <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"> 2014-03-26-17-20 download http://www.youtube.com/watch?v=Qnk7lTM20mI 2015-04-02-16-14 watch YouTube Qnk7lTM20mI 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">
2014-03-26-17-20 download
http://www.youtube.com/watch?v=Qnk7lTM20mI
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
http://www.google.com/advanced_search
sin(α+(n-1)*β)
https://www.google.com/search?as_q=&as_epq=sin(%CE%B1%2B(n-1)*%CE%B2)&as_oq=&as_eq=&as_nlo=&as_nhi=&lr=&cr=&as_qdr=all&as_sitesearch=&as_occt=any&safe=images&tbs=&as_filetype=&as_rights=&gws_rd=ssl

Doubt in a formula cos(α)+cos(α+β)+cos(α+2β)+___+cos(α+{n-1}β)  Yahoo Answers
2015-04-05-14-23
https://in.answers.yahoo.com/question/index?qid=20130113041031AAEdQvY
Below follow in.answers.yahoo.com
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">
Liu,Hsinhan follow 
https://in.answers.yahoo.com/question/index?qid=20130113041031AAEdQvY
discussion. 
LiuHH notes may contain error!
Please verify. Above solved eq.LAm1
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



<a name="docA999"> 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*e --- 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〉 ∈∀ ∂ ⊥ ∃ ∋ ∆ ∇ ∟ ∠ ∫ ∬ ∭ ∮ ∥ ○ ● ◎  ∧ ∨ ∩ ∪ ∴ ∵ ∶ ∷ ⊂ ⊃ ⊄ ⊅ ⊆ ⊇ ⊿ + - * / 2014-12-09-13-16 add next for hollow 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 倒數;互反;互逆 reciprocal 象限 quadrant 畢氏定理 Pythagoras theorem a403251521 &#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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