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[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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Hi there! I hope this message finds everyone well. I have recently completed Proficiency in Complex Analysis Certification. I have happily posted my certificate credentials at LinkedIn, however, since I'm a newbie, I do not have much people around at LinkedIn, who can really understand how interesting of work I've accomplished. Thus, I humbly request my kind fellow members, who took up this course, to kindly check out my Certification post on LinkedIn and show some humble love in the comment section. If you can, I highly appreciate endorsing "Complex Analysis" as one of the skillsets that I posted on LinkedIn as learning outcome of this course. You can access in scroling down below on my LinkedIn profile. Your efforts mean alot to me and it will make my day better. Thanks in advance

Link to the post: https://www.linkedin.com/posts/syed-ramis-hussain-zaidi-272209247_wolfram-complexanalysis-mathematics-activity-7346631892537901056-6pee?utm_source=social_share_send&utm_medium=member_desktop_web&rcm=ACoAAD0TxjYBcRn80gLEaLhVjfYem-vraAkeous
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

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

I posted this message below and never got a response from anyone at Wolfram. So, I am reposting.

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

Question About Deforming Contours
POSTED BY: Joseph Smith
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

I completed the exam for "Introduction to Complex Analysis" awhile ago. The download link for the level 1 certification in the online course is active, but it doesn't work. Can this be fixed, or is there an alternative way to get the certificate?
POSTED BY: Mike Melko
POSTED BY: Charles Maplethorpe
POSTED BY: Joseph Smith
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

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Attachments:
Screenshot 2024-...png
Screenshot 2024-...png
POSTED BY: Charles Glover
Murray Wolinsky
Murray Wolinsky
Posted 8 months ago
POSTED BY: Murray Wolinsky
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
Murray Wolinsky
Murray Wolinsky
Posted 8 months ago
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
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
Attachments:
ZeroPolesWork.nb
POSTED BY: Mitchell Sandlin

I ask for an explanation.

I successfully plotted 5 points on the complex function f[z]=Sqrt[(z+1)(z-1)], with Show[ComplexPlot3D, Graphics3D].

If I use the same procedure and plot the point p1=(1, 1, Abs[g[z1]] ), with z1=1+I, on the function g[z]=Log[z] the program does not work. Why? I attach my notebook.

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
Attachments:
Example24.1_Problem.nb
POSTED BY: Mitchell Sandlin
POSTED BY: Michael Rogers
POSTED BY: Mitchell Sandlin
POSTED BY: Michael Rogers
Donald Durack
Donald Durack
Posted 9 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
POSTED BY: Taiboo Song
POSTED BY: Taiboo Song
POSTED BY: Marco Saragnese
POSTED BY: LORIS LORI
POSTED BY: Marco Saragnese
Matthew Mawson
Matthew Mawson
Posted 8 months ago

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

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

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

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
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
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
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
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
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
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 9 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
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

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

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

Wow. That needs to find itself into some SciFi story about space travel. Love it. Thanks.
POSTED BY: Carl Hahn

I was also going to recommend Penrose's book; it's a good read!!
POSTED BY: Charles Glover

Hi Joseph,

The Riemann sphere is a way to understand the role of infinity in complex analysis. To understand functions of a complex variable is a mathematical application, but it is important and the one for which the Riemann sphere was invented. To fully understand it, one needs to know some (differential) topology, which I will not attempt to explain. I will state some facts and hope they will be intelligible enough that they will be motivating. The upshot is this: The Riemann sphere is a really good, accurate model for understanding analytic, complex-valued functions. Every point on a sphere looks the same by symmetry. Via the Riemann sphere we can define limits, derivatives etc. at infinity and treat infinity just like any other complex number.

Every nonconstant rational function $R(z)=P(z)/Q(z)$, with $P$ and $Q$ polynomials, maps every point on the Riemann sphere to another point on the sphere; and further, every point is the value of $R(z)$ for at least one point $z$.

  • $R(z)=1/z$ maps $0$ to $R(0)=\infty$ and $\infty$ to $R(\infty)=0$.
  • ${1 + 2z^2 \over 12 - 3 z^2}$ maps both $\pm2$ to $R(\pm2)=\infty$ and $\infty$ to $R(\infty)=-2/3$.

To get to another part of your question, it turns out that the only differentiable symmetries (a.k.a. "diffeomorphisms" in differential topology) of the Riemann sphere as a model of complex analysis are fractional linear transformations, a.k.a Möbius transformations: $$z \mapsto {az+b \over cz + d}$$ Fractional linear transformations map circles and lines in the complex plane to circles and lines. Circles and lines in the complex plane correspond to circles on the Riemann sphere, and so the Riemann sphere gives a unified look to them. It's not incorrect to think of a line as a circle with an infinite radius in this context.

Symmetries play an important role in understanding how things are related. For instance, $1/(z-2)$ is just the function $1/z$ translated over two units. What one knows about $1/z$ corresponds to the same facts about $1/(z-2)$ translated over. Likewise, knowledge about $1/z$ corresponds to the same facts about $(1+i)/z$, but with function values scaled by $1+i$. Fractional linear transformations, then, are the ways that one can translate facts about the derivatives etc. of a given function $f(z)$ to other functions. If $f(z)$ and $g(z)$ are related by such symmetries, then they are essentially the same function, with locations and values connected by the symmetries.

A particular class of functions that is important in this context is the class of functions that can be represented by power series. You should get to that later in the course.

You will also see other ways that infinity is important in this course (poles, residues, to name two). Going beyond the course, in topology I learned why the first type of integral below is easy and taught in first-year calculus and why the second has books written about them: $$(1)\ \int {dx \over \sqrt{\text{quadratic polynomial}}} \qquad (2)\ \int{dx \over \sqrt{\text{cubic polynomial}}}$$

The characterization of fractional linear transformations in terms of lines and circles makes even more geometric (and visual) the projection connecting the Riemann sphere and the complex-number plane. That possibility of visualization also makes WL/Mathematica a particularly good tool to use to explore complex analysis.
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

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 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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