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Wolfram U

[WSG24] Daily Study Group: Introduction to Complex Analysis

Marco Saragnese, Wolfram
Posted 9 months ago
POSTED BY: Marco Saragnese
142 Replies
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POSTED BY: Syed Ramis Hussain Zaidi

I finally finished all the lessons, quizzes, and final exam. This was a useful and interesting course. I was surprised that no mathematics teacher of mine ever mentioned the story of how complex numbers were first used as a means of dealing with cubic equations.
In question 20 for the final exam, there did not seem to be a zero at z=2. Based on the formula, it appears that there is a pole instead of a zero at z=2.

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

Thanks
POSTED BY: Joseph Smith

You are right... Thank you!
POSTED BY: Marco Saragnese

Thanks for your response! It took me a long time to work through all the lessons but I enjoyed this course.

POSTED BY: Joseph Smith
Attachment

Attachments:
Captura de tela ...png
b7ca7e05-2a39-46...pdf
POSTED BY: Ricardo Rovere de Santi
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Attachment

Attachments:
Screenshot 2025-...png
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POSTED BY: Charles Glover

Question About Deforming Contours
POSTED BY: Joseph Smith

In order to deform C and turn the integral over C into a sum of integrals over C1, C2 and C3 you can take an approach similar to that of fig. 3 of lesson 14. Try "connecting" C to (-C1) by a line and (-C1) to C by another almost-overlapping line, traversed in the opposite direction. The contributions of the two lines cancel. Then do the same with C to (-C2) and C to (-C3). The result is one continuous path inside of which f is analytic, and so the integral is zero by Cauchy's theorem. The integrals over the "double lines" cancel and the result is that integral(C-C1-C2-C3)=0. So integral(C) = integral(C1+C2+C3).
POSTED BY: Marco Saragnese

Thanks for your response. I will need to read this over carefully.

POSTED BY: Joseph Smith

POSTED BY: Joseph Smith
Mike Melko
Mike Melko
Posted 7 months 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

For Lesson 13 Problem 4, I can't see how the path for the horizontal section is rho(t)=1-2t with 1>t>-1 as stated in the solution. The path has a nonzero imaginary component. Isn't the path best described by rho(1)=1+it with t going from1 to -1?

POSTED BY: Joseph Smith

There is a typo in the exercises (thank you!). But I would propose rho(t) = 1+i-2t with 0<t<1. Your proposed path would, instead, be a vertical line.
POSTED BY: Marco Saragnese

Thanks so much for your reply
POSTED BY: Joseph Smith

Thanks so much for your reply
POSTED BY: Joseph Smith
Gerald Dorfman
Gerald Dorfman
Posted 8 months ago
POSTED BY: Gerald Dorfman
Attachment

Attachment

Attachments:
Screenshot 2024-...png
Screenshot 2024-...png
POSTED BY: Charles Glover
Murray Wolinsky
Murray Wolinsky
Posted 8 months ago
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
POSTED BY: Joseph Smith
Murray Wolinsky
Murray Wolinsky
Posted 8 months ago

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
POSTED BY: Michael Rogers
Phil Earnhardt
Phil Earnhardt
Posted 8 months ago

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
Murray Wolinsky
Murray Wolinsky
Posted 8 months ago

You might enjoy "The origin of the refractive index," Chapter 31 of Volume 1 of the Feynman Lectures on Physics. You can find it online. It seems Derek Muller is following Feynman's treatment.

Note also that Rayleigh scattering is treated in the following chapter.

Feynman's lectures are amazing. I find this part of his work especially so.
POSTED BY: Murray Wolinsky
POSTED BY: Mitchell Sandlin
POSTED BY: Michael Rogers

Hi Michael;

Thanks so much for the information you provided. I find it useful to get my hands a little dirty and work with the mathematics, which I did with the information you provided - see attached notebook. In the notebook, I have questions about the plotting of the function that you provided along with using the Limit[] function to determine the order of the function. Additionally, I included all my calculations, so if you find anything that I am doing incorrectly or not understanding, please let me know.

I certainly appreciate your help.

Thanks,

Mitch Sandlin

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

You integrate on the boundary (Circle[]) and sum the residues in the interior (Disk[]):

ResidueSum[{z(z - 1)/((z - 2) (z - 3)), {Re[z], Im[z]} \[Element] 
   Disk[{0, 0}, 5/2]}, z]

By the Residue Theorem, you need to multiply the residue sum by $2 \pi i$ to get the value of the integral.
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
POSTED BY: Michael Rogers
Donald Durack
Donald Durack
Posted 8 months 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

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

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

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

You are right, thank you. I'll have it fixed...
POSTED BY: Marco Saragnese
Matthew Mawson
Matthew Mawson
Posted 8 months ago

my mistake. please ignore this post.
POSTED BY: Matthew Mawson
POSTED BY: Michael O'Connor

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

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

Thank you....
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

You are right... thanks again... 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

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

You are right, I'll get it fixed
POSTED BY: Marco Saragnese
Updating Name
Updating Name
Posted 9 months 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

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
Murray Wolinsky
Murray Wolinsky
Posted 9 months ago
POSTED BY: Murray Wolinsky

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

Hello,

I have some questions about quizzes 8 and 9, specifically problems 5 and 2. For problem 5, I selected option C, and for problem 2, I chose option B. I cross-checked my answers using mathematical references and other reliable sources, which confirm that my answers are correct. However, the official answer key lists different alternatives. Could you please clarify if there might be any errors in the answer key?

Thank you for your assistance.

Best regards, Ricardo
POSTED BY: Ricardo Rovere de Santi

Hi, Quiz 8 problem 5 has a mistake and will be corrected soon. Thank you. But Quiz 9 problem 2 has no mistake: the correct answer should be A as you can check with Integrate[Exp[-I x]/(x + I)^2, {x, -Infinity, Infinity}]

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

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

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
POSTED BY: Mitchell Sandlin
Phil Earnhardt
Phil Earnhardt
Posted 9 months ago

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
Phil Earnhardt
Phil Earnhardt
Posted 9 months ago

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

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
POSTED BY: Michael Rogers

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

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
Tingting Zhao
Tingting Zhao
Posted 9 months 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 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
Tingting Zhao
Tingting Zhao
Posted 9 months 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

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
Phil Earnhardt
Phil Earnhardt
Posted 9 months ago
POSTED BY: Phil Earnhardt

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
Updating Name
Updating Name
Posted 9 months 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

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
Gerald Dorfman
Gerald Dorfman
Posted 9 months ago

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
POSTED BY: Michael Rogers
Gerald Dorfman
Gerald Dorfman
Posted 9 months ago

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
Gerald Dorfman
Gerald Dorfman
Posted 9 months ago

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
Gerald Dorfman
Gerald Dorfman
Posted 9 months ago

Michael, I took a look at the function f(z)=(z−z∗)(z∗+z)^2 / ( (z−z∗)^2 − (z∗+z)^4 ). I assume that you intend z* to denote the complex conjugate of z. f(z) is a real valued function because ( z - z* ) = 2 * Im(z) and ( z* + z ) = 2 * Re(z). For that reason, I don't understand why i (the square root of -1) appears in the value of the function f along either the straight line ( f((a+ib)t) ) or the parabola ( f(ta+ibt^2) ).

On a side note, if a real valued function of a complex variable has a derivative at any point, that derivative must be zero. So, if it has a derivative everywhere in an open set (region), the function must be a constant. To see that, if we approach a complex point z0 along a vertical line, then the derivative is the limit as h approaches 0 of ( f(a + i h) - f(a) ) / ( i h ). The numerator is real and the denominator is complex but f is real. So the derivative must be zero. A slightly more detailed description of this is provided on page 23 of "Complex Analysis" second edition by Lars V. Ahlfors, copyright McGraw-Hill, 1966.
POSTED BY: Gerald Dorfman

No, (z - z*) = 2 i Im(z)
POSTED BY: Marco Saragnese
Gerald Dorfman
Gerald Dorfman
Posted 9 months ago

Marco, My mistake. Thanks for the correction.
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
Gerald Dorfman
Gerald Dorfman
Posted 9 months ago

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
Phil Earnhardt
Phil Earnhardt
Posted 9 months ago

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 8 months 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

Notebook attached.

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
Gerald Dorfman
Gerald Dorfman
Posted 9 months ago

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
Gerald Dorfman
Gerald Dorfman
Posted 9 months ago

I think I see the answer to my question. The Postfix code is just a function; no Map (/@) is specified.
POSTED BY: Gerald Dorfman
Gerald Dorfman
Gerald Dorfman
Posted 9 months ago

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

The idea behind that plot is to look at the function f(z)=z^2 as a complex function of the complex variable z. The horizontal plane is the z plane, so if you write z = x + I * y, the x axis corresponds to Re[z] and the y axis to Im[z]. Remember that Re[f(z)]=Re[z^2] is real-valued. So, it will be a surface in the plot, constructed as follows: the height of the surface at coordinates (x,y) is Re[(x+I*y)^2]. Don't confuse the meaning of z (a complex number) with the vertical axis of the plot. For example, try calculating the height of the yellow curve at the point (1,2) which corresponds to z=(1+2* I) . It will be Re[(1+2* I)^2] = -3.
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
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

I was also going to recommend Penrose's book; it's a good read!!
POSTED BY: Charles Glover
POSTED BY: Michael Rogers
Gerald Oberg
Gerald Oberg
Posted 9 months 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
POSTED BY: Michael Rogers

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

You are right, I'll have to fix the exercise. Thank you
POSTED BY: Marco Saragnese
Ulf Schmidt
Ulf Schmidt
Posted 9 months 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 9 months 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
POSTED BY: Carl Hahn
Tingting Zhao
Tingting Zhao
Posted 9 months ago

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

This method is used to produce a map from a sphere to a cartesian plane. Say for example to show a flat map of the spherical Earth.
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 9 months ago

Oooo, bro set x = s + t so that he can cancel the imaginary numbers in certain conditions. Noice!
POSTED BY: Tingting Zhao
Tingting Zhao
Tingting Zhao
Posted 9 months 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
Murray Wolinsky
Murray Wolinsky
Posted 9 months ago

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 9 months 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

You are right, I'll have the exercise fixed. Thanks
POSTED BY: Marco Saragnese
Henry Ward
Henry Ward
Posted 9 months ago

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

POSTED BY: Henry Ward

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
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