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Wolfram U

[WSG24] Daily Study Group: Introduction to Complex Analysis

Marco Saragnese, Wolfram
Posted 1 year ago
POSTED BY: Marco Saragnese
142 Replies
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POSTED BY: Syed Ramis Hussain Zaidi
Attachments:
FinalExamQuestion20.nb
POSTED BY: Joseph Smith

Hello Joseph,

Thank you for reaching out. I will follow up with you outide this post.

Best,

Christine Owens Wolfram U Project Manager Wolfram U
POSTED BY: Christine Owens
POSTED BY: Joseph Smith
POSTED BY: Marco Saragnese
POSTED BY: Joseph Smith
Attachment

Attachments:
Captura de tela ...png
b7ca7e05-2a39-46...pdf
POSTED BY: Ricardo Rovere de Santi
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Attachments:
Screenshot 2025-...png
Screenshot 2025-...png
POSTED BY: Charles Glover
POSTED BY: Joseph Smith
POSTED BY: Marco Saragnese
POSTED BY: Joseph Smith

POSTED BY: Joseph Smith
Mike Melko
Mike Melko
Posted 1 year ago
POSTED BY: Mike Melko

FYI, there were questions about the use of stereographic representation. This recent post on Medium about the absence of antimatter in our universe may be of interest to some.

https://medium.com/the-infinite-universe/the-universes-missing-antimatter-may-be-all-around-us-c461c77f83ef
POSTED BY: Charles Maplethorpe
POSTED BY: Joseph Smith
POSTED BY: Marco Saragnese

Thanks so much for your reply
POSTED BY: Joseph Smith
POSTED BY: Joseph Smith
POSTED BY: Gerald Dorfman

I just finished Quiz 3, and I found no answer was correct. The final two problems are binary answers, so I changed my answers to both, but they were still graded as wrong. I believe there was an error in the grading. Please take a look at the attached screenshots.

Attachment

Attachment

Attachments:
Screenshot 2024-...png
Screenshot 2024-...png
POSTED BY: Charles Glover
POSTED BY: Murray Wolinsky

I have not heard from anyone from Wolfram U about this issue. In my Quiz 3, all answers are still being graded wrong. Please see the original post. Answers to questions 4 & 5 are binary (Yes, No). NO MATTER what is chosen for the answer, it is graded as wrong.
POSTED BY: Charles Glover

In the solution to Chapter 9 Exercise 5 it is stated "The function is a conformal transformation at all points where it is analytic and where f'(z)!=0. " What does the notation f'(z)! mean?

I don't see how the quotient rule is being applied here to get the derivative of (z^2-1)/((3+z)(z-2i)).

Can anyone suggest what I am missing? screen shot of solution .
POSTED BY: Joseph Smith

In the c programming language (perhaps others) != means "not equal."

So "where f'(z) != 0" means "where f'(z) is not equal to 0."
POSTED BY: Murray Wolinsky

Thanks! Makes sense in context.

POSTED BY: Joseph Smith

Thanks
POSTED BY: Joseph Smith

The code in the D[] operator omits the (z+3) factor in the denominator of the function in the question. The solution is correct, if they meant the function (z^2-1)/(z-2i).
POSTED BY: Michael Rogers

The latest Veritasium video You're Probably Wrong About Rainbows dropped this morning. In his video, Derek Muller shines a laser of various frequencies on a glass sphere. He uses the laser to demonstrate the critical angle of the water droplets -- approximately 48 degrees (4π/15 radians). Aha! I immediately noticed the similarity of his angle-varying physical experiment to an abstract Riemann Sphere.

At ~9:45 in the video, Derek models the particles in the sphere as masses with springs. The excitation of those masses stores energy in those springs -- pushing the phase of the light (he calls it a "phase kick"). The influence of the light waves on the medium is highly dependent on its frequency. I'm familiar with alterations of the phase in capacitors and inductors, but I'd never contemplated its application to refraction. I have long advocated for an impedance model to understand the dynamics of musculoskeletal movement -- "rigid body dynamics is flat earth thinking about our posture and movement". This is a shining example of the role that energy storage and release plays in everything.

My brain is screaming, "What about Rayleigh Scattering!!!"; I'll save that exploration for another day.

I highly recommend the video. The kicker at the end about CTR Wilson's Nobel Prize is a great example of scientific curiosity going off on a remarkable tangent.

POSTED BY: Phil Earnhardt
POSTED BY: Murray Wolinsky
POSTED BY: Mitchell Sandlin
POSTED BY: Michael Rogers
Attachments:
ZeroPolesWork.nb
POSTED BY: Mitchell Sandlin
Attachments:
points onto Comp...nb
POSTED BY: LORIS LORI
POSTED BY: Michael Rogers

A thousand thanks. I wouldn't have solved it without your help.
POSTED BY: LORIS LORI

In Example 24.1 the problem is solved using the function ContourIntegrate[], but also states that the problem can be solved using the function ResidueSum[]. However, the two functions give two different answers, and I don't understand why. No doubt, it is something that I am doing incorrectly, but I am not seeing it, so would someone please explain what I am doing incorrectly - See Attached Notebook.

Thanks,

Mitch Sandlin

Attachments:
Example24.1_Problem.nb
POSTED BY: Mitchell Sandlin
POSTED BY: Michael Rogers

Hi Michael; Thanks so much, your modification worked great. However, I do not understand why Circle[] did not work and Disk[] did. Can you tell me why Disk[] worked and Circle[] didn't? Also, I noted that in your modification, you defined the imaginary and real parts of "z" separately before executing Disk[], which also appears to be necessary.

Thanks,

Mitch Sandlin
POSTED BY: Mitchell Sandlin

Hi Mitchell,

A Circle[] represents the boundary of a Disk[]. If {x, y} is a in point "in" the set represented by the Cirlce[], then the point is on the circumference and not in the interior of the circle. A Disk[] represents the region consisting of the circle and its interior. Since we want the sum of the residues over the interior, we need to sum over the Disk[].

This distinction between a circle and disk was not made when I took geometry in school. In Euclid a circle is "a region contained by a [curved] line called the periphery." That is, it was the same as a disk. It was a little strange when the distinction was made, I think around the time I got to line integrals and Green's theorem. When we got to surface integrals a bit later, I ran into a similar distinction between sphere (hollow surface) and ball (solid, equal to the sphere plus its interior). (They are Sphere[] and Ball[] in WL.)

I think I tried Element[z, Disk[{0, 0}, 5/2]], but it didn't work. I guessed I needed to specify the coordinates since Disk[] is two-dimensional in the two-dimensional real plane. One could specify the complex disk with Abs[z] < 5/2, which is how the examples in the docs are set up.
POSTED BY: Michael Rogers
Donald Durack
Donald Durack
Posted 1 year ago
POSTED BY: Donald Durack
POSTED BY: Taiboo Song

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

POSTED BY: Taiboo Song

The section "Bonus materials" of the course is a super-shortened condensate of the course with all the main formulas.
POSTED BY: Marco Saragnese
POSTED BY: LORIS LORI

You are right, thank you. I'll have it fixed...
POSTED BY: Marco Saragnese

my mistake. please ignore this post.
POSTED BY: Matthew Mawson

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
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
POSTED BY: Marco Saragnese
POSTED BY: Joseph Smith
POSTED BY: Marco Saragnese

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
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

You are right, much appreciated. I'll get it fixed.
POSTED BY: Marco Saragnese

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

You are right, I'll get it fixed
POSTED BY: Marco Saragnese
Updating Name
Updating Name
Posted 1 year ago
POSTED BY: Updating Name

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

I've worked the quiz problems and they all make sense to me, except for one.The answer I get for Quiz 8, Problem 5 is present among the offered choices, but it's marked wrong by the grader.

I just tried the other three choices and found the one the grader likes, but I don't understand why.

Anyone else experience this -- or know why the answer marked correct is correct? I can get the answer it marks correct if I integrate over only half the unit circle.

As I said, the answers to all the other quiz problems make sense to me.
POSTED BY: Murray Wolinsky

You are right, I'll have it fixed. Thank you
POSTED BY: Marco Saragnese
POSTED BY: Ricardo Rovere de Santi
POSTED BY: Marco Saragnese

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
POSTED BY: Marco Saragnese

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

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.
POSTED BY: Charles Maplethorpe

Thank you. I'll have it fixed.
POSTED BY: Marco Saragnese

The same mistake is in Fig. 10.
POSTED BY: Charles Maplethorpe

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

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

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: Paul Tikotin

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
POSTED BY: Michael O'Connor
POSTED BY: Michael Rogers
Tingting Zhao
Tingting Zhao
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
POSTED BY: Michael Rogers
Tingting Zhao
Tingting Zhao
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
POSTED BY: Michael Rogers
POSTED BY: Phil Earnhardt
POSTED BY: Paul Tikotin
Updating Name
Updating Name
Posted 1 year ago
POSTED BY: Updating Name

You are correct, I'll have it fixed. Thank you!
POSTED BY: Marco Saragnese

The same problem comes up in the following figure 10
POSTED BY: Charles Maplethorpe

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 don't think that the existence of the limit along all straight lines is sufficient for the existence of the complex limit. After all, it is not so in two-variable calculus. I will try to look for an explicit counterexample.
POSTED BY: Marco Saragnese

I think @Marco is right. If we are considering arbitrary functions of a complex $z$, here is an example based on a real two-variable example:

$$f(z)=\frac{\left(z-z^*\right) \left(z^*+z\right)^2}{\left(z-z^*\right)^2-\left(z^*+z\right)^4}$$

Along lines through $z=0$,

$$f\bigl((a+ib)\,t\bigr)=-\frac{2 i a^2 b}{4 a^4 t^2+b^2} \,t$$

approaches $0$. Along certain parabolas through $z=0$,

$$f(t a + i b t^2)=-\frac{2 i a^2 b}{4 a^4+b^2}$$

is constant and so has a nonzero limit if $a$ and $b$ are nonzero.

Here is a visual:

ComplexPlot3D[
 ((z + Conjugate[z])^2 (z - Conjugate[z])) /
  ((z - Conjugate[z])^2 - (z + Conjugate[z])^4)
 , {z, -1 - I, 1 + I}, PlotRange -> {0, 1}, 
 MaxRecursion -> 6]

enter image description here
POSTED BY: Michael Rogers

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

Attachments:
ComplexLimitCoun...nb
POSTED BY: Gerald Dorfman

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:
ComplexLimitCoun...nb
POSTED BY: Gerald Dorfman
POSTED BY: Gerald Dorfman

No, (z - z*) = 2 i Im(z)
POSTED BY: Marco Saragnese

Marco, My mistake. Thanks for the correction.
POSTED BY: Gerald Dorfman
POSTED BY: Michael Rogers
POSTED BY: Gerald Dorfman

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
Graham Gyatt
Graham Gyatt
Posted 1 year ago

I think Phil's question might have got posted in the wrong spot and missed, but nice I'm interested in the answer too, I thought I would add this reply to try to re-highlight it
POSTED BY: Graham Gyatt
POSTED BY: Michael Rogers
Attachments:
A Modest Proposa...nb
POSTED BY: Joseph Smith

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

Thanks so much for your response!

POSTED BY: Joseph Smith

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

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

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

Marco, Thanks. Our replies crossed. I came to the same conclusion as indicated in your reply.
POSTED BY: Gerald Dorfman

For figure 2 lesson 3, is the caption correct?
POSTED BY: Joseph Smith

You are right, I'll have the caption fixed.
POSTED BY: Marco Saragnese

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:
14381729.nb
POSTED BY: Joseph Smith
POSTED BY: Marco Saragnese

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

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

Thxs, Marco.

I am wondering how you got into complex number analysis. What trigger made you enter this field?
POSTED BY: Taiboo Song

I studied physics at the university and complex analysis was a required course.
POSTED BY: Marco Saragnese

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

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

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
POSTED BY: Marco Saragnese
POSTED BY: Carl Hahn
POSTED BY: Charles Glover
POSTED BY: Michael Rogers
Gerald Oberg
Gerald Oberg
Posted 1 year ago
POSTED BY: Gerald Oberg
POSTED BY: Michael Rogers
POSTED BY: Joseph Smith

You are right, I'll have to fix the exercise. Thank you
POSTED BY: Marco Saragnese
Ulf Schmidt
Ulf Schmidt
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

You are correct, thank you!
POSTED BY: Marco Saragnese
Gerald Oberg
Gerald Oberg
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:
Updated References.pdf
POSTED BY: Gerald Oberg

Thxs, Gerald for the updated references. Do any books cover aerospace applications that use complex numbers?
POSTED BY: Taiboo Song

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

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
Tingting Zhao
Tingting Zhao
Posted 1 year ago

I think maybe complex vectors on a unit sphere?
POSTED BY: Tingting Zhao
POSTED BY: Paul Warburton

I tried to describe the only application I know of in a reply to Joseph Smith, above.
POSTED BY: Marco Saragnese
Tingting Zhao
Tingting Zhao
Posted 1 year ago
POSTED BY: Tingting Zhao
Tingting Zhao
Tingting Zhao
Posted 1 year ago
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
Tingting Zhao
Tingting Zhao
Posted 1 year ago

Yes, I have the same answer as you.
POSTED BY: Tingting Zhao

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
POSTED BY: Marco Saragnese
Henry Ward
Henry Ward
Posted 1 year ago
POSTED BY: Henry Ward
POSTED BY: Devendra Kapadia
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