Group Abstract Group Abstract

Message Boards Message Boards

10
|
19.7K Views
|
142 Replies
|
38 Total Likes
View groups...
Share
Share this post:
GROUPS:

[WSG24] Daily Study Group: Introduction to Complex Analysis

Posted 1 year ago
POSTED BY: Marco Saragnese
142 Replies

Marco has worked hard to create this superb introduction to complex analysis which is one of the most beautiful and useful branches of mathematics.

I strongly recommend this study group to everyone!

POSTED BY: Devendra Kapadia

About applications of the Riemann sphere and the stereographic projection, others in this forum have mentioned cartography, but that does not require complex numbers. The only "practical" application of the Riemann sphere I know of is to Lorentz transformations. The situation is this: imagine two observers moving at very large relative velocities. By a relativistic effect, the two observers will see the stars in different positions. How to relate these positions? In this problem it can actually be useful to think of the sky as a Riemann sphere for the purpose of calculation. Then, the position of a star will be given by a complex number and the problem becomes how to relate the position of the same star as seen by the two observers, or to relate the two complex numbers. More precisely, the Lorentz transformations of the "celestial sphere" can be described by conformal transformations of the Riemann sphere. So the theorems about the stereographic projection (circles mapped to circles etc.) become useful to derive properties about Lorentz transformations. For example, a perfectly circular constellation should remain circular as seen by the second observer. Anyway I am no expert, so I refer you to Chapter 1 of Vol. 1 of "Spinors and Spacetime" by Penrose and Rindler, or perhaps to Chapter 18 of "The road to reality" by Penrose (and I apologize for any wrong statements).

POSTED BY: Marco Saragnese

You are right, I'll have to fix the exercise. Thank you

POSTED BY: Marco Saragnese

All code can already be seen, including that for images and animations. You just have to double click on the bracket that hides the input, identifiable by the little upward-pointing arrow:

Input code hidden

Double clicking will expand the code: Input code visible after double clicking

POSTED BY: Marco Saragnese

No, (z - z*) = 2 i Im(z)

POSTED BY: Marco Saragnese

Hi,

I'm not exactly sure what you want, but here are formulas for the various elements of the projection. You can put them together in graphics as you see fit:

(* elements *)
pole = {0, 0, 2};
numberPt = {Re[z], Im[z], 0};
projectionPt = { (* main formula: proj. onto R. sphere *)
  (4 Re[z])/(4 + Abs[z]^2), 
  (4 Im[z])/(4 + Abs[z]^2), 
  (2 Abs[z]^2)/(4 + Abs[z]^2)};
riemann = Sphere[{0, 0, 1}, 1];
plane = InfinitePlane[{{0, 0, 0}, {1, 0, 0}, {0, 1, 0}}];

 z = 2 - I; (* need to give z a value *)
 Graphics3D[{
   Opacity[0.6],(* change style as desired *)
   Point[{pole, numberPt, projectionPt}],
   Line[{pole, numberPt}],
   riemann,
   plane},
  PlotRange -> {{-3, 3}, {-3, 3}, {0, 2}},
  FaceGrids -> { (* may omit *)
    {{0, 0, -1}, {Range[-3, 3], Range[-3, 3]}}}
  ]

enter image description here

Changing FaceGrids to the following will highlight the real and imaginary axes:

FaceGrids -> { (* may omit *)
  {{0, 0, -1},
   {Range[-3, 3], Range[-3, 3]} /.
     0 -> {0, AbsoluteThickness[1]}}}

Omitting FaceGrids gives a plain plane below the sphere. Omitting plane and using FaceGrids gives a nice look, too.

POSTED BY: Michael Rogers

Michael, Thanks. In the attached notebook ( file ComplexLimitCounterExample_15nov24.nb ), I've provided code to present your example.

Attachments:
POSTED BY: Gerald Dorfman

Perhaps the contributed function RiemannSphereComplexPlot[] -- a 3D rotatable Riemann sphere version of ComplexPlot -- is what you are seeking: https://resources.wolframcloud.com/FunctionRepository/resources/RiemannSphereComplexPlot/

POSTED BY: Phil Earnhardt
POSTED BY: Michael Rogers
Posted 1 year ago

In Lesson 13, in the proof of Cauchy's Theorem, I think there is a sign error in the first (left) integrand. I think partial of u with respect to y (du/dy) should have a negative sign in accordance with Green's Theorem. Then, the Cauchy-Riemann equations yield that the first integrand is 0. Is that the case?

POSTED BY: Updating Name
POSTED BY: Marco Saragnese
POSTED BY: Michael Rogers
POSTED BY: Marco Saragnese

You are correct, thank you!

POSTED BY: Marco Saragnese
Posted 1 year ago

A still from Star Wars: Revenge of the Sith with the character Count Dooku saying:

POSTED BY: Henry Ward
Posted 1 year ago
POSTED BY: Graham Gyatt
Posted 1 year ago

In the exercises for lesson 23 on calculating Residues, the solutions for Exercises 3 and 4 are done for double poles, but in both cases the poles appear to be single poles, as verified with FunctionPole.

POSTED BY: Donald Durack

The section "Bonus materials" of the course is a super-shortened condensate of the course with all the main formulas.

POSTED BY: Marco Saragnese

You are right, thank you. I'll have it fixed...

POSTED BY: Marco Saragnese

Please emphasize what formula to remember and what formula to derive will be better to understand the calculus.

POSTED BY: Taiboo Song

This calculus class included complex analysis introduced us to numerous formulas and theories. Are there any comprehensive collections of all the formulas and theorems related to calculus? It can be challenging to remember everything, so having a resource that compiles them all would be incredibly helpful.

POSTED BY: Taiboo Song

in Lesson 21, Example 21.3. Laurent series of f(z)=e^(1/z) around z=0 is reported Sum z^(-n)/n! for n from 0 to infinitive. in Lesson 23, Example 23.2. Laurent series of f(z)=e^(1/z) around z=0 is reported Sum n! z^(-n) for n from 0 to infinitive. I think there is an error in example 23.2

POSTED BY: LORIS LORI

What can you say about physical analogues to Residue Theory -- how it applies to physics, electromagnetism, etc.? I found myself completely disconnected to what a "residue" means -- how a number of singularities has physical significance. What is the significance of the orthogonality of harmonics? As a professor, how to you create the engagement/pertinence of these abstract concepts, @Michael? I can search and ask the AIs, but I wanted to know what you and @Marco can say personally. Thank you.

POSTED BY: Phil Earnhardt

Hi Michael, Thanks for the simple and clear analysis that shows that the only 2 arguments of f(z) are Pi/2 and -Pi/2, where

f[z_] := ((z + Conjugate[z])^2 (z - 
      Conjugate[z]))/((z - Conjugate[z])^2 - (z + Conjugate[z])^4)

I had not seen previously the Option TargetFunctions that you used to quickly resolve my question in the instruction

ComplexExpand[f[z], z, TargetFunctions -> {Re, Im}] // Simplify
POSTED BY: Gerald Dorfman

Hi Gerald,

For some reason, I wasn't notified of your replies. Since Study Group threads get long, I didn't happen to see them until today. Nice notebooks, btw. You asked:

From the plots, it appears that the values of function f have only two arguments (angles with the real axis) and that they are determined by whether Im[z} is positive or negative. I have not analyzed f to confirm this. Am I correct? Do you see a simple analysis to determine this?

I think the following shows it:

ClearAll[f]
f[z_] := ((z + Conjugate[z])^2 (z - 
      Conjugate[z]))/((z - Conjugate[z])^2 - (z + Conjugate[z])^4)

ComplexExpand[f[z], z, TargetFunctions -> {Re, Im}] // Simplify

(*  -(2 I Im[z] Re[z]^2)/(Im[z]^2 + 4 Re[z]^4)  *)

Note the squares on Re[z] and Im[z] except for the factor of Im[z] in the numerator. The argument of f[z] is the same as the argument of -I Im[z], which is -Pi/2 when Im[z] > 0 and Pi/2 when Im[z] < 0.

In a different reply, you remarked:

On a side note, if a real valued function of a complex variable has a derivative at any point, that derivative must be zero.

Right. It was this sort of thing I had in mind when I prefaced my example with, "If we are considering arbitrary functions of a complex variable $z$..." [emphasis added]. The function f[z] is not complex-differentiable.

POSTED BY: Michael Rogers

I intended C2 to be defined clockwise in that specific figure (because I drew the arrow that way) So the integral over (C1+C2) is zero by Cauchy. So int_C1 = - int_C2 or int_C1 = int_(-C2) . So I don't think there are typos there. In this example it is (-C2) which is counterclockwise. An exception to our regular convention since normally one always takes paths to be counterclockwise. But it is marked by an arrow so I hope it's clear.

POSTED BY: Marco Saragnese

Thank you....

POSTED BY: Marco Saragnese

Question on notation in Section 12. Should the equation read the contour integral over C1 equals the contour integral over C2 (it says it equals the contour integral over negative C2)? I believe that C2 is defined as being in the clockwise direction, so the negative sign will naturally arise when the integral is done in the counterclockwise direction.

enter image description here

POSTED BY: Michael O'Connor

The discontinuities are in half-lines parallel to the real axis, not the imaginary axis. And they are not for any half-lines, only for half-lines departing at points with coordinates (x,y) = (0, (2k+1)*pi ), which are the branch points of the function. At each branch point, a branch cut starts: the branch cuts here are half-lines in the direction of increasing real coordinate.

POSTED BY: Marco Saragnese

You are right... thanks again... I'll get it fixed

POSTED BY: Marco Saragnese

There is a typo in Problem 6 on Quiz 4. The first answer should read: "The integral does not exist." (and not "The integral does not").

enter image description here

POSTED BY: Michael O'Connor

Are branch cuts in a unique location? In Lesson 5 exercise 3 we are asked to "describe the branch cuts" of the function Sqrt[(e^2+1)]. Given the definition from Wolfram Mathworld "A branch cut is a curve (with ends possibly open, closed, or half-open) in the complex plane across which an analytic multivalued function is discontinuous. For convenience, branch cuts are often taken as lines or line segments." The solution states: enter image description here

I can see from the figure that there are discontinuities for any line parallel to the Im axis for x>0. Is the branch cut where these discontinuities start or can the branch cut be anywhere x>0?

POSTED BY: Joseph Smith

In Section 11 of the Lesson, shouldn't the first definition of a complex integral be the integral of f(z)dz over the contour (and not f(x)dx over the contour).

enter image description here

POSTED BY: Michael O'Connor

You are right, much appreciated. I'll get it fixed.

POSTED BY: Marco Saragnese

You are right, I'll get it fixed

POSTED BY: Marco Saragnese

In Example 5.3, shouldn't the text read: "In other words, the branch cut corresponds to either z purely imaginary or to z real between -1 and 1."

enter image description here

POSTED BY: Michael O'Connor

Shouldn't the Caption to Fig 9 in the book be: "Fig. 9 Vector Plot of z^2" and not "Fig. 9 Vector Plot of z^2 + 1 = z^2"

enter image description here

POSTED BY: Michael O'Connor

It refers to all other possibilities for n, so n<0. So there are no negative-power terms in the series.

POSTED BY: Marco Saragnese

This whole discussion of locating Riemann Spheres has been very entertaining and educational! I will file it away. I'm not sure there's an unresolved question, but I do have a suggestion. This clearly calls for a demonstration with Manipulate[] that shows different locations for the Riemann Sphere: a variety of locations, and a variety of equations to show the various tradeoffs. Michael has a bunch of examples in the demonstrations project; this would be a good addition. Professors who do Wolfram U continuing education programs are inspiring.

The sphere reminds me of a Dome Magnifier -- a hemisphere placed in contact with type or an image to improve its visibility: enter image description here

Magnifying is not the only property. As this item's description notes:

Bright field magnifiers have a calculated light guidance that directs all possible illumination onto the object. This provides a brighter field of view without an additional light source.

Was Riemann thinking of an optical magnifier when he invented the thought-spheres?

POSTED BY: Phil Earnhardt
Posted 1 year ago

If f(z)=2+2z+ Sum(z^n), where n = 2,3, ..., Infinity, the output of SeriesCoefficient[f[z],{z,0,n}] is:

1 n>1

2 n==0||n==1

0 True

What does "0 True" mean?

POSTED BY: Updating Name

If you can post a specific example I'll look at it. Otherwise, I would suggest following the steps in example 18.1 in the course materials and see if that helps.

POSTED BY: Marco Saragnese

You are right, I'll have it fixed. Thank you

POSTED BY: Marco Saragnese
POSTED BY: Murray Wolinsky

Hi;

In working with the properties of Harmonic Functions, how is the mean value calculated when you are given the formula of (x,y) along with the radius and center of a circle? I have gone through the course materials up to include unit 18 and have not found anything to help me set-up the problem. Thanks, Mitch Sandlin

POSTED BY: Mitchell Sandlin

The same mistake is in Fig. 10.

Marco, My mistake. Thanks for the correction.

POSTED BY: Gerald Dorfman

Right! the projected points end up on the other side of the sphere. And you were right before: If you use rays from the north pole $N$, then you get what said in your first reply, and the north pole has to be above the $z$ plane to get a second intersection point. (And if one uses the line segment from $N$ to $z$, then the whole sphere has to be above, or tangent to, the $z$ plane. I was used to lines, which avoids concerns about betweenness.)

POSTED BY: Michael Rogers
Posted 1 year ago

I suppose if we choose to use lines instead of rays for projection then after the sphere moved below the complex plane the projection will be an inverse image of the one from above?

POSTED BY: Tingting Zhao

Thank you. I'll have it fixed.

POSTED BY: Marco Saragnese
POSTED BY: Gerald Dorfman

Thanks to another student, I found this article in the Mathematica Journal: "Domain Coloring on the Riemann Sphere. I would like to use the code. But simply downloading the article as a notebook fails to give me anything other than the ability to look at the pictures.

I think I need to download and instal a package called complexVisualize.m.

Are you able to tell me where to find that file?

I am using this course to both extend my Mathematica knowledge and refresh my Complex Analysis.

POSTED BY: Paul Tikotin

Thank you, This is nice and will, I hope, give me some basis for experimentation. I would like to define my own shapes on the sphere and see the mapping not the plane.

POSTED BY: Paul Tikotin

Nice!

Even better (?) is the package pointed to in the reference.

BUT - it needs a package complexVisualize.m which I can't find. So I cannot do anything but look at the pretty pictures.

My knowledge of Mathematica is not up to the task of downloading and using packages...yet.

POSTED BY: Paul Tikotin

The same problem comes up in the following figure 10

Hi Tingting,

The only problem is if the "north pole" $N$, the point that you connect with $z$, lies in the complex plane. If it's not in the complex plane, then the line through $N$ and $z$ intersects the sphere in another point $P(z)$, which we may define as the projection of $z$ onto the sphere.

If the sphere is tangent to the complex plane at $N$, then all lines through $N$ and $z$ are tangent to the sphere at $N$. The projection is undefined. (Or all numbers $z$ are projected onto $N$, which is pretty useless.)

POSTED BY: Michael Rogers
Posted 1 year ago

What happens if one keeps moving the sphere down the y axis? At some point the north pole would overlap the origin and infinity becomes 0 and then nothingness, yes?

POSTED BY: Tingting Zhao

Hi Michael,

In my Complex Analysis textbook (Bak & Newman), the Riemann Sphere has unit diameter and sits above the complex plane tangent at $z=0$ (the origin), just as you say. It has the nice property that the equator maps to the unit circle.

In my code, I used a sphere with a unit radius, since this seemed to be quite common in a google search. I thought maybe things had changed since I took the course. The equator in this model is nothing special, though.

I have seen before the Riemann Sphere presented as the standard unit sphere (unit radius centered at the origin) so that the complex plane cuts it in two. This model seems less common to me. It does have the property that the equator is the unit circle.

All three models are effectively equivalent: Any proof or explanation based on one can be translated to another, changing projection formulas if necessary. In all three, the south pole corresponds to $z=0$ and the north pole to $z=\infty$, which is convenient.

Because of this, it seems more a matter of convention than correctness to me. I personally prefer the tangent-sphere models to the sphere centered at the origin. (And I like the unit diameter one better than the unit radius one, but I went with the peer-pressure of google in my previous reply.) Possibly the one centered at the origin is easier to deal with pedagogically.

POSTED BY: Michael Rogers

I am reading the book. The legend for chapter 3 Fig.9 in the book says “Fig. 9 Vector Plot of z2+1=z2”. This puzzled me, so I checked notebook 3, and it says “Fig. 9 Vector Plot of f(z)=z2”, which I think is correct.

In the Bonus Material we encounter the following figure: enter image description here

Shouldn't the Riemann Sphere sit on the complex plane and only touch it at point {0,0}? If I am reading the figure correctly, it suggests the Riemann Sphere is bisected by the complex plane, which I believe is incorrect.

POSTED BY: Michael O'Connor

Michael, I've written some code that enhances the ComplexPlot3D display of the function

f[z_] := ((z + Conjugate[z])^2 (z - 
      Conjugate[z]))/((z - Conjugate[z])^2 - (z + Conjugate[z])^4)

The code is in the attached Notebook, file ComplexLimitCounterExample_16nov24.nb .

From the plots, it appears that the values of function f have only two arguments (angles with the real axis) and that they are determined by whether Im[z} is positive or negative. I have not analyzed f to confirm this. Am I correct? Do you see a simple analysis to determine this?

BTW, "Calculus Volume II" by Tom M. Apostol, Copyright 1962 by Blaisdell Publishing Company has on page 70 a simple real-valued function of two real variables that is not continuous at 0 even though approaching 0 along either axis has limit 0. However, when approaching 0 along the line x = y, its limit is 1/2. Here is the function:
f(x,y) = x*y / (x^2 + y^2) if (x,y) != (0,0) and f(0,0) = 0.

Attachments:
POSTED BY: Gerald Dorfman

In the European tradition of philosophy, Plato famously asked (demanded?) a knowledge of mathematics from his students. At that time, I think, mathematics was considered as part of Logic which was an essential component of philosophy.

The Pythagorean School (much earlier than Plato) also took mathematics very seriously, but the school looks quasi religious by today's (western) standards.

POSTED BY: Paul Tikotin

This is not a specific answer to your question...but

The Schaum's Outline text gives some physical examples relating to fluid flow in the chapter on Conformal Transformations.

POSTED BY: Paul Tikotin

The Wolfram Function Repository has a function StereographicProjection[]. Documentation at https://resources.wolframcloud.com/FunctionRepository/resources/StereographicProjection/ . Does that do what you are hoping for?

Check out the neat example at the bottom of the function's documentation: enter image description here

POSTED BY: Phil Earnhardt

Hi;

Is there a straight-Forward method of producing a Stereographic Projection of a complex number? The examples given in our discussion materials seems extremely convoluted with several complex plots joined together with possibly some primitives added in.

Thanks,

Mitch Sandlin

POSTED BY: Mitchell Sandlin

Thanks so much for your response!

POSTED BY: Joseph Smith

You are correct, I'll have it fixed. Thank you!

POSTED BY: Marco Saragnese

Lesson 9 states "for the limit of a complex function to exist, it must exist no matter the direction by which it is approached." Is a necessary and sufficient condition for this that the limit exists and is the same for all approaches along straight lines? Is it possible for the limit to exist and be the same for all approaches along straight lines but for that not to be the case when approaching along some curve that is not a straight line?

POSTED BY: Gerald Dorfman

I studied physics at the university and complex analysis was a required course.

POSTED BY: Marco Saragnese

Notebook attached.

Attachments:
POSTED BY: Joseph Smith

I was also going to recommend Penrose's book; it's a good read!!

POSTED BY: Charles Glover

Thxs, Marco.

I am wondering how you got into complex number analysis. What trigger made you enter this field?

POSTED BY: Taiboo Song

Marco, Thanks. Our replies crossed. I came to the same conclusion as indicated in your reply.

POSTED BY: Gerald Dorfman

I think I see the answer to my question. The Postfix code is just a function; no Map (/@) is specified.

POSTED BY: Gerald Dorfman

In that code, Labeled applies to the whole Row object,

(Row[...]) // Labeled[#, "......"]&

Consider the difference between:

Row[{a, b, c}] // Labeled[#, "mylabel"] &

and

Map[Labeled[#, "mylabel"] &, Row[{a, b, c}], {2}]

Hope this helps

POSTED BY: Marco Saragnese

In Lesson 3, why does the Postfix instruction //Labeled[#,Text[ ...]& result in just one Text for the entire row rather than a separate text for each of the 3 ComplexPlot3D[...] entries in the Row[...] for the following code?

Row[{
ComplexPlot3D[z,{z,3},PlotRange->{{-3,3},{-3,3},{0,6}},BoxRatios->{1,1,1},PlotStyle->Directive[Opacity[0.8`]],AxesLabel->{Style[ToString[Re[z],TraditionalForm],12],Style[ToString[Im[z],TraditionalForm],12],Style[ToString[Abs[z],TraditionalForm],12]},ImageSize->160],
ComplexPlot3D[Conjugate[z],{z,3},PlotRange->{{-3,3},{-3,3},{0,6}},BoxRatios->{1,1,1},PlotStyle->Directive[Opacity[0.8`]],AxesLabel->{Style[ToString[Re[z],TraditionalForm],12],Style[ToString[Im[z],TraditionalForm],12],Style[ToString[Abs[Overscript[z, _]],TraditionalForm],12]},ImageSize->160],
ComplexPlot3D[1/z,{z,3},PlotRange->{{-3,3},{-3,3},{0,6}},BoxRatios->{1,1,1},PlotStyle->Directive[Opacity[0.8`]],AxesLabel->{Style[ToString[Re[z],TraditionalForm],12],Style[ToString[Im[z],TraditionalForm],12],Style[ToString[HoldForm[Abs[1/z]],TraditionalForm],12]},ImageSize->160,PlotLegends->Automatic]
}]//Labeled[#,Text[StringJoin["Fig. 4. Plots of the functions ",ToString[z,TraditionalForm],", ",ToString[Overscript[z, _],TraditionalForm],", ",ToString[1/z,TraditionalForm],"."]]]&
POSTED BY: Gerald Dorfman
POSTED BY: Marco Saragnese
POSTED BY: Marco Saragnese

For figure 2 lesson 3, is the caption correct?

POSTED BY: Joseph Smith

Trying to catch up here. Surely this is a simple question. When plotting Re[z] and Im[z], shouldn't these plots be confined to either the {z, Re[z]} or the {z,Im[z]} planes? I don't see why these plots appear to be surfaces in 3 dimensions.

Thanks!

Attachments:
POSTED BY: Joseph Smith

They found the need when studying cubic equations. From Wikipedia, it looks like Girolamo Cardano had a tortured life, which he described in an autobiography. The front page of the autobiography on Wikipedia describes him as a "medical philosopher and man of letters". He was a medical doctor, engineer, mathematician and philosopher. The "Cardanic joint" bears his name. Less is known about Rafael Bombelli's life; he was an architect and civil engineer. I don't think that the career of professional mathematicians in the modern sense existed at the time. People studied multiple disciplines which today we treat as distinct. They would perhaps use terms like "natural philosopher" or "scholar" to describe themselves, even if today we remember them for their mathematical contributions. Often, mathematics was a hobby and they had a practical profession in commerce, law or other fields. Fermat was famously a lawyer.

POSTED BY: Marco Saragnese
Posted 1 year ago

Hi Michael, What references (texts) would you recommend for studying the topics "(differential) topology" and integrals of the form (2) that you referred to?

POSTED BY: Gerald Oberg

Thxs, Gerald for the updated references. Do any books cover aerospace applications that use complex numbers?

POSTED BY: Taiboo Song

I am wondering about the history of complex numbers, Girolamo and Rafel, in 16C. How did they find the need, and what profession both were in? Did mathematician careers exist in 16C?

POSTED BY: Taiboo Song
POSTED BY: Michael Rogers

Although I have encountered and used complex numbers over the decades, I never understood what motivated their introduction until your first lecture. Thanks.

POSTED BY: Joseph Smith

Wow. That needs to find itself into some SciFi story about space travel. Love it. Thanks.

POSTED BY: Carl Hahn

You are right, I'll have the exercise fixed. Thanks

POSTED BY: Marco Saragnese

Can you comment on the significance or the application of the Riemann sphere concept? Why should it matter how lines or circles in the complex plane map to the sphere?

POSTED BY: Joseph Smith

Quiz 1 Question 3: Is the stereo graphic representation of z1=1/2+i*(3^1/2)/2 in the northern (upper?) or southern (lower?) hemisphere? The text says: "Numbers with absolute value less than 1 are mapped to the southern hemisphere, and 0 to the south pole. Numbers with absolute value greater than 1 are mapped to the northern hemisphere."

Abs[z1] = 1, but the answer "in the upper hemisphere" comes back wrong. Can you explain?

POSTED BY: Joseph Smith
POSTED BY: Paul Warburton
Posted 1 year ago

Lesson 2 The Complex Plane

There might be an error in the proof for the formulas of the stereographic projection: "... And because the triangles (0,z,N) and (z',z,Overscript[z, ^]) are similar, then |z'|/|z|=1/(1-Z)...."

correct: |z|/|z'|=1/(1-Z) <--- This formula was used for the rest of the proof. The formulas itself are valid.

POSTED BY: Ulf Schmidt
Posted 1 year ago

I think maybe complex vectors on a unit sphere?

POSTED BY: Tingting Zhao
Posted 1 year ago

Attached is an updated list of the References provided, with a link to the publisher’s page first and then a link to the corresponding Amazon page.

Attachments:
POSTED BY: Gerald Oberg

yeah, I noticed that too. I picked North since the equator was not available. It said it was the wrong answer.

POSTED BY: Carl Hahn

So, I asked this question during the lecture and Marco did not have an answer for it. I had never heard of the stereographic projection of complex numbers. I was wondering if anybody knows of a practical application for it. It seemed to me that it might be used in modulation/coding theory or maybe image recognition problems but I'm just shooting in the dark. Anybody know?

POSTED BY: Carl Hahn
Posted 1 year ago

Oooo, bro set x = s + t so that he can cancel the imaginary numbers in certain conditions. Noice!

POSTED BY: Tingting Zhao
Posted 1 year ago

I find the projection of the Reimann sphere is extremely distorted. I prefer that both sides of the sphere be projected on the inner disk separately so that the two projections have symmetry.

POSTED BY: Tingting Zhao
Posted 1 year ago

Yes, I have the same answer as you.

POSTED BY: Tingting Zhao

Please look at the 3rd question in Quiz 1. The question asks whether a given point is stereographically projected into the lower or the upper hemisphere of the Riemann sphere. In fact, the magnitude of the point is exactly 1, so that the given point lies on the equator. But that's not an available answer.

So, what's the right response?

POSTED BY: Murray Wolinsky
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Reply Preview
Attachments
Remove
or Discard