<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.
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
n^{s}
=
p=∞
∏
p=prime
p^{s}
p^{s}-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
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,∞]{-1^{n+1}/n^s}/(1-2^{1-s}) --- eq_cz02a
or
ζ(s)={[1-2^(1-s)]^(-1)}*∑[n=1,∞]{-1^{n+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
1-2^{1-s}
*
n=∞
∑
n=1
(-1)^{n+1}
n^{s}
--- 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)=2^{s}*PI^{s-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)
=
2^{s}
*
π^{s-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
x^{2}=m*x+b --- eq.AC04
● Solutions are
x
=
m
2
±
√
[
m^{2}
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=x^{2} --- eq.AC06
and
y=m*x+b --- eq.AC07
In y=x^{2}, x^{2} 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=x^{2} 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
x^{2}=m*x+b --- eq.AC04
Its solution is
x=(m/2) ± √[(m^{2}/4)+b] --- eq.AC05
If you look at eq.AC04 by itself, it really
represent the intersection of graph y=x^{2} and
another graph y=m*x+b this is just a line.
eq.AC04 says curve y=x^{2} equal line y=m*x+b
<a name="docA016">
Where curve y=x^{2} 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) ± √[(m^{2}/4)+b] --- eq.AC05
● What if (m^{2}/4)+b < 0 ?
● In particular
x^{2}=-1 --- eq.AC08
has no real solution.
● It is often argued that x^{2}=-1 led to i=√-1 .
● But ... Historically, no interest in non-real
solutions. Since the graphs of y=x^{2} 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 x^{2}=-1 it has no real solution.
It is often argued that this x^{2}=-1 led to invention
of complex number i=√-1 . But, historically, there
is no interest in non-real solutions.
<a name="docA019">
Because in x^{2}=-1 case, simply curve y=x^{2} and
line y=m*x+b do not intersect.
y=x^{2} 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
x^{3}=p*x+q --- eq.AC09
● Represents intersection of
y=x^{3} --- 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 x^{2} term here. It is just an
example of cubic equation. For eq.AC09 we look at
y=x^{3} --- eq.AC10
and
y=p*x+q --- eq.AC11
Solution of eq.AC09 represent the intersection of
cubic curve y=x^{3} and line y=p*x+q . You can see
that cubic curve y=x^{3} go all the way from plus
infinity to minus infinity. No matter what line
you draw, y=p*x+q always intersect with y=x^{3}.
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
x^{3}=p*x+q --- eq.AC09
has a solution given by
x
=
∛
{
√
[
q^{2}
4
－
p^{3}
27
]
＋
q
2
}
－
∛
{
√
[
q^{2}
4
－
p^{3}
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
x^{3}=-6*x+20 --- eq.AC13
Del Ferro and Tartaglia followed by Cardano showed
x^{3}=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
q^{2}/4 = 20*20/4=100
p^{3}/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, 2^{3} 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
x^{3}=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
x^{3}=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} = a^{3}+3*a^{2}*b+3*a*b^{2}+b^{3} --- eq.AC20
To use eq.AC20 right here in eq.AC18
(2+√-1)^{3} = 2+√-121 --- eq.AC18
(2+√-1)^{3} = 2^{3}+3*2^{2}*√-1+3*2*√-1^{2}+√-1^{3} --- eq.AC21
Because
√-1^{2}=-1 --- eq.AC22
and
√-1^{3} = √-1^{2}*√-1 = -√-1 --- eq.AC23
<a name="docA031">
eq.AC21 change to eq.AC24 below
(2+√-1)^{3} = 2^{3}+3*2^{2}*√-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
x^{3}=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 ℝ^{2}<a name="docA058">Set of real numbers: 'ℝ', both x-axis and
y-axis are real is ℝ^{2} 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}=x^{2}+y^{2} --- 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+i^{2}yv --- 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 i^{2} has the property -1, then i^{2} is just -1.
<a name="docA067">
"Suppose all the regular rules held" is the
motivation and i^{2}=-1 is the complex rule
(i^{2}=-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 i^{2}yv 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
(z_{1}z_{2})z_{3}=z_{1}(z_{2}z_{3}) --- eq.AC41
commutative
z_{1}z_{2}=z_{2}z_{1} --- eq.AC42
distributive
z_{1}(z_{2}+z_{3})=z_{1}z_{2}+z_{1}z_{3} --- 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
i^{2}=(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 i^{2}
Use the rule of multiplication find i^{2}=-1 as
shown in eq.AC45 . i squared is indeed -1.
Multiplication we defined has the property
that we wanted it to have i^{2}=-1.
<a name="docA071">
What is i cubed?
i^{3}=i^{2}*i=-1*i=-i --- eq.AC46
From associative law
i^{3}=i*i^{2}=i*(-1)=-i --- eq.AC47
i to the fourth power
i^{4}=i^{2}*i^{2}=(-1)*(-1)=+1 --- eq.AC48
i to the fifth power
i^{5}=i^{4}*i=1*i=i --- eq.AC49
i to the sixth power
i^{6}=i^{4}*i^{2}=1*(-1)=-1 --- eq.AC50
.....
keeps going, i=i, i^{2}=-1, i^{3}=-i, i^{4}=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)
u^{2}+v^{2}+i(-uv+vu)
=
xu+yv
u^{2}+v^{2}
+
i*
yu-xv
u^{2}+v^{2}
--- eq.AC52
u^{2}+v^{2} 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="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
x^{2}+y^{2}
--- 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)=x^{2}+y^{2}=|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 conjugatez.
<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)=x^{2}+y^{2}=|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
<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
<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}=x^{2}+y^{2} --- eq.AC68
x and y both are real numbers. (video 25:05/30:39)
x^{2}+y^{2} 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}＞2^{2} is OK.
In complex |2i|＞|1| ok, (2i)^{2}＞1^{2} 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 a_{0}, a_{1} ●●● a_{n} are complex numbers with a_{n}≠0,
then the polynomial
p(z)=a_{n}z^{n}+a_{n-1}z^{n-1}+●●●+a_{1}z+a_{0} --- eq.AC73
has n roots z_{1}, z_{2} ●●●, z_{n} in ℂ.
It can be factored as
p(z)=a_{n}(z-z_{1})(z-z_{2})●●●(z-z_{n}) --- eq.AC74
<a name="docA092">
For example
p(z)=5*z^{3}+4*z^{2}-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)=a_{3}(z-z_{1})(z-z_{2})(z-z_{3}) --- 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)=x^{2}+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=x^{2}+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
z^{2}+1=(z+i)*(z-i) --- eq.AC78
i and -i let z^{2}+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*z^{3}+4*z^{2}-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 1Acer2 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^{iθ}=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^{iθ} is a shortcut notation.
e^{iθ} is a place holder for cos(θ)+i*sin(θ) Later
on we will understand why we choose e^{iθ} 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^{iθ} as a symbol.
● z=r*[cos(θ)+i*sin(θ)] --- eq.AC86
become
z=r*e^{iθ} --- eq.AC92
z=r*e^{iθ} is the polar form of z=x+iy.
z=r*e^{iθ} is much shorter than z=r*[cos(θ)+i*sin(θ)]
● Note:
e^{i(θ+2π)}=e^{iθ}=e^{i(θ+4π)}=...=e^{i(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
e^{iπ/2}=i --- eq.AC94
What e^{iπ/2} really stand for?
e^{iπ/2} stand for cos(π/2)+i*sin(π/2)
Since cos(π/2)=0 and sin(π/2)=1, then
e^{iπ/2}=cos(π/2)+i*sin(π/2)=0+1*i=i
Another way to look at e^{iπ/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^{iπ} ?
e^{iπ}=-1 --- eq.AC95
Again, I can write e^{iπ} 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
e^{iπ/4}=(1+i)/√2 --- eq.AC96
radius is one. Angle is π/4. This is (1/√2,1/√2)
point in complex plane.
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<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 e^{iπ/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 e^{iπ/4} is 1/√2+i/√2
Another form is √2/2+i*√2/2 .
<a name="docA122">
Finally,
e^{2πi}=1 --- eq.AC97
What is that. By definition
e^{2πi}=1*[cos(2π)+i*sin(2π)]=1*[1+0i]=1 --- eq.AC98
You possibly see e^{2πi}=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
e^{2πi}=1 combine five most important constants
in mathematics into one equation.
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<a name="docA123">
2015-03-20-14-57 start
Properties of the Exponential Notation
● |e^{iθ}|=1 --- eq.AC99
From eq.Eul1 e^{iθ}=cos(θ)+i*sin(θ) --- eq.Eul1
Therefore
|e^{iθ}|=|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^{iθ}|=√cos(θ)*cos(θ)+sin(θ)*sin(θ)=√1=1 --- eq.ACa1
e^{iθ} can be viewed as a complex number whose
length r=1 and angle from positive x-axis is θ.
<a name="docA124">
___
● e^{iθ}=e^{-iθ} --- eq.ACa2
What is the complex conjugate of e^{iθ} ?
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^{iθ}=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^{iθ}=cos(θ)+i*sin(θ) --- eq.Eul1
[e^{iθ} ≠ 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 e^{i(-θ)}=e^{-iθ}
Here we showed eq.ACa3 is true.
<a name="docA127">
● 1/e^{iθ}=e^{-iθ} --- eq.ACa7
What is one over e^{iθ} ?
We have talked about the reciprocal of a complex
number. see eq.AC57
<a name="docA129">
2015-03-20-16-18 here
Multiply top and bottom with complex conjugate
of e^{iθ} 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^{iθ}*e^{-iθ}=[cos(θ)]^{2}+[sin(θ)]^{2}=1 --- eq.ACa9
This calculation end up to
1/e^{iθ}=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
● e^{i(θ+φ)}=e^{iθ}*e^{iφ} --- eq.ACb0
Here we meet another Greek symbol φ for angle.
From eq.Eul1 e^{iθ}=cos(θ)+i*sin(θ) --- eq.Eul1
Change θ to θ+φ we have
e^{i(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
e^{i(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
e^{i(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^{iθ}*e^{iφ} --- 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(z_{1}z_{2})=arg(z_{1})+arg(z_{2}) --- 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(z_{1}z_{2})=arg(z_{1})+arg(z_{2}) --- eq.ACb8
Left side and right side may differ by multiple
of 2*π.
<a name="docA136">
● Consider
z_{1}=r_{1}*e^{iφ1} --- eq.ACc3
z_{2}=r_{2}*e^{iφ2} --- eq.ACc4
What is the polar form of z_{1}z_{2}?
Just plug in get
z_{1}z_{2}=r_{1}*e^{iφ1} * r_{2}*e^{iφ2} --- eq.ACc5
z_{1}z_{2}=(r_{1}r_{2})*e^{i(φ1+φ2)} --- eq.ACc6
eq.ACc6 is the answer.
Video 26:16/32:31 has picture.
<a name="docA137">
De Moivre's Formula
● e^{iθ}*e^{iθ}=e^{i(θ+θ)}=e^{i*2θ} --- eq.ACc7
We can use eq.ACb0
e^{i(θ+φ)}=e^{iθ}*e^{iφ} --- eq.ACb0
set θ=φ to get eq.ACc7
eq.ACc7 relate to De Moivre's Formula
eq.ACc7 say e^{iθ} squared is same as e^{i*2θ}.
How about e^{iθ} cubed?
[e^{iθ}]^{3}=[e^{iθ}]^{2} * [e^{iθ}] --- eq.ACc8
<a name="docA138">
In eq.ACc8 e^{iθ} cubed apply e^{iθ} square eq.ACc7
get
● [e^{iθ}]^{3}=e^{i*2θ} * e^{iθ} --- eq.ACc9
By the rule we already established, find
[e^{iθ}]^{3}=e^{i*3θ} --- eq.ACd0
Similarly, you find e^{iθ} to power n is
● [e^{iθ}]^{n}=e^{i*nθ} --- eq.ACd1
eq.ACd1 is also true for negative n.
[e^{iθ}]^{-1}=1/e^{iθ}=e^{-iθ} --- eq.ACd2
<a name="docA139">
Recall e^{iθ} 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^{iθ}]^{n}=e^{i*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)=cos^{3}(θ)-3*cos(θ)*sin^{2}(θ) --- eq.ACd4
sin(3θ)=3*cos^{2}(θ)*sin(θ)-sin^{3}(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} = a^{3}+3*a^{2}*b+3*a*b^{2}+b^{3} --- 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}=
cos^{3}(θ)+3*cos^{2}(θ)*[i*sin(θ)]
+3*cos(θ)*[i*sin(θ)]^{2}+[i*sin(θ)]^{3} --- eq.ACd9
[cos(θ)+i*sin(θ)]^{3}=
cos^{3}(θ)+3*i*cos^{2}(θ)*[sin(θ)]
-3*cos(θ)*[sin^{2}(θ)]-[i*sin^{3}(θ)] --- eq.ACe0
//remind: i*i=-1 ; i*i*i=-i ;
<a name="docA142">
[cos(θ)+i*sin(θ)]^{3}=
cos^{3}(θ)-3*cos(θ)*[sin^{2}(θ)]
3*i*cos^{2}(θ)*[sin(θ)]-[i*sin^{3}(θ)] --- eq.ACe1
[cos(θ)+i*sin(θ)]^{3}=
cos^{3}(θ)-3*cos(θ)*[sin^{2}(θ)]
i*[3*cos^{2}(θ)*sin(θ)-sin^{3}(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)=cos^{3}(θ)-3*cos(θ)*sin^{2}(θ) --- eq.ACe3
eq.ACe3 and eq.ACd4 are identical
cos(3θ)=cos^{3}(θ)-3*cos(θ)*sin^{2}(θ) --- eq.ACd4
<a name="docA144">
imag(eq.ACe2) = imag(eq.ACd6) get
sin(3θ)=3*cos^{2}(θ)*sin(θ)-sin^{3}(θ) --- eq.ACe4
eq.ACe4 and eq.ACd5 are identical
sin(3θ)=3*cos^{2}(θ)*sin(θ)-sin^{3}(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)
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2015-03-22-16-38 done second proofread
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<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
z^{n}=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^{iφ} --- eq.ACe6
z=r*e^{iθ} --- eq.ACe7
w=ρ*e^{iφ} is given, one value.
z=r*e^{iθ} is answer, n values.
Here introduce another Greek symbol ρ for
radius of w. Because r is used for radius of z.
The equation
z^{n}=w --- eq.ACe5
becomes
<a name="docA155">
r^{n}*e^{inθ}=ρ*e^{iφ} --- eq.ACe8
Compare eq.ACe8 left side with right side.
Both side are the same, then
both modulus are equal
r^{n}=ρ --- eq.ACe9
and both arguments are equal
e^{inθ}=e^{iφ} --- 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 = ^{n}√ρ --- eq.ACf2
ρ is radius which is a positive real number,
therefore ^{n}√ρ is taking regular n-th root of rho.
No complex calculation involved in eq.ACf2.<a name="docA157">
from eq.ACf1 nθ=φ+2kπ , k∈ℤ (ℤ =integer set) so
θ = φ/n + 2kπ/n --- eq.ACf3
In eq.ACf3 look k values from 0,1,2,... up to n-1
I have exhausted all possibilities. After k=n-1
k values start repeat (k=0 and k=n are identical)
When I look at k=n for example. When k=n, what
would I get? I get θ = φ/n + 2kπ/n = φ/n + 2nπ/n
n cancel out, so get θ = φ/n + 2π. But I already
look at θ = φ/n when k=0. φ/n and φ/n + 2π are
same angle. I get different angle if k=0,1,2,...
up to k=n-1 . I get all of my different solutions
We write //2015-03-22-20-36 here
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
w^{1/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^{iφ} --- eq.ACe6
We need find out what is ρ and what is φ.
4i=4*e^{iπ/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
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*e^{iπ/4} = √2+i*√2 --- eq.ACf7
If k=1, 2kπ/2=π, eq.ACf6 become second answer:
√4i = 2*e^{i(π+π/4)} =-√2-i*√2 --- eq.ACf8
In eq.ACf8 factor e^{iπ} rotate 180 degree
change sign to both real and imaginary.
2*e^{i(π+π/4)} is actually negative of 2*e^{i(π/4)}<a name="docA163">
Second answer is negative of first answer.
Square root has only two answers.
Remember e^{iπ/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^{iφ} --- 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^{iπ} --- eq.ACf9
Here e^{iπ}=-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.
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*e^{iπ/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^{iπ}=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*e^{i5π/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*e^{i0} --- 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.
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:
1^{1/8}=e^{iπ/4}=cos(π/4)+i*sin(π/4) --- eq.ACg7
If k=2, 2kπ/8=π/2, eq.ACg6 become second answer:
1^{1/8}=e^{iπ/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:
1^{1/8}=e^{i3π/4}=cos(3π/4)+i*sin(3π/4) --- eq.ACg9
If k=4, 2kπ/8=π, eq.ACg6 become fourth answer:
1^{1/8}=e^{iπ}=cos(π)+i*sin(π)=-1 --- eq.ACh0
<a name="docA173">
If k=5, 2kπ/8=5π/4, eq.ACg6 become fifth answer:
1^{1/8}=e^{i5π/4}=cos(5π/4)+i*sin(5π/4) --- eq.ACh1
If k=6, 2kπ/8=3π/2, eq.ACg6 become sixth answer:
1^{1/8}=e^{i3π/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:
1^{1/8}=e^{i7π/4}=cos(7π/4)+i*sin(7π/4) --- eq.ACh3
If k=0, 2kπ/8=0, eq.ACg6 become zeroth answer:
1^{1/8}=e^{0}=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^{iθ}--- 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 tableeq.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 z_{0} 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
z_{0}=x_{0}+i*y_{0} --- eq.ACh5
B_{r}(z_{0})={z∈ℂ: z has distance less than r from z_{0}} --- eq.ACh6
B_{r}(z_{0}) is disk of radius r centered at z_{0}.
K_{r}(z_{0})={z∈ℂ: z has distance r from z_{0}} --- eq.ACh7
K_{r}(z_{0}) is circle of radius r centered at z_{0}.
B_{r}(z_{0}) use inequality distance less than r
K_{r}(z_{0}) 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 z_{0}=x_{0}+i*y_{0}LiuHH guess z_{0}=3+4i
Another point z=x+i*y LiuHH guess z=7+7i
On x-axis, draw x_{0} and x two points.
On y-axis, draw y_{0} and y two points.
Draw a line from z_{0}=x_{0}+i*y_{0} to z=x+i*y
Draw a right triangle from z_{0} to z as hypotenuse.
Draw horizontal side x-x_{0} and vertical side y-y_{0}<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-x_{0}
The length of leg parallel to y-axis is y-y_{0}<a name="docA207">
I can use Pythagoras theorem to find
● d=√[(x-x_{0})^{2}+(y-y_{0})^{2}] --- eq.ACh8
d=|(x-x_{0})+i*(y-y_{0})| --- eq.ACh9
d=|z-z_{0}| --- eq.ACi0
Pythagoras theorem give us eq.ACh8 . eq.ACh8 is
same as eq.ACh9 the length of (x-x_{0})+i*(y-y_{0}).
Complex number in eq.ACh9 is really z-z_{0}.
<a name="docA208">
This distance d is actually the modulus of
complex number z-z_{0}. With that notation I can
write down disk expression
B_{r}(z_{0})={z∈ℂ: |z-z_{0}|＜r} --- eq.ACi1
write down circle expression
K_{r}(z_{0})={z∈ℂ: |z-z_{0}|=r} --- eq.ACi2
View from geometry, B_{r}(z_{0}) use inequality and
B_{r}(z_{0}) has two-dimension area. On the other
hand, K_{r}(z_{0}) use equality, K_{r}(z_{0}) is a one
dimensional curve. a403261520<a name="docA209">
Definition (interior point)
Let E⊂ℂ. A point z_{0} is an interior point of E
if there is some r＞0 such that B_{r}(z_{0})⊂E
E⊂ℂ say set E is within set ℂ.
c∈ℂ say element c is within set ℂ.
Let E be a complex number set //E⊂ℂ
A point z_{0} is interior point of E ? or not an
interior point of E ? This is determined by
next condition.
Take point z_{0} as center, if we can find a disk
B_{r}(z_{0}) centered at z_{0} and with radius r＞0 such
that disk B_{r}(z_{0}) 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 B_{r}(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-z_{0}|＜r} and
{z∈ℂ :|z-z_{0}|＞r} are open
A ball centered at z_{0} 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-z_{0}|≤r} and
{z∈ℂ :|z-z_{0}|=r} are closed set
They are closed set because these set definition
include all of its boundary points.
A circle {z∈ℂ :|z-z_{0}|=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-z_{0}|＜r}∪
{z∈ℂ :|z-z_{0}|＝r and imag(z-z_{0})＞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.
______
● B_{r}(z_{0})=B_{r}(z_{0})∪K_{r}(z_{0}) --- eq.ACi4
______
B_{r}(z_{0})={z∈ℂ:|z-z_{0}|≦r} --- eq.ACi5
The closure of disk of radius r centered at z_{0}
is disk of radius r centered at z_{0} together
with its circle. Adding all the boundary points.
<a name="docA225">
______
● K_{r}(z_{0})=K_{r}(z_{0}) --- 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">
___________
● B_{r}(z_{0})\{z_{0}}={z∈ℂ:|z-z_{0}|≦r} --- eq.ACi7
If a disk B_{r}(z_{0}) take its center away "\{z_{0}}"
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-z_{0}|≦r} --- eq.ACi8
。
E=B_{r}(z_{0}) --- 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 B_{r}(z_{0}) only and throw away
all boundary points.
<a name="docA228">
On the other hand, if E is circle K_{r}(z_{0}) only
● With E=K_{r}(z_{0}) --- 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=B_{1}(0) a disk center at 0 and radius=1
and
V=B_{1}(2) a disk center at 2 and radius=1
X⊂B_{1}(0) and Y⊂B_{1}(2). We also know
B_{1}(0)∩B_{1}(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=B_{1}(0) and choose V=B_{1}(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⊂B_{R}(0). If no such R exists then
A is called unbounded.
If there exist a disk B_{R}(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 θ_{1}=30 degree.
If second ray from (0,0) to infinity at
phase angle θ_{2}=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.
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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)=z^{2}+c --- eq.ACj2
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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)=x^{2}+1 --- eq.ACj3
● The graph helps us understand the function.
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A function f:A→B is a rule that assigns each
element of A exactly one element of B. For example
f(x)=x^{2}+1 --- eq.ACj3
If input x=+2, eq.ACj3 output f( 2)=2^{2}+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)=0^{2}+1=1
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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)=z^{2}+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)=(x^{2}-y^{2}+1)+i*2xy --- eq.ACj7
w=f(z)=u(x,y)+i*v(x,y) --- eq.ACj8
where u and v are ℝ^{2}→ℝ
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To graph complex functions need complex input
and complex output. For example
f(z)=z^{2}+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.
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Write input as z=x+i*y --- eq.ACj5
For complex function f(z)=z^{2}+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)=(x^{2}-y^{2}+1) --- eq.ACj9
assign imag part as v(x,y)=2*x*y --- eq.ACk0
Both u(x,y) and v(x,y) are ℝ^{2}→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 ℝ^{2}→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
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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
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Example //at video 06:12/26:38
f(z)=z^{2} --- eq.ACk1
w=(x+i*y)^{2}=(x^{2}-y^{2})+2ixy --- eq.ACk2
Not so useful?
More useful in this case: polar coordinates!
z=r*e^{iθ} --- eq.ACe7
then
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w=r^{2}*e^{2iθ} --- 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)=z^{2} --- eq.ACk1
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see video 07:28/26:38 for f(z)=z^{2} graph.
As z moves around a circle of radius r once,
w moves around a circle of radius r^{2} at double
speed twice.
see video 09:55/26:38 for f(z)=z^{2} movie.
see video 10:08/26:38 for f(z)=z^{2} mesh mapping.
see video 11:25/26:38 for f(z)=z^{2}+1 graph.
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More complicated Functions
● How do we understand more complicated functions
such as
f(z)=z^{2}+1 --- eq.ACk6
● Same idea!
Video graph show first step change from z to z^{2}.
First step is square operation.
Second step change from z^{2} to z^{2}+1
Second step is parallel moving operation (add 1).
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at video 11:25/26:38 for f(z)=z^{2}+1 graph.
for f(z)=z^{2}+1 idea is the same, except take
this function apart into f(z)=z^{2} and separate
function just add one to the result. We first
look at w1 plane for f(z)=z^{2} (not add one)
We already know how this function work. See
|w|=|z|^{2} --- eq.ACk4
and
arg(w)=2*arg(z) --- eq.ACk5
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Then we compose f(z)=z^{2} 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)=z^{2}+1
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How about ... //video 12:45/26:38
● How about
f(z)=z^{2}+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
● f^{2}(z)=f(f(z))=(z+1)+1=z+2 --- eq.ACk9
● f^{3}(z)=f(f^{2}(z))=f(z+2)=(z+2)+1=z+3 --- eq.ACL0
● ●●●
● f^{n}(z)=z+n --- eq.ACL1
● f^{n} (read "Eff n") is called the n-th iterate
of f. (Not to be confused with n-th power of f)
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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).
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This period computer auto start external storage
hard drice e:\. //video 13:56/26:38
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on and off not in control
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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
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In general
● f^{2}(z)=f(f(z))=(z+1)+1=z+2 --- eq.ACk9
● f^{3}(z)=f(f^{2}(z))=f(z+2)=(z+2)+1=z+3 --- eq.ACL0
You get the idea keep doing this n times, get
● f^{n}(z)=z+n --- eq.ACL1
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f two is also written as f compose with f
f^{2} is also written as f◦f
f^{3} is also written as f◦f◦f
and so forth. In other words,
f^{n}=f◦f◦f...f◦f◦f n times.
We read f^{n} as "Eff n". Some time people write
f^{n} as f^{○n} with a little circle before n.
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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 f^{n}(z) or [f(z)]^{n}
are easily confused.
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For now we use f^{n}(z) as n-th iteration.
If we are interested at n-th power of f(z) we
write as (z+1)^{n} that is z^{n}+n*z^{n-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).
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2015-03-27-20-02 upload file to
http://freeman2.com/calcity0.pdf
http://freeman2.com/calcity1.jpg
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Another Example
Let
f(z)=3z --- eq.ACL4
Then
● f^{2}(z)=f(f(z))=f(3z)=3*3z=3^{2}z --- eq.ACL5
● f^{3}(z)=f(f^{2}(z))=f(3^{2}z)=3*3^{2}z=3^{3}z --- eq.ACL6
● ●●●
● f^{n}(z)=3^{n}z --- eq.ACL7
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Two more examples
Let
f(z)=z^{d} --- eq.ACL8
Then
● f^{2}(z)=f(f(z))=f(z^{d})=(z^{d})^{d}=z^{d*d} --- eq.ACL9
● f^{3}(z)=f(f^{2}(z))=f(z^{d*d})=(z^{d*d})^{d}=z^{d*d*d} --- eq.ACm0
● ●●●
● f^{n}(z)=z^{(dn)} --- eq.ACm1
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Last example. Now let
f(z)=z^{2}+2 --- eq.ACm2
Then
● f^{2}(z)=f(f(z))=f(z^{2}+2)=(z^{2}+2)^{2}+2=z^{4}+4z^{2}+6 --- eq.ACm3
● f^{3}(z)=(z^{4}+4z^{2}+6)^{2}+2=z^{8}+... --- eq.ACm4
● f^{n}(z) is a polynomial of degree 2^{n}.
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Julia Sets
● To study the Julia set of the polynomial
f(z)=z^{2}+c --- eq.ACm5
We will study the behavior of the iterates
f,f^{2},f^{3},f^{4} ... f^{n} ... 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!
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To study the Julia set we study quadratic
polynomial z^{2}+c We do not need other type
quadratic polynomial. For example we do
NOT study
3*z^{2}-15z+4 --- eq.ACm6
z^{2}+c does not allow any linear term and
not allow any number before z^{2} Later we study
Julia set we will see why z^{2}+c is enough.
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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)=z^{2}+c , the sequence is
0, c, c^{2}+c, [c^{2}+c]^{2}+c ●●●
it keeps going like that. You can multiply
through to get a sequence of complex numbers.
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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.
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2015-03-28-13-03 done first proofread
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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
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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
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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"