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[WSG24] Daily Study Group: Introduction to Complex Analysis

Posted 5 months ago

A Wolfram U Daily Study Group on "Introduction to Complex Analysis" begins on November 11, 2024.

Join me and a group of fellow learners to learn about the fundamentals of complex analysis using the powerful tools for symbolic computation and visualization in Wolfram Language. Our topics for the study group include elementary complex functions, the Cauchy-Riemann equations, complex integration, Cauchy's theorem and the residue theorem.

No prior Wolfram Language experience is required.

Please feel free to use this thread to collaborate and share ideas, materials and links to other resources with fellow learners.

Dates: November 11-22, 2024

Register here: https://wolfr.am/1qGinMguv

Wolfram U banner image

POSTED BY: Marco Saragnese
136 Replies

Hello I recently completed the final exam for the "Daily Cause of Complex Analysis" course, and I successfully passed. However, when I downloaded my Level 1 certificate, I noticed that it appears to have some kind of defect.

I have attached both the certificate and a screenshot of my progress window for you reference . Could you kindly investigate this issue and provide assistance in resolving it?

Thank you for your support.

Attachment

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

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
Posted 3 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

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
Posted 4 months ago

I've written a large Notebook. The output and input cells are interspersed. I'd like to have all the output together in 3 Sections with Section names, say "General Results", "Case A Findings", and "Case B Findings". I'd like the 3 Sections to have Subsections, and the Subsections to have Subsubsections. All the output cells can be either after the end of all the input cells or in a separate Notebook. How can that be coded?

I know how to generate a new Notebook with Sections, etc. For example:

nb = CreateDocument[{ipSection = TextCell["Input Data", "Section"], 

   aSection = TextCell["General Results", "Section"], 

  bSection = TextCell["Case A Findings", "Section"]},

  WindowSelected -> True, 

  WindowTitle -> 
   "Analysis Created on " <> DateString[Today],

  NotebookFileName -> "Analysis_" <> ToString[Today]]

However, I don't know how to "Print" output into the new notebook (nb in the example). For example, if I have an output plot generated using ComplexPlot3D[...], how do I get that into the Notebook nb in a specified section (like bSection)? I found ResourceFunction["PrintAsCellObject"] but I've been unable to use it to do what I just described. I've also looked at Write, etc. and have not been able to develop any code that works. The simplest thing would be to direct all the output to "Print" at the end of all the input in designated Sections and Subsections. Are there some settings in Format > Option Inspector ... that might help?

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

POSTED BY: Charles Glover
Posted 4 months ago

I agree with your answers to questions 5 and 6. I think you intended to show that the questions were misgraded regardless of how you answered them, but I think you ended up just repeating your answers.

Regardless, I agree with your answers as shown.

The grader must have been changed and broken.

Here are my reasons for agreeing with you. Question 6 presented before Question 5.

enter image description here

So the answer to Question 6 should be "Yes," as you chose, and the answer to Question 5 should be "No,", also as you chose.

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
Posted 4 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

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
Posted 4 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
Posted 4 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

Hi Mitchell,

Here are some connections. Connections to other fields may require some background for a full understanding, such as a course in the field. First a definition:

Def: $f(z)$ has a pole of integer order $k>0$ at $c$ if $(z-c)^kf(z)$ is holomorphic at $c$ but is not holomorphic for smaller powers $k$. Example: The following has poles of orders $1,2,3$ at $z=10,20,30$ respectively: $$f(z)={(z-40)^4 \over (z-10)(z-20)^2(z-30)^3} \,.$$

A number $c$ is a pole of order $k$ of $f(z)$ if it is a zero of multiplicity $k$ of $1/f(z)$. If we express the example above as a product, zeros and poles are distinguished by the sign of the exponent; otherwise, they are algebraically similar: $$f(z)={(z-40)^4 (z-10)^{-1}(z-20)^{-2}(z-30)^{-3}} \,.$$

Generalized mathematical significance (hope this isn't too abstract): Suppose we have a continuous object, such as a function $f_a(z)$ that varies continuously with one parameter/coefficient (or more) represented by $a$. And suppose there is a discrete quantity that is calculated from $f_a(z)$, such as the number of zeros or number of poles. Finally suppose that the discrete quantity can be proved to vary continuously with $a$. Then the discrete quantity is an invariant of $f_a(z)$. That is, it's constant.

Contrariwise, if the discrete quantity changes (is discontinuous), something interesting is probably happening with the functions $f_a(z)$ at the value for $a$ at which there is a discontinuity.

For instance, a meromorphic function on the Riemann sphere has a finite number of zeros and the same number of zeros ( $Z$) as poles ( $P$) counted with multiplicity and including poles/zeros at infinity. Thus $Z-P=0$. It is constant. You cannot change the number of zeros of $f_a(z)$ without also changing the number of poles in the same way (assuming $f_a(z)$ is meromorphic for all $a$). Example: $f_a(z)=z^2+a$ has two zeros because it has a pole of order $2$ at infinity (and vice versa).

More examples:

I suppose it's clear that solving equations has many applications, and the number of solutions has implications for the problem in which the equation arises. A solution to $f(z)=c$ is a zero of $g(z)=f(c)-c$.

I suppose it's also clear from the complex analysis course that the poles themselves are important in integration. Perhaps the number of them is not so significant, but it might sometimes be helpful to know how many there are.

An application: I suppose periodicity has been an important topic in mathematical physics and therefore in mathematics since humans noticed things like days, months, and years recur periodically. A periodic function has infinitely solutions to $f(z)=c$ if it has one. If $f(z)$ has infinitely many zeros, then it must have an essential singularity on the Riemann sphere (examples: $e^z$, $\sin z$, $\sin(1/z)$). Periodic functions must have this feature.

In control theory, the zeros and poles of the transfer function determine the behavior of a linear system. See MIT notes (read $2{du\over dt} + u$ for the RHS of the example at the top of page 2) or CalTech notes (read $e^{st}$ in the RHS of the equation between (6.6) and (6.7) on page 142).

POSTED BY: Michael Rogers
Attachments:
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:
POSTED BY: LORIS LORI

u is Labeled[Legended[Graphics3D[...],...],...]
v is Graphics3D[...]

Show[] can combine legended graphics and non-legended graphics. It doesn't seem to be programmed to combine labeled graphics.

My feeling is that this is plausibly a bug. If Show[] isn't supposed to handle labeled graphics, then it should say so. If it is supposed to handle them, then it failed. Further, the Rule[] error does not come from a user-specified rule, but from Mathematica. It suggests it has not been programmed carefully enough.

I'm not sure of the best workaround, but either of these is fairly simple:

u /. p_Graphics3D :> Show[p, v]

u /. p_Legended :> Show[p, v]
POSTED BY: Michael Rogers

A thousand thanks. I wouldn't have solved it without your help.

POSTED BY: LORIS LORI

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
Posted 4 months ago
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
Posted 4 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
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

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

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

POSTED BY: Joseph Smith

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

POSTED BY: Marco Saragnese
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
Posted 4 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
Posted 4 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

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
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.

Thank you. I'll have it fixed.

POSTED BY: Marco Saragnese

The same mistake is in Fig. 10.

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
Posted 4 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
Posted 4 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

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
Posted 4 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
Posted 4 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
Posted 4 months ago

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

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

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

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

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

POSTED BY: Phil Earnhardt

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
Posted 4 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 4 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

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
Posted 4 months ago

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

Attachments:
POSTED BY: Gerald Dorfman
Posted 4 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:
POSTED BY: Gerald Dorfman
Posted 4 months ago
POSTED BY: Gerald Dorfman

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

POSTED BY: Marco Saragnese
Posted 4 months ago

Marco, My mistake. Thanks for the correction.

POSTED BY: Gerald Dorfman
POSTED BY: Michael Rogers
Posted 4 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
Posted 4 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
Posted 4 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:
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
Posted 5 months ago
POSTED BY: Gerald Dorfman
POSTED BY: Marco Saragnese
Posted 5 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
Posted 5 months ago

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

POSTED BY: Gerald Dorfman
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:
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

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
POSTED BY: Michael Rogers
Posted 5 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
Posted 5 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
Posted 5 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.

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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
Posted 5 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
Posted 5 months ago

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

POSTED BY: Tingting Zhao
Posted 5 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
Posted 5 months ago
POSTED BY: Murray Wolinsky
Posted 5 months ago
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
Posted 5 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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