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[WSG24] Daily Study Group: Introduction to Complex Analysis

Posted 1 month 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
118 Replies

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

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

Can anyone suggest what I am missing? screen shot of solution .

POSTED BY: Joseph Smith

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

So "where f'(z) != 0" means "where f'(z) is not equal to 0."

POSTED BY: Murray Wolinsky

Thanks! Makes sense in context.

POSTED BY: Joseph Smith

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

POSTED BY: Michael Rogers

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

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

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

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

POSTED BY: Phil Earnhardt

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

Hi;

A good amount of the material presented in this webinar involves calculating information about poles and the number of zeros (or zero crossings); however (even after completing the webinar), I don’t believe that I have a good understanding of the importance of these values and what to do with them - the why. I can understand that knowing about the poles and number of zeros in a function gives a better understanding of the function, but I am not sure why that is important and how it is used in the end. My best guess is that these areas are problem areas in the function and need to be somehow removed or worked around to allow calculus to be performed on the function.

Furthermore, I am not sure that I know exactly what a pole is. Using Mathematica’s function ComplexPlot3D they appear to be 3 dimensional objects but aren’t we working in a 2 dimensional plane with complex numbers – its confusion.

Additionally, the “number of zeros” calculation appears to be an extension of Calculus’s critical points, and they didn’t present a problem in non-complex calculus.

In any event, any information you could give that would give me a better understanding of why poles and number of zeros are important and how they are used would be greatly appreciated. I have some new tools but are not sure how to use them or what they mean.

Thanks,

Mitch Sandlin

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

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

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:
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

Hi Mitchell,

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

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

I think I tried Element[z, Disk[{0, 0}, 5/2]], but it didn't work. I guessed I needed to specify the coordinates since Disk[] is two-dimensional in the two-dimensional real plane. One could specify the complex disk with Abs[z] < 5/2, which is how the examples in the docs are set up.

POSTED BY: Michael Rogers
Posted 12 days 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

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

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

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

POSTED BY: Joseph Smith

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

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

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
Posted 15 days 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

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

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.

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 19 days 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 19 days 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

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
Posted 17 days 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 16 days 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 15 days 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 19 days 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 19 days 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 19 days 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 18 days 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 17 days 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
Posted 16 days 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
Posted 14 days 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 14 days 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 12 days 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

I haven't taught Complex Analysis yet. So I haven't had to think about how to present the subject to students. You might be more disappointed in this opinion: Not every mathematical concept needs to anchored in, or even connected to, a physical phenomenon or application. When I was a physics major in college, my interest in complex analysis and in math in general was in understanding how functions work and to have better skills for solving the math in scientific problems. It was clear from my physics professors that such skills would be of value in physics.

Residues let you calculate integrals; indeed, even integrals from real analysis, which is a surprise. Calculating integrals was a skill valued by me and my fellow physics majors. You might think Mathematica would take care of all that. But it doesn't. Sometimes the integrand contains a general function that has no explicit formula but about which there are only some assumptions. Residues, branch cuts, integration by parts, and other tools may help say something about the integral. Other times, the understanding I have obtained from calculus, complex analysis, and differential equations has allowed me to help people (or myself) when Integrate[] has trouble. Sometimes I can figure out workarounds. Understanding complex integration sometimes has helped solve issues with NIntegrate[].

So: How do I create engagement? That invites a long answer. I'll boil it down to a quote (from faulty memory). Eva Brann, basing a question on the root meaning of "to teach," which is "to show," asked what is the best thing a teacher can show?: "How they think."

What residues mean to me: - The integral of $(z-a)^k$ around a simple closed path is zero unless $k=-1$ and the path goes around $a$, in which case the integral is $2\pi i$. - This fact is due to Green's theorem, the Cauchy-Riemann equations, and a simple calculation for the path $z=a+e^{it}$. - This means that if $f$ is analytic or has a Laurent expansion $f(z) = \sum\limits_{k=n}^\infty {c_k (z-a)^k}$, and we integrate around a path that contains $a$ and stays within the radius of convergence, then the integral of each term of the series vanishes except for the $k=-1$ term; and the integral is determined by the residue (times $2\pi i$). This is the key for me. - If the path goes around several poles, the integral is determined by the sum of the residues. - Dimensional analysis: Given that the integral $\int (z-a)^k\;dz$ is path-independent, the value is constant for paths that go around $a$. The integral also scales by like the $k+1$ power: If the path is a circle centered $a$, then the integral around a circle of radius $2r$ is equal to $2^{k+1}$ times the integral around a circle of radius $r$. Since the integrals are equal, the integral must be zero unless $k=-1$. This is one way to look at why the $-1$ power is the only significant one. It also suggests why expanding a function in a power series is revealing. - The coefficient $c_k$ of the series expansion of $f(z)$ at $a$ may be computed from ${1 \over 2\pi i} \int f(z)(z-a)^{-k-1}\,dz$. (This may be done efficiently in NIntegrate[] with the "Trapezoidal" strategy.) Multiplying by $k!$ yields the value of the derivative $f^{(k)}(a)$. In other words, dividing the Laurent series of $f(z)$ by $(z-a)^{k+1}$ shifts the coefficients of the series so that $c_k$ becomes the residue. - The radius of convergence is equal to the distance from the center to the nearest pole (other than $a$). So the path staying within the radius of convergence means it does not go around another pole and pick up another residue.

I'm not sure of the context for "orthogonality of harmonics." In AC power, the average power is the inner product of the voltage and current, which can be broken down into a linear combination of harmonics. If they are composed of different harmonics, then the average power is zero. If they have nonzero components for the same harmonic, then the average power will be nonzero. (If you're talking about Fourier series. In general, an orthogonal basis is golden, in many different topics in mathematics; just as square is in carpentry and woodworking.)

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 20 days 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
Posted 20 days 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 20 days 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!

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
Posted 21 days 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

Hi Gerald,

I learned these things back in the 1980s, some facts as an undergraduate but mainly in graduate school from lectures. The professors sometimes recommended books as ancillary materials, but only a few used books as textbooks and followed them. Further, I've been at an institution teaching only 1st and 2nd year students for 30 years, so I haven't taught this or related material for a long time. Nonetheless, I can give you some recommendations, but I personally have not field-tested them, so to speak.

Integrals of the form (2) first showed up for me as an undergraduate in physics. Integrals with the square of 3rd and 4th degree polynomials are called elliptic integrals. The arc length of an ellipse can be expressed with such an integral, hence the name, and they first came to the attention of mathematicians in the mid 18th century. They were studied somewhat intensely in the early 19th century by Gauss, Jacobi, Legendre, and in particular by Abel, who extended the scope to include integrals involving square roots of higher degree polynomials. These are called hyperelliptic or "Abelian* integrals. Much knowledge of them was developed before the Riemann sphere or topology was even invented. These facts give you some keywords to google, but they also warn you that you will find a lot of information about (2) not relating to my remarks above.

I came to know more about elliptic integrals from a sideways direction. I was studying number theory and algebraic geometry in grad school; in particular, "elliptic curves" was a hot topic. They come from the square-roots in the elliptic integrals, that is, curves of the form $y^2 = \text{cubic or quartic}$. In the 1980s, they were thought to hold a key to proving Fermat's Last Theorem (which turned out to be the case). They also yielded a powerful factorization algorithm, which was important because of the then relatively new RSA encryption algorithm. It's truly amazing the connections the integral (2) has to seemingly unrelated branches of mathematics.

It turns out there is a book that "does it all", or almost:

Elliptic Curves: Function Theory, Geometry, Arithmetic by McKean and Moll

The book was recommended to me by a friend, and on that basis, I am recommending it here. It gives an explicit construction of the differential manifold structure on the Riemann sphere, which is fairly simple, considering. It also gives a definition of manifold. So it covers everything that is necessary. However, it is question in my mind whether, having given someone the needed tool, they will know why, when, and how to use it. But it's less likely to be a problem than the general case, because the goal is to understand this manifold, namely the Riemann sphere, not manifolds in general. That aside, it covers elliptic integrals, elliptic functions, elliptic curves (though not the connections to Fermat's Last Theorem and factorization), and a few other things.

For differential topology, it's a bit harder to suggest a book. Many assume the reader is familiar with topology, especially manifolds. General introductions to topology include manifolds and much else that isn't strictly necessary for the question at hand. The following was my undergraduate textbook. I remember liking it and using it as a reference in grad school from time to time. And it seems to have the right scope, at least through, say, Chapter 5, which covers differentiable manifolds:

Singer and Thorpe, Lecture Notes on Elementary Topology and Geometry.

This Math.SE Q&A contains some good recommendations:

https://math.stackexchange.com/questions/46482/introductory-texts-on-manifolds

These two have a good reputation:

  • L. Tu, An Introduction to Manifolds (includes differentiable a.k.a. smooth manifolds†)
  • John M. Lee, Introduction to Smooth Manifolds

†Note: Usually there is a slight difference between these two; however, every smooth manifold is a differentiable manifold.

I hope that helps.

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 23 days 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 23 days ago

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

Attachments:
POSTED BY: Gerald Oberg

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 23 days 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 23 days ago

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

POSTED BY: Tingting Zhao
Posted 23 days 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

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
Posted 23 days 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
Posted 23 days 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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